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495 | OPTICS | In 1845, Faraday studied the influence of electricity and magnetism on polarized light and discovered that heavy glass, originally non-optically active, exhibited optical activity under the influence of a strong magnetic field, causing the plane of polarization of polarized light to rotate. This was the first time humanity recognized the relationship between electromagnetic phenomena and light phenomena. The magneto-optical rotation effect later became known as the Faraday Effect.
In the magneto-optical rotation effect, the rotation angle of the vibration plane \(\beta\) is proportional to the distance \(d\) that light travels through the medium and is also proportional to the magnetic induction intensity \(B\) within the medium, expressed as:
\[
\beta = vBd
\]
where \(v\) is the proportionality constant, known as the Verdet constant, which depends on the properties of the medium and is also related to the wavelength of the incident light. Assume that the medium can be modeled using the Lorentz model, where the medium is composed of atoms, with \(N\) atoms per unit volume, and each atom contains \(Z\) electrons in the outer shell. The outer electrons deviate from their equilibrium positions and are subject to linear restoring forces: \(\vec{f} = -k\vec{r}\), with the harmonic resonance frequency of the outer shell electrons being \(\omega_0\). The interaction between the incident light and the medium exclusively involves the outer shell electrons.
Circularly polarized light can be regarded as the result of the synthesis of two linearly polarized light waves with the same frequency and direction, mutually perpendicular vibration axes, and a stable phase relationship. The two linearly polarized light waves have the same amplitude and a phase difference of \(\pm\pi/2\), representing left-handed and right-handed circularly polarized light, respectively. Similarly, a linearly polarized light wave can be decomposed into left-handed and right-handed circularly polarized components.
Based on these models, calculate the Verdet constant \(\beta\) for incident light with an angular frequency of \(\omega\). Assume that the medium does not absorb the incident light and its refractive index near the angular frequency \(\omega\) is \(n\). | Solution: The propagation of light waves in a medium under an applied magnetic field involves the motion of electrons:
\[m\ddot{\vec{r}} = -k\vec{r} - g\vec{v} - e(\vec{E} + \vec{v} \times \vec{B})\]
\(\vec{E}\) is the electric field intensity of the incident light wave; \(\vec{B}\) is the magnetic induction intensity applied to the medium; \(g\) is the damping term, and since the medium does not absorb the incident light, it exhibits weak damping and low loss, thus neglecting the damping term:
\[m\ddot{\vec{r}} = -k\vec{r} - e(\vec{E} + \vec{v} \times \vec{B})\]
Let \(\vec{B}\) be along the positive \(z\)-axis, which is also the propagation direction of the light wave, thus:
\[
\begin{cases}
m\ddot{x} + e\dot{y}B + kx = -eE_x \\
m\ddot{y} - e\dot{x}B + ky = -eE_y
\end{cases}
\]
Assuming the incident wave is a steady-state wave, the relationship between the electric field and time: \(e^{-i\omega t}\)
The relationship between the perturbation term of the forced vibration of the outer electrons and time: \(e^{-i\omega t}\)
Thus: \(\frac{\partial}{\partial t} \equiv -i\omega\), \(\frac{\partial^2}{\partial t^2} \equiv -\omega^2\), therefore:
\[
\begin{cases}
(\omega_0^2 - \omega^2)r_x - i\omega\Omega r_y = -\frac{e}{m}E_x \\
(\omega_0^2 - \omega^2)r_y + i\omega\Omega r_x = -\frac{e}{m}E_y
\end{cases}
\]
Where:
\(\Omega = \frac{eB}{m}\), \(\omega_0 = \sqrt{\frac{k}{m}}\)
For left-hand circularly polarized light, \(E_+ = E_x + iE_y\), then: \(r_+ = r_x + ir_y\)
\[(\omega_0^2 - \omega^2 - \omega\Omega)r_+ = -\frac{e}{m}E_+\]
Solving this, we get:
\[r_+ = -\frac{e}{m(\omega_0^2 - \omega^2 - \omega\Omega)}E_+\]
The polarization vector: The atomic number density of the medium is \(N(1/m^3)\), with each atom providing \(Z\) electrons weakly bound in the outer layer;
\[\vec{P}_+ = NZ(-e)\vec{r}_+ = \frac{NZ\frac{e^2}{m}}{\omega_0^2 - \omega^2 - \omega\Omega}E_+\]
Since \(P = \varepsilon_0\chi E\), thus:
\[\chi_+ = \frac{NZ\frac{e^2}{\varepsilon_0 m}}{\omega_0^2 - \omega^2 - \omega\Omega}\]
The refractive index for left-hand circularly polarized light is:
\[n_+^2 = \varepsilon_+ = (1 + \chi_+) = 1 + \frac{NZ\frac{e^2}{\varepsilon_0 m}}{\omega_0^2 - \omega^2 - \omega\Omega}\]
For right-hand circularly polarized light \(E_- = E_x - iE_y\), then: \(r_- = r_x - ir_y\)
\[(\omega_0^2 - \omega^2 + \omega\Omega)r_- = -\frac{e}{m}E_-\]
Solving this, we get:
\[r_- = -\frac{e}{m(\omega_0^2 - \omega^2 + \omega\Omega)}E_-\]
\[\vec{P}_- = NZ(-e)\vec{r}_- = \frac{NZ\frac{e^2}{m}}{\omega_0^2 - \omega^2 + \omega\Omega}E_-\]
Thus:
\[\chi_- = \frac{NZ\frac{e^2}{\varepsilon_0 m}}{\omega_0^2 - \omega^2 + \omega\Omega}\]
The refractive index for right-hand circularly polarized light is:
\[n_-^2 = \varepsilon_- = (1 + \chi_-) = 1 + \frac{NZ\frac{e^2}{\varepsilon_0 m}}{\omega_0^2 - \omega^2 + \omega\Omega}\]
That is, the refractive indices for left-hand and right-hand circularly polarized light are different:
\[n_{\pm}^2 = \varepsilon_{\pm} = (1 + \chi_{\pm}) = 1 + \frac{NZ\frac{e^2}{\varepsilon_0 m}}{\omega_0^2 - \omega^2 \mp \omega\Omega}\]
Any linearly polarized light can be decomposed into two circularly polarized lights with the same frequency, equal amplitude, and a stable phase relationship. When linearly polarized light passes through a medium with an external magnetic field aligned with the direction of propagation, the rotation angle of the polarization plane is:
\[\beta = \frac{1}{2}(\alpha_+ - \alpha_-) = \frac{\pi}{\lambda}(n_+ - n_-)d\]
where \(d\) is the propagation length of light in the medium.
Find \((n_+ - n_-)\)
\[
\begin{align*}
n_+^2 - n_-^2 &= (n_+ - n_-)(n_+ + n_-)\\
&= NZ\frac{e^2}{\varepsilon_0 m} \cdot \frac{2\omega\Omega}{(\omega_0^2 - \omega^2)^2 - (\omega\Omega)^2}
\end{align*}
\]
Generally, the cyclotron frequency \(\Omega\) is much smaller than the optical frequency \(\omega\), and the optical frequency \(\omega\) is not near the resonance frequency \(\omega_0\), hence:
\[n_+ + n_- \approx 2n\]
\[n^2 \approx 1 + NZ\frac{e^2}{\varepsilon_0 m} \cdot \frac{1}{\omega_0^2 - \omega^2}\]
Therefore:
\[n_+ - n_- = NZ\frac{e^2}{\varepsilon_0 mn} \cdot \frac{\omega\Omega}{(\omega_0^2 - \omega^2)^2}\]
Thus:
\[
\begin{align*}
\beta &= \frac{\pi}{\lambda}(n_+ - n_-)d = \frac{\omega}{2c}NZ\frac{e^2}{\varepsilon_0 mn} \cdot \frac{\omega\Omega}{(\omega_0^2 - \omega^2)^2}Bd = \frac{NZ e^3}{2c\varepsilon_0 m^2 n} \cdot \frac{\omega^2}{(\omega_0^2 - \omega^2)^2} \cdot Bd\\
&= vBd
\end{align*}
\]
The Verdet constant is obtained as:
\[v = \frac{NZ e^3}{2c\varepsilon_0 m^2 n} \cdot \frac{\omega^2}{(\omega_0^2 - \omega^2)^2}\] | \[v = \frac{NZ e^3}{2c\varepsilon_0 m^2 n} \cdot \frac{\omega^2}{(\omega_0^2 - \omega^2)^2}\] |
600 | OPTICS | The Yang's double slit interference experiment consists of three parts: light source, double slit, and receiving screen. In the experiment, we generated a line light source by placing a point light source behind the slit $S_0$, and the light was then projected onto the receiving screen through the double slits $S_1$ and $S_2$, forming interference fringes. Now we will perform the following actions on this device:
Place polarizer $P_0$ next to slit $S_0$, and polarizers $P_1$ and $P_2$ next to slits $S_1$ and $S_2$, respectively, so that the transmission direction of $P_0$ is parallel to $P_1$ and forms a $\theta=60\degree$ angle with $P_2$.
Try to determine the contrast of the stripes on the receiving screen at this time. | The magnitude of electric field intensity is set to A after passing through polarizer P0.
The optical amplitude through P1 is A, and the optical amplitude through P2 is $\ frac 12 $A.
Let the direction parallel to P0 be the x direction and the direction perpendicular to it be the y direction.
standard
$$
E_{1x}=A,E_{2x}=\frac{1}{4} A,E_{2y}=\frac {\sqrt{3}}{4} A
$$
$E2 {1x} $and $E2 {2x} $are coherently superimposed, with a total light intensity of
$$
I=I_{0}\left(\frac 54 +\frac 12 \cos\delta\right)
$$
The lining ratio is
$$
\gamma=\frac 25
$$ | $$\gamma = \frac{2}{5}$$ |
285 | MECHANICS | A small bug with a mass of $m$ crawls on a disk with a radius of $2R$. Relative to the disk, its crawling trajectory is a circle of radius $R$ that passes through the center of the disk. The disk rotates with a constant angular velocity $\omega$ about an axis passing through its center and perpendicular to the plane of the disk. The bug's angular crawling velocity relative to the disk is in the same direction as the disk's angular velocity and has the same magnitude. Solve for the maximum force $F_{max}$ between the bug and the disk required to maintain this motion (neglecting gravity). | The problem can use the acceleration transformation formula for rotating reference frames, where the actual acceleration of the small insect
$$
\overrightarrow{a}=\overrightarrow{a}+2\overrightarrow{\omega}\times\overrightarrow{v}+\dot{\overrightarrow{\omega}}\times\overrightarrow{r}-\omega^{2}\overrightarrow{r}
$$
Here, $\vec{a}^{\prime}$ is the acceleration relative to the rotating system, and since the rotating system is a uniform circular motion,
$$
\dot{a_{\mathrm{\scriptsize~=~}}}\omega^{2}R
$$
is the speed relative to the rotating system, with a magnitude of $\omega R$. Thus, the magnitude of the Coriolis acceleration is:
$$
2\omega v^{'}=2\omega^{2}R
$$
The last term is the centripetal acceleration, which is directly proportional to the distance from the insect to the center of the disk,
$$
\omega^{2}r=2\omega^{2}R\cos\left(\frac{\theta}{2}\right)
$$
(Analysis of each acceleration, 7 points)
These three acceleration components are combined to derive the total acceleration magnitude as:
$$
a=\omega^{2}R{\sqrt{\left(3+2\cos\left({\frac{\theta}{2}}\right)^{2}\right)^{2}+4\cos\left({\frac{\theta}{2}}\right)^{2}\sin\left({\frac{\theta}{2}}\right)^{2}}}=\omega^{2}R{\sqrt{9+16\cos\left({\frac{\theta}{2}}\right)^{2}}}
$$
which is a constant, so the maximum force is
$$
F_{m a x}=5m\omega^{2}R
$$ | $$F_{\max} = 5m\omega^2R$$ |
321 | MECHANICS | The civilization of *Three-Body* changes the values of fundamental physical constants, such as the Planck constant, inside the "water drop" to alter the range of strong forces.
What kind of effects would this produce? This problem provides a suitable discussion about this question.
Hideki Yukawa pointed out that the propagator of nuclear forces is the $\pi$ meson. One can equivalently simplify the effect by considering that protons and protons, protons and neutrons, and neutrons and neutrons have an identical strong-force potential energy. For a force field with a propagator of mass, Yukawa potential and its range can be expressed as:
$$
U = -\frac{A\mathrm{e}^{-\frac{2\pi m c r}{h}}}{r}\quad,\quad\lambda = \frac{h}{2\pi m c}
$$
For the deuteron (the strong-force binding between a proton and a neutron), consider two particles rapidly rotating around the center of mass. If the distance between the proton and the neutron is precisely $r$ and they do not collide, find the angle rotated around the center of mass during one complete cycle of small radial oscillation. Express the answer in terms of $\lambda$ and $r$. | $$
F=-\frac{A(1+\frac{2\pi m c r}{h})e^{-\frac{2\pi m c r}{h}}}{r^{2}}
$$
Newton's Law
$$
m\omega^{2}\frac{r}{2}=\frac{A(1+\frac{2\pi m c r}{h})e^{-\frac{2\pi m c r}{h}}}{r^{2}}
$$
Effective Potential Energy
$$
V_{eff}=-\frac{A e^{-\frac{2\pi m c r}{h}}}{r}+\frac{L^{2}}{2\cdot m/2\cdot r^{2}}
$$
Solving the Second Derivative
$$
\frac{d^{2}V_{eff}}{d r^{2}}=\frac{A e^{-\frac{r}{\lambda}}}{r^{3}}\left(1+\frac{r}{\lambda}-\frac{r^{2}}{\lambda^{2}}\right)
$$
Angular Frequency
$$
\omega_{0}=\sqrt{\frac{2\frac{d^{2}V_{eff}}{d r^{2}}}{m}}
$$
Geometric Relation
$$
\varphi=2\pi\cdot\frac{\omega}{\omega_{0}}
$$
Resulting
$$
\varphi=2\pi\sqrt{\frac{1+r/\lambda}{1+r/\lambda-r^2/\lambda^2}}
$$ | $$\varphi = 2\pi \sqrt{\frac{1 + \frac{r}{\lambda}}{1 + \frac{r}{\lambda} - \frac{r^2}{\lambda^2}}}$$ |
674 | MECHANICS | A coin is placed at rest on the edge of a smooth tabletop, with only a small portion of its right side extending beyond the edge. The coin can be considered as a uniform disk with mass $m$, radius $r$, and gravitational acceleration $g$. A vertical impulse $I$ is applied to the right side of the coin. During its subsequent motion after the impact, the coin might fly off the table at some point. Determine the minimum initial velocity of the center of mass, $v_{0, \min}$, required for the coin to leave the table, expressed in terms of $g$ and $r$.",
| The moment of inertia of the coin about the axis is \[J=\frac14mr^2\] According to the impulse and impulse-momentum theorem, about the center of mass we have \[I+I_N=m v_0, Ir-I_Nr=J\omega_0\] There is also the velocity relation \[v_0=\omega_0 r\] Therefore, we have \[v_0=\frac{8I}{5m}, \omega_0=\frac{8I}{5mr}\] After being struck, assume the angle between the coin and the horizontal direction is \(\phi\). Because the tabletop is smooth, the horizontal coordinate of the center of mass remains unchanged, so \[\vec r_c=(0,r\sin\phi)\] Thus, we have \[E_k=\frac 12m\dot y_c^2+\frac12 J\dot\phi ^2=\frac12mr^2\left(\cos^2\phi+\frac 14\right)\dot\phi^2\] \[E_p=mgy_c=mgr\sin\phi\] Thus, mechanical energy is conserved \[\frac 18mr^2(4\cos^2\phi+1)\dot\phi^2+mgr\sin\phi=\frac{8I^2}{5m}\] From this, we can obtain \[\dot\phi^2=\frac{64I^2-40m^2gr\sin\phi}{5m^2r^2(4\cos^2\phi+1)}\] Since \(\frac{\mathrm d\dot\phi^2}{\mathrm d\phi}=2\ddot\phi\), we have \[\ddot\phi = \frac{4\cos\phi}{5r^2(4\cos^2\phi+1)^2}\left(64\frac{I^2}{m^2}\sin\phi-5gr(5+4\sin^2\phi)\right)\] Hence, we can obtain \[\begin{align*}\ddot y_c&=r(\ddot\phi\cos\phi-\dot\phi^2\sin\phi)\\ &=\frac{4\cos^2\phi}{5r(4\cos^2\phi+1)^2}\left(64\frac{I^2}{m^2}\sin\phi-5gr(5+4\sin^2\phi)\right)-\frac{(64 I^2-40 m^2gr\sin\phi)\sin\phi}{5m^2r(4\cos^2\phi+1)}\end{align*}\] When there is no support force from the tabletop, the coin leaves the table, at that moment \[m\ddot y_c=-mg\] At this point \(\phi=\theta\), \[20\sin^2\theta-64\frac{I^2}{m^2gr}\sin\theta+25=0\] It can be seen that the solution for \(\sin\theta\) can take all values in \((0,1]\), which means for it to jump, we have \[I=m\sqrt{gr}\left(\sqrt{\frac{20\sin^2\theta+25}{64\sin\theta}}\right)_{\min}=\frac{3\sqrt5}{8}m\sqrt{gr}\] Thus we have \[v_{0,\min}=\frac{3\sqrt 5}{5}\sqrt{gr}\]
| $$\frac{3\sqrt{5}}{5}\sqrt{gr}$$ |
83 | ELECTRICITY |
---
A concentric spherical shell with inner and outer radii of $R_{1}=R$ and $R_{2}=2^{1/3}R$, and magnetic permeability $\mu=3\mu_{0}$, is placed in an external uniform magnetic field with magnetic flux density $B_{0}$ and the magnetic field generated by a fixed magnetic dipole located at the center of the sphere. The magnetic dipole has a magnetic moment $m$ that is aligned with the external magnetic field. The system reaches magnetostatic equilibrium. To solve this problem, we can use the method of magnetic scalar potential. The concept is to express the magnetic field intensity as ${\pmb{H}}=-\nabla\varphi$, where $\varphi$ is referred to as the magnetic scalar potential. This approach applies to all three regions: the space inside the shell, the space outside the shell, and the region within the shell itself. The relationship between magnetic flux density and magnetic field intensity is $B=\mu H$.
If the magnetic moment is rotated by an angle $\alpha$, calculate the torque it experiences. The answer should be expressed in terms of $m$, $B_0$, and $\alpha$. Please verify the result before outputting.
--- | (1)
Let:
$$
\varphi={\left\{\begin{array}{l l}{A_{1}{\frac{r}{R}}\cos\theta+B_{1}{\frac{R^{2}}{r^{2}}}\cos\theta}&{(r<R)}
{A_{2}{\frac{r}{R}}\cos\theta+B_{2}{\frac{R^{2}}{r^{2}}}\cos\theta}&{(R<r<k R)}\ {A_{3}{\frac{r}{R}}\cos\theta+B_{3}{\frac{R^{2}}{r^{2}}}\cos\theta}&{(r>k R)}\end{array}\right.}
$$
(Tangential) continuity $(k=2^{1/3})$
$$
\begin{array}{c}{{A_{1}+B_{1}=A_{2}+B_{2}}}\ {{\qquad}}\ {{k A_{2}+k^{-2}B_{2}=k A_{3}+k^{-2}B_{3}}}\end{array}
$$
Magnetic induction normal continuity:
$$
\mu_{0}A_{1}-2\mu_{0}B_{1}=\mu A_{2}-2\mu B_{2}
$$
$$
k\mu A_{2}-2k^{-2}\mu B_{2}=k\mu_{0}A_{3}-2k^{-2}\mu_{0}B_{3}
$$
According to the boundary condition at infinity:
$$
A_{3}=-\frac{B_{0}R}{\mu_{0}}
$$
According to the boundary condition at the magnetic dipole:
$$
B_{1}=\frac{m}{4\pi R^{2}}
$$
The solution is:
$$
\begin{array}{c}{{A_{2}=\displaystyle\frac{21}{31}A_{3}+\displaystyle\frac{6}{31}B_{1}}}\ {{{}}}\ {{B_{2}=\displaystyle\frac{6}{31}A_{3}+\displaystyle\frac{15}{31}B_{1}}}\ {{{}}}\ {{A_{1}=\displaystyle\frac{27}{31}A_{3}-\displaystyle\frac{10}{31}B_{1}}}\ {{{}}}\ {{B_{3}=\displaystyle-\frac{14}{31}A_{3}+\displaystyle\frac{27}{31}B_{1}}}\end{array}
$$
Substituting:
$$
\varphi=\left\{\begin{array}{l l}{\left(-\frac{27}{31}\frac{B_{0}R}{\mu_{0}}-\frac{10}{31}\frac{m}{4\pi R^{2}}\right)\frac{r}{R}\cos\theta+\frac{m}{4\pi R^{2}}\frac{R^{2}}{r^{2}}\cos\theta}&{(r<R)}\ {\left(-\frac{21}{31}\frac{B_{0}R}{\mu_{0}}+\frac{6}{31}\frac{m}{4\pi R^{2}}\right)\frac{r}{R}\cos\theta+\left(-\frac{6}{31}\frac{B_{0}R}{\mu_{0}}+\frac{15}{31}\frac{m}{4\pi R^{2}}\right)\frac{R^{2}}{r^{2}}\cos\theta}&{(R<r<k R)}\ {-\frac{B_{0}R}{\mu_{0}}\frac{r}{R}\cos\theta+\left(\frac{14}{31}\frac{B_{0}R}{\mu_{0}}+\frac{27}{31}\frac{m}{4\pi R^{2}}\right)\frac{R^{2}}{r^{2}}\cos\theta}&{(r>k R)}\end{array}\right.
$$
(2)
Magnetic field induced by external magnetic field:
$$
B^{\prime}=\frac{27}{31}B_{0}
$$
Thus, torque:
$$
{\cal M}=\underline{{{m\times B}}}
$$
Result:
$$
M=-{\frac{27}{31}}m B_{0}\sin\alpha
$$ | $$ M = -\frac{27}{31} m B_0 \sin \alpha $$ |
54 | MECHANICS | Three small balls are connected in series with three light strings to form a line, and the end of one of the strings is hung from the ceiling. The strings are non-extensible, with a length of $l$, and the mass of each small ball is $m$.
Initially, the system is stationary and vertical. A hammer strikes one of the small balls in a horizontal direction, causing the ball to acquire an instantaneous velocity of $v_0$. Determine the instantaneous tension in the middle string when the topmost ball is struck. (The gravitational acceleration is $g$.) | Strike the top ball, the acceleration of the top ball is:
$$a=\frac{v_0^2}{l}$$
Switching to the reference frame of the top ball, let the tensions from top to bottom be $T_1, T_2, T_3$. Applying Newton's second law to the middle ball:
$$T_2-T_3-mg-m\frac{v_0^2}{l}=m\frac{v_0^2}{l}$$
The acceleration of the bottom ball is: $$a=2\frac{v_0^2}{l}$$
Applying Newton's second law to the bottom ball: $$T_3-mg=2m\frac{v_0^2}{l}$$
Thus, we have: $$T_2=2mg+4m\frac{v_0^2}{l}$$ | $$T_2 = 2mg + 4m \frac{v_0^2}{l}$$ |
10 | ELECTRICITY | The region in space where $x > 0$ and $y > 0$ is a vacuum, while the remaining region is a conductor. The surfaces of the conductor are the $xOz$ plane and the $yOz$ plane. A point charge $q$ is fixed at the point $(a, b, c)$ in the vacuum, and the system has reached electrostatic equilibrium. Find the magnitude of electric field intensity on the surface of the conductor at the $xOz$ plane, ${E}(x, +0, z)$. | The infinite conductor and the point at infinity are equipotential, $U{=}0$, with the boundary condition that the tangential electric field $E_{\tau}{=}0$. By the uniqueness theorem of electrostatics, the induced charges on the conductor can be equivalently replaced by image charges to determine the contribution to the electric potential in the external space of the conductor. The top view of the image charges is shown below.
Therefore, the answer is
\(\vec{E}\left(x,+0,z\right)=\left(\frac{-2q}{4\pi\varepsilon_{0}}\frac{b}{r_{1}^{3}}{+}\frac{2q}{4\pi\varepsilon_{0}}\frac{b}{r_{2}^{3}}\right)\hat{y}=\frac{q b}{2\pi\varepsilon_{0}}\biggl[\bigl[(a+x)^{2}+(z-c)^{2}+b^{2}\bigr]^{\frac{-3}{2}}-\bigl[(x-a)^{2}+(z-c)^{2}+b^{2}\bigr]^{\frac{-3}{2}}\biggr]\hat{y},\) \(\vec{E}\left(+0,y,z\right)=(-\frac{2q}{4\pi\varepsilon_{0}}\frac{a}{r_{1}^{3}}+\frac{2q}{4\pi\varepsilon_{0}}\frac{a}{r_{2}^{3}})\hat{x}=\frac{q a}{2\pi\varepsilon_{0}}\Bigg[\Big[(y+b)^{2}+\big(z-c\big)^{2}+a^{2}\Big]^{\frac{-3}{2}}-\Big[\big(y-b\big)^{2}+\big(z-c\big)^{2}+a^{2}\Big]^{\frac{-3}{2}}\Bigg]\hat{x}\) | $$\frac{q b}{2\pi \varepsilon_0} \left[ ((a+x)^2 + (z-c)^2 + b^2)^{-3/2} - ((x-a)^2 + (z-c)^2 + b^2)^{-3/2} \right]$$ |
33 | THERMODYNAMICS | Consider an infinite-length black body with inner and outer cylinders, which are in contact with heat sources at temperatures $T_{1}$ and $T_{2}$, respectively; assume that the temperature of the heat sources remains constant. Let the inner cylinder have a radius $r$, the outer cylinder have a radius $R$, and the distance between the axes of the inner and outer cylinders be $b$, with $r < b < R$ and $r + b < R$. Find the power $p(\theta)$ absorbed per unit area from the heat source at angle $\theta$ on the surface of the outer cylinder (i.e., the power density at $\theta$), where $\theta$ is the angle between the line connecting a point on the surface of the outer cylinder and the center of the outer cylinder, and the line connecting the centers of the inner and outer cylinders. The Stefan-Boltzmann constant is denoted as $\sigma$. | In the case of non-coaxial, we examine the long strip-shaped area element at the outer wall reaching $\theta + \mathrm{d} \theta$:
The heat flux projected onto the inner axis is:
$$
\frac{d S}{2}\sigma T_{2}^{4}\int_{\mathrm{arcsin}\left({\frac{b \sin\theta}{D}}\right)-\mathrm{arcsin}({\frac{r}{D}})}^{\mathrm{arcsin}\left({\frac{b \sin\theta}{D}}\right)+\mathrm{arcsin}({\frac{r}{D}})}\cos\theta \, d\theta
$$
The above equation equals $\sigma T_{2}^{4}\frac{r(R-b \cos \theta)}{R^{2}+b^{2}-2R b \cos \theta}\mathrm{d}S$, where dS is the area of the strip-shaped area element. According to the zeroth law of thermodynamics, the heat flux sent from the inner cylinder to the strip-shaped area element at $\theta + \mathrm{d} \theta$ on the outer cylinder is $\begin{array}{r}{\sigma T_{1}^{4}\frac{r(R-b \cos\theta)}{R^{2}+b^{2}-2R b \cos\theta}\mathrm{d} S}\end{array}$: Similarly, according to the zeroth law of thermodynamics,
It is known that the heat flux sent from other places on the outer cylinder to this area element is
$$
\begin{array}{r}{(\sigma T_{2}^{4}-\sigma T_{2}^{4}\frac{r(R-b}{R^{2}+b^{2}-2R b \cos \theta})\mathrm{d} S}\end{array}
$$
It can be seen that the heat absorbed by the unit area element at $\theta$ from the heat source with temperature $T_{2}$ is equal to
$$
\begin{array}{r}{(\sigma T_{2}^{4}-\sigma T_{1}^{4})\frac{r(R-b)}{R^{2}+b^{2}-2R b \cos \theta}}\end{array}
$$ | $$
p(\theta) = (\sigma T_2^4 - \sigma T_1^4) \frac{r(R - b \cos \theta)}{R^2 + b^2 - 2Rb \cos \theta}
$$ |
41 | MECHANICS | The rocket ascends vertically from the Earth's surface with a constant acceleration $a = k_1g_0$, where $g_0$ is the gravitational acceleration at the Earth's surface. Inside the rocket, there is a smooth groove containing a testing apparatus. When the rocket reaches a height $h$ above the ground, the pressure exerted by the instrument on the bottom of the groove is $k_2$ times what it was at the time of liftoff. Given that the radius of the Earth is $R$, find $h$. | At the moment of takeoff, $N_1=(1+k_1)mg_0$, where $m$ is the mass of the tester.
When ascending to a height $h$, the gravitational acceleration changes to $g=g_0(\dfrac{R}{R+h})^2$
The pressure at this time is $N_2=mg_0(\dfrac{R}{R+h})^2+k_1mg_0$
Since $N_2=k_2N_1$, solving the equations together gives
$h=R(\dfrac{1}{\sqrt{k_2(1+k_1)-k_1}}-1)$ | $$R\left(\frac{1}{\sqrt{k_2(1+k_1)-k_1}}-1\right)$$ |
453 | ELECTRICITY | Most media exhibit absorption of light, where the intensity of light decreases as it penetrates deeper into the medium.
Let monochromatic parallel light (with angular frequency $\omega$) pass through a uniform medium with a refractive index of $n$. Experimental results show that, over a reasonably wide range of light intensities, after passing through a small thickness, the decrease in light intensity is proportional to both the light intensity itself and this small thickness. The proportionality coefficient is denoted as $\alpha$.
Next, we will estimate the coefficient $\alpha$ based on the classical oscillator model of atoms. In this model, the atomic nucleus can be considered stationary, while the electron, when in motion, is bound by the nucleus. This binding force can be approximated as a linear restoring force: $-m\omega_{0}^{2}x$. Here, $\mathsf{m}$ is the mass of the electron, $\omega_{0}$ is the natural angular frequency (describing the strength of the restoring force), and $x$ is the distance between the electron and the nucleus. Simultaneously, when the electron is in motion, there often exists a damping force: $-m\gamma v$. However, since the damping is usually quite weak, it can be approximately assumed that the electron undergoes simple harmonic motion within any given period.
Additionally, it is important to know that the power of radiation emitted by a charged particle in accelerated motion is given by:
$$
P = {\frac{e^{2}{\overline{{{\dot{v}}^{2}}}}}{6\pi\varepsilon_{0}c^{3}}}
$$
Based on this model, derive the expression for $\alpha$.
**Hint**: The primary interaction between electromagnetic waves and the medium originates from the contribution of the electric field component. $e$ is the magnitude of the electron charge, $\overline{{{\dot{v}}^{2}}}$ represents the average value of the square of the acceleration, $\gamma$ is a positive constant, $\varepsilon_{0}$ is the vacuum permittivity, $v$ is the velocity of the electron, $c$ is the speed of light in vacuum, and $N$ is the number density of electrons. | The dynamics equation for electrons:
$$
m\ddot{x}=-m\gamma\dot{x}-m\omega_{0}^2x-eE_0e^{i\omega t}
$$
Substituting $x=\tilde{x}e^{i\omega t}$ yields $\tilde{x}=\frac{-eE_0/m}{-\omega^2+\omega_0^2+i\gamma \omega}$
$$
\tilde{v}=i\omega \tilde{x}=\frac{-i\omega eE_0/m}{-\omega^2+\omega_0^2+i\gamma \omega}
$$
$$
\dot{\tilde{v}}=\frac{-\omega^2 eE_0/m}{-\omega^2+\omega_0^2+i\gamma \omega}
$$
$$
\overline{\dot{v^2}}=\frac{\frac{1}{2}(\omega^2 eE_0/m)^2}{(\omega_0^2-\omega^2)^2+\gamma^2 \omega^2}
$$
$$
P=-\frac{e^2\overline{\dot{v^2}}}{6\pi\epsilon_0 c^3}=-\frac{1}{12\pi\epsilon_0 c^3}\frac{(\omega^2 e^2 E_0/m)^2}{(\omega_0^2-\omega^2)^2+\gamma^2 \omega^2}
$$
The energy flux density of the electromagnetic wave:
$$
I=\frac{1}{2}nc\epsilon_0E_0^2
$$
Assuming the number density of electrons is \(N\), then the energy loss per unit area in a length \(dx\) is \(NPdx=dI\).
Also, from $-dI=\alpha Idx$, we have
$$
\alpha=-\frac{dI}{Idx}=-\frac{NP}{I}
$$
Thus,
$$
\alpha=\frac{N}{6\pi n\epsilon_0^2c^4}\frac{(\frac{\omega^2 e^2}{m})^2}{(\omega_0^2-\omega^2)^2+\gamma^2 \omega^2}
$$ | $$
\alpha=\frac{N}{6\pi n\epsilon_0^2c^4}\frac{(\frac{\omega^2 e^2}{m})^2}{(\omega_0^2-\omega^2)^2+\gamma^2 \omega^2}
$$
|
281 | THERMODYNAMICS | Due to the imbalance of molecular forces on the surface layer of a liquid, tension will arise along any boundary of the surface layer. In this context, the so-called "capillary phenomenon" occurs, affecting the geometric shape of a free liquid surface. Assume a static large container exists in a uniform atmosphere. The container holds a liquid in static equilibrium, and its side walls are vertical planes. The contact angle between the liquid and the side wall is $\theta$ $(0 < \theta < \frac{\pi}{2})$, the density of the liquid is $\rho$, its surface tension coefficient is $\sigma$, atmospheric pressure is $p_{0}$, and the gravitational acceleration is $g$.
Find the additional height $h$ of the liquid surface rise at the junction between the container wall and the liquid surface due to the capillary phenomenon, expressing the result using $\theta, \rho, \sigma, g$.
Hint: The formula for the radius of curvature in a two-dimensional Cartesian coordinate system is: $R = {\frac{(1+z^{\prime}(x)^{2})^{\frac{3}{2}}}{\mid z^{\prime\prime}(x)\mid}}$. | Choose $x\sim x+\mathrm{d}x.$ , analyze the surface microelement of the liquid located at a height between $z\sim z+\mathrm{d}z$, considering the liquid pressure $p_{i}$ at that point, it is evident that:
$$
p_{0}=p_{i}+\rho g z
$$
Next, consider the forces perpendicular to the liquid surface. From the Laplace's additional pressure formula:
$$
p_{i}+\frac{\sigma}{R}=p_{0}
$$
$R$ is the radius of curvature of the liquid surface equation at that point. By combining equations (5) and (6), we obtain:
$$
z=z(x)=\cfrac{\sigma}{\rho g R}=\cfrac{\sigma z^{\"v}(x)}{\rho g(1+z^{\prime}(x)^{2})^{\frac{3}{2}}}
$$
Note that here, due to the concave shape of the liquid surface, $z"(x)$ is positive.
Integrate equation (7):
$$
\int_{h}^{0}z\mathrm{d}z=\frac{\sigma}{\rho g}\int_{h}^{0}\frac{z^{\prime}(x)}{\left(1+z^{\prime}(x)^{2}\right)^{\frac{3}{2}}}\mathrm{d}z=\frac{\sigma}{\rho g}\int_{z^{\prime}(0)}^{z^{\prime}(\infty)}\frac{\mathrm{d}z^{\prime}(x)}{\left(1+z^{\prime}(x)^{2}\right)^{\frac{3}{2}}}=\frac{\sigma}{\rho g}\int_{\frac{1}{\sqrt{1+z\left(\left(x\right)^{2}\right)}}}^{\frac{1}{\sqrt{1+z\left(0\right)^{2}}}}\mathrm{d}\left(\frac{1}{\sqrt{1+z^{\prime}(x)^{2}}}\right)
$$
Here, the boundary conditions are specifically pointed out: $z(0)=h,z(\infty)=0,z^{\prime}(\infty)=0$ (the liquid surface approaches horizontal infinitely far away). Thus, $h$ can be expressed as:
$$
\frac{1}{2}h^{2}=\frac{\sigma}{\rho g}[1-\frac{1}{\sqrt{1+z^{\prime}(0)^{2}}}]
$$
The contact angle $\theta$ satisfies:
$$
z^{\prime}(0)=\tan(\frac{\pi}{2}-\theta)=-\cot\theta
$$
Substitute to calculate, and the final result is:
$$
h={\sqrt{\frac{2\sigma}{\rho g}(1-\sin\theta)}}
$$ | $$\sqrt{\frac{2\sigma}{\rho g}(1-\sin \theta)}$$ |
95 | MECHANICS | At the North Pole, the gravitational acceleration is considered a constant vector of magnitude $g$, and the Earth's rotation angular velocity $\Omega$ is along the $z$-axis pointing upward (the $z$-axis direction is vertically upward). A mass $M$ carriage has a mass $m=0.1M$ particle suspended from its ceiling by a light string of length $l$. It is known that $\frac{g}{l} = \omega_{0}^{2} = 10\Omega^{2}$. Initially, the particle is at rest, hanging vertically. The carriage slides frictionlessly along a track in the $x$-direction. Suddenly, the carriage is acted upon by a constant forward force $F$ in the $x$-direction. Determine the expression for the $x$-coordinate of the particle relative to the carriage's reference frame as a function of time $t$. Use the small-angle approximation. At $t=0$, $x=0$. Express the answer in terms of $M$ and $\Omega$. Consider the effects of the Coriolis force caused by the Earth's rotation in this scenario. | The two components of the force exerted on the small ball can be approximated as:
$$
(F_{x},F_{y})=\left(-\frac{m g}{l}x,-\frac{m g}{l}y\right)
$$
Assume the coordinate of the carriage is $x$. Then the Coriolis force acting on the small ball is:
$$
(C_{x},C_{y})=\Big(2m\Omega\dot{y},-2m\Omega(\dot{X}+\dot{x})\Big)
$$
Newton's laws:
$$
F_{\pi}+C_{x}=m\left(\ddot{X}+\ddot{x}\right)
$$
$$
F_{y}+C_{y}=m{\ddot{y}}
$$
$$
F-F_{x}=M\ddot{X}
$$
Rearranging:
$$
\ddot{X}-\frac{m g}{M l}x=\frac{F}{M}
$$
$$
\ddot{X}+\ddot{x}+\frac{g}{l}x-2\Omega\dot{y}=0
$$
$$
\ddot{y}+\frac{g}{l}y+2\Omega(\dot{X}+\dot{x})=0
$$
We find the following particular solutions $\scriptstyle X={\frac{1}{2}}a t^{2}$ ${\pmb y}=b t$ ${\pmb x}={\pmb c}$
$$
\begin{array}{c}{{a-\displaystyle\frac{m}{M}\omega_{0}^{2}c=\displaystyle\frac{F}{M}}}\ {{}}\ {{a+\omega_{0}^{2}c-2\Omega b=0}}\ {{}}\ {{\omega_{0}^{2}b+2\Omega a=0}}\end{array}
$$
From this, we obtain:
$$
X_{0}=\frac{25F}{57M}t^{2}\quad,\quad y_{0}=-\frac{10F}{57M\Omega}t\quad,\quad x_{0}=-\frac{7F}{57M\Omega^{2}}
$$
The final solution should be $X=X_{0}+X_{1}$, ${\mathfrak{y}}={\mathfrak{y}}_{0}+{\mathfrak{y}}_{1}$ ${\boldsymbol{x}}={\boldsymbol{x}}_{0}+{\boldsymbol{x}}_{1}$, where $X_{1}, y_{1}, x_{1}$ satisfy the corresponding homogeneous equations. Let:
$$
X_{1}, x_{1}, y_{1}=A, B, C\mathrm{e}^{i\omega t}
$$
Substituting these into the equations results in the matrix equation:
$$
\left[\begin{array}{c c c}{{-\omega^{2}}}&{{-m\omega_{0}^{2}/M}}&{{0}}\ {{-\omega^{2}}}&{{-\omega^{2}+\omega_{0}^{2}}}&{{-2\mathrm{i}\Omega\omega}}\ {{2\mathrm{i}\Omega\omega}}&{{2\mathrm{i}\Omega\omega}}&{{-\omega^{2}+\omega_{0}^{2}}}\end{array}\right]\left[\begin{array}{c}{{A}}\ {{B}}\ {{C}}\end{array}\right]=\left[\begin{array}{l}{{0}}\ {{0}}\ {{0}}\ {{0}}\end{array}\right]
$$
Taking the determinant of the coefficient matrix equal to zero allows it to be factored:
$$
\omega^{2}\left(\omega^{2}-6\Omega^{2}\right)\left(\omega^{2}-19\Omega^{2}\right)=0
$$
The case $\omega^{2}=0$ corresponds to the trivial solution:
$$
X_{1}, x_{1}, y_{1}=C_{1}, 0, 0
$$
The solution for $\omega=\sqrt{6}\Omega$ is:
$$
X_{1}, x_{1}, y_{1}=C_{2}, -6C_{2}, \frac{5\sqrt{6}}{2}\mathrm{i}C_{2}\mathrm{e}^{\mathrm{i}\omega t}
$$
The solution for $\omega=19\Omega$ is:
$$
X_{1}, x_{1}, y_{1}=C_{3}, -19C_{3}, -4\sqrt{19}\mathrm{i}C_{3}\mathrm{e}^{\mathrm{i}\omega t}
$$
By superimposing these three solutions and considering the initial conditions:
$$
C_{1}+C_{2}+C_{3}=0
$$
$$
-6C_{2}-19C_{3}+c=0
$$
$$
-\frac{5\sqrt{6}}{2}\cdot\sqrt{6}\Omega\cdot C_{2}+4\sqrt{19}\cdot\sqrt{19}\Omega\cdot C_{3}+b=0
$$
We obtain:
$$
C_{1}=\frac{59F}{3249M\Omega^{2}}\quad,\quad C_{2}=-\frac{2F}{117M\Omega^{2}}\quad,\quad C_{3}=-\frac{5F}{4693M\Omega^{2}}
$$
Finally, substituting these into the equations and taking the real part:
$$
\boxed{x=-\frac{7F}{57M\Omega^{2}}+\frac{4F}{39M\Omega^{2}}\cos{\sqrt{6}}\Omega t+\frac{5F}{247M\Omega^{2}}\cos{\sqrt{19}}\Omega t}
$$ | $$
\boxed{x=-\frac{7F}{57M\Omega^2}+\frac{4F}{39M\Omega^2}\cos(\sqrt{6}\Omega t)+\frac{5F}{247M\Omega^2}\cos(\sqrt{19}\Omega t)}
$$ |
101 | MECHANICS | The problem discusses a stick figure model. The stick figure's head is a uniform sphere with a radius of $r$ and a mass of $m$, while the rest of the body consists of uniform rods with negligible thickness and a mass per unit length of $\lambda$. All parts are connected by hinges. Specifically, the torso and both arms have a length of $l$, while the legs have a length of $1.2l$. $\theta$ represents the angle between the arms and the torso, and $\varphi$ represents the angle between the legs and the extended torso line (not an obtuse angle). The stick figure is composed of a head, a torso, two arms, and two legs. Initially, the stick figure is in a state where $\theta=\textstyle{\frac{\pi}{2}}$ and $\varphi={\frac{\pi}{4}}$. The head has a mass of $m=0.6\lambda l$, with the result expressed in terms of $\lambda$.
The stick figure is a simulation of a real biological body, so if necessary, the connections between the torso and the arms/legs, in addition to providing interaction forces, can also exert a couple (i.e., a torque), even if there is no interaction force between them. This net torque is independent of the choice of reference point. In the following discussion, this will be referred to simply as a "couple."
During a certain free-fall motion, the stick figure, overwhelmed with panic, forgets to bring its legs together and swings its arms and torso with an amplitude of $\theta_{0}$ (small angle) and an angular frequency of $\omega$. As a result, its legs passively oscillate near $\varphi={\frac{\pi}{4}}$. Find the amplitude of resonance for the legs (seek the steady-state solution starting from the initial position). Assume that the stick figure can actively generate a couple for the arms but not for the legs. |
(1) The forces are already marked on the diagram. List Newton's laws and rotational equations, assuming the displacement of the driven object is $\pmb{x}$, positive upwards.
Vertical direction of the leg
$$
T_{3}=1.2\lambda l\times\left(0.6l\times{\frac{d^{2}}{d t^{2}}}\left({\mathrm{cos}}\varphi\right)-{\ddot{x}}\right)
$$
Vertical direction of the arm
$$
T_{1}=\lambda l\left(\ddot{x}-0.5l\times\frac{d^{2}}{d t^{2}}\left(\mathrm{cos}\theta\right)\right)
$$
Vertical direction of the torso (head)
$$
2\left(T_{3}-T_{1}\right)=(m+\lambda l)\ddot{x}
$$
Horizontal direction of the leg
$$
-T_{4}=1.2\lambda l\times0.6l\times\frac{d^{2}}{d t^{2}}\left(\mathrm{sin}\varphi\right)
$$
Horizontal direction of the arm
$$
-T_{2}=\lambda l\times0.5l\times{\frac{d^{2}}{d t^{2}}}\left(\mathrm{sin}\theta\right)
$$
Angular momentum theorem for the leg
$$
T_{3}\times0.6l\mathrm{sin}\varphi+T_{4}\times0.6l\mathrm{cos}\varphi-M=\frac{1}{12}\times1.2\lambda l\times(1.2l)^{2}\ddot{\varphi}
$$
Angular momentum theorem for the arm
$$
T_{2}\times0.5l\mathrm{cos}\theta-T_{1}\times0.5l\mathrm{sin}\theta=\frac{1}{12}\times\lambda l\times l^{2}\ddot{\theta}
$$
Substitute in, we have
$$
\frac{1}{12}l^{2}{\ddot{\theta}}+0.25l^{2}{\mathrm{cos}}\theta\frac{d^{2}}{d t^{2}}{\mathrm{sin}}\theta+\frac{T_{1}}{\lambda}\times0.5{\mathrm{sin}}\theta=0
$$
By combining the previous equations, we get
$$
T_{1}=\left({\frac{6}{25}}{\frac{d^{2}}{d t^{2}}}\left(\mathrm{cos}\varphi\right)-{\frac{1}{3}}{\frac{d^{2}}{d t^{2}}}\left(\mathrm{cos}\theta\right)\right)\lambda l^{2}
$$
By substituting, we obtain the final expression
$$
\left(\frac{1}{3}-\frac{1}{12}\mathrm{sin}^{2}\theta\right)\ddot{\theta}-\frac{1}{12}\mathrm{sin}\theta\mathrm{cos}\theta\dot{\theta}^{2}=\frac{3}{25}\omega^{2}\mathrm{sin}\theta\mathrm{cos}\varphi
$$
Therefore,
$$
\begin{array}{c}{{A(\theta)=\displaystyle\frac{1}{3}-\frac{1}{12}\mathrm{sin}^{2}\theta}}\ {{{}}}\ {{B(\theta)=-\displaystyle\frac{1}{12}\mathrm{sin}\theta\mathrm{cos}\theta}}\ {{{}}}\ {{C(\theta)=\displaystyle\frac{3}{25}\omega^{2}\mathrm{sin}\theta}}\end{array}
$$
Substituting back into the previous angular momentum theorem, we have
$$
\begin{array}{r l r}{M=\frac{9}{25}\mathrm{sin}\varphi\lambda l^{3}\left(\frac{12}{25}\mathrm{cos}\varphi\omega^{2}+\frac{1}{3}\mathrm{cos}\theta\dot{\theta}^{2}+\frac{1}{3}\mathrm{sin}\theta\ddot{\theta}\right)}&{{}=\frac{12}{25}\frac{\mathrm{sin}\varphi}{\mathrm{sin}\theta}\ddot{\theta}}\end{array}
$$
There still remain the first five equations from the first question and the modified angular momentum theorem
$$
T_{3}\times0.6l\mathrm{sin}\varphi+T_{4}\times0.6l\mathrm{cos}\varphi=\frac{1}{12}\times1.2\lambda l\times(1.2l)^{2}\ddot{\varphi}
$$
Substitute and combine, we get
$$
T_{3}\times0.6l\mathrm{sin}\varphi-1.2\times0.6\lambda l^{2}{\frac{d^{2}}{d t^{2}}}(\mathrm{sin}\varphi)\times0.6l\mathrm{cos}\varphi={\frac{1}{12}}\times1.2\lambda l\times(1.2l)^{2}\ddot{\varphi}
$$
$$
T_{3}=\lambda l\left(\ddot{x}-0.5l\frac{d^{2}}{d t^{2}}\left(\mathrm{cos}\theta\right)\right)+\frac{1}{2}(m+\lambda l)\ddot{x}
$$
Combining, we get
$$
T_{3}=-\frac{1}{5}\lambda l^{2}\frac{d^{2}}{d t^{2}}\left(\mathrm{cos}\theta\right)+\frac{54}{125}\lambda l^{2}\frac{d^{2}}{d t^{2}}\left(\mathrm{cos}\varphi\right)
$$
Substituting the above equation, we obtain the final expression
$$
\left({\frac{54}{125}}{\frac{d^{2}}{d t^{2}}}\left(\mathrm{cos}\varphi\right)-{\frac{1}{5}}{\frac{d^{2}}{d t^{2}}}\left(\mathrm{cos}\theta\right)\right)\mathrm{sin}\varphi-{\frac{18}{25}}{\frac{d^{2}}{d t^{2}}}\left(\mathrm{sin}\varphi\right)\mathrm{cos}\varphi={\frac{6}{25}}{\ddot{\varphi}}
$$
Let = +, then retain terms up to the first order of and θ, we get
$$
\ddot{\delta}=-\frac{25\sqrt{2}}{204}\omega^{2}\theta_{0}\mathrm{cos}\omega t
$$
That is
$$
A={\frac{25{\sqrt{2}}}{204}}\theta_{0}
$$ | $$
A = \frac{25 \sqrt{2}}{204} \theta_0
$$ |
630 | THERMODYNAMICS | In this problem, we study a simple \"gas-fueled rocket.\" The main body of the rocket is a plastic bottle, which can take off after adding a certain amount of fuel gas and igniting it. It is known that the external atmospheric pressure is $P_{0}$, and the initial pressure of the gas in the bottle is $P_{0}$, with a temperature of $T_{0}$. The proportion of the molar amount of fuel gas is $\alpha$, and the rest is air. Alcohol is chosen as the fuel, with the reaction equation in air as: $C_{2}H_{5}OH(g) + 3O_{2}(g) \rightarrow 3H_{2}O(g) + 2CO_{2}(g)$. Assume that the air contains only $N_{2}$ and $O_{2}$, all gases are considered as ideal gases, and none of the vibrational degrees of freedom are excited. It is known that under conditions of pressure $P_{0}$ and temperature $T_{0}$, the reaction heat is $-\lambda$ (enthalpy change for $1\mathrm{mol}$ of ethanol reacting under isothermal and isobaric conditions). Ignore heat loss, assume the reaction is complete, and that there is air remaining, with the gas volume remaining constant throughout. Find the pressure $P_{1}$ of the gas inside the bottle after the gas has completely reacted upon ignition. | Let the initial moles of air be $n_{0}$, and the moles of alcohol be $$ n_{x}=\frac{\alpha n_{0}}{1-\alpha} $$ The ideal gas equation of state is $$ {P_{0}V_{0}=\left(n_{x}+n_{0}\right)R T_{0}} \\ {P_{0}V_{1}=\left(2n_{x}+n_{0}\right)R T_{0}} $$ According to the definition of enthalpy change, we have $$ \Delta{{H}_{0}}=\Delta{{U}_{0}}+\Delta\left(P V\right)=\Delta{{U}_{0}}+{{P}_{0}}\left({{V}_{1}}-{{V}_{0}}\right) $$ It follows that $$ \Delta U_{0}=-n_{x}\lambda-n_{x}RT_{0} $$ Since heat loss is neglected and no gas escapes from the bottle, the reaction is adiabatic and the volume does not change, giving us $$ \Delta U=\Delta U_{\mathrm{0}}+\Big[\frac{5}{2}R\cdot\big(n_{0}-n_{x}\big)+3R\cdot 3n_{x}\Big]\big(T_{1}-T_{0}\big)=0 $$ We obtain $$ T_{1}=T_{0}+\frac{2n_{x}\left(\lambda+R T_{0}\right)}{R\left(5n_{0}+13n_{x}\right)}=T_{0}+\frac{2\alpha\left(\lambda+R T_{0}\right)}{R\left(5+8\alpha\right)} $$ The ideal gas equation of state is $$ {P_{0}V_{0}=\left(n_{x}+n_{0}\right)R T_{0}} \\ {P_{1}V_{0}=\left(2n_{x}+n_{0}\right)R T_{1}} $$ We find $$ P_{1}=P_{0}\frac{(1+\alpha)(5+10\alpha+2\alpha(\frac{\lambda}{R T_{0}}))}{5+8\alpha} $$ | $$P_0 \frac{(1+\alpha)(5+10\alpha+2\alpha \frac{\lambda}{R T_0})}{5+8\alpha}$$ |
134 | MECHANICS | A fox is escaping along a straight line $AB$ at a constant speed $v_{1}$. A hound is pursuing it at a constant speed $v_{2}$, always aimed at the fox. At a certain moment, the fox is at $F$ and the hound is at $D$. $FD \bot AB$ and $FD = L$. Find the magnitude of the hound's acceleration at this moment. | To solve $\textcircled{1}$, the hunting dog performs uniform circular motion. Thus, the tangential acceleration is zero, while the normal acceleration changes. Assume that during a very short period of time $\mathrm{d}t$ after the given moment, the curvature radius of the trajectory of the hunting dog is $\rho,$ then the normal acceleration is
$$
a_{n}={\frac{v_{2}^{2}}{\rho}}.
$$
During the time interval $\mathrm{d}t$, the fox and the hunting dog respectively reach positions $F^{\prime}$ and $D^{\prime}$ (see Figure 1.7(a) in the problem). The angle turned by the hunting dog's direction of motion is
$$
\alpha={\frac{\widehat{D D^{\prime}}}{\rho}}={\frac{v_{2}\mathrm{d}t}{\rho}}.
$$
However, the distance traveled by the fox is
$$
\begin{array}{r}{\overline{{F F^{\prime}}}=v_{1}\mathrm{d}t\approx L\tan\alpha\approx\alpha L,}\end{array}
$$
which implies
$$
\alpha={\frac{v_{1}\mathrm{d}t}{L}}.
$$
Thus,
$$
\frac{\upsilon_{2}\mathrm{d}t}{\rho}=\frac{\upsilon_{1}\mathrm{d}t}{L},
$$
and from this, the curvature radius can be obtained as
$$
\rho=\frac{L v_{2}}{v_{1}}.
$$
Therefore, the magnitude of the acceleration is
$$
a_{n}={\frac{v_{2}^{2}}{\rho}}={\frac{v_{1}v_{2}}{L}}.
$$
It should be noted that the acceleration obtained this way corresponds only to the specific moment mentioned in the problem, not to the acceleration at any arbitrary position during the chasing process. | $$\frac{v_1 v_2}{L}$$ |
650 | MECHANICS | Two parallel light strings, each with length $L$, are horizontally separated by a distance $d$. Their upper ends are connected to the ceiling, and the lower ends are symmetrically attached to a uniformly distributed smooth semicircular arc. The radius of the semicircle is $R$, and its mass is $m$. The gravitational acceleration is $g$. A small ball with the same mass $m$ is placed at the bottom of the semicircle. The entire system undergoes small oscillations in the plane in some manner, with a total energy of $E_{0}$. The potential energy zero point is chosen such that the energy of the system is zero when it is vertically stable. Find the maximum $d$, denoted as $d_{max}$, that allows the system to have zero tension in one of the strings at some moment.
| Let the angle between the rope and the vertical be $\theta$, and the angle between the line connecting the small ball and the center of the circle and the vertical be $\varphi$. First, consider the kinetic equations. The kinetic energy $$ T={\frac{1}{2}}m L^{2}{\dot{\theta}}^{2}+{\frac{1}{2}}m(L{\dot{\theta}}+R{\dot{\varphi}})^{2} $$ The potential energy $$ V=m g L\theta^{2}+{\frac{1}{2}}m g R\varphi^{2} $$ Thus, the equations of motion are $$ \begin{array}{c}{{2m L^{2}\ddot{\theta}+m L R\ddot{\varphi}+2m g L\theta=0}}{{\phantom{-}}}{{m R^{2}\ddot{\varphi}+m L R\ddot{\theta}+m g R\varphi=0}}\end{array} $$ Consider the torque at the midpoint of the endpoints of the two ropes on the ceiling $$ \tau=2m g L\theta+m g R\varphi+m L^{2}\ddot{\theta}+m L(L-\left(1-\frac{2}{\pi}\right)R)\ddot{\theta}+m L R\ddot{\varphi} $$ Substituting the equations of motion, simplifying gives $$ \tau=2m g\left(1-\frac{2}{\pi}\right)R\theta+\frac{2}{\pi}m g R\varphi $$ To require that one rope has no tension, then $\tau=m g d$. Keeping $E_{0}$ constant and maximizing $d$ is equivalent to minimizing $E_{0}$ under the condition. Obviously, at this time one should have $$ \dot{\theta}=\dot{\varphi}=0 $$ because the velocity terms do not contribute to the torque and only increase the energy. Therefore, at the critical point $$ E_{0}=m g L\theta^{2}+{\frac{1}{2}}m g R\varphi^{2} $$ According to the Cauchy inequality $$ \theta^{2}+\frac{1}{2}m g R\varphi^{2}\Bigg)\left(\left(\frac{2m g\left(1-\frac{2}{\pi}\right)R}{\sqrt{m g L}}\right)^{2}+\left(\frac{\frac{2}{\pi}m g R}{\sqrt{\frac{1}{2}m g R}}\right)^{2}\right)\geq\left(2m g\left(1-\frac{2}{\pi}\right)R\theta+\frac{2}{\pi}m g R\varphi\right)^{2} $$ Substituting the torque, the energy has: $$ d_{\operatorname*{max}}=\frac{2}{\pi}\sqrt{\frac{E_{0}R}{m g}}\sqrt{(\pi-2)^{2}\frac{R}{L}+2} $$
| $$\frac{2}{\pi}\sqrt{\frac{E_0 R}{m g}}\sqrt{(\pi-2)^2\frac{R}{L}+2}$$ |
332 | MECHANICS | There is a uniform thin spherical shell, with its center fixed on a horizontal axis and able to rotate freely around this axis. The spherical shell has a mass of $M$ and a radius of $R$. There is also a uniform thin rod, with one end smoothly hinged to the axis at a distance $d$ from the center of the sphere, and the other end resting on the spherical shell. The thin rod has a mass of $m$ and a length of $L$ ($d-R<L<\sqrt{d^{2}-R^{2}}$). Initially, the point of contact and the line connecting the sphere's center lie in the same vertical plane as the rod. Now, a small perturbation is applied, causing the system to begin rotating as a whole without relative sliding around the horizontal axis. When the entire system rotates by an angle $\varphi$, the spherical shell and the rod just begin to experience relative sliding. Determine the static friction coefficient $\mu$ between the spherical shell and the rod, expressed as a function of $\varphi$. | Introducing simplified parameters $$ \begin{array}{l}{\displaystyle{\beta=\cos^{-1}\frac{L^{2}+d^{2}-R^{2}}{2L d}}}{\displaystyle{\gamma=\cos^{-1}\frac{R^{2}+d^{2}-L^{2}}{2R d}}}\end{array} $$ The coordinates of the contact point \( B \) are written as $$ \overrightarrow{O B}=(\cos\beta,\sin\beta\sin\varphi,\sin\beta\cos\varphi)L $$ Thus, the rod direction vector is written as $$ \hat{\boldsymbol{l}}=(\cos\beta,\sin\beta\sin\varphi,\sin\beta\cos\varphi) $$ Support force direction vector Assume the direction vector of the friction force is $$ \pmb{\hat{n}}=(-\cos\gamma,\sin\gamma\sin\varphi,\sin\gamma\cos\varphi) $$ $$ {\hat{f}}=\cos\theta{\frac{{\hat{l}}\times{\widehat{\pmb{n}}}}{\left|{\hat{l}}\times{\widehat{\pmb{n}}}\right|}}+\sin\theta{\frac{{\widehat{\pmb{n}}}\times({\hat{l}}\times{\widehat{\pmb{n}}})}{\left|{\widehat{\pmb{n}}}\times({\hat{l}}\times{\widehat{\pmb{n}}})\right|}} $$ That is $$ \hat{\boldsymbol f}=\cos\theta\left(0,-\cos\varphi,\sin\varphi\right)+\sin\theta\left(\sin\gamma,\cos\gamma\sin\varphi,\cos\gamma\cos\varphi\right)\qquad\mathrm{(}||\boldsymbol\varphi||\overset{\leq}{\sin\theta}|\leq\sin\gamma\sin\varphi\mathrm{(}||\boldsymbol\varphi||,\cos\varphi| $$ For the spherical shell, by the angular momentum theorem $$ {\frac{2}{3}}M R^{2}{\frac{d^{2}\varphi}{d t^{2}}}=R\cos\theta\sin\gamma f $$ For the rod, by the angular momentum theorem $$ \left\{\begin{array}{c}{{\frac{1}{3}m L^{2}\sin\beta^{2}\frac{d^{2}\varphi}{d t^{2}}=\frac{1}{2}m g L\sin\beta\sin\varphi-L\cos\theta\sin\beta f\qquad\quad\bigtriangledown}}{{\frac{1}{3}m L^{2}\sin\beta\cos\beta(\frac{d\varphi}{d t})^{2}=\frac{1}{2}m g L\cos\beta\cos\varphi-L\sin(\beta+\gamma)N-L\sin\theta\cos(\beta+\gamma)f\qquad\quad\bigtriangledown}}\end{array}\right. $$ From Equation $\textcircled{1}$ It follows that $$ {\begin{array}{r l r}{f\cos\theta={\frac{M m g R^{2}\sin\varphi}{2M R^{2}+m L^{2}\sin\beta^{2}}}}&{\quad{\textcircled{14}}}{{\frac{d^{2}\varphi}{d t^{2}}}={\frac{3m g L\sin\beta\sin\varphi}{4M R^{2}+2m L^{2}\sin\beta^{2}}}}&{\quad{\textcircled{15}}}\end{array}} $$ From Equation $\textcircled{1}$ $$ (\frac{d\varphi}{d t})^{2}=\frac{3m g L\sin\beta(1-\cos\varphi)}{2M R^{2}+m L^{2}\sin\beta^{2}} $$ $$ N=\frac{m g\left[M R^{2}(\cos\beta\cos\varphi-\tan\theta\sin\varphi\cos(\beta+\gamma))+m L^{2}\sin\beta^{2}\cos\beta(\frac{3}{2}\cos\varphi-1)\right]}{\sin(\beta+\gamma)(2M R^{2}+m L^{2}\sin\beta^{2})} $$ Thus $$ \mu=\frac{f}{N}=\frac{M R^{2}\sin\varphi\sin(\beta+\gamma)}{\cos\theta\left[M R^{2}(\cos\beta\cos\varphi-\tan\theta\sin\varphi\cos(\beta+\gamma))+m L^{2}\sin\beta^{2}\cos\beta(\frac{3}{2}\cos\varphi-1)\right]} $$ To happen relative slipping, the above expression should take the minimum value $$ \mu={\frac{M R^{2}\sin\varphi\sin(\beta+\gamma)}{\sqrt{\cos\beta^{2}\left[M R^{2}\cos\varphi+m L^{2}\sin\beta^{2}({\frac{3}{2}}\cos\varphi-1)\right]^{2}+[M R^{2}\sin\varphi\cos(\beta+\gamma)]^{2}}}} $$ | $$
\mu = \frac{MR^2 \sin\varphi \sin\left(\arccos\frac{L^2 + d^2 - R^2}{2L d} + \arccos\frac{R^2 + d^2 - L^2}{2R d}\right)}{\sqrt{\left(\frac{L^2 + d^2 - R^2}{2 L d}\right)^2 \left[ M R^2 \cos\varphi + m L^2 \sin^2\left(\arccos\frac{L^2 + d^2 - R^2}{2L d}\right) \left(\frac{3}{2} \cos\varphi - 1\right)\right]^2 + \left[ M R^2 \sin\varphi \cos\left(\arccos\frac{L^2 + d^2 - R^2}{2L d} + \arccos\frac{R^2 + d^2 - L^2}{2R d}\right)\right]^2 }}
$$ |
250 | ELECTRICITY |
There is now an electrolyte with thickness $L$ in the $z$ direction, infinite in the $x$ direction, and infinite in the $y$ direction. The region where $y > 0$ is electrolyte 1, and the region where $y < 0$ is electrolyte 2. The conductivities of the two dielectrics are $\sigma_{1}, \sigma_{2}$, and the dielectric constants are $\varepsilon_{1}, \varepsilon_{2}$, respectively. On the $xOz$ interface of the two dielectrics, two cylindrical holes with a radius $R$ are drilled in the $z$ direction, spaced $2D (D > R, R, D \ll L)$ apart, with centers located on the interface as long straight cylindrical holes. Two cylindrical bodies $\pm$ are inserted into the holes, with the type of the cylinders given by the problem text below.
The cylindrical bodies $\pm$ are metal electrodes filling the entire cylinder. Initially, the system is uncharged, and at $t=0$, a power source with an electromotive force $U$ and internal resistance $r_{0}$ is used to connect the electrodes. Find the relationship between the current through the power source and time, denoted as $i(t)$. | Given the potential difference $u$, it can be seen:
$$
\varphi_{+}=u/2,\varphi_{-}=-u/2,\lambda=\frac{2\pi(\varepsilon_{1}+\varepsilon_{2})\varphi_{+}}{2\xi_{+}}=\frac{\pi(\varepsilon_{1}+\varepsilon_{2})u}{\operatorname{arccosh}(\mathrm{D/R})}
$$
Select a surface encapsulating the cylindrical surface and examine Gauss's theorem. For the positive electrode, it is easy to see:
$$
\iint\vec{E}\cdot d\vec{S}=L\oint\vec{E}\cdot\hat{n}d l=\frac{\lambda L}{(\varepsilon_{1}+\varepsilon_{2})/2}=\frac{2\pi u L}{2\mathrm{arccosh}(D/R)}
$$
Since the above potential distribution is deemed directly applicable for the calculation of current, the total current flowing out of the positive electrode is:
$$
I=\iint\sigma\vec{E}\cdot d\vec{S}=\frac{\sigma_{1}+\sigma_{2}}{2}\times\frac{2\pi u L}{2\mathrm{arccosh}(D/R)}
$$
Given the current $i$ passing through the power source, this current changes the net charge:
$$
{\frac{d(\lambda L)}{d t}}=i-I=i-{\frac{2\pi u L}{2\mathrm{arccosh}(D/R)}}={\frac{\pi(\varepsilon_{1}+\varepsilon_{2})L}{2\mathrm{arccosh}(D/R)}}{\frac{d u}{d t}}
$$
According to the loop voltage drop equation:
$$
\begin{array}{l l}{{U=r_{0}i+u\rightarrow u=U-r_{0}i}}\ {{\rightarrow i-{\displaystyle\frac{\pi(\sigma_{1}+\sigma_{2})L}{2\mathrm{arccosh}(D/R)}}}(U-r_{0}i)=-{\displaystyle\frac{\pi(\varepsilon_{1}+\varepsilon_{2})L}{2\mathrm{arccosh}(D/R)}}r_{0}{\displaystyle\frac{d i}{d t}}}\ {{\rightarrow{\displaystyle\frac{d i}{d t}}={\displaystyle\frac{(\sigma_{1}+\sigma_{2})U}{r_{0}(\varepsilon_{1}+\varepsilon_{2})}}-\left(\frac{\sigma_{1}+\sigma_{2}}{\varepsilon_{1}+\varepsilon_{2}}+{\displaystyle\frac{2\mathrm{arccosh}(D/R)}{\pi r_{0}L(\varepsilon_{1}+\varepsilon_{2})}}\right)i}}\end{array}
$$
At time $t=0$, all current should preferentially enter the capacitor. At this time, the initial current is $U/r_{0}$, and this differential equation yields:
$$
i(t)=\frac{U}{r_{0}\left(1+\frac{2\mathrm{arccosh}(D/R)}{\pi r_{0}L(\sigma_{1}+\sigma_{2})}\right)}\left\{2+\frac{2\mathrm{arccosh}(D/R)}{\pi r_{0}L(\sigma_{1}+\sigma_{2})}-\exp{\left[-\left(\frac{\sigma_{1}+\sigma_{2}}{\varepsilon_{1}+\varepsilon_{2}}+\frac{2\mathrm{arccosh}(D/R)}{\pi r_{0}L(\varepsilon_{1}+\sigma_{2})}\right)\right]}\right\},
$$ | $$
i(t)=\frac{U}{r_0\left(1+\frac{2\arccosh(D/R)}{\pi r_0 L(\sigma_1+\sigma_2)}\right)}\left(2+\frac{2\arccosh(D/R)}{\pi r_0 L(\sigma_1+\sigma_2)}-\exp\left[-\left(\frac{\sigma_1+\sigma_2}{\varepsilon_1+\varepsilon_2}+\frac{2\arccosh(D/R)}{\pi r_0 L(\varepsilon_1+\varepsilon_2)}\right)t\right]\right)
$$ |
409 | ELECTRICITY | In electromagnetism, we often study the problem of electromagnetic field distribution in a region without charge or current distribution. In such cases, the electromagnetic field will become a tubular field. The so-called tubular field is named because of the nature of the velocity field of an incompressible fluid at every instant. Its characteristic can be described using the language of vector analysis as the field's divergence being equal to zero. Another perspective is that if we take the field lines of a vector field $\pmb{F}$ (the tangent line at each point being the direction of field strength) and select a flux tube along a set of field lines passing through a certain cross-section, then at any cross-section of the flux tube, the field flux:
$$
\int F\cdot\mathrm{d}S=\Phi
$$
will be a conserved quantity. This problem will investigate a special case of a tubular field that is rotationally symmetric around the $z$ axis: using oblate spheroidal coordinates to construct a unique tubular field as an electric field or magnetic field. The so-called oblate spheroidal coordinate system is similar to spherical coordinates, and it is generated by the following coordinate transformations:
$$
x= a \mathrm{ch}\mu \cos \nu \cos \varphi
$$
$$
y= a \mathrm{ch}\mu \cos \nu \sin \varphi
$$
$$
z=a\mathrm{sh}\mu\sin\nu
$$
We want the shape of the field lines of an electric field or magnetic field to precisely follow its generatrix direction, i.e., the tangent direction of the curve where field strength changes with $\mu$ while $\nu,\varphi$ remain fixed (upwards when $\relax z>0$).
Find the charge distribution in the $xy$ plane that can produce this electric field distribution. Assume the field strength near the origin is $E_0$. The dielectric constant is $\varepsilon_{0}$. | First, we determine the equation of the constant-$\mu$ line. Now, using the trigonometric identity $\sin^{2}\nu+\cos^{2}\nu=1$, we eliminate the parameter $\nu$:
$$
{\frac{x^{2}}{a^{2}{\mathrm{ch}}^{2}\mu}}+{\frac{z^{2}}{a^{2}{\mathrm{sh}}^{2}\mu}}=1
$$
Its physical significance is that, for an electric field, it forms a surface of equal potential. This is based on the fact that equipotential surfaces should always be perpendicular to the electric field lines. The constant-$\mu$ lines, being ellipses, satisfy this condition: their tangent directions are aligned with the external bisectors of angles formed by lines connecting the two foci, and they are perpendicular to the tangent directions of the hyperbolic electric field lines, which correspond to the internal bisectors.
Next, we determine an inverse transformation. For a point in the first quadrant with coordinates $(x, z)$, the distances to the two foci are:
$$
r_{+}=\sqrt{(x+a)^{2}+z^{2}}\quad,\quad r_{-}=\sqrt{(x-a)^{2}+z^{2}}
$$
Thus, the semi-major axis lengths of the ellipse and hyperbola are respectively:
$$
r_{+}+r_{-}=2a\mathrm{ch}\mu
$$
$$
r_{+}-r_{-}=2a\cos\nu
$$
These two equations help convert Cartesian coordinates into generalized coordinates $\mu$ and $\nu$. Next, we utilize the following property: the infinitesimal displacement caused by an increase in $\mu$ between equipotential surfaces is the derivative of the coordinates with respect to $\mu$:
$$
\mathrm{d}x=a\mathrm{sh}\mu\cos\nu\mathrm{d}\mu\quad,\quad\mathrm{d}z=a\mathrm{ch}\mu\sin\nu\mathrm{d}\mu
$$
We proceed to calculate the length of this displacement:
$$
\mathrm{d}s={\sqrt{\mathrm{d}x^{2}+\mathrm{d}z^{2}}}=a{\sqrt{\mathrm{ch}^{2}\mu-\cos^{2}\nu}}\mathrm{d}\mu
$$
Next, we consider the electric potential difference caused by displacements between one ellipse and another. This difference is equal to the electric field strength multiplied by the displacement length. Because the ellipses represent equipotential surfaces, this product is constant everywhere. In particular, taking $\nu=90^{\circ}$ places this point exactly on the $z$-axis, and the electric field strength at this location has already been computed in (2). We obtain the following equation:
$$
{\frac{E_{0}a^{2}}{a^{2}+a^{2}\mathrm{{sh}}^{2}\mu\sin^{2}{90^{\circ}}}}\cdot a\mathrm{{ch}}\mu\mathrm{{d}}\mu=E\cdot a{\sqrt{\mathrm{{ch}}^{2}\mu-\cos^{2}\nu}}\mathrm{{d}}\mu
$$
This gives the distribution of electric field strength. To determine the direction of electric field lines, we only need to locate the bisector of the angle formed by lines connecting the two foci:
$$
\theta={\frac{\theta_{+}+\theta_{-}}{2}}
$$
$$
\tan\theta_{\pm}={\frac{z}{x\pm a}}
$$
We organize the result:
$$
f(x,z)={\frac{2a^{2}}{\left[{\sqrt{(x+a)^{2}+z^{2}}}+{\sqrt{(x-a)^{2}+z^{2}}}\right]^{2}}}\left[\left[{\frac{(x+a)^{2}+z^{2}}{(x-a)^{2}+z^{2}}}\right]^{\frac{1}{4}}+\left[{\frac{(x-a)^{2}+z^{2}}{(x+a)^{2}+z^{2}}}\right]^{\frac{1}{4}}\right]~.
$$
With $z=0$, $x=r$, we obtain:
$$
f(r)=\frac{2a^{2}}{(r+a+|r-a|)\sqrt{|r^{2}-a^{2}|}}
$$
For the electric field, a charged circular disk with radius $a$ is required to generate it. In this case, the field strength at $r<a$ is:
$$
E=E_{0}f(r)={\frac{E_{0}a}{\sqrt{a^{2}-r^{2}}}}
$$
By symmetry and Gauss's law:
$$
E=\frac{\sigma}{2\varepsilon_{0}}
$$
Thus, the surface charge density on the disk is:
$$
\boxed{\sigma=\frac{2\varepsilon_{0}E_{0}a}{\sqrt{a^{2}-r^{2}}}}
$$ | $$
\boxed{\sigma=\frac{2\varepsilon_{0}E_{0}a}{\sqrt{a^{2}-r^{2}}}}
$$ |
133 | MECHANICS | A particle undergoes planar motion, where the $x$ component of its velocity $v_{x}$ remains constant. The radius of curvature at this moment is $R$. Determine the acceleration at this moment. | To solve for the radius of curvature, it's more appropriate to use the natural coordinate system. In this coordinate system, the tangential and normal components of acceleration are represented as:
$$
a_{\tau} = {\frac{\mathrm{d}v}{\mathrm{d}t}}, \quad a_{n} = {\frac{v^{2}}{R}}.
$$
Also,
$$
v_{x} = v \cos\theta,
$$
where $\theta$ can be referenced in Figure 1.5. If we take the time derivative of equation (2) and consider that $v_{x}$ is a constant, we get:
$$
{\frac{\mathrm{d}v}{\mathrm{d}t}} = v {\dot{\theta}} \tan\theta.
$$
Since $\dot{\theta} = \frac{\mathrm{d}\theta}{\mathrm{d}s} \frac{\mathrm{d}s}{\mathrm{d}t} = \frac{v}{R}$, where $s$ is the arc length coordinate, substituting this relation into equation (3) gives:
$$
a_{\tau} = {\frac{\mathrm{d}v}{\mathrm{d}t}} = {\frac{v^{2}}{R}} \tan\theta.
$$
From equations (1), (2), and (4), the magnitude of the particle's acceleration can be expressed as:
$$
a = \sqrt{a_{n}^{2} + a_{\tau}^{2}} = \frac{v^{2}}{R} \frac{1}{\cos\theta} = \frac{v^{3}}{R v_{x}}.
$$ | $$\frac{v^3}{R v_x}$$ |
336 | MECHANICS | A homogeneous solid small elliptical cylinder is placed inside a thin circular cylinder. The semi-major axis of the elliptical cylinder's cross section is $a$, the semi-minor axis is $b$, and its mass is $m$. The inner radius of the circular cylinder is $R$, and its mass is $M$. The central axes of both the circular cylinder and the elliptical cylinder are horizontal. The magnitude of gravitational acceleration is $g$. The circular cylinder is fixed and cannot rotate. Initially, the elliptical cylinder is in equilibrium at the bottom of the circular cylinder with its minor axis aligned vertically. Assuming the elliptical cylinder undergoes rolling without slipping inside the circular cylinder, derive the frequency $f$ of the small oscillations of the elliptical cylinder's center of mass near this equilibrium position. | The radius of curvature at the contact point is
$$
\rho_{1}=\frac{a^{2}}{b}
$$
Thus, in geometric terms, it can be equivalent to a cylinder whose center of mass is offset by $(\rho_{1}-b)$. Let the angle between the contact point and the cylinder center line be $\theta$, and the self-rotation angle of the elliptical cylinder be $\varphi$.
No-slip rolling gives
$$
R\theta=\rho_{1}(\theta+\varphi)
$$
That is
$$
\varphi=({\frac{b R}{a^{2}}}-1)\theta
$$
Thus the potential energy of the system is
$$
E_{p}=m g[(R-\rho_{1})(1-\cos\theta)+(\rho_{1}-b)(1-\cos\varphi)]
$$
Under the small angle approximation
$$
E_{p}={\frac{1}{2}}m g\left[(R-{\frac{a^{2}}{b}})+({\frac{a^{2}}{b}}-b)({\frac{b R}{a^{2}}}-1)^{2}\right]\theta^{2}={\frac{1}{2}}m g(b+R-{\frac{b^{2}R}{a^{2}}})({\frac{b R}{a^{2}}}-1)\theta^{2}
$$
The stability condition for a single degree of freedom conservative system:
$$
\frac{d^{2}E_{p}}{d\theta^{2}}>0
$$
That is
$$
R>{\frac{a^{2}}{b}}
$$
The moment of inertia of the elliptical cylinder about its central axis is
$$
I={\frac{1}{4}}m(a^{2}+b^{2})
$$
Thus, the kinetic energy of the system is
$$
E_{k}=\frac{1}{2}(I+m b^{2})\dot{\varphi}^{2}=\frac{1}{2}\frac{m(a^{2}+5b^{2})}{4}(\frac{b R}{a^{2}}-1)^{2}\dot{\theta}^{2}
$$
Therefore, the frequency of oscillation is
$$
f={\frac{1}{2\pi}}{\sqrt{\frac{4g(b+R-{\frac{b^{2}R}{a^{2}}})}{(a^{2}+5b^{2})({\frac{b R}{a^{2}}}-1)}}}
$$ | $$
f=\frac{1}{2\pi}\sqrt{\frac{4g(b+R-\frac{b^2R}{a^2})}{(a^2+5b^2)\left(\frac{bR}{a^2}-1\right)}}
$$ |
88 | MODERN | In a certain atom $A$, there are only two energy levels: the lower energy level $A_{0}$ is called the ground state, and the higher energy level $A^{\star}$ is called the excited state. The energy difference between the excited state and the ground state is $E_{0}$. When the atom is in the ground state, its rest mass is $m_{0}$; when it is in the excited state, due to its intrinsic instability, it will transition to the ground state while emitting photons externally. It is known that the probability of transition from the excited state to the ground state in unit time in the atom's rest reference frame is $\lambda$, and the probability of emitting photons in all directions is equal. At the initial moment, the total number of atoms is $N_{0}(N_{0}\gg1)$, and the experimental reference frame $S$ and the atom's proper reference frame $S^{\prime}$ are time-synchronized. All $N_{0}$ atoms are in the excited state and have a common velocity $\pmb{v}$ in the $+\hat{\pmb{x}}$ direction relative to the $S$ frame.
Considering $m_{0}c^{2}\gg E_{0}$, at this time the recoil effect on the atom due to photon emission can be ignored. Try to solve the angular distribution of the light emission power $w(\theta)$ from the atom at time $\pmb{t}$ in the laboratory reference frame $S$. (Calculate the emission power per unit solid angle in the direction of angle $\pmb{\theta}$ relative to the $+\hat{\pmb{x}}$ axis, rather than the received power at an infinite distance. The speed of light in vacuum is $c$ (known). The answer should be expressed using $\lambda, N_0, E_0, v, \theta, \pi, t, c$. Please check and output the final answer. | Consider the reference frame $S^{\prime}$:
$$
h\nu_{0}=E_{0}
$$
Let the remaining number of atoms at time $t^{\prime}$ be $N^{\prime}$:
$$
-\mathrm{d}N^{\prime}=\lambda N^{\prime}\mathrm{d}t^{\prime}
$$
Solving for $N^{\prime}$, we get:
$$
N^{\prime}=N_{0}e^{-\lambda t^{\prime}}
$$
The total power emitted by photons:
$$
P^{\prime}=-\dot{N}^{\prime}h\nu_{0}=\lambda N_{0}E_{0}e^{-\lambda t^{\prime}}
$$
Consider the relativistic transformation of energy, momentum, and time between $s^{\prime}$ and $s$:
$$
\mathrm{d}t=\gamma\mathrm{d}t^{\prime}
$$
$$
\mathrm{d}E_{l i g h t}=\gamma(\mathrm{d}E_{l i g h t}^{\prime}+v\mathrm{d}p_{l i g h t}^{\prime})=\gamma\mathrm{d}E_{l i g h t}^{\prime}
$$
Using the relations:
$$
\begin{array}{r}{\mathrm{d}E_{l i g h t}=P\mathrm{d}t}\ {\mathrm{d}E_{l i g h t}^{\prime}=P^{\prime}\mathrm{d}t^{\prime}}\end{array}
$$
We obtain:
$$
P=P^{\prime}=\lambda N_{0}E_{0}e^{-\lambda\frac{\epsilon}{\gamma}}
$$
Where:
$$
\gamma={\frac{1}{\sqrt{1-{\frac{v^{2}}{c^{2}}}}}}
$$
Considering that number density is Lorentz invariant, let the number of photons emitted in $S$, within $\theta \sim \theta + \mathrm{d}\theta$, be $\dot{n}$.
From the transformation of the photon energy and momentum, the relationship between $\pmb\theta^{\prime}$ and $\pmb\theta$ can be expressed as:
$$
\dot{n}\mathrm{d}t\times2\pi\sin\theta\mathrm{d}\theta=\frac{\dot{N}^{\prime}}{4\pi}\mathrm{d}t^{\prime}\times2\pi\sin\theta^{\prime}\mathrm{d}\theta^{\prime}
$$
$$
\cos\theta^{\prime}=\frac{\cos\theta-\beta}{1-\beta\cos\theta}
$$
And the transformation of time:
$$
\mathrm{d}t^{\prime}=\sqrt{1-\beta^{2}}\mathrm{d}t
$$
Substituting into the equations, we get:
$$
\dot{n}=\frac{\dot{N}^{\prime}}{4\pi}\cdot\frac{\mathrm{d}\cos\theta^{\prime}}{\mathrm{d}\cos\theta}\cdot\frac{\mathrm{d}t^{\prime}}{\mathrm{d}t}=\frac{\dot{N}^{\prime}}{4\pi}\cdot\frac{(1-\beta^{2})^{\frac{3}{2}}}{(1-\beta\cos\theta)^{2}}
$$
Considering the frequency transformation of photons:
$$
\nu=\frac{\sqrt{1-\beta^{2}}}{1-\beta\cos\theta}\nu_{0}
$$
We further obtain the power distribution of the emitted light:
$$
w={\frac{\mathrm{d}P}{\mathrm{d}\Omega}}={\dot{n}h\nu}={\frac{P}{4\pi}}{\frac{(1-\beta^{2})^{2}}{(1-\beta\cos\theta)^{3}}}={\frac{\lambda N_{0}E_{0}}{4\pi}}{\frac{(1-\beta^{2})^{2}}{(1-\beta\cos\theta)^{3}}}\cdot e^{-\lambda{\frac{4}{7}}}
$$
Rewriting $\beta$ in terms of velocity:
In the relativistic framework, $\beta = \frac{v}{c}$ (where $v$ is the velocity and $c$ is the speed of light in a vacuum). Replacing $\beta$ and $\lambda$ with velocity-related quantities, the formula becomes:
$$
w=\frac{\mathrm{d}P}{\mathrm{d}\Omega}=\dot{n}h\nu=\frac{P}{4\pi}\frac{(1 - (\frac{v}{c})^{2})^{2}}{(1 - \frac{v}{c}\cos\theta)^{3}}=\frac{\lambda N_{0}E_{0}}{4\pi}\frac{(1 - (\frac{v}{c})^{2})^{2}}{(1 - \frac{v}{c}\cos\theta)^{3}}\cdot e^{-\lambda\frac{4}{7}}
$$ | $$
w=\frac{\lambda N_0 E_0}{4\pi}\frac{\left(1-\left(\frac{v}{c}\right)^2\right)^2}{\left(1-\frac{v}{c}\cos\theta\right)^3}e^{-\frac{4\lambda}{7}}
$$ |
105 | ELECTRICITY | This problem aims to guide everyone to discover a very interesting way to understand electromagnetic fields. It is known that the Maxwell equations in vacuum are
$$
\nabla\cdot{\pmb{E}}=\frac{\rho}{\varepsilon_{0}}
$$
$$
\nabla\times{\pmb{{E}}}=-\frac{\partial{\pmb{{B}}}}{\partial t}
$$
$$
\nabla\cdot\pmb{B}=0
$$
$$
\nabla\times\boldsymbol{B}=\mu_{0}\boldsymbol{j}+\varepsilon_{0}\mu_{0}\frac{\partial\boldsymbol{E}}{\partial t}
$$
To make the equations of electromagnetic fields more symmetrical, we introduce magnetic charge. Magnetic charge $\pmb{g}$ will generate a static magnetic field, and the motion of magnetic charge will generate a static electric field. It takes the following form:
$$
\boldsymbol{B}={\frac{g \boldsymbol e_{r}}{4\pi r^{2}}}
$$
$$
\pmb{{E}}=\frac{\lambda g\pmb{{v}}\times\pmb{\boldsymbol{e}}_{r}}{4\pi c r^{2}}
$$
Find $\lambda$ by conservation of magnetic charge. | $$
\pmb{\varepsilon}\rightarrow\pmb{\varepsilon}
$$
It can be obtained:
$$
{\pmb B}\rightarrow\mu_{0}{\pmb B}c
$$
$$
\nabla\cdot\pmb{\cal E}=\rho
$$
$$
\nabla\times\pmb{\mathcal{E}}=-\frac{1}{c}\frac{\partial\pmb{\mathcal{B}}}{\partial t}
$$
$$
\nabla\cdot\pmb{B}=0
$$
$$
\nabla\times\pmb{B}=\frac{1}{c}\pmb{j}+\frac{1}{c}\frac{\partial\pmb{E}}{\partial t}
$$
Following the position where the charge density term appears, add the divergence term of the magnetic field and the curl term of the electric field, we obtain:
$$
\begin{array}{c}{\boldsymbol{\nabla\cdot E}=\rho}\ {\boldsymbol{\nabla\times E}=\lambda\frac{k}{c}-\frac{1}{c}\frac{\partial\boldsymbol{B}}{\partial t}}\ {\boldsymbol{\nabla\cdot B}=\omega}\ {\boldsymbol{\nabla\times B}=\frac{1}{c}\boldsymbol{j}+\frac{1}{c}\frac{\partial E}{\partial t}}\end{array}
$$
The conservation of magnetic charge implies:
$$
\frac{\partial\omega}{\partial t}+\nabla\cdot\boldsymbol{k}=0
$$
Since the curl of the electric field does not have divergence, the identity is obtained:
$$
\lambda\frac{\nabla\cdot\pmb{k}}{c}-\frac{1}{c}\frac{\partial\nabla\cdot\pmb{B}}{\partial t}=0
$$
Substituting the divergence expression of the magnetic field and comparing it with the conservation of magnetic charge, we find:
$$
\lambda=-1
$$ | $$\lambda = -c$$ |
617 | MECHANICS | Consider each domino as a uniform, smooth slender rod with height $h$ and mass $m$ (neglecting thickness and width), connected to the ground with a smooth hinge. All rods are initially vertical to the ground and arranged evenly in a straight line, with the distance between adjacent rods being $l$, where $l \ll h$. The coefficient of restitution between the rods is $e = 0$. For convenience, assume that forces between the rods only occur at the instant of collision, and are otherwise negligible. Assume there are infinitely many rods, all initially standing upright. At a certain moment, a disturbance is applied to the first rod, causing it to fall with a certain initial angular velocity. The angular velocity of each rod at the instant it is hit by the previous rod may vary, but as the dominoes collectively fall forward, this post-collision angular velocity will tend toward a certain value. Determine this value $\omega_{\infty}$.
| Suppose the initial angular velocity of the $n$th rod when it falls is $\omega _{n}$. From the beginning of the fall until it collides with the next rod, the angle through which the rod rotates is denoted as $\theta$. Since $l\ll h$, we have $$\theta\approx\tan\theta\approx\sin\theta=\frac lh$$ Rigid body dynamics: $$\frac12mgh\theta=\frac13mh^2\ddot{\theta}$$ Thus: $\ddot{\theta}-\rho^2\theta=0$ where $\rho=\sqrt{\frac{3g}{2h}}$ The general solution is: $\theta(t)=C_1\cosh\rho t+C_2\sinh\rho t$ With initial conditions $$\theta(0)=0,\quad\dot{\theta}(0)=\omega_n$$ Therefore, $$\theta(t)=\frac{\omega_n}{\rho}\sinh\rho t$$ The angle rotated by the $n$th rod before colliding with the $n+1$th rod: $$\theta_0=\frac{l}{h}$$ Time required: $$t_n=\dfrac{1}{\rho}\sinh^{-1}\dfrac{\rho l}{h\omega_n}$$ Using properties of hyperbolic functions $$\cosh^2x-\sinh^2x=1$$ We obtain $$\dot{\theta}(t_n)=\sqrt{\omega_n^2+\frac{\rho^2l^2}{h^2}}$$ At the moment of collision between the $n$th rod and the $n+1$th rod, the force and impulse must be along the horizontal direction. Let $\omega_n^{\prime}$ be the angular velocity of the $n$th rod after the collision. Angular momentum: $$\dot{\theta}(t_n)=\omega_n^{\prime}+\omega_{n+1}$$ Inelastic collision: $\sqrt{h^2-l^2}\omega_{n+1}-h\omega^{\prime}\cdot\frac{\sqrt{h^2-l^2}}h=e\left[h\dot{\theta}(t_n)\cdot\frac{\sqrt{h^2-l^2}}h-0\right]=0$ Solving yields $$\omega_{n+1}^{2}=\frac14\omega_{n}^{2}+\frac14\frac{\rho^{2}l^{2}}{h^{2}}\\\omega_{n+1}^{2}-\frac13\frac{\rho^{2}l^{2}}{h^{2}}=\frac14\left(\omega_{n}^{2}-\frac13\frac{\rho^{2}l^{2}}{h^{2}}\right)$$ Solving this sequence gives the angular velocity of the $n$th rod just after collision $$\omega_n=\sqrt{\frac{\omega_0^2}{4^n}+\frac{\rho^2l^2}{3h^2}\left(1-\frac1{4^n}\right)}$$ In this formula, $\omega_0$ is the initial angular velocity when the first rod starts to fall. Let $n\to\infty$ to obtain $$\omega_\infty=\frac{\rho l}{\sqrt{3}h}=\frac lh\sqrt{\frac g{2h}}$$ | $$\omega_\infty = \frac{l}{h} \sqrt{\frac{g}{2h}}$$ |
702 | ELECTRICITY | To establish a rectangular coordinate system in an infinitely large three-dimensional space, there are two uniformly positively charged infinite lines at $(a,0)$ and $(-a,0)$, parallel to the $z$-axis, with a charge line density of $\lambda$; and two uniformly negatively charged lines at $(0,a)$ and $(0,-a)$, also parallel to the $z$-axis, with a charge line density of $-\lambda$. For each line, the point at a distance $a$ from the line can be taken as where the electric potential is zero. If a positive ion with a mass of $m$ and a charge of $q$ is placed near the origin, and it is restricted to move only within the $xy$ plane, and somehow all four charged lines are made to rotate counterclockwise around the origin with an angular velocity $\Omega$, with the angular velocity vector directed along the $z$-axis, given that the vacuum permittivity is $\varepsilon_{0}$, find the minimum value of $\Omega$, denoted as $\Omega_{\min}$, required for the ion to maintain stable equilibrium near the origin. | Near the origin, the electric potential can be expressed as $$ \begin{array}{r l} U&=-\displaystyle\frac\lambda{2\pi\varepsilon_{0}}\ln\displaystyle\frac{\sqrt{(a-x)^{2}+y^{2}}}{a}-\displaystyle\frac\lambda{2\pi\varepsilon_{0}}\ln\displaystyle\frac{\sqrt{(a+x)^{2}+y^{2}}}{a}\\ &+\displaystyle\frac\lambda{2\pi\varepsilon_{0}}\ln\displaystyle\frac{\sqrt{x^{2}+(a-y)^{2}}}{a}+\displaystyle\frac\lambda{2\pi\varepsilon_{0}}\ln\displaystyle\frac{\sqrt{x^{2}+(a+y)^{2}}}{a}\end{array} $$ Using the formula given in the problem, it can be considered that $x\cdot y$ are both small quantities, thus we have $$ \begin{array}{r l} \frac{\lambda}{2\pi\varepsilon_{0}}\ln{\frac{\sqrt{(a-x)^{2}+y^{2}}}{a}} & =\frac{\lambda}{4\pi\varepsilon_{0}}\ln\frac{a^{2}-2a x+x^{2}+y^{2}}{a^{2}} \\ & =\frac{\lambda}{4\pi\varepsilon_{0}}\ln\left(1+{\frac{x^{2}+y^{2}-2a x}{a^{2}}}\right) \\ & =\frac{\lambda}{4\pi\varepsilon_{0}}\left[{\frac{x^{2}+y^{2}-2a x}{a^{2}}}-{\frac{1}{2}}\left({\frac{x^{2}+y^{2}-2a x}{a^{2}}}\right)^{2}\right] \\ & =\frac{\lambda}{4\pi\varepsilon_{0}}\left({\frac{x^{2}+y^{2}-2a x}{a^{2}}}-{\frac{2x^{2}}{a^{2}}}\right) \\ & =\frac{\lambda\left(-x^{2}+y^{2}-2a x\right)}{4\pi\varepsilon_{0}a^{2}} \end{array} $$ By proceeding similarly, we obtain the approximate results of the other terms, thus we have $$ {\begin{array}{r l}\\&{U(x,y)=-{\frac{\lambda\left(-x^{2}+y^{2}-2a x\right)}{4\pi\varepsilon_{0}a^{2}}}-{\frac{\lambda\left(-x^{2}+y^{2}+2a x\right)}{4\pi\varepsilon_{0}a^{2}}}}\\ &{\qquad+{\frac{\lambda\left(x^{2}-y^{2}-2a y\right)}{4\pi\varepsilon_{0}a^{2}}}+{\frac{\lambda\left(x^{2}-y^{2}+2a y\right)}{4\pi\varepsilon_{0}a^{2}}}}\\ &{\qquad={\frac{\lambda\left(x^{2}-y^{2}\right)}{\pi\varepsilon_{0}a^{2}}}}\end{array}} $$ After obtaining the electric potential, we can find the components of the electric field intensity along the $x$ and $y$ directions, obtaining $$ {\begin{array}{r}{E_{x}=-{\cfrac{\partial U(x,y)}{\partial x}}=-{\cfrac{2\lambda x}{\pi\varepsilon_{0}a^{2}}}}\ {E_{y}=-{\cfrac{\partial U(x,y)}{\partial y}}={\cfrac{2\lambda y}{\pi\varepsilon_{0}a^{2}}}}\end{array}} $$ Therefore, the equation of motion for the positive ion is $$ \begin{array}{l}{{m{\ddot{x}}+\displaystyle\frac{2q\lambda}{\pi\varepsilon_{0}a^{2}}x=0}}\ {{m{\ddot{y}}-\displaystyle\frac{2q\lambda}{\pi\varepsilon_{0}a^{2}}y=0}}\end{array} $$ It is obvious that in the $x$ direction it is stable, whereas in the $y$ direction it is unstable. Choose the $\Omega$ system that rotates with the four lines, in this system, the four lines are stationary, so the form of the electric force near the origin remains unchanged. Parameterize the electric force, letting $$ \omega_{0}^{2}={\frac{2q\lambda}{\pi\varepsilon_{0}a^{2}m}} $$ Considering the Coriolis force and inertial centrifugal force, the particle's dynamic equation in this system is $$ \begin{array}{c}{m\ddot{x}=m\left(-\omega_{0}^{2}+\Omega^{2}\right)x+2m\Omega\dot{y}}\ {m\ddot{y}=m\left(\omega_{0}^{2}+\Omega^{2}\right)x-2m\Omega\dot{x}}\end{array} $$ Which can be simplified to $$ \begin{array}{r}{\ddot{x}-2\Omega\dot{y}+\left({\omega_{0}^{2}-\Omega^{2}}\right)x=0}\ {\ddot{y}+2\Omega\dot{x}-\left({\omega_{0}^{2}+\Omega^{2}}\right)x=0}\end{array} $$ To discuss the stability of this two-degree-of-freedom system, consider the normal solution, thus we have $$ \begin{array}{r}{x=A e^{i\omega t}}\ {y=B e^{i\omega t}}\end{array} $$ Substitute into the above equations to obtain $$ \begin{array}{r}{\left(\omega_{0}^{2}-\Omega^{2}-\omega^{2}\right)A-2i\omega\Omega B=0}\ {2i\omega\Omega A-\left(\omega_{0}^{2}+\Omega^{2}+\omega^{2}\right)B=0}\end{array} $$ The characteristic equation is $$ -\left(\omega_{0}^{2}-\Omega^{2}-\omega^{2}\right)\left(\omega_{0}^{2}+\Omega^{2}+\omega^{2}\right)-4\omega^{2}\Omega^{2}=0 $$ Solving for the angular frequencies of the normal modes gives $$ \begin{array}{r}{\omega_{1}=\sqrt{\Omega^{2}-\omega_{0}^{2}}}\ {\omega_{2}=\sqrt{\Omega^{2}+\omega_{0}^{2}}}\end{array} $$ The existing normal modes of the system must all be stable for the system to be stable. Therefore, the condition is $$ \Omega>\omega_{0}=\sqrt{\frac{2q\lambda}{\pi\varepsilon_{0}a^{2}m}} $$ | $$\sqrt{\frac{2q\lambda}{\pi\varepsilon_0 a^2 m}}$$ |
367 | MECHANICS | There is a smooth elliptical track, and the equation of the track satisfies $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b > 0$). On the track, there is a small charged object that can move freely along the track, with mass $m$ and charge $q$. Now, a point charge with charge $Q$ (same sign as $q$) is placed at the origin. Find the period $T$ of the small oscillations of the charged object around its stable equilibrium position. | It is evident that at this point, the small object is in a stable equilibrium position at the two vertices of the major axis of the elliptical orbit. When the small object experiences a slight deviation along the orbit from the stable equilibrium position, as shown in the figure below, the distance from the small object to the origin O is \( r \), and the angle relative to the x-axis is \( \theta \) (\( \theta << 1 \)).
From the elliptical orbit equation, differentiating both sides, we have:
\[ \frac{2x dx}{a^2} + \frac{2y dy}{b^2} = 0 \]
This leads to the elliptic tangent slope at the point:
\[ \tan \alpha = \frac{a^2}{b^2} \tan \theta \]
The angle between the direction of the electrostatic force on the small object and the normal is:
\[ \beta = \alpha - \theta \]
Since the distance of the small object deviating from the stable equilibrium position is very small, we have \(\theta, \alpha, \beta << 1\), which implies:
\[ \beta = \alpha - \theta \approx \left( \frac{a^2}{b^2} - 1 \right) \theta \]
Magnitude of the electrostatic force on the small object:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{Qq}{r^2} \]
Where \( r \) satisfies:
\[ \frac{r^2 \cos^2 \theta}{a^2} + \frac{r^2 \sin^2 \theta}{b^2} = 1 \]
Thus, we obtain:
\[ r = \sqrt{\frac{a^2 b^2}{b^2 + (a^2 - b^2) \sin^2 \theta}} = a \left[ 1 + \frac{a^2 - b^2}{b^2} \sin^2 \theta \right]^{-\frac{1}{2}} \approx a \left[ 1 + \frac{a^2 - b^2}{b^2} \theta^2 \right]^{\frac{1}{2}} \]
\[ \approx a \left( 1 - \frac{a^2 - b^2}{2b^2} \theta^2 \right) \]
And the tangential component of the electrostatic force:
\[ F_t = -F \sin \beta \approx -\frac{Qq}{4\pi \epsilon_0 a^2} \left( 1 - \frac{a^2 - b^2}{2b^2} \theta^2 \right)^{-2} \cdot \left( \frac{a^2}{b^2} - 1 \right) \theta \]
\[ \approx -\frac{Qq (a^2 - b^2)}{4\pi \epsilon_0 a^2 b^2} \theta \]
Velocity of the small object:
\[ v = \dot{r} + r \dot{\theta} = -\frac{a (a^2 - b^2)}{2b^2} 2 \theta \dot{\theta} + a \left( 1 - \frac{a^2 - b^2}{2b^2} \theta^2 \right) \dot{\theta} \approx a \dot{\theta} \]
\[ F_t = m a_t = m \dot{v} = m a \ddot{\theta} \]
\[ ma \ddot{\theta} = -\frac{Qq (a^2 - b^2)}{4\pi \epsilon_0 a^2 b^2} \theta \]
This is the standard equation of simple harmonic motion, with the angular frequency of motion being:
\[ \omega = \sqrt{\frac{Qq (a^2 - b^2)}{4\pi \epsilon_0 m a^3 b^2}} \]
Period of the small vibrations:
\[ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{4\pi \epsilon_0 m a^3 b^2}{Qq (a^2 - b^2)}} \] | $$ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{4\pi \epsilon_0 m a^3 b^2}{Qq (a^2 - b^2)}} $$ |
497 | MECHANICS |
---
There is a uniform spring with a spring constant of \(k\) and an original length of \(L\), with a mass of \(m\). One end is connected to a wall, and the other end is connected to a mass \(M\) block, placed on a horizontal smooth surface for simple harmonic motion. It is known that \(m \ll M\).
The spring, after being used for a long time, experiences corrosion, causing the spring constant to decrease and the mass to increase. It is approximately assumed that the spring constant and mass uniformly increase or decrease throughout the spring.
A new spring initially has an amplitude of \(A_0\). After a period of time, the spring mass becomes \(\gamma_1\) times its original mass, and the spring constant becomes \((1 - \gamma_2\frac{m}{M})\) times its original value. Under the following model, find the change in amplitude \(\Delta A\).
Assume that the additional mass added to the spring during corrosion moves at the same speed as the spring, meaning the added mass has the same velocity as the point where it is added.
--- | First, based on the analysis of the change in the system's mechanical energy, through the integration of energy relationships and separation of variables, the relationship between the energy ratio and the stiffness coefficient as well as mass is obtained:
\(\ln\frac{E}{E_0} = \frac{1}{2}\ln\frac{k'}{k} + \frac{1}{2}\ln\frac{M'}{M_0'}\), taking the exponential on both sides gives \(\frac{E}{E_0} = \sqrt{\frac{k'}{k}\frac{M'}{M_0'}}\).
From \(E = \frac{1}{2}k'A^2\) and \(E_0 = \frac{1}{2}kA_0^2\), it follows that \(\frac{\frac{1}{2}k'A^2}{\frac{1}{2}kA_0^2} = \sqrt{\frac{k'}{k}\frac{M'}{M_0'}}\).
Substituting \(k' = (1 - \gamma_2\frac{m}{M})k\), \(M' = M + \frac{1}{3}\gamma_1 m\), \(M_0' = M + \frac{1}{3}m\), and approximating using \(m \ll M\):
\(\frac{A^4}{A_0^4} = \frac{M + \frac{1}{3}\gamma_1 m}{M + \frac{1}{3}m} \cdot \frac{1}{1 - \gamma_2\frac{m}{M}} \approx \left(1 + \frac{\gamma_1 - 1}{3}\frac{m}{M}\right)\left(1 + \gamma_2\frac{m}{M}\right) \approx 1 + \left(\frac{\gamma_1 - 1}{3} + \gamma_2\right)\frac{m}{M}\).
Expanding \(A^4 \approx A_0^4\left[1 + \left(\frac{\gamma_1 - 1}{3} + \gamma_2\right)\frac{m}{M}\right]\) to the fourth root gives \(A \approx A_0\left[1 + \frac{1}{4}\left(\frac{\gamma_1 - 1}{3} + \gamma_2\right)\frac{m}{M}\right]\).
Thus, the change in amplitude \(\Delta A = A - A_0\) is:
\[
\Delta A = \frac{m A_0}{4M}\left[\frac{1}{3}(\gamma_1 - 1) + \gamma_2\right]
\] | \[
\Delta A = \frac{m A_0}{4M}\left[\frac{1}{3}(\gamma_1 - 1) + \gamma_2\right]
\] |
457 | MECHANICS | A regular solid uniform $N$-sided polygonal prism, with a mass of $m$ and the distance from the center of its end face to a vertex as $l$, is resting on a horizontal table. The axis of the prism is horizontal and points forward. A constant horizontal force $F$ acts on the center of the prism, perpendicular to its axis and large enough to cause the prism to start rotating. As a result, the prism will roll to the right while undergoing completely inelastic collisions with the table. It is known that the coefficient of static friction with the ground is sufficiently large. After a sufficiently long time, the angular velocity after each collision becomes constant. Find this angular velocity. | For a regular $N$-sided polygon with the distance from its center to a vertex being $l$, the moment of inertia about the centroid is:
$$
I_{o}=\frac{m l^{2}}{2}(\frac{1}{3}\sin^{2}\frac{\pi}{N}+\cos^{2}\frac{\pi}{N})
$$
The moment of inertia about a vertex is:
$$
I=\frac{m l^{2}}{2}(\frac{1}{3}\sin^{2}\frac{\pi}{N}+\cos^{2}\frac{\pi}{N})+m l^{2}
$$
After sufficient time has passed:
$$
\frac{1}{2}I(\Omega^{2}-\omega^{2})=F\cdot 2l\sin\theta
$$
$$
I_{o}\Omega+m\Omega a\cos{2\theta} a=\mathbf{I}\omega
$$
Where:
$$
\theta = \frac{\pi}{N}
$$
After collision,
$$
{\omega}=\sqrt{\frac{8F\sin\displaystyle\frac{\pi}{N}(9-5\tan^{2}\displaystyle\frac{\pi}{N})^{2}}{m l(2+\displaystyle\frac{1}{3}\sin^{2}\displaystyle\frac{\pi}{N}+\cos^{2}\displaystyle\frac{\pi}{N})((9+7\tan^{2}\displaystyle\frac{\pi}{N})^{2}-(9-5\tan^{2}\displaystyle\frac{\pi}{N})^{2})}}
$$ | $$
{\omega}=\sqrt{\frac{8F\sin\displaystyle\frac{\pi}{N}(9-5\tan^{2}\displaystyle\frac{\pi}{N})^{2}}{m l(2+\displaystyle\frac{1}{3}\sin^{2}\displaystyle\frac{\pi}{N}+\cos^{2}\displaystyle\frac{\pi}{N})((9+7\tan^{2}\displaystyle\frac{\pi}{N})^{2}-(9-5\tan^{2}\displaystyle\frac{\pi}{N})^{2})}}
$$ |
432 | ELECTRICITY | A homogeneous sphere with a mass of $m$ and a radius of $R$ carries a uniform charge of $Q$ and rotates around the $z$-axis (passing through the center of the sphere) with a constant angular velocity $\omega$. The formula for calculating the magnetic moment is $\sum_{i} I_{i} S_{i}$, where $I_{i}$ represents the current in a current loop and $S_{i}$ is the area vector of the current loop. The sphere is placed on an infinitely large superconducting plate, with the $z$-axis oriented vertically and perpendicular to the plane of the plate. The gravitational acceleration is $g$. When the system reaches a stable state (the sphere rotates but its center remains stationary), determine the distance $h$ from the center of the sphere to the surface of the plate. Assume the parameters satisfy the conditions for stable levitation, and neglect dissipative effects such as air resistance.
The vacuum permittivity is given as $\epsilon_0$, and the vacuum permeability is given as $\mu_0$. | Using the method of electric imaging and magnetic imaging, the electric force is given by
$$
F_{e}={\frac{-Q^{2}}{4\pi\varepsilon_{0}(2h)^{2}}}={\frac{-Q^{2}}{16\pi\varepsilon_{0}h^{2}}}
$$
Magnetic force is given by
$$
F_{m}=-{\frac{d}{d x}}\left({\frac{2\mu_{0}\mu^{2}}{4\pi x^{3}}}\right)\mid_{x=2h}={\frac{3\mu_{0}Q^{2}R^{4}\omega^{2}}{800\pi h^{4}}}
$$
Force balance condition is
$$
F_{e}+F_{m}=m g
$$
That is
$$
\frac{3\mu_{0}Q^{2}R^{4}\omega^{2}}{800\pi h^{4}}=m g+\frac{Q^{2}}{16\pi\varepsilon_{0}h^{2}}
$$
Solving for \( h \), we get
$$
h={\sqrt{\frac{-Q^{2}+{\sqrt{Q^{4}+{\frac{96\pi\varepsilon_{0}^{2}m g\mu_{0}Q^{2}R^{4}\omega^{2}}{25}}}}}{32\pi\varepsilon_{0}m g}}}
$$ | $$
h={\sqrt{\frac{-Q^{2}+{\sqrt{Q^{4}+{\frac{96\pi\varepsilon_{0}^{2}m g\mu_{0}Q^{2}R^{4}\omega^{2}}{25}}}}}{32\pi\varepsilon_{0}m g}}}
$$
|
284 | OPTICS | Fiber-optic communication has greatly advanced our information technology development. Below, we will briefly calculate how the light carrying information propagates through a rectangular optical fiber. Consider a two-dimensional waveguide (uniform in the direction perpendicular to the paper), extending along the $x$ direction, with variations in refractive index in the $y$ direction. The middle layer is called the core, with a thickness of $b$ and refractive index $n$; the top and bottom layers are called the cladding, with a refractive index of $n+\delta n$. Assume the light entering the fiber is a monochromatic wave with a wavelength of $\lambda$.
When reflecting at the waveguide interface, the light transitioning from $n_{1}$ to $n_{2}$ has a reflection coefficient:
$$
r={\frac{n_{1}\cos\theta_{1}-n_{2}\cos\theta_{2}}{n_{1}\cos\theta_{1}+n_{2}\cos\theta_{2}}}
$$
where $\theta_1$ is the angle of incidence and $\theta_2$ is the angle of refraction. According to theoretical calculations, determine the phase shift $\varphi$ of the light after its first reflection, given that the angle of incidence is $\theta$, assuming total internal reflection occurs. Assume the plane wave phase is $\omega t-kx$. | Considering that the optical signal should propagate without loss, the light transmission in the optical fiber should involve total internal reflection at the interface. Assume the incident electric field is represented by ${\widetilde{E}}_{i}$, and the reflected electric field by $\widetilde{E}_{r}$; we have $\widetilde{E_{r}}=r\cdot\widetilde{E_{i}}$. By substituting Snell's law
$$
n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}
$$
into the expression for the reflection coefficient, we obtain:
$$
r={\frac{n_{1}\cos\theta_{1}-{\sqrt{{n_{2}}^{2}-{n_{1}}^{2}\sin^{2}\theta_{1}}}}{n_{1}\cos\theta_{1}+{\sqrt{{n_{2}}^{2}-{n_{1}}^{2}\sin^{2}\theta_{1}}}}}
$$
Total internal reflection condition (no loss) requires:
$$
n_{2}\leqslant n_{1}\sin\theta_{1}
$$
Thus, the reflection coefficient can be expressed in the form of a complex number:
$$
r={\frac{n_{1}\cos\theta_{1}-j{\sqrt{{n_{1}}^{2}\sin^{2}\theta_{1}-{n_{2}}^{2}}}}{n_{1}\cos\theta_{1}+j{\sqrt{{n_{1}}^{2}\sin^{2}\theta_{1}-{n_{2}}^{2}}}}}={\frac{{n_{1}}^{2}\cos2\theta+{n_{2}}^{2}}{{n_{1}}^{2}-{n_{2}}^{2}}}-j{\frac{2n_{1}\cos\theta_{1}{\sqrt{{n_{1}}^{2}\sin^{2}\theta_{1}-{n_{2}}^{2}}}}{{n_{1}}^{2}-{n_{2}}^{2}}}
$$
Using the double angle formula of tangent, the phase shift of the reflection coefficient $\widetilde{r}=r_{0}\cdot e^{j\varphi}$ can be obtained:
$$
\varphi=2\arctan\left[{\frac{\sqrt{\sin^{2}\theta_{1}-\left(n_{2}/n_{1}\right)^{2}}}{\cos\theta_{1}}}\right]
$$
Substituting the refractive index distribution of the waveguide, the phase shift resulting from total internal reflection at the upper and lower interfaces can be obtained as:
$$
\varphi=2\arctan\left[\frac{\sqrt{\sin^{2}\theta-\left(1+\delta n/n\right)^{2}}}{\cos\theta}\right]
$$ | $$\varphi=2\arctan\left(\frac{\sqrt{\sin^2\theta-\left(1+\frac{\delta n}{n}\right)^2}}{\cos\theta}\right)$$ |
39 | MECHANICS | A rectangular wooden block of height $H$ and density $\rho_{1}$ is gently placed on the water surface. The density of the water is $\rho_{2}$, where $\rho_{1} < \rho_{2}$. The gravitational acceleration is $g$. Consider only the translational motion of the wooden block in the vertical direction, neglecting all resistance in the direction of motion and assuming the height of the water surface remains unchanged. Gently press down on the wooden block and then release it. Afterward, the wooden block is neither fully submerged nor does it detach from the water surface. Determine the period of motion of the wooden block $T$; | Taking the top at equilibrium as the origin, establish the coordinate system as shown.
Let the position of the top of the block be $y$. The net force on the block is $-\rho_{1}H S g + \rho_{2}(H-h_{0}-y)S g = -\rho_{2}S g y$. The net force on the block is a linear restoring force, and its motion is simple harmonic motion. The equation of motion for the block is
$$
\rho_{1}H S{\ddot{y}}+\rho_{2}S g y=0
$$
Therefore, the period of the block is
$$
T=2\pi{\sqrt{\frac{\rho_{1}H}{\rho_{2}g}}}
$$ | $$
T = 2\pi \sqrt{\frac{\rho_1 H}{\rho_2 g}}
$$ |
124 | MECHANICS | A wedge-shaped block with an inclination angle of $\theta$, having a mass of $M$, is placed on a smooth tabletop. Another small block of mass $m$ is attached to the top of the wedge using a spring with a spring constant $k$. The natural length of the spring is $L_{0}$, and the surfaces between the two blocks are frictionless. Now, the small block is released from rest at a position $L_{0}$ away from the top of the wedge, allowing it to slide freely downward. Find the oscillation period of the small block $m$. | Let's set the vertical and horizontal accelerations of the small wooden block $m$ as $a_{y}$ and $a\_x$, respectively. Assume that the horizontal acceleration of the wedge-shaped wooden block $M$ is $A_{x}$. We obtain:
$$
\begin{array}
{r l}{m a_{x}+M A_{x}=0.}&{(1)}\\
\ \frac{a_{x}-A_{x}}{a_{y}}=\cot \theta&{(2)}\\
\ {-N\sin\theta+F\cos\theta=M A_{x}}&{(3)}\\
\ {m g-N\cos\theta-F\sin\theta=m a_{y}}&{(4)}\\
\ {F=k(l-l_{0})=\frac{k x}{\cos\theta}\Bigl(1+\frac{m}{M}\Bigr)}&{(5)}
\end{array}
$$
From equations (1) and (2), we get:
$$
a_{y}=(a_{x}-A_{x})\tan\theta=a_{x}\left(1+\frac{m}{M}\right)\tan\theta
$$
Substitute into equation (4),
$$
\begin{array}
{r l}m g-N\cos\theta-F\sin\theta=m a_{x}\left(1+\frac{m}{M}\right)\tan\theta &{(6)}
\end{array}
$$
From equations (3) and (6), we can eliminate $N$, resulting in:
$$
m g\sin\theta-F=m a_{x}\left(1+\frac{m}{M}\right)\frac{\sin^{2}\theta}{\cos\theta}-M A_{x}\cos\theta=m a_{x}\left[\left(1+\frac{m}{M}\right)\frac{\sin^{2}\theta}{\cos\theta}+\cos\theta\right]
$$
Substitute $F$ from equation (5),
$$
m g\sin\theta-\frac{k x}{\cos\theta}\Bigl(1+\frac{m}{M}\Bigr)=m a_{x}\left[\Bigl(1+\frac{m}{M}\Bigr)\frac{\sin^{2}\theta}{\cos\theta}+\cos\theta\right]
$$
We find that there is a term $a_{x}\propto-x$ in the expression. Therefore, the angular frequency is:
$$
\begin{array}{c c c}{\displaystyle{\omega^{2}=\frac{\frac{k}{\cos\theta}\left(1+\frac{m}{M}\right)}{m\left[\left(1+\frac{m}{M}\right)\frac{\sin^{2}\theta}{\cos\theta}+\cos\theta\right]}=\frac{k\left(1+\frac{m}{M}\right)}{m[\left(1+\frac{m}{M}\right)\sin^{2}\theta+\cos^{2}\theta]}}}\ {\displaystyle{\Rightarrow\omega^{2}=\frac{k(M+m)}{m[M+m\sin^{2}\theta]}}}\
\end{array}
$$
Finally, we get the period:
$$
T=2\pi\sqrt{\frac{m(M+m\sin^{2}\theta)}{k(M+m)}}
$$ | $$
T=2\pi\sqrt{\frac{m(M+m\sin^2\theta)}{k(M+m)}}
$$ |
55 | ELECTRICITY | In a vacuum, there is an infinitely long, uniformly charged straight line fixed in place, with a charge line density of $\lambda$. Additionally, there is a dust particle with mass $m$, which can be considered as an isotropic, uniform dielectric sphere with a volume $V$, and a relative permittivity $\varepsilon_{r}$. It is given that the volume $V$ of the dielectric sphere is very small, the permittivity of vacuum is $\varepsilon_{0}$, and the following factors can be neglected: gravity, electromagnetic radiation caused by the motion of charges, and relativistic effects. Study the motion of the dust particle under the influence of the charged straight line:
Find the force acting on the particle and provide its magnitude. | Since the volume of the medium sphere $V$ is very small, we can approximate the external electric field $E$ at the medium sphere as a uniform field. Therefore, the medium sphere is uniformly polarized, and let the polarization intensity inside the sphere be $P$. The polarization surface charge density follows the cosine distribution as below (where $\alpha=0$ corresponds to the direction of the external field $E$):
$$
\sigma_{p}(\alpha)=P\cos{\alpha}
$$
The depolarization field generated by the polarized charges inside the medium sphere is a uniform field (the negative sign indicates the opposite direction to the external field $E$):
$$
E^{\prime}=-{\frac{P}{3\varepsilon_{0}}}
$$
The total electric field inside the sphere is $E+E^{\prime}$, and according to the polarization law, we have
$$
P=(\varepsilon_{r}-1)\varepsilon_{0}(E+E^{\prime})=(\varepsilon_{r}-1)\varepsilon_{0}\left(E-\frac{P}{3\varepsilon_{0}}\right)
$$
According to Gauss's theorem, the field strength generated by a charged line at the medium sphere is
$$
E(r)=\frac{\lambda}{2\pi\varepsilon_{0}r}
$$
By combining the above equations, we can solve for the required polarization intensity
$$
P(r)=\frac{\varepsilon_{r}-1}{\varepsilon_{r}+2}\frac{3\lambda}{2\pi r}
$$
The electric dipole moment of the medium sphere is
$$
p(r)=P(r)\cdot V=\frac{\varepsilon_{r}-1}{\varepsilon_{r}+2}\frac{3\lambda V}{2\pi r}
$$
Thus, the force on the particle is
$$
F(r)=p(r){\frac{\mathrm{d}E}{\mathrm{d}r}}=-{\frac{3\left(\varepsilon_{r}-1\right)V\lambda^{2}}{4\pi^{2}\left(\varepsilon_{r}+2\right)}}{\frac{1}{r^{3}}}
$$ | $$
-\frac{3(\varepsilon_r-1)V\lambda^2}{4\pi^2(\varepsilon_r+2)r^3}
$$ |
419 | OPTICS | Consider a thin layer with refractive index $ n_1 $ and thickness $ d $, sandwiched between a medium with refractive index $ n_2 $ on both sides (for simplicity, let $ n = \frac{n_1}{n_2} < 1 $). Now, suppose a beam of light with wavelength $ \lambda $ (wavelength inside $ n_2 $) is incident at an angle $ i_2 $, with a refraction angle $ i_1 $. Through multiple reflections and refractions, and given that the surface of the thin layer exhibits total internal reflection, determine the total reflectance $ R_s $ of the light intensity in this scenario, considering only the s-polarized (perpendicular to the plane of incidence) light intensity. Express the result using $ \lambda $, $ d $, $ i_2 $, and $ n$. | Calculate the amplitude after infinite reflections:
\[
\begin{align*}
E_{r}&=E_{0}\left(r + tr't\mathrm{e}^{\mathrm{j}2\delta}+t(r')^{3}t\mathrm{e}^{\mathrm{j}4\delta}+\cdots +t(r')^{2n - 1}t\mathrm{e}^{\mathrm{j}(2n - 2)\delta}+\cdots\right)\\
&=E_{0}\left(r+tr't\mathrm{e}^{\mathrm{j}2\delta}\left(1 + r'^{2}\mathrm{e}^{\mathrm{j}2\delta}+\cdots\right)\right)\\
&=E_{0}\left(r + tr't\mathrm{e}^{\mathrm{j}2\delta}\frac{1}{1 - r'^{2}\mathrm{e}^{\mathrm{j}2\delta}}\right)\\
&=E_{0}\left(r-\frac{(1 - r'^{2})r\mathrm{e}^{\mathrm{j}2\delta}}{1 - r'^{2}\mathrm{e}^{\mathrm{j}2\delta}}\right)\\
&=E_{0}\left(\frac{r(1 - \mathrm{e}^{\mathrm{j}2\delta})}{1 - r'^{2}\mathrm{e}^{\mathrm{j}2\delta}}\right)
\end{align*}
\]
Substitute the complex refractive index to obtain:
\[
R_{s}=\left(1+\left(\frac{2\mathrm{e}^{-\frac{2\pi d}{\lambda}\sqrt{\mathrm{sin}i_2^2-n^2}}}{1 - \mathrm{e}^{-\frac{4\pi d}{\lambda}\sqrt{\mathrm{sin}i_2^2-n^2}}}\frac{2\sqrt{(1 - \sin^{2}i_{2})(\sin^{2}i_{2}-n^{2})}}{1 - n^{2}}\right)^{2}\right)^{-1}
\] | \[
R_{s}=\left(1+\left(\frac{2\mathrm{e}^{-\frac{2\pi d}{\lambda}\sqrt{\mathrm{sin}i_2^2-n^2}}}{1 - \mathrm{e}^{-\frac{4\pi d}{\lambda}\sqrt{\mathrm{sin}i_2^2-n^2}}}\frac{2\sqrt{(1 - \sin^{2}i_{2})(\sin^{2}i_{2}-n^{2})}}{1 - n^{2}}\right)^{2}\right)^{-1}
\] |
458 | ADVANCED | The incompressible viscous fluid satisfies the Navier-Stokes equations:
$$
\frac{\partial \vec{v}}{\partial t} + (\vec{v} \cdot \nabla) \vec{v} = -\frac{1}{\rho} \nabla p + \frac{\mu}{\rho} \Delta \vec{v}
$$
where $\eta$ is the viscosity of the viscous fluid, $\rho$ is the density of the viscous fluid, and:
Using the Navier-Stokes equations, solve the following problem: An incompressible viscous fluid flows through a regular triangular pipe with a side length of $a$ and a length of $l$, with a pressure difference of $\Delta p$ between the two ends. Determine the volumetric flow rate $Q$. | Hypothesis:
$ v = \frac{\Delta p}{l} \frac{2}{\sqrt{30\eta}} h_1 h_2 h_3 $
It can be proved that:
$ \Delta(h_1 h_2 h_3) = -(h_1 + h_2 + h_3) = -\frac{\sqrt{3}}{2} a $
Therefore, the hypothesis satisfies the Navier-Stokes equations and boundary conditions and constitutes a solution. It is evident that the solution is unique, thus $ v = \frac{\Delta p}{l} \frac{2}{\sqrt{30\eta}} h_1 h_2 h_3 $ is the only correct solution.
Integrating yields the flow rate:
$ Q = \frac{\sqrt{3} a^4 \Delta p}{320 l} $ | $ Q = \frac{\sqrt{3} a^4 \Delta p}{320\eta l} $ |
206 | ELECTRICITY | In a zero-gravity space, two coaxial cone surfaces $A$ and $B$ are placed. Assume their common vertex is located at the origin of the coordinate system. The cylindrical coordinate equations are given as:
$$
A: r = z \tan\alpha_{1} \quad,\quad B: r = z \tan\alpha_{2}
$$
where $\alpha_{2} > \alpha_{1} (\alpha$ is the angle between the line connecting a point in space to the origin and the positive direction of the common axis $\hat{z}$).
Analysis of the electric potential in the space: It is known that the electric potentials on surfaces $A$ and $B$ satisfy $V_{A} = 0$, $V_{B} = V_{0}$. Solve for the potential distribution $V(\alpha)$ between the cone surfaces. | **Electric Field Distribution Analysis:**
Based on symmetry, select the cone vertex as the origin, and establish spherical coordinates $(R,\alpha,\theta)$ to solve the spatial electric field potential distribution. Here, $R=\sqrt{r^{2}+z^{2}}$ and $R\sin\alpha=r$, which correspond to cylindrical coordinates. The spherical coordinate form of Laplace's equation $\nabla^{2}V=0$ is written as:
$$
\frac{1}{R^{2}}\frac{\partial}{\partial R}(R^{2}\frac{\partial V}{\partial R})+\frac{1}{R^{2}\sin\alpha}\frac{\partial}{\partial\alpha}(\sin\alpha\frac{\partial V}{\partial\alpha})+\frac{1}{R^{2}\sin^{2}\alpha}\frac{\partial^{2}V}{\partial\theta^{2}}=0
$$
Considering symmetry, ${\mathrm{{.}}}V=V(\alpha)$, the equation simplifies to:
$$
\frac{1}{R^{2}\sin\alpha}\frac{\partial}{\partial\alpha}(\sin\alpha\frac{\partial V}{\partial\alpha})=0
$$
That is:
$$
\sin\alpha\frac{\partial V}{\partial\alpha}=A
$$
Where $A$ is a constant. Integrating both sides:
$$
V=\int{\frac{\mathrm{d}\alpha}{\sin\alpha}}=A\ln\tan{\frac{\alpha}{2}}+C
$$
Considering boundary conditions:
$$
V(\alpha_{1})=0,V(\alpha_{2})=V_{0}
$$
Substituting these into the solution yields:
$$
V=\frac{V_{0}}{\ln\frac{\tan\frac{\alpha_{2}}{2}}{\tan\frac{\alpha_{1}}{2}}}\mathrm{ln}\frac{\tan\frac{\alpha}{2}}{\tan\frac{\alpha_{1}}{2}}
$$ | $$
\frac{V_0}{\ln\left(\frac{\tan\left(\frac{\alpha_2}{2}\right)}{\tan\left(\frac{\alpha_1}{2}\right)}\right)}\ln\left(\frac{\tan\left(\frac{\alpha}{2}\right)}{\tan\left(\frac{\alpha_1}{2}\right)}\right)
$$ |
204 | ELECTRICITY | In modern plasma physics experiments, two methods are commonly used to confine negatively charged particles. In the following discussion, relativistic effects and contributions such as delayed potentials are not considered.
In space, uniformly charged rings with a radius of $R$ and a charge $Q_{0}$ are placed on planes $z=l$ and $z=-l$, respectively. Additionally, a particle with a charge of $-q$ and a mass of $m$ is located at the origin of the coordinate system. The permittivity of vacuum is given as $\varepsilon_{0}$.
Find the angular frequency $\omega_{z}$ corresponding to the perturbation stability of the particle in the $\hat{z}$ direction. | Analyzing the electric field at a small displacement $z$ from the equilibrium position along the $z$ axis
we obtain
$$
\vec{E_{z}}=\frac{Q}{4\pi\varepsilon_{0}}\left(\frac{l+z}{[R^{2}+(l+z)^{2}]^{\frac{3}{2}}}-\frac{l-z}{[R^{2}+(l-z)^{2}]^{\frac{3}{2}}}\right)
$$
$$
\vec{E_{z}}=\frac{Q}{2\pi\varepsilon_{0}}\frac{\left(R^{2}-2l^{2}\right)}{\left(R^{2}+l^{2}\right)^{\frac{5}{2}}}\vec{z}
$$
The equation of motion is
$$
m{\ddot{z}}=-{\frac{Q q}{2\pi\varepsilon_{0}}}{\frac{(R^{2}-2l^{2})}{(R^{2}+l^{2})^{\frac{5}{2}}}}z
$$
For stability, the condition of the simple harmonic oscillator equation must be satisfied
$$
R^{2}>2l^{2}
$$
Solving gives
$$
\omega_{z}=\sqrt{\frac{Q q}{2\pi\varepsilon_{0}m}\frac{(R^{2}-2l^{2})}{(R^{2}+l^{2})^{\frac{5}{2}}}}
$$ | $$
\sqrt{\frac{Q q}{2 \pi \varepsilon_0 m} \frac{R^2 - 2l^2}{(R^2+l^2)^{5/2}}}
$$ |
478 | ELECTRICITY | In an infinitely large, isotropic, linear dielectric medium, there exists a uniform external electric field $\vec{E}_{0}$. The vacuum permittivity is given as $\varepsilon_{0}$.
The dielectric medium is a liquid dielectric with a relative permittivity of $\varepsilon_{r}$. A solid, ideal conducting sphere with a radius of $R$ is placed inside the dielectric medium. The net charge carried by the conducting sphere is $Q$. It is assumed that the conducting sphere and the dielectric medium are in close contact, such that when considering forces, the free charges on the surface of the conducting sphere and the polarization charges at the interface of the dielectric medium must be regarded as a whole.
Determine the magnitude of the electrostatic force $F$ acting on the conducting sphere. | Spatial Electric Field Distribution
$$
\vec{E}=0,~0<r<R
$$
$$
\vec{E}=\vec{E}_{0}+R^{3}\frac{3\hat{n}(\hat{n}\cdot \vec{E}_{0})-\vec{E}_{0}}{r^{3}}+\frac{Q \hat{n}}{4\pi\varepsilon_{0}\varepsilon_{r}r^{2}},~r>R
$$
Total Surface Charge Density on the Conductor Sphere (including free charge and polarization charge):
$$
\sigma(\theta)=\frac{Q}{4\pi\varepsilon_{r} R^2}+3\varepsilon_{0}E_{0}\cos\theta
$$
Utilizing the technique where the electric field experienced by the surface charge is the average of the electric fields on both sides, the total electrostatic force obtained by integrating over the sphere's surface is:
$$
F=\frac{1}{\varepsilon_{r}}QE_{0}
$$ | $$
F=\frac{1}{\varepsilon_{r}}QE_{0}
$$ |
71 | ELECTRICITY | Given a particle with charge $q$ and mass $m$ moving in an electric field $\pmb{E}=E_{x}\pmb{x}+E_{z}\pmb{z}$ and a magnetic field $\pmb{B}=B\pmb{z}$. The initial conditions are: position $(x_{0},y_{0},z_{0})$ and velocity $(v_{\perp}\cos\delta,v_{\perp}\sin\delta,v_{z})$.
We know that a particle in a uniform magnetic field undergoes circular Larmor gyration, and the center of this rotation is called the guiding center. The drift velocity of this guiding center due to the electric field can be expressed using $\pmb{E},\pmb{B}$ and their magnitudes.
Problem: We discuss the case where the magnetic field is uniform, but the electric field is non-uniform. For simplicity, we assume $\pmb{E}$ is in the $\pmb{x}$ direction and varies sinusoidally in the $\pmb{y}$ direction.
$$
\pmb{E} \equiv E_{0}\cos(k y)\hat{\pmb{x}}
$$
In reality, such a charge distribution can occur in a plasma during wave propagation. Find the corresponding guiding center drift velocity. It is known that the electric field is very weak, and the particle's initial position is $(x_{0},y_{0},z_{0})$. Approximate to the lowest order that can distinguish from the uniform electric field case. | This problem has been modified; the original problem had four questions.
If there is an electric field present, we find that the motion of the particle will be a combination of two movements: the normal circular Larmor gyration and a drift towards the center of guidance. We can choose the $\pmb{x}$ axis along the direction of $\pmb{E}$, so $\pmb{E}_{y}=0$, and the velocity components related to transverse components can be treated separately. The equation of motion is
$$
m{\frac{d v}{d t}}=q({\pmb{E}}+v\times{\pmb{B}})
$$
Its $\pmb{z}$ component is
$$
\frac{d v_{z}}{d t}=\frac{q}{m}E_{z}
$$
Integrating yields
$$
v_{\tau}=\frac{q E_{z}}{m}t+v_{z0}
$$
This is a simple motion along the direction of $\pmb{B}$. The transverse components of the previous equation are
$$
\frac{d\upsilon_{x}}{d t}=\frac{q}{m}E_{x}\pm\omega_{c}\upsilon_{y}
$$
$$
\frac{d\upsilon_{y}}{d t}=0\pm\omega_{c}\upsilon_{x}
$$
where $\omega_{c}\equiv{\frac{|g|B}{m}}$ is the Larmor gyration angular frequency. Differentiating these two equations gives
$$
\ddot{v}_{x}=-\omega_{c}^{2}v_{x}
$$
$$
\ddot{v}_{y}=-\omega_{c}^{2}(v_{y}+\frac{E_{x}}{B})
$$
We can rewrite the above equations as
$$
\frac{d^{2}}{d t^{2}}(v_{y}+\frac{E_{x}}{B})=-\omega_{c}^{2}(v_{y}+\frac{E_{x}}{B})
$$
Therefore, if we use $\begin{array}{r}{v_{y}+\frac{E_{x}}{B}}\end{array}$ instead of ${\pmb v_{\pmb y}}$, the equation simplifies to the situation when the electric field is zero. Thus, the two equations can be replaced by
$$
v_{x}=v_{\perp}e^{i(\omega_{\mathrm{e}}t+\delta)}
$$
$$
\upsilon_{y}=-i\upsilon_{\perp}e^{i(\omega_{c}t+\delta)}-\frac{E_{x}}{B}
$$
where $\delta$ is the angle between the component of the particle's initial velocity in the $_{xy}$ plane ${\pmb v}_{\bot}$ and the $\pmb{x}$ axis (counter-clockwise direction). Further integration and taking the real part yields the equations of motion
$$
x=x_{0}+\frac{v_{\perp}}{\omega_{c}}\sin(\omega_{c}t+\delta)
$$
$$
y=y_{0}-\frac{E_{x}}{B}t+\frac{v_{\perp}}{\omega_{c}}[1-\cos(\omega_{c}t+\delta)]
$$
$$
z=\frac{q E_{z}}{2m}t^{2}+v_{z0}t+z_{0}
$$
From the equations of motion, it is observed that the Larmor motion of the particle is identical to the case without an electric field, but there is a drift superimposed in the $-y$ direction towards the center of guidance ${\pmb v}_{{\pmb g}{\pmb c}}$ (for $E_{z}>0)$.
To obtain the general formula for ${\pmb v}_{{\pmb g}{\pmb c}}$, we can solve the equation (1) using vector form. Since we already know that the term $\textstyle m{\frac{d v}{d t}}$ only gives rise to circular motion with frequency $\omega_{c}$, this term can be ignored in equation (1). Thus, equation (1) becomes
$$
\pmb{{ E}}+\pmb{{\upsilon}}\times\pmb{{ B}}=0
$$
By taking the cross product of $\pmb{B}$ with the above expression, we get
$$
\pmb{E}\times\pmb{B}=\pmb{B}\times(\pmb{v}\times\pmb{B})=\pmb{v}\pmb{B}^{2}-\pmb{B}(\pmb{v}\cdot\pmb{B})
$$
The transverse component of this equation is
$$
v_{\perp q c}=E\times B/B^{2}\equiv v_{E}
$$
We define this transverse component as $\pmb{v}_{E}$, which is the electric field drift towards the center of guidance.
The equation of motion for the particle is
$$
m{\frac{d{\boldsymbol{v}}}{d t}}=q[{\boldsymbol{E}}({\boldsymbol{y}})+{\boldsymbol{v}}\times{\boldsymbol{B}}]
$$
Decomposing into scalar form, we have
$$
\dot{v}_{x}=\frac{q B}{m}v_{y}+\frac{q}{m}E_{x}(y)
$$
$$
\dot{v}_{y}=-\frac{q B}{m}v_{x}
$$
Simplifying yields
$$
\ddot{\upsilon}_{x}=-\omega_{c}^{2}\upsilon_{x}\pm\omega_{c}\frac{\dot{E}_{x}(y)}{B}
$$
$$
\ddot{v}_{y}=-\omega_{c}^{2}v_{y}-\omega_{c}^{2}\frac{E_{x}(y)}{B}
$$
Here, $\pmb{E_{x}}(\pmb{y})$ is the electric field at the particle's location. To compute this value, we need to know the particle's trajectory, which is precisely what we attempt to solve initially. If the electric field is weak, as a first approximation, we can estimate $E_{x}(y)$ using the undisturbed trajectory.
The trajectory in the absence of an electric field is given by
$$
y=y_{0}\pm r_{L}\cos\omega_{c}t
$$
where $\begin{array}{r}{r_{L}=\frac{m v_{\perp}}{|q|B}}\end{array}$ is the Larmor gyration radius, and we obtain
$$
\ddot{v}_{y}=-\omega_{c}^{2}v_{y}-\omega_{c}^{2}\frac{E_{0}}{B}\cos k(y_{0}\pm r_{L}\cos\omega_{c}t)
$$
We seek a solution that is the sum of the gyration at $\omega_{c}$ and a stationary drift $_{v_{E}}$. Since we are interested in expressing $v_{E}$, we can average over one cycle to eliminate the gyration motion. Thus, the equation yields
$$
\bar{v}_{x}=0
$$
In the above equation, the average of the oscillating term $\ddot{v}_{y}$ is clearly zero, giving us
$$
\overline{{\ddot{v}}}_{y}=0=-\omega_{c}^{2}\overline{{v}}_{y}-\omega_{c}^{2}\frac{E_{0}}{B}\overline{{\cos k(y_{0}\pm r_{L}\cos\omega_{c}t)}}
$$
Upon expanding the small approximation and averaging, we find
$$
\overline{{v}}_{y}=-\frac{E_{x}(y_{0})}{B}(1-\frac{1}{4}k^{2}r_{L}^{2})
$$
Thus, due to non-uniformity, the usual $\pmb{{E}}\times\pmb{{B}}$ drift is modified to
$$
\boxed{v_{E}={\frac{E\times B}{B^{2}}}(1-{\frac{1}{4}}{(k \frac{m v_{\perp}}{\vert{q}\vert B})}^2)}
$$
The correction term represents the finite Larmor radius effect under the sinusoidal electric field distribution. | $$
v_E = \frac{E \times B}{B^2} \left(1 - \frac{1}{4} \left(k \frac{m v_\perp}{|q| B}\right)^2 \right)
$$ |
259 | MODERN | In the inertial frame \(S\), at time \(t = 0\), four particles simultaneously start from the origin and move in the directions of \(+x, -x, +y, -y\), respectively, with velocity \(v\). Consider another inertial frame \(S'\), which moves relative to \(S\) along the positive x-axis with velocity \(u\). At the initial moment, the two reference frames satisfy \(t = t' = 0\), where \(t'\) represents the time in the \(S'\) reference frame, and at the initial moment, the origins of the two reference frames coincide. Derive the relationship between the area of the quadrilateral formed by connecting the four particles in reference frame \(S'\) and time \(t'\). Consider the effects of special relativity, with the speed of light given as \(c\). | In the inertial frame \( S \), four particles start from the origin at \( t = 0 \), moving with velocity \( v \) in the \( +x, -x, +y, -y \) directions, respectively. Another inertial frame \( S' \) moves relative to \( S \) in the positive \( x \)-direction with velocity \( u \). At the initial moment, the origins of the two reference frames coincide, and their clocks are synchronized. We need to find the relationship between the area of the quadrilateral formed by connecting the four particles' positions and time \( t' \) in frame \( S' \), taking into account relativistic effects as per special relativity.
---
1. **Lorentz Transformation**:
- The coordinate transformation equations:
\[
x' = \gamma \left( x - ut \right), \quad t' = \gamma \left( t - \frac{ux}{c^2} \right)
\]
where \( \gamma = \frac{1}{\sqrt{1 - u^2/c^2}} \).
---
2. **Coordinate Transformation of Particles**:
- **Particle 1 (in the \( +x \) direction)**:
\[
x_1' = \gamma \frac{(v - u)t}{1 - uv/c^2}, \quad y_1' = 0
\]
Time \( t_1' = \gamma t \left(1 - \frac{uv}{c^2}\right) \)
- **Particle 2 (in the \( -x \) direction)**:
\[
x_2' = \gamma \frac{-(v + u)t}{1 + uv/c^2}, \quad y_2' = 0
\]
Time \( t_2' = \gamma t \left(1 + \frac{uv}{c^2}\right) \)
- **Particle 3 (in the \( +y \) direction)**:
\[
x_3' = -\gamma ut, \quad y_3' = \frac{vt}{\gamma}
\]
Time \( t_3' = \gamma t \)
- **Particle 4 (in the \( -y \) direction)**:
\[
x_4' = -\gamma ut, \quad y_4' = -\frac{vt}{\gamma}
\]
Time \( t_4' = \gamma t \)
---
3. **Coordinates at the Same Time \( t' \)**:
- Coordinates of Particle 1 at time \( t' \) in frame \( S' \):
\[
x_1' = \frac{(v - u)t'}{1 - uv/c^2}, \quad y_1' = 0
\]
- Coordinates of Particle 2 at time \( t' \) in frame \( S' \):
\[
x_2' = \frac{-(v + u)t'}{1 + uv/c^2}, \quad y_2' = 0
\]
- Coordinates of Particles 3 and 4 at time \( t' \) in frame \( S' \):
\[
x_3' = -ut', \quad y_3' = \frac{vt'}{\gamma}
\]
\[
x_4' = -ut', \quad y_4' = -\frac{vt'}{\gamma}
\]
---
4. **Calculation of the Area**:
- Coordinates of the four vertices of the quadrilateral:
- \( A \left( \frac{(v - u)t'}{1 - uv/c^2}, 0 \right) \)
- \( B \left( -ut', \frac{vt'}{\gamma} \right) \)
- \( C \left( -ut', -\frac{vt'}{\gamma} \right) \)
- \( D \left( \frac{-(v + u)t'}{1 + uv/c^2}, 0 \right) \)
- Length of the base (distance from \( A \) to \( D \)):
\[
\text{Base} = \left| \frac{(v - u)t'}{1 - uv/c^2} - \frac{-(v + u)t'}{1 + uv/c^2} \right| = 2vt' \frac{(1 - u^2/c^2)}{1 - (uv/c)^2}
\]
- Height (distance from \( B \) to the \( x \)-axis):
\[
\text{Height} = \frac{vt'}{\gamma}
\]
- Area formula:
\[
\text{Area} = \text{Base} \times \text{Height} = 2vt' \frac{(1 - u^2/c^2)}{1 - (uv/c)^2} \times \frac{vt'}{\gamma}
\]
Substituting \( \gamma = \frac{1}{\sqrt{1 - u^2/c^2}} \) and simplifying:
\[
\text{Area} = \frac{2v^2 t'^2 (1 - u^2/c^2)^{3/2}}{1 - (uv/c)^2}
\] | $$\frac{2 v^2 t'^2 (1 - u^2/c^2)^{3/2}}{1 - (uv/c)^2}$$ |
766 | MECHANICS | A small ring $A$ with mass $m$ is placed on a smooth horizontal fixed rod and connected to a small ball $B$ with mass $m$ by a thin string of length $l$. Initially, the string is pulled to a horizontal position, and then the system is released from rest. Find: When the angle between the string and the horizontal rod is $\theta$, what is the tension in the string?
| 【Solution】Let the angle between the rope and the rod be $\theta$, the velocity of the small ring A be $\boldsymbol{v}_{A}$, and the velocity of the small ball B relative to the small ring be $v'$. Then the velocity of the small ball B relative to the bottom surface $\boldsymbol{v}_{B}$ is given by the relative motion formula, that is $$ \boldsymbol{v}_{B} = \boldsymbol{v'} + \boldsymbol{v}_{A} \tag{1} $$ $$ \upsilon_{Br} = \upsilon_{A} - \upsilon^{\prime} \sin\theta \tag{2} $$ $$ \upsilon_{B} = \sqrt{(\upsilon_{A} - \upsilon^{\prime} \sin\theta)^2 + (\upsilon^{\prime} \cos\theta)^2} \tag{3} $$ By horizontal momentum conservation and mechanical energy conservation for the system, we have $$ m \upsilon_{A} + m \upsilon_{Br} = 0 \tag{4} $$ $$ \frac{1}{2} m \upsilon_{A}^2 + \frac{1}{2} m \upsilon_{B}^2 - m g l \sin\theta = 0 \tag{5} $$ Where the zero point of gravitational potential energy is taken at the release point of ball B. Substitute equations (2) and (3) into equations (4) and (5) respectively to obtain $$ m\upsilon_{A}+ m (\upsilon_{A} - \upsilon^{\prime} \sin\theta) = 0 \tag{6} $$ $$ \frac{1}{2} m \upsilon_{A}^2 + \frac{1}{2} m \left[(\upsilon_{A} - \upsilon^{\prime} \sin\theta)^2 + (\upsilon^{\prime} \cos\theta)^2\right] - m g l \sin\theta = 0 \tag{7} $$ Jointly solving equations (6) and (7), we obtain $$ \upsilon^{\prime 2} = \frac{4g l \sin\theta}{1 + \cos^2\theta} \tag{8} $$ $$ \upsilon_{A} = \frac{1}{2} \sin\theta \sqrt{\frac{4g l \sin\theta}{1 + \cos^2\theta}} \tag{9} $$ Let the angle between $ u_{B}$ and $ u^{\prime}$ be $a$. According to the geometric relationship shown in the figure, using the sine theorem, we have $$ \frac{\upsilon_{A}}{\sin a} = \frac{\upsilon^{\prime}}{\sin\left(\frac{\pi}{2} + \theta - a\right)} \tag{10} $$ Substituting equations (8) and (9) into equation (10), we can obtain $$ \frac{\upsilon^{\prime}}{\upsilon_{A}} = \frac{\cos(\theta - a)}{\sin a} = \frac{\cos\theta \cos a + \sin\theta \sin a}{\sin a} = \frac{2}{\sin\theta} \tag{11} $$ Therefore, we can solve to get $$ \cot a = \frac{1 + \cos^2\theta}{\sin\theta \cos\theta} $$ That is $$ \left\{ \begin{array}{l} \sin a = \frac{\sin\theta \cos\theta}{\sqrt{1 + 3\cos^2\theta}} \\ \cos a = \frac{1 + \cos^2\theta}{\sqrt{1 + 3\cos^2\theta}} \end{array} \right. \tag{12} $$ Let the tension in the rope at this time be $T$ and the acceleration of the small ring A be $a$, then we have $ T \cos\theta = m a \tag{13} $ Taking the small ring A as the reference frame, in this non-inertial frame, the forces acting on the small ball B are shown in the figure. In the figure $f_i = m a$, from the figure, we obtain the normal equation for the circular motion of the small ball B as $$ T + m a \cos\theta - m g \sin\theta = m \frac{\upsilon^{\prime 2}}{l} \tag{14} $$ Substituting equation (7) into equation (13), we obtain $$ T = \frac{5 + \cos^2\theta}{1 + \cos^2\theta} m g \sin\theta - m a \cos\theta \tag{15} $$ Jointly solving equations (12) and (14), we obtain $$ T = \frac{5 + \cos^2\theta}{(1 + \cos^2\theta)^2} m g \sin\theta \tag{16} $$
| $$
T = \frac{5 + \cos^2 \theta}{(1 + \cos^2 \theta)^2} m g \sin \theta
$$ |
716 | MECHANICS | The principle of a rotational speed measurement and control device is as follows. At point O, there is a positive charge with an electric quantity of Q. A lightweight, smooth-walled insulating thin tube can rotate around a vertical axis through point O in the horizontal plane. At a distance L from point O inside the tube, there is a photoelectric trigger control switch A. A lightweight insulating spring with a free length of L/4 is fixed at the O end, and the other end of the spring is connected to a small ball with mass m and positive charge q. Initially, the system is in static equilibrium. The thin tube rotates about a fixed axis under the action of an external torque, allowing the small ball to move within the thin tube. When the rotational speed $\omega$ of the thin tube gradually increases, the small ball reaches point A in the thin tube and just achieves radial equilibrium relative to the thin tube, triggering the control switch. The external torque instantaneously becomes zero, thus limiting excessive rotational speed; at the same time, the charge at point O becomes an equal amount of negative charge -Q. By measuring the position B of the radial equilibrium point of the small ball relative to the thin tube thereafter, the rotational speed can be determined. If the distance OB is measured to be $L/2$, determine the rotational speed $\omega$ of the thin tube when the ball is at point B. Express the result using the following physical quantities: Electric charge $Q$, ball's electric charge $q$, mass $m$, length $L$, and Coulomb's constant $k$.
| Let the angular velocity of the thin tube be $\omega_A$. When the small ball is in equilibrium at point A relative to the thin tube, we have: $$ k_0 \cdot \frac{3}{4}L - \frac{k q Q}{L^2} = m L \omega_A^2 \tag{1} $$ When the small ball is in equilibrium at point B ($OB = L/2$), with angular velocity $\omega_B$, we have: $$ k_0 \cdot \frac{1}{4}L + \frac{k q Q}{L^2 / 4} = \frac{L}{2} m \omega_B^2 \tag{2} $$ Conservation of angular momentum gives: $$ m L^2 \omega_A = m \cdot \frac{L^2}{4} \omega_B \tag{3} $$ From this, we obtain: $$ \omega_B = 4 \omega_A $$ Substituting the above into equation (2): $$ k_0 \cdot \frac{1}{4}L + \frac{k q Q}{L^2 / 4} = 8 L m \omega_A^2 \tag{4} $$ Simultaneously solving equations (1) and (4), we find the spring constant: $$ k_0 = \frac{48 k q Q}{23 L^3} \tag{5} $$ Substituting equation (5) into equation (2), we can find the angular velocity of the thin tube at this time: $$ \omega_B = 4 \sqrt{ \frac{13 k q Q}{23 m L^3} } \tag{6} $$
| $$\omega_B = 4 \sqrt{\frac{13 k q Q}{23 m L^3}}$$ |
103 | THERMODYNAMICS | Solving physics problems involves many techniques and methods: analogy, equivalence, diagrams, and so on. A smart person like you can definitely use these techniques and methods to solve the following problem:
In space, there is an infinite series of nodes, numbered in order as $\cdots -3, -2, -1, 0, 1, 2, 3, \cdots.$ From each node, there are three thermal resistances connected. The connection method is: for node ${\pmb n}$, there are thermal resistances $R$ to nodes ${\pmb n} \pm 1$ respectively. Additionally, if $m$ is even, ${n}$ is odd, and $n = m + 3$, there is a thermal resistance $R$ between node $m$ and node $n$. Find the equivalent thermal resistance $R_{04}$ between node 0 and node 4 when all other nodes remain adiabatic. | Noticing that it can be arranged in a zigzag pattern, the diagram is shown above:
First, consider $R_{02}$, which is relatively simple. Due to symmetry, the two nodes divide the network into two parts, each side being equivalent to $\scriptstyle{R^{\prime}}$. The self-similarity of $\scriptstyle{R^{\prime}}$ provides the equivalence shown on the right of the diagram above. Therefore, the 35 resistors are also $\scriptstyle{R^{\prime}}$, leading to:
$$
R^{\prime}=R+R//(R+R^{\prime})
$$
Discarding the negative root gives:
$$
\smash{R^{\prime}=\sqrt{3}R}
$$
Thus:
$$
R_{02}=\frac{\sqrt{3}}{2}R
$$
By treating $\scriptstyle{R^{\prime}}$ as equivalent to the series connection of $\scriptstyle{R^{\prime}-R}$ and $\scriptstyle{R}$, the equivalent network between 0123 is shown in the lower-left diagram above. Consequently:
$$
R_{01}=(R^{\prime}-R)//(R^{\prime}-R+R+R)
$$
$$
R_{03}=R//(R^{\prime}-R+R^{\prime}-R+R)
$$
The calculations yield:
$$
R_{01}=\frac{\sqrt{3}}{3}R
$$
$$
R_{03}=\left(1-\frac{\sqrt{3}}{6}\right)R
$$
Finally, the equivalent circuit diagram for $R_{04}$ is shown in the lower-right diagram above, which is a balanced bridge. Thus:
$$
R_{04}=2R//2R^{\prime}
$$
Resulting in:
$$
R_{04}=\left(3-\sqrt{3}\right)R
$$ | $$
R_{04} = (3 - \sqrt{3})R
$$ |
310 | MODERN | Consider an ideal mirror moving at relativistic velocity, with mass $m$ and area $S_{\circ}$. (The direction of photon incidence is the same as the direction of the mirror's motion.)
Now consider the case where the mirror is moving with an initial velocity $\beta_{0}c$. In this situation, the mirror is unconstrained by external forces, and photons are incident on it with constant power for a certain period of time, with energy $E$. Assuming the mirror's velocity after irradiation is $\beta_{1}\mathfrak{c}$, find the expression for $\beta_{1}$. | List the conservation of energy and momentum:
$$
E+{\frac{m c^{2}}{\sqrt{1-{\beta_{0}}^{2}}}}=E^{\prime}+{\frac{m c^{2}}{\sqrt{1-{\beta_{1}}^{2}}}}
$$
$$
\frac{E}{c}+\frac{m c\beta_{0}}{\sqrt{1-\beta_{0}{}^{2}}}=\frac{m c\beta_{1}}{\sqrt{1-\beta_{1}{}^{2}}}-\frac{E^{\prime}}{c}
$$
Solving, we get:
$$
\beta_{1}=\frac{\left(\sqrt{\displaystyle\frac{1+\beta_{0}}{1-\beta_{0}}}+\frac{2E}{m c^{2}}\right)^{2}-1}{\left(\sqrt{\displaystyle\frac{1+\beta_{0}}{1-\beta_{0}}}+\frac{2E}{m c^{2}}\right)^{2}+1}
$$ | $$\frac{\left(\sqrt{\frac{1+\beta_0}{1-\beta_0}}+\frac{2E}{mc^2}\right)^2 - 1}{\left(\sqrt{\frac{1+\beta_0}{1-\beta_0}}+\frac{2E}{mc^2}\right)^2 + 1}$$ |
110 | MECHANICS | A homogeneous picture frame with a light string, string length $2a$, frame mass $m$, length $2c$, and width $2d$, is hanging on a nail. Ignoring friction, with gravitational acceleration $g$. The mass of the light string is negligible, and it is inextensible, with its ends connected to the two vertices of one long side of the picture frame. Objects other than the picture frame, nail, and light string are not considered. Find the angle $\alpha$ between the long side of the picture frame and the horizontal line when in equilibrium, where $\alpha \in (0, \frac{\pi}{2}], c^4 > a^2d^2 - c^2d^2$. | As shown in the figure, the trajectory of the nail is an ellipse $\begin{array}{r}{\frac{x^{2}}{\alpha^{2}}+\frac{y^{2}}{b^{2}}=1}\end{array}$ where $b={\sqrt{a^{2}-c^{2}}}$. It is easy to know from geometric relationships that the angle between the line connecting the nail and the center of mass of the picture frame and the $y$-axis is $\alpha$.
Write the parametric equations of the ellipse:
$$
\begin{array}{r}{x=a\sin\theta}\ {}\ {y=b\cos\theta}\end{array}
$$
Obtain the distance between the center of mass of the picture frame and the nail, and $\pmb{\alpha}$:
$$
h={\sqrt{x^{2}+(y+d)^{2}}}={\sqrt{(a\sin\theta)^{2}+(b\cos\theta+d)^{2}}}
$$
$$
\alpha=\arctan({\frac{x}{y+d}})=\arctan({\frac{a\sin\theta}{b\cos\theta+d}})
$$
By the equilibrium of moments, the line connecting the center of mass of the picture frame and the nail is vertical, so the potential energy is:
$$
E_{p}=-m g h=-m g\sqrt{(a\sin\theta)^{2}+(b\cos\theta+d)^{2}}
$$
Condition for extremum of potential energy:
$$
{\frac{d E_{p}}{d\theta}}=-m g{\frac{c^{2}\sin\theta\cos\theta-b d\sin\theta}{\sqrt{(a\sin\theta)^{2}+(b\cos\theta+d)^{2}}}}=0
$$
Solve to obtain:
$$
\theta_{2}=\operatorname{arccos}(\frac{b d}{c^{2}})
$$
The condition for the existence of ${\theta_{2}}$ is:
$$
c^{2}>{\sqrt{a^{2}-c^{2}}}d
$$
Substitute into the previous equation to obtain:
$$
{{\alpha=\arctan({\frac{\sqrt{c^{4}-d^{2}(a^{2}-c^{2})}}{a d}})}}
$$ | $$
\alpha = \arctan\left(\frac{\sqrt{c^4 - d^2(a^2 - c^2)}}{ad}\right)
$$ |
256 | MODERN | Two relativistic particles X, each with rest mass $M$, experience a short-range attractive force $F(r) = \alpha/r^2$ (where $\alpha$ is a positive constant) in the zero momentum reference frame C, and are bound by this short-range attractive force to form a pair $\mathrm{X_{2}}$. The speed of light in a vacuum is $c$, and the reduced Planck constant is $\hbar = h / (2\pi)$.
If a bound pair is stable in the ground state $n=1$, the maximum value of $\alpha$, $\alpha_{c}$, can be calculated.
To bombard $\mathrm{X_{2}}$ and separate $\mathrm{X}$, neglecting the rest mass and utilizing a particle with ultra-high speed and energy $E$, the experimental setup involves bombarding $\mathrm{X_{2}}$ at rest in the laboratory frame $\mathrm{L}$. Set $\alpha = \alpha_{c}/2$. If the ultra-high-speed particle causes the transition of $\mathrm{X_{2}}$ and emit a photon, find the minimum value $E_{1}$ of $E$. | In the center-of-mass system, the momentum of the two particles X is the same, denoted as $p$, and at this time, there exists the angular momentum quantization condition:
$$
p\times{\frac{r}{2}}\times2=p r=n\hbar,n\in\mathbb{N}
$$
According to the dynamics equation, the angular velocity $\omega=2v/r$. Based on Newton's second law:
$$
p\omega={\frac{\alpha}{r^{2}}}\rightarrow\alpha=p\times{\frac{2v}{r}}\times r^{2}=2p r v=2n\hbar v
$$
Note that the attractive potential energy is $V(r)=-\alpha/r$. Introducing the kinetic energy $T$, and noting that the dynamic mass is $T/c^{2}$, it is straightforward to write the system's energy as:
$$
E=-{\frac{\alpha}{r}}+2T=-{\frac{\alpha p}{n\hbar}}+2T=-{\frac{\alpha T v}{n\hbar c^{2}}}+2T=2T\left(1-\left({\frac{\alpha}{2n\hbar c}}\right)^{2}\right)
$$
By calculating the system energy, we obtain:
$$
T={\frac{M c^{2}}{\sqrt{1-\left({\frac{\alpha}{2n\hbar c}}\right)^{2}}}}\to E_{n}=2M c^{2}{\sqrt{1-\left({\frac{\alpha}{2n\hbar c}}\right)^{2}}}
$$
Since the square root must be meaningful, we establish:
$$
\left({\frac{\alpha}{2\hbar c}}\right)^{2}\leq1\rightarrow\alpha_{c}=2\hbar c
$$
Writing out the binding energy for the ground states with $n=1,2$:
$$
\begin{array}{r}{E_{n=1}=2M c^{2}\sqrt{1-\left(\cfrac{\hbar c}{2\hbar c}\right)^{2}}=\sqrt{3}M c^{2}}\ {E_{n=2}=2M c^{2}\sqrt{1-\left(\cfrac{\hbar c}{4\hbar c}\right)^{2}}=\cfrac{\sqrt{15}}{2}M c^{2}}\end{array}
$$
The invariant modulus squared of the initial state is:
$$
\left(\sum E\right)^{2}-\left(\sum P c\right)^{2}=\left(E+{\sqrt{3}}M c^{2}\right)^{2}-E^{2}=3\left(M c^{2}\right)^{2}+2{\sqrt{3}}M c^{2}E
$$
If the system successfully ensures a transition, then in the critical state, the system's total energy in the zero-momentum frame must be no less than the binding energy of the transition to $n=2$:
$$
3\left(M c^{2}\right)^{2}+2{\sqrt{3}}M c^{2}E\geq\left({\sqrt{15}}M c^{2}\right)^{2}/4
$$
Solving this gives:
$$
E_{1}={\frac{15/4-3}{2{\sqrt{3}}}}M c^{2}={\frac{\sqrt{3}}{8}}M c^{2}
$$ | $$
E_1 = \frac{\sqrt{3}}{8}Mc^2
$$ |
112 | ELECTRICITY | Initially, a conductive dielectric sphere with a free charge of 0 is placed in a vacuum. It is known that the radius of the conducting sphere is $R$, its relative permittivity is $\varepsilon_{r}$, and its conductivity is $\sigma$. At the moment $t=0$, a uniform external field ${{\vec{E}}_{0}}$ is applied around the conducting sphere, and free charge begins to accumulate on the surface of the conductor. The permittivity of the vacuum is $\varepsilon_0$. Try to determine the total Joule heat generated by the system from the initial state to the steady state. | Considering the moment when $\ell\to0^{+}$, the polarization properties of the dielectric are prioritized over the conductive properties. At this time, the potential distribution is equivalent to the polarization of a dielectric sphere with a relative dielectric constant of $\varepsilon_{\mathsf{r}}$ in a uniform external field.
$$
\sigma_{i}=P\cos\theta
$$
Further,
$$
\bar{P}=(\varepsilon_{\mathrm{r}}-1)\varepsilon_{0}\left(E_{0}-\frac{P}{3\varepsilon_{0}}\right)
$$
Solving,
$$
P=3\varepsilon_{0}\frac{\varepsilon_{r}-1}{\varepsilon_{r}+2}E_{0}
$$
Effective electric dipole outside the sphere,
$$
\mathscr{P}=\frac{4}{3}\pi\mathscr{R}^{3}P=\frac{4\pi\varepsilon_{0}(\varepsilon_{r}-1)}{\varepsilon_{r}+2}E_{0}R^{3}
$$
Solving,
$$
U_{i n.}(t\rightarrow0^{+})=-{\frac{3}{\varepsilon_{r}+2}}E_{0}r\cos\theta
$$
$$
\zeta_{c x}(t\rightarrow0^{+})=-E_{0}r\cos\theta+\frac{\varepsilon_{r}-1}{\varepsilon_{r}+2}E_{0}\frac{R^{3}}{r^{2}}\cos\theta
$$
Considering the moment when it reaches a steady state, the conductive dielectric sphere fully exhibits the properties of an ideal conductor $(\varepsilon_{r}\to+\infty)$, solving,
$$
\bar{U}_{i n}(t;\to\infty)=0
$$
$$
U_{\mathrm{cx}}(t\rightarrow\infty)=-E_{0}\tau\cos\theta+E_{0}\frac{R^{3}}{r^{2}}\cos\theta
$$
Considering the moment $t$, when the interior of the conductive dielectric sphere is free from charge, its free charges should be distributed sinusoidally on the surface, assuming the spatial potential distribution satisfies,
$$
U=\left\{\begin{array}{l l}{-E_{0}r\cos\theta+A(t)\frac{\cos\theta}{r^{2}},}&{\mathrm{if~}R<r} \\ {-E_{0}r\cos\theta+B(t)r\cos\theta,}&{\mathrm{if~}0<r<R}\end{array}\right.
$$
Considering the continuity of potential on $\mathcal{R}$,
$$
-E_{0}R\cos{\theta}+A(t)\frac{\cos{\theta}}{R^{2}}=-E_{0}R\cos{\theta}+B(t)R\cos{\theta}
$$
Simplifying to get,
$$
B(t)=A(t)\frac{1}{R^{3}}
$$
Further considering the distribution of space electric field along the $\hat{r}$ direction,
$$
E_{\mathrm{r}}=-\frac{\partial U}{\partial r}=\left\{\begin{array}{l l}{E_{0}\cos\theta+A(t)\frac{2\cos\theta}{r^{3}},}&{\mathrm{if~}R<r} \\ {E_{0}\cos\theta-B(t)\cos\theta,}&{\mathrm{if~}0<r<R}\end{array}\right.
$$
Considering the charging process at the surface,
$$
j_{n}=\sigma\vec{E}_{i n}\cdot\hat{r}=\frac{\mathrm{d}\sigma_{f}}{\mathrm{d}t}
$$
Where,
$$
\boldsymbol{\sigma}_{f}=(\vec{D}_{e x}-\vec{D}_{i n})\cdot\boldsymbol{\hat{r}}=(\varepsilon_{0}\vec{E}_{c x}-\varepsilon_{0}\varepsilon_{r}\vec{E}_{i n})\cdot\boldsymbol{\hat{r}}
$$
Substituting into the previous equation,
$$
\sigma B(t)+\varepsilon_{0}\frac{2}{R^{3}}\frac{\mathrm{d}A(t)}{\mathrm{d}t}+\varepsilon_{0}\varepsilon_{r}\frac{\mathrm{d}B(t)}{\mathrm{d}t}=0
$$
Further considering the previous equation,
$$
\sigma\left(B(t)-E_{0}\right)+\varepsilon_{0}(2+\varepsilon_{r})\frac{\mathrm{d}B(t)}{\mathrm{d}t}=0
$$
Considering the initial condition,
$$
B(t\rightarrow0^{+})=\frac{\varepsilon_{r}-1}{\varepsilon_{r}+2}E_{0}
$$
Solving,
$$
B(t)=E_{0}-\frac{3}{\varepsilon_{r}+2}E_{0}e^{-\frac{\sigma}{\varepsilon_{0}(2+\varepsilon_{r})}t}
$$
Further obtaining the spatial potential distribution,
$$
\begin{array}{r}{U=\left\{\begin{array}{l l}{-E_{0}r\cos\theta+\left(1-\frac{3}{\epsilon_{\mathrm{r}}+2}e^{-\frac{\epsilon}{\epsilon_{0}(2+\epsilon_{\mathrm{r}})}t}\right)E_{0}\frac{R^{3}}{r^{2}}\cos\theta,}&{\mathrm{if~}R<r} \\ {-\frac{3}{\epsilon_{\mathrm{r}}+2}e^{-\frac{\sigma}{\epsilon_{0}(2+\epsilon_{\mathrm{r}})}t}E_{0}r\cos\theta,}&{\mathrm{if~}0<r<R}\end{array}\right.}\end{array}
$$
Utilizing the previous equation, it is easy to obtain the radial component distribution of the electric displacement vector inside and outside the surface of the conductive dielectric sphere,
$$
D_{\mathsf{r}}=\left\{\begin{array}{l l}{3\varepsilon_{0}\left[1-\frac{2}{\epsilon_{r}+2}e^{-\frac{\sigma}{\varepsilon_{0}(2+\varepsilon_{r})}t}\right]E_{0}\cos\theta,}&{\mathrm{if~}r\to R^{+}} \\ {-\frac{3\varepsilon_{0}\varepsilon_{\mathrm{r}}}{\varepsilon_{r}+2}e^{-\frac{\sigma}{\varepsilon_{0}(2+\varepsilon_{r})}t}E_{0}\cos\theta,}&{\mathrm{if~}r\to R^{-}}\end{array}\right.
$$
Obtaining the surface free charge distribution,
$$
\sigma_{f}=(\overrightarrow{\vec{D}}_{e x}-\overrightarrow{\vec{D}}_{i n})\cdot\hat{r}=3\epsilon_{0}\left(1-e^{-\frac{\sigma}{\epsilon_{0}(2+\epsilon_{r})}t}\right)E_{0}\cos\theta
$$
Considering the Joule heat theorem, the Joule heat power density inside the conductive dielectric sphere is,
$$
P_{J}=\stackrel{\rightarrow}{\vec{j}}\cdot\vec{E}_{i n}^{\prime}=\frac{9\sigma}{(\varepsilon_{r}+2)^{2}}e^{-2\frac{\sigma}{\varepsilon_{0}(2+\varepsilon_{r})}t}E_{0}^{2}
$$
Total Joule heat power in the entire space,
$$
W_{J}=\underbrace{P_{J}\mathrm{d}V}_{V_{\Theta\:\sharp\wedge\Theta}}=\frac{12\pi\sigma}{(\varepsilon_{r}+2)^{2}}e^{-2\frac{\sigma}{\varepsilon_{0}(2+\varepsilon_{r})}t}R^{3}E_{0}^{2}
$$
Further obtaining the total Joule heat,
$$
Q_{t o t}=\int_{0}^{\infty}W_{J}\mathrm{d}t=\int_{0}^{\infty}\frac{12\pi\sigma}{(\varepsilon_{r}+2)^{2}}e^{-2\frac{\sigma}{\varepsilon_{0}(2+\varepsilon_{r})}t}R^{3}E_{0}^{2}\mathrm{d}t=\frac{6\pi\varepsilon_{0}}{(\varepsilon_{r}+2)}R^{3}E_{0}^{2}
$$ | $$
Q_{tot} = \frac{6 \pi \varepsilon_0}{\varepsilon_r + 2} R^3 E_0^2
$$ |
505 | MECHANICS | It is often observed that when a puppy gets wet, it will vigorously shake its body, and with just a few quick shakes, it can rid itself of most of the water. In fact, the loose skin and longer fur of dogs provide a biological rationale for this behavior. However, this might drive you crazy when you are giving the dog a bath!
To address this, consider the following kinematics problem: treat the puppy as a sphere with radius $R$ in a uniform gravitational field with gravity $g$. The center of the sphere is suspended at a height $H>R$ above the ground. Establish a Cartesian coordinate system where the $z$-axis is vertical, the ground is at $z=0$, and the sphere's center is located at $(0,0,H)$. The sphere rotates uniformly with an angular velocity vector along the $x$ direction, and the magnitude is $\omega$. Consider a droplet of water being flung off the surface of the sphere, and then follows a trajectory under the influence of gravity.
If the angular velocity is too small, water droplets cannot be flung off the top, so there's a minimum angular velocity $\omega_{1}$.
When the angular velocity $\omega>\omega_{1}$, there is still a region on the sphere where droplets cannot be normally flung off. Find the equation of the curve projecting the boundary of this region onto the $xy$-plane (express this as an equation involving $y^2$ in terms of $x$). | Take the $x$-axis as the polar axis, and the $z$-axis as the reference axis for the azimuthal angle, i.e.:
$$
\begin{aligned}{x}&={R\cos{\theta}}\\ {y}&={R\sin{\theta}\sin{\varphi}}\\{z}&={H+R\sin{\theta}\cos{\varphi}}\end{aligned}
$$
Then the velocity of the circular motion at the point:
$$
v=\omega r=\omega R\sin\theta
$$
The radius of curvature of the parabolic path of the water droplet follows:
$$
g\cos\varphi={\frac{v^{2}}{\rho}}
$$
In order to eject the water drop, it needs to satisfy:
$$
\rho>r=R\sin\theta
$$
Setting the equality gives the boundary condition:
$$
\omega^{2}R\sin\theta=g\cos\varphi
$$
Expressing $\theta, \varphi$ in terms of $x, y$ and simplifying:
$$
\boxed{y^{2}=\frac{\omega^{4}}{g^{2}}(R^{2}-x^{2})\left(x^{2}-R^{2}+\frac{g^{2}}{\omega^{4}}\right)}
$$ | $$
\boxed{y^{2}=\frac{\omega^{4}}{g^{2}}(R^{2}-x^{2})\left(x^{2}-R^{2}+\frac{g^{2}}{\omega^{4}}\right)}
$$ |
331 | OPTICS | Building the experimental setup for diffraction with a steel ruler: Establish a spatial Cartesian coordinate system, with the positive $x$ direction pointing vertically downward and the positive $z$ direction pointing perpendicularly towards the wall. The angle between the steel ruler and the $z$-axis in the horizontal plane is $\phi$, and the angle between the steel ruler and the horizontal plane is $\xi$. A laser is directed along the positive $z$ direction onto the surface of the steel ruler, with the horizontal distance from the point of contact to the wall being $L$. The graduations on the steel ruler are evenly distributed with a spacing of $d$, oriented perpendicular to the direction of the ruler.
By considering the wave vector continuous along the direction of the scale markings, one can derive the equation of the curve where the diffraction pattern appears (general property); by considering the strong interference between the graduations, one can determine the locations of the primary maxima in the diffraction pattern (specific property).
For simplicity, assume the steel ruler is completely horizontal, setting $\xi=0$. Find: the spacing of the primary maxima near the zeroth-order maximum $\Delta y$. | Let the coordinates of the diffraction point be $(x,y)$, we can derive the incident wave vector and the reflected wave vector
$$
\left\{\begin{array}{c}{\displaystyle\boldsymbol{k}=\frac{2\pi}{\lambda}\hat{\boldsymbol{z}}}\ {\displaystyle\boldsymbol{k}^{\prime}=\frac{2\pi}{\lambda}\frac{x\hat{\boldsymbol{x}}+y\hat{\boldsymbol{y}}+L\hat{\boldsymbol{z}}}{\sqrt{x^{2}+y^{2}+L^{2}}}}\end{array}\right.
$$
The wave vector is continuous along the direction of the scale mark
$$
({\pmb k}^{\prime}-{\pmb k})\cdot{\widehat{\pmb x}}=0
$$
This yields
$$
x=0
$$
The interference between scales is extremely strong (assume the order is $n$)
$$
(k^{\prime}-k)\cdot d(\sin\phi\widehat{\pmb{y}}+\cos\phi\widehat{\pmb{z}})=2n\pi
$$
Solving, we obtain
$$
(\frac{n\lambda}{d}+\cos\phi)^{2}x^{2}+\left[(\frac{n\lambda}{d}+\cos\phi)^{2}-\sin\phi^{2}\right]y^{2}-2\sin\phi\cos\phi L y+\left[(\frac{n\lambda}{d})^{2}+2\frac{n\lambda}{d}\cos\phi\right]L^{2}=0
$$
The solution for $y$ is
$$
y_{n}=\frac{\sin\phi\cos\phi+\sqrt{\sin\phi^{2}\cos\phi^{2}-\left[(\frac{n\lambda}{d}+\cos\phi)^{2}-\sin\phi^{2}\right]\left[(\frac{n\lambda}{d})^{2}+2\frac{n\lambda}{d}\cos\phi\right]}}{(\frac{n\lambda}{d}+\cos\phi)^{2}-\sin\phi^{2}}L
$$
According to the problem statement
$$
\frac{n\lambda}{d}\ll1
$$
Then
$$
y_{n}\approx\tan2\phi L-\frac{1+\tan2\phi^{2}}{\sin\phi}\frac{n\lambda}{d}L
$$
The difference in $y$ is
$$
\Delta y=\frac{1+\tan2\phi^{2}}{\sin\phi}\frac{\lambda L}{d}
$$ | $$
\Delta y = \frac{1 + \tan^2(2\phi)}{\sin\phi} \frac{\lambda L}{d}
$$ |
498 | OPTICS | The interference phenomenon in variable refractive index systems is a new issue of concern in the field of optics in recent years. There is a thin film of non-dispersive variable refractive index medium with a constant thickness \(d\), where the refractive index within the film changes linearly with the distance from the film's surface. The refractive index at the lower surface is \(n_a\), and at the upper surface is \(n_b > n_a\). A beam of parallel light enters the film from the lower surface at an incident angle \(i\). Taking the incident point as the origin, the positive direction of the \(x\)-axis is along the lower surface of the film to the right, and the positive direction of the \(y\)-axis is perpendicular to the film surface upward.
If the incident light is monochromatic light with wavelength \(\lambda\), and the refractive index of air for this wavelength of light is \(n_0 = 1\), the distance between the incident points of two adjacent beams of incident light is \(2S\). The incident point of the first beam on the lower surface is denoted as Q, and the point from which it exits from the lower surface again after its first reflection from the upper surface is denoted as P, with OP = \(2S\). It is important to note that \(S\) is unknown and can be determined through other physical quantities; the final answer should not contain \(S\). Ignoring the phase change caused by reflection, find the phase difference \(\varphi\) between the two beams of light. | The phase difference is \(\varphi = \frac{2\pi}{\lambda} \Delta L\), where \(\Delta L = L - L_0\).
\(L\) is the total optical path from the incident point Q through reflection to the exit point P in the medium.
\(L_0 = 2S \sin i\) is the optical path difference in air for two adjacent incident beams, which can also be understood as: if the light travels a horizontal distance \(2S\) in air (with a refractive index of 1), it is the projection of the optical path along the direction parallel to the incident light.
This definition of \(\Delta L\) describes the additional optical path difference caused by light traveling in the medium relative to traveling the same equivalent distance in air, which is often used to describe the formation of interference patterns.
1. **Calculate the optical path \(L\) in the medium**:
We have already calculated:
\(L = 2 \int_0^d \frac{n(y)^2}{\sqrt{n(y)^2 - \sin^2 i}} dy\)
Substituting variables \(u = n(y) = n_a + ky\), \(dy = \frac{d}{n_b - n_a} du\).
\(L = 2 \int_{n_a}^{n_b} \frac{u^2}{\sqrt{u^2 - \sin^2 i}} \frac{d}{n_b - n_a} du = \frac{2d}{n_b - n_a} \int_{n_a}^{n_b} \frac{u^2}{\sqrt{u^2 - \sin^2 i}} du\)
Using the integration formula \(\int \frac{x^2 dx}{\sqrt{x^2 - A^2}} = \frac{x}{2} \sqrt{x^2 - A^2} + \frac{A^2}{2} \ln|x + \sqrt{x^2 - A^2}| + C\), where \(A = \sin i\).
\(L = \frac{2d}{n_b - n_a} \left[ \frac{u}{2} \sqrt{u^2 - \sin^2 i} + \frac{\sin^2 i}{2} \ln(u + \sqrt{u^2 - \sin^2 i}) \right]_{n_a}^{n_b}\)
\(L = \frac{d}{n_b - n_a} \left[ u \sqrt{u^2 - \sin^2 i} + \sin^2 i \ln(u + \sqrt{u^2 - \sin^2 i}) \right]_{n_a}^{n_b}\)
\(L = \frac{d}{n_b - n_a} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} + \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right]\)
2. **Calculate the horizontal displacement \(2S\)**:
We have already calculated:
\(2S = 2 \int_0^d \frac{\sin i}{\sqrt{n(y)^2 - \sin^2 i}} dy\)
\(2S = 2 \sin i \int_0^d \frac{dy}{\sqrt{n(y)^2 - \sin^2 i}}\)
\(2S = 2 \sin i \left[ \frac{d}{n_b - n_a} \int_{n_a}^{n_b} \frac{du}{\sqrt{u^2 - \sin^2 i}} \right]\)
\(2S = \frac{2d \sin i}{n_b - n_a} \left[ \ln(u + \sqrt{u^2 - \sin^2 i}) \right]_{n_a}^{n_b}\)
\(2S = \frac{2d \sin i}{n_b - n_a} \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right)\)
3. **Calculate the optical path difference \(L_0\) in air**:
\(L_0 = 2S \sin i\)
\(L_0 = \left[ \frac{2d \sin i}{n_b - n_a} \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right] \sin i\)
\(L_0 = \frac{2d \sin^2 i}{n_b - n_a} \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right)\)
4. **Calculate the total optical path difference \(\Delta L\)**:
\(\Delta L = L - L_0\)
\(\Delta L = \frac{d}{n_b - n_a} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} + \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right] - \frac{2d \sin^2 i}{n_b - n_a} \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right)\)
\(\Delta L = \frac{d}{n_b - n_a} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} + (\sin^2 i - 2\sin^2 i) \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right]\)
\(\Delta L = \frac{d}{n_b - n_a} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} - \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right]\)
5. **Calculate the phase difference \(\varphi\)**:
\(\varphi = \frac{2\pi}{\lambda} \Delta L\)
\[ \varphi = \frac{2\pi d}{\lambda (n_b - n_a)} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} - \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right] \]
Or written in the form of arccosh:
\[ \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) = \text{arccosh}\left(\frac{n_b}{\sin i}\right) - \text{arccosh}\left(\frac{n_a}{\sin i}\right) \]
Therefore:
\[ \varphi = \frac{2\pi d}{\lambda (n_b - n_a)} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} - \sin^2 i \left( \text{arccosh}\left(\frac{n_b}{\sin i}\right) - \text{arccosh}\left(\frac{n_a}{\sin i}\right) \right) \right] \]
Final answer:
The phase difference \(\varphi\) between the two beams is:
\[ \varphi = \frac{2\pi d}{\lambda (n_b - n_a)} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} - \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right] \] | \[ \varphi = \frac{2\pi d}{\lambda (n_b - n_a)} \left[ n_b\sqrt{n_b^2 - \sin^2 i} - n_a\sqrt{n_a^2 - \sin^2 i} - \sin^2 i \ln \left( \frac{n_b + \sqrt{n_b^2 - \sin^2 i}}{n_a + \sqrt{n_a^2 - \sin^2 i}} \right) \right] \] |
737 | ELECTRICITY | There is a centrally symmetric magnetic field in space that is directed inward perpendicular to the paper. The magnitude of the magnetic field varies with distance $r$ from the center O, and is given by the formula ${\mathbf{B}(\mathbf{r})=\mathbf{B}_{0}\left(\frac{r}{R}\right)^{n}}$. A charged particle with charge $q$ and mass $m$ moves in uniform circular motion of radius $R$ around O in the plane perpendicular to the magnetic field. If the particle is given a small radial disturbance, it will oscillate slightly along the radial direction. Find the period of these small oscillations. | (1) According to Newton's second law: $$ {\mathfrak{q B}}_{0}v_{0}={\frac{m v_{0}^{2}}{R}} $$ We obtain: $$ \mathrm{v}_{0}={\frac{q B_{0}R}{m}} $$ (2) The principle of angular momentum: $$ \frac{d\mathrm{L}}{d t}=\mathrm{B}_{0}\Big(\frac{r}{R}\Big)^{n}r q\dot{r} $$ Rearrange terms: $$ dL-\frac{B_0q}{R^n}r^{n+1}dr=0 $$ Integrate and substitute the result from question (1): $$ \mathrm{L}-\frac{\mathrm{B}_{0}q}{(n+2)R^{n}}r^{n+2}=q B_{0}R^{2}(1-\frac{1}{n+2}) $$ (3) Conservation of energy: $$ {\frac{{\mathbf{L}}^{2}}{2m r^{2}}}+{\frac{1}{2}}m{\dot{r}}^{2}=c $$ That is: $$ \frac{(q B_{0}R^{2}\Big(1-\frac{1}{n+2}\Big)+\frac{{\bf B}_{0}q}{(n+2)R^{n}}r^{n+2})^{2}}{2m r^{2}}+\frac{1}{2}m{\dot{r}}^{2}=c $$ Differentiate the above equation with respect to time: $$ \mathrm{m}\ddot{r}+\frac{(n+1)B_{0}^{2}q^{2}r^{2n+1}}{m(n+2)^{2}R^{2n}}+\frac{q^{2}B_{0}^{2}n(n+1)r^{n-1}}{(n+2)^{2}R^{n-2}m}-\frac{q^{2}B_{0}^{2}R^{4}(n+1)^{2}}{m(n+2)^{2}r^{3}}=0 $$ Let: $$ {\bf{r}}={\bf{R}}+\delta{\bf{r}} $$ Expand the above equation for small quantities and get: $$ \mathrm{m}\ddot{\delta{r}}+\frac{(n+1)B_{0}^{2}q^{2}(2n+1)\delta r}{m(n+2)^{2}}+\frac{q^{2}B_{0}^{2}n(n+1)(n-1)\delta r}{(n+2)^{2}m}+\frac{3 q^{2}B_{0}^{2}R^{4}(n+1)^{2}\delta r}{m(n+2)^{2}}=0 $$ That is: $$ \mathrm{m}\ddot{\delta{r}}+(n+1)\frac{q^2B_{0}^2\delta r}{m}=0 $$ The above equation is the standard formula for simple harmonic motion, yielding: $$ \displaystyle\mathbb{T}=\frac{2\pi}{\sqrt{n+1}}\frac{m}{q B_{0}} $$ | $$
\frac{2\pi}{\sqrt{n+1}} \frac{m}{q B_0}
$$ |
636 | ELECTRICITY | Establish a Cartesian coordinate system Oxyz, with a hypothetical sphere of radius $R$ at the origin. Place $n$ rings of radius $R$ along the meridional circles, all passing through the points (0, 0, R) and (0, 0, -R). The angle between any two adjacent rings is $\frac{\pi}{n}$, and each ring is uniformly charged with positive charge $Q$. The setup is stationary, and a negative charge $−q$ with mass $m$ is performing circular motion on the equatorial plane at a distance $r_{0}$ from the center of the sphere ($r_0\gg R$). Now, give the charge a radial disturbance, and attempt to find the difference between its radial oscillation period and angular revolution period. The vacuum permittivity is known as $\epsilon_0$. | Consider the case of a single circular loop, establishing a spherical coordinate system $(r, \alpha)$ with the loop axis as the polar axis. Then: $$ V = {\frac{1}{4\pi\varepsilon_{0}}}\int_{0}^{2\pi}{\frac{Q d\theta}{2\pi}}{\frac{1}{\sqrt{r^{2}+R^{2}-2R r \sin\alpha \cos\Theta}}}={\frac{Q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}+{\frac{R^{2}}{r^{3}}}{\frac{1-3\cos^{2}\alpha}{2}}\right)\#\left({\frac{1}{r}}+{\frac{R^{2}}{r^{2}}}\right). $$ Next, consider the superposition of multiple circular loops. Based on geometric relations: $$ V_{i}={\frac{Q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}+{\frac{R^{2}}{r^{3}}}{\frac{1-3\sin^{2}\theta\cos^{2}\left(\varphi-{\frac{\mathrm{i}\pi}{\mathrm{n}}}\right)}{2}}\right) $$ Summing results in: $$ V={\frac{n Q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}-{\frac{R^{2}}{r^{3}}}{\frac{1-3\cos^{2}\theta}{4}}\right) $$ On the equatorial plane, $\textstyle{\theta={\frac{\pi}{2}}}$ $$ V={\frac{n Q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}-{\frac{R^{2}}{4r^{3}}}\right) $$ Assuming the angular momentum is $L=m\omega_{\theta}r_{0}^{2}$, the effective potential energy is $$ \mathrm{V}_{\mathrm{eff}}=-{\frac{n Q q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}-{\frac{R^{2}}{4r^{3}}}\right)+{\frac{L^{2}}{2m r^{2}}} $$ The first derivative set to zero: $$ -{\frac{n Q q}{4\pi\varepsilon_{0}}}{\left({\frac{-1}{r^{2}}}+{\frac{3R^{2}}{4r^{4}}}\right)}-{\frac{L^{2}}{m r^{3}}}=0 $$ Solving gives: $$ \omega_{\theta}^{2}=\frac{n Q q}{4\pi\varepsilon_{0}m r_{0}}\left(\frac{1}{r_{0}^{2}}-\frac{3R^{2}}{4r_{0}^{4}}\right) $$ The second derivative: $$ -{\frac{n Q q}{4\pi\varepsilon_{0}}}\left({\frac{2}{r^{3}}}-{\frac{3R^{2}}{r^{5}}}\right)+{\frac{3L^{2}}{m r^{4}}}=m\omega_{r}^{2} $$ Solving gives: $$ \omega_{r}^{2}=\frac{n Q q}{4\pi\varepsilon_{0}m r_{0}}\left(\frac{1}{r_{0}^{2}}+\frac{3R^{2}}{4r_{0}^{4}}\right) $$ $$ T_{r}-T_{\oplus}=\frac{2\pi}{\omega^{2}}(\omega_{\theta}-\omega_{r})=-\frac{3\pi R^{2}}{2r_{0}^{2}}\sqrt{\frac{4\pi\varepsilon_{0}m r_{0}^{3}}{n Q q}} $$ | $$-\frac{3\pi R^2}{2r_0^2}\sqrt{\frac{4\pi\varepsilon_0 m r_0^3}{n Q q}}$$ |
420 | MECHANICS | B and C are two smooth fixed pulleys with negligible size, positioned on the same horizontal line. A and D are two objects both with mass $m$, connected by a light and thin rope that passes over the fixed pulleys. Initially, the system is stationary, and the distances between AB and CD are both in the direction of gravity, where the distance between AB is $x_{0}$ and the distance between CD is $L$, which is sufficiently large such that D will not touch C within the time frame discussed in the problem. At this moment, ball A is pulled so that the line AB deviates from the vertical line by a small angle $\theta_{0}$ (without changing the length of AB), and the system begins to move. Taking gravitational acceleration as $\pmb{g}$, and assuming that A descends very slowly so that we can approximately consider the length of AB remains unchanged, A moves around B with a pendulum-like motion. Try to solve for the amplitude of the oscillation angle $\theta$ of A when $AB=x$.
| Let the tension in the rope be $\intercal$. According to Newton's second law for block D, $T-mg=m{\ddot{x}}$ (1). Using the given assumptions, write the relation of the pendulum's oscillation angle with time: $$ \theta_{t} = \theta \cos(\sqrt{\frac{g}{x}}t + \phi) $$ For $\mathsf{A}$, write down the expression of Newton's second law: $$ mg-T+m\dot{\theta}_{t}^{2}x = m\ddot{x} $$ Combining (2) and (3), we get: $$ mg-T+{\textstyle{\frac{1}{2}}}mg\theta^{2}=m\ddot{x} $$ Since the work done on A by the tension in the rope and the gravity of A is equal to the increment in A's kinetic energy, and A's kinetic energy can be separately written as the kinetic energy of vertical motion and the kinetic energy of oscillation (as the oscillation angle is small, the two can be considered independent): $$ (mg-T)\dot{x}=\frac{d}{dt}(\frac{1}{2}m\dot{x}^{2}+\frac{1}{2}mg x\theta^{2}) $$ Combining (1), (4), and (5), we get: $$ g\theta^{2}=4\ddot{x} \quad -\ddot{x}\dot{x}=\frac{d}{dt}(\frac{1}{2}\dot{x}^{2}+2x\ddot{x}) $$ $\textstyle{\bar{\mathcal{F}}}\psi/{\bar{\mathcal{H}}}{\ddot{x}}dx={\frac{1}{2}}d{\dot{x}}^{2}$, simplifying the above equation, we get: $$ \dot{x}^{2}+2x\ddot{x}=const $$ Using the initial conditions at $\mathtt{t}{=}0$: $$ x=x_{0} \quad \ddot{x}=g\theta_{0}^{2}/4 $$ We can rewrite equation (8) as: $$ \dot{x}^{2}+2x\ddot{x}=g\theta_{0}^{2}x_{0}/2 $$ Multiplying both sides of the above equation by ${\mathsf{dx}}$, and using $\ddot{\bf x}dx=\frac{1}{2}d\dot{x}^{2}$, we get: $$ \dot{x}^{2}dx+xd\dot{x}^{2}=d(x\dot{x}^{2})=\frac{1}{2}g\theta_{0}^{2}x_{0}dx $$ Integrating, we obtain: $$ \dot{x}^{2}=\frac{1}{2}g\theta_{0}^{2}x_{0}(x-x_{0})/x $$ Differentiating both sides with respect to time, eliminating the first-order derivative of $\mathsf{x}$ with respect to $\mathtt{t}$, we get: $$ \ddot{x}=\frac{1}{4}g\theta_{0}^{2}\frac{x_{0}^{2}}{x^{2}} $$ Combining with equation (6), we can solve for the amplitude of the oscillation angle $\theta$: $$ \theta=\theta_{0}\frac{x_{0}}{x} $$
| $$
\theta=\theta_{0}\frac{x_{0}}{x}
$$ |
228 | MECHANICS | AB is a uniform thin rod with mass $m$ and length $l_{2}$. The upper end B of the rod is suspended from a fixed point O by an inextensible soft and light string, which has a length of $l_{1}$. Initially, both the string and the rod are hanging vertically and at rest. Subsequently, all motion occurs in the same vertical plane, with all angles rotating counterclockwise being positive.
Assume at a certain moment, the angles between the string, the rod, and the vertical direction are $\theta_{1}$ and $\theta_{2}$ respectively. The angular velocities of the string around the fixed point O and the rod around its center of mass are $\omega_{1}$ and $\omega_{2}$ respectively. Determine the angular acceleration $\alpha_{2}$ of the rod around its center of mass, expressing the answer only in terms of the physical quantities given in the problem and the gravitational acceleration $g$. | In the vertical plane where points O, B, and A are located, establish a plane coordinate system with point O as the origin, the horizontal ray to the right as the $\mathbf{X}$ axis, and the vertical ray upward as the $\mathbf{y}$ axis. The acceleration of the pole's center of mass C, denoted as $(\boldsymbol{a}_{\mathrm{G}x},\boldsymbol{a}_{\mathrm{G}y})$, satisfies the center of mass motion theorem
$$
m a_{\mathrm{c}x}=-T\sin\theta_{1},m a_{\mathrm{c}y}=-m g+T\cos\theta_{1}\textcircled{7}
$$
In this equation, $T$ is the magnitude of the tension in the rope. At the same time, the pole rotates around its center of mass under the moment caused by the rope tension $T$ relative to the center of mass. By the rotational theorem, we have
$$
\frac{1}{12}m l_{2}^{2}\alpha_{2}=-T\frac{1}{2}l_{2}\sin(\theta_{2}-\theta_{1})\textcircled{8}
$$
From the geometric relationship, we have
$$
x_{\mathrm{{B}}}(t)=x_{\mathrm{{C}}}(t)-{\frac{1}{2}}l_{2}\sin\theta_{2}(t),y_{\mathrm{{B}}}(t)=y_{\mathrm{{C}}}(t)+{\frac{1}{2}}l_{2}\cos\theta_{2}(t)
$$
Differentiating both sides of the above equations with respect to time $t$, the velocity of point B satisfies the conditions
$$
{\boldsymbol{v}}_{\mathrm{{B}}x}(t)={\boldsymbol{v}}_{\mathrm{{C}}x}(t)-{\frac{1}{2}}{\boldsymbol{\omega}}_{2}(t){\boldsymbol{l}}_{2}\cos\theta_{2}(t),\quad{\boldsymbol{v}}_{\mathrm{{B}}y}(t)={\boldsymbol{v}}_{\mathrm{{C}}y}(t)-{\frac{1}{2}}{\boldsymbol{\omega}}_{2}(t){\boldsymbol{l}}_{2}\sin\theta_{2}(t),
$$
Differentiating both sides of the above equations with respect to time $t$ again, the acceleration of point B satisfies the conditions
$$
a_{\mathrm{B}x}=a_{\mathrm{C}x}-\frac{1}{2}\alpha_{2}l_{2}\cos\theta_{2}+\frac{1}{2}\omega_{2}^{2}l_{2}\sin\theta_{2},a_{\mathrm{B}y}=a_{\mathrm{C}y}-\frac{1}{2}\alpha_{2}l_{2}\sin\theta_{2}-\frac{1}{2}\omega_{2}^{2}l_{2}\cos\theta_{2}
$$
Meanwhile, point B rotates around point O with a fixed axis under the condition of an inextensible rope, thus
$$
x_{\mathrm{{B}}}(t)=l_{\mathrm{{1}}}\sin\theta_{\mathrm{{1}}}(t),y_{\mathrm{{B}}}(t)=-l_{\mathrm{{1}}}\cos\theta_{\mathrm{{1}}}(t)
$$
Differentiating both sides of the above equations with respect to time $t$, the velocity of point B also satisfies the conditions
$$
V_{_{\mathrm{B}x}}(t)=\omega_{1}(t)l_{1}\cos\theta_{1}(t),~V_{_{\mathrm{B}y}}(t)=\omega_{1}(t)l_{1}\sin\theta_{1}(t)
$$
Differentiating both sides of the above equations with respect to time $t$ again, the acceleration of point B also satisfies the conditions
$$
a_{\scriptscriptstyle{\mathrm{B}x}}=\alpha_{1}l_{1}\cos\theta_{1}-\alpha_{1}^{2}l_{1}\sin\theta_{1},a_{{\scriptscriptstyle{\mathrm{B}y}}}=\alpha_{1}l_{1}\sin\theta_{1}+\omega_{1}^{2}l_{1}\cos\theta_{1}
$$
The angular accelerations of the rope around the suspension point and the pole around the center of mass can be solved as
$$
\begin{array}{r l}&{\alpha_{2}=\cfrac{3\sin(\theta_{1}-\theta_{2})\left[2g\cos\theta_{1}+2l_{1}\omega_{1}^{2}+l_{2}\omega_{2}^{2}\cos(\theta_{1}-\theta_{2})\right]}{l_{2}\left[1+3\sin^{2}(\theta_{1}-\theta_{2})\right]}}\end{array}
$$ | $$
\frac{3\sin(\theta_1-\theta_2)\left[2g\cos\theta_1+2l_1\omega_1^2+l_2\omega_2^2\cos(\theta_1-\theta_2)\right]}{l_2\left[1+3\sin^2(\theta_1-\theta_2)\right]}
$$ |
450 | MECHANICS | Xiao Ming discovered an elliptical plate at home with semi-major and semi-minor axes of \( A \) and \( B \), respectively. Using one focus \( F \) as the origin, a polar coordinate system was established such that the line connecting the focus and the vertex closest to the focus defines the polar axis direction. Through extremely precise measurements, it was found that the mass surface density satisfies the equation \(\sigma = \sigma_{0}(1 + e\cos\varphi)^3\), where \( e \) is the eccentricity of the ellipse. When the plate's major axis is positioned vertically and released from rest on a sufficiently rough tabletop, the plate moves only within the plane it lies in. Find the angular acceleration \(\beta\) of the plate when the major axis becomes horizontal. | At this time, the angular velocity is $\omega$, the angular acceleration is $\beta$, and the acceleration of the center of mass is $a_{x}, a_{y}$.
The moment of inertia about the instantaneous center $\mathrm{P}$ is $I_{P}=I+m A^{2}$.
From the conservation of energy, the contact condition at point P gives the acceleration
$a = \omega^2 \rho = \omega^2 \frac{A^2}{B}$.
Considering the rotational theorem about the instantaneous center:
$I_{P} \vec{\beta} = \vec{M} - m \vec{r}_{c} \times \vec{a}$
We obtain:
$$
I_{P} \beta = m \sqrt{A^2 - B^2} \bigg(g + \omega^{2} \frac{A^{2}}{B} \bigg)
$$
$$
\beta = 4A \sqrt{A^2 - B^2} g \frac{2A^{4} + 2A^{3} \sqrt{A^2 - B^2} - A^{3}B + B^{4}}{(2A^{3} + B^{3})^{2}B}
$$ | $$
\beta=4A \sqrt{A^2 - B^2} g\frac{2A^{4}+2A^{3}\sqrt{A^2 - B^2}-A^{3}B+B^{4}}{(2A^{3}+B^{3})^{2}B}
$$ |
708 | OPTICS | \"Choose a sodium lamp for Young's double-slit interference experiment. The wavelengths of the sodium lamp's double yellow lines are $\lambda_{1}$ and $\lambda_{2}$ respectively. Due to a very small wavelength difference, the higher-order interference fringes on the screen will become blurred. Neglecting the width of the slits themselves, introduce the contrast function $\gamma \equiv\frac{I_{max}-I_{min}}{I_{max}+I_{min}}$ to measure the clarity of the interference. When $\gamma = 0$, the fringes will become blurred. Additionally, if the background light intensity becomes $1/e$ of the system's maximum possible light intensity, it will also cause the observer to subjectively perceive the interference fringes as too dim, leading to blurriness. If the width of the double slits is $d$ and the horizontal distance from the screen to the double slits is $L$. Use a filter to filter out the light with a wavelength of $\lambda_{1}$. Due to the broadening of the spectral line caused by the thermal motion of molecules in the sodium lamp, if the temperature of the sodium lamp is $T$ and the mass of the sodium atom is $m$, try to calculate the position where the fringes become blurred for the first time. Assume that the sodium atoms follow the Maxwell velocity distribution law (and that the problem can be solved using the far-field condition). | Examine the effect of monochromatic light on the screen: $$ U = U_0 \left( e^{i k \cdot \frac{d x}{L}} + 1 \right) $$ The resulting light intensity is: $$ I = U \cdot U^* = U_0^2 \left(2 + 2\cos\left(k \cdot \frac{d x}{L}\right)\right) = I_0 \left(1 + \cos\left(k \cdot \frac{d x}{L}\right)\right) $$ For Doppler frequency shift due to thermal motion, examine the Doppler effect of electromagnetic waves: $$ \frac{k}{k_0} = \frac{\sqrt{1 - \beta^2}}{1 - \beta \cos\theta} \approx 1 + \beta \cos\theta = 1 + \frac{v_x}{c} \quad $$ For particles with horizontal velocity in the interval $v - v + \mathrm{d}v$ (in terms of emitted light intensity), we have: $$ \frac{\mathrm{d}I}{I_0} = f_x(v) \, \mathrm{d}v \quad $$ For light intensity on the screen: $$ I \propto \int f_x(v)\,\mathrm{d}v \cos\frac{k_0 \left(1 + \frac{v_x}{c}\right) d x}{L}\ = \int f_x(v)\,\mathrm{d}v \left( \cos\frac{k_0 d x}{L}\cos\frac{v k_0 d x}{L c}\ - \sin\frac{k_0 d x}{L}\sin\frac{v k_0 d x}{L c} \right) \quad $$ Notice that the second term is an odd function, and its integral equals 0. The integral of the first term is calculated as: $$ s_1 = \operatorname{Re} \int_{-\infty}^{\infty} \sqrt{\frac{m}{2\pi k T}} e^{- \frac{m v^2}{2k T} + i \frac{v k_0 d x}{L c}} \, \mathrm{d}v = e^{\frac{\frac{k_0 d x^2}{L c}}{\frac{2m}{k T}}} \quad $$ When it decreases to $\frac{1}{e}$, the background light will become blurred, thus: $$ \frac{\frac{k_0 d x^2}{L c}}{\frac{2m}{k T}} = 1 \quad $$ Solve for: $$ x = \sqrt{\frac{2m}{k T}} \cdot \frac{L c \lambda_2}{2\pi d} \quad $$ | $$
x = \sqrt{\frac{2m}{k T}} \cdot \frac{L c \lambda_2}{2\pi d}
$$ |
649 | ELECTRICITY | In modern plasma physics experiments, negative particles are often constrained in two ways. In the following discussion, we do not consider relativistic effects or retarded potentials. Uniformly charged rings with radius $R$ are placed on planes $z=l$ and $z=-l$ in space, respectively. The rings are perpendicular to the $z$ axis, and each carries a charge of $Q_{0}$. A particle with charge $-q$ and mass $m$ is placed at the origin of the coordinate system. Given that $Q_0, q > 0$, the vacuum permittivity is $\epsilon_{0}$ and the vacuum permeability is $\mu_{0}$. To keep the point charge stable in both the $\hat{z}$ and $\hat{r}$ directions, we consider rotating the two charged rings in the $\hat{z}$ direction with a constant angular velocity $\Omega$. Given a small disturbance to the point charge at the origin, provide the minimum $\Omega$ required for the particle to remain stable in all directions. | Analyzing the magnetic field at a small displacement $z$ along the $z$-axis away from the coordinate origin $$ \vec{B_{z}}={\frac{\mu_{0}Q\Omega R}{4\pi}}\left({\frac{R}{[R^{2}+(l+z)^{2}]^{\frac{3}{2}}}}+{\frac{R}{[R^{2}+(l-z)^{2}]^{\frac{3}{2}}}}\right) $$ yields $$ \vec{B_{z}}=\frac{\mu_{0}Q\Omega}{2\pi}\frac{R^{2}}{(R^{2}+l^{2})^{\frac{3}{2}}}\hat{z} $$ Observing that near the origin, $\vec{B}$ can be considered as a uniform field along the $\hat{z}$ direction, denoted as $B_{0}$, consider a particle starting from the origin. By the angular momentum theorem $$ \frac{\mathrm{d}\vec{L}}{\mathrm{d}t}=-q B_{0}r\dot{r}\vec{\varphi} $$ $$ L+\frac{1}{2}q B_{0}r^{2}=0 $$ $$ v_{\varphi}=-\frac{q B_{0}r}{2m} $$ Substitute into the conservation of energy equation to obtain the system's effective potential energy $$ V_{e f f}=[{\frac{\mu_{0}^{2}q^{2}Q^{2}\Omega^{2}}{32m\pi^{2}}}{\frac{R^{4}}{(R^{2}+l^{2})^{3}}}-{\frac{q Q}{8\pi\varepsilon_{0}}}{\frac{(R^{2}-2l^{2})}{(R^{2}+l^{2})^{\frac{5}{2}}}}]r^{2}+{\frac{q Q}{4\pi\varepsilon_{0}}}{\frac{(R^{2}-2l^{2})}{(R^{2}+l^{2})^{\frac{5}{2}}}}z^{2} $$ For the system to be stable along $\hat{r},\hat{z}$, the coefficients should both be greater than 0, solving gives $$ R^{2}>2l^{2} $$ $$ \Omega>\frac{2\pi}{\mu_{0}R^{2}}\sqrt{\frac{m(R^{2}-2l^{2})(R^{2}+l^{2})^{\frac{1}{2}}}{\pi\varepsilon_{0}Q q}} $$ That is, the minimum value is $$ \frac{2\pi}{\mu_{0}R^{2}}\sqrt{\frac{m(R^{2}-2l^{2})(R^{2}+l^{2})^{\frac{1}{2}}}{\pi\varepsilon_{0}Q q}} $$ | $$
\frac{2\pi}{\mu_0 R^2} \sqrt{\frac{m(R^2 - 2l^2)(R^2 + l^2)^{1/2}}{\pi \varepsilon_0 Q q}}
$$ |
688 | ELECTRICITY | In space, there is an axisymmetric magnetic field, with the direction of the magnetic field pointing outward perpendicular to the plane, and its magnitude depends only on the distance from the center of symmetry, $B(r) = B_{0} \left(\frac{r}{r_{0}}\right)^{n}$. A particle with mass $m$ and charge $q$ moves in a circular motion with radius $r_{0}$ around the center of symmetry under the influence of the magnetic field. Find: if the charged particle is given a small radial disturbance, determine the period $T$ of the small radial oscillations. | According to symmetry, the system in this problem has conserved canonical angular momentum with respect to the magnetic field's center of symmetry. To find the canonical angular momentum, we list the corresponding rate of change equation: $$ \frac{d L}{d t} = q v_{r} B_{0} \Big(\frac{r}{r_{0}}\Big)^{n} r $$ Rearranging terms and integrating yields: $$ d L - \frac{q B_{0}}{r_{0}^{n}} r^{n+1} d r = 0 \to L - \frac{q B_{0}}{(n+2) r_{0}^{n}} r^{n+2} = C = \frac{n+1}{n+2} q B_{0} r_{0}^{2} $$ (Conservation quantity 6 points) Thus, we obtain the function of angular momentum and $r$: $$ L = L(r) = \frac{q B_{0}}{n+2} \biggl(\frac{r^{n+2}}{r_{0}^{n}} + (n+1) r_{0}^{2}\biggr) $$ The energy conservation of the charged particle gives: $$ \frac{1}{2} m \dot{r}^{2} + \frac{1}{2} m v_{\theta}^{2} = \frac{1}{2} m \dot{r}^{2} + \frac{L(r)^{2}}{2m r^{2}} = E $$ (Energy conservation 6 points) Give the particle a radial perturbation, let $r = r_{0} + \delta r$, $\dot{r} = (\dot{\delta r})$, resulting in: $$ \frac{1}{2} m (\dot{\delta r})^{2} + \frac{q^{2} B_{0}^{2}}{2m(n+2)^{2}} \frac{\left(\frac{(r_{0}+\delta r)^{n+2}}{r_{0}^{n}} + (n+1) r_{0}^{2}\right)^{2}}{(r_{0}+\delta r)^{2}} = E $$ To analyze the oscillation, we need to expand the potential energy and retain terms up to the second order, so we first look at the denominator: $$ \begin{array}{r l} &\left(\frac{(r_{0}+\delta r)^{n+2}}{r_{0}^{n}} + (n+1) r_{0}^{2}\right)^{2} = \left(r_{0}^{2}\Bigl(1+\frac{\delta r}{r_{0}}\Bigr)^{n+2} + (n+1) r_{0}^{2}\right)^{2} \\ &\qquad = r_{0}^{4} \Biggl(1 + (n+2) \frac{\delta r}{r_{0}} + \frac{(n+2)(n+1)}{2} \Bigl(\frac{\delta r}{r_{0}}\Bigr)^{2} + (n+1) \Biggr)^{2} \\ &\qquad = r_{0}^{4} (n+2)^{2} \Biggl(1 + \frac{\delta r}{r_{0}} + \frac{n+1}{2} \Bigl(\frac{\delta r}{r_{0}}\Bigr)^{2} \Biggr)^{2} = r_{0}^{4} (n+2)^{2} \left(1 + 2 \frac{\delta r}{r_{0}} + (n+2) \frac{\delta r}{r_{0}}\right)^{2} \end{array} $$ In the above equation, we repeatedly used the small quantity approximation formula $\begin{array}{r} (1+x)^{n} = 1 + n x + \frac{n(n-1)}{2}x^{2} + \ldots, \end{array}$ so the potential energy term can be written as: $$ \frac{\left(\displaystyle\frac{(r_{0}+\delta r)^{n+2}}{r_{0}^{n}} + (n+1) r_{0}^{2}\right)^{2}}{(r_{0}+\delta r)^{2}} = r_{0}^{2} (n+2)^{2} \left(1 + 2 \frac{\delta r}{r_{0}} + (n+2) \Big(\frac{\delta r}{r_{0}}\Big)^{2}\right) \left(1 + \frac{\delta r}{r_{0}}\right)^{-2} $$ $$ = r_{0}^{2} (n+2)^{2} \left(1 + 2 \frac{\delta r}{r_{0}} + (n+2) \Big(\frac{\delta r}{r_{0}}\Big)^{2}\right) \left(1 - 2 \frac{\delta r}{r_{0}} + 3 \Big(\frac{\delta r}{r_{0}}\Big)^{2}\right) $$ Thus the second-order terms of potential energy are: $$ \frac{q^{2} B_{0}^{2}}{2m(n+2)^{2}} r_{0}^{2} (n+2)^{2} (n+1) \bigg(\frac{\delta r}{r_{0}}\bigg)^{2} = \frac{q^{2} B_{0}^{2}}{2m} (n+1) \delta r^{2} $$ (Correctly expanding small terms 12 points) Thus, the energy conservation equation is: $$ \frac{1}{2} m (\dot{\delta r})^{2} + \frac{q^{2} B_{0}^{2}}{2m} (n+1) \delta r^{2} = C $$ According to the formula for simple harmonic oscillation, we know this represents an oscillation period of: $$ T = \frac{2 \pi m}{q B_{0} \sqrt{n+1}} $$ | $$T = \frac{2\pi m}{q B_0 \sqrt{n+1}}$$ |
118 | ELECTRICITY | A regular dodecahedron resistor network is given. Except for $R_{BC} = 2r$, the resistance between all other adjacent vertices is $r$. Points $B$ and $C$ are the two endpoints of one edge of the dodecahedron. Find the resistance between points $B$ and $C$. | Boldly introducing negative resistance, $BC$ is equivalent to a parallel combination of $\pmb{r}$ and $-2r$:
$$
\frac{1}{r}+\frac{1}{-2r}=\frac{1}{2r}
$$
After extracting $-2r$, the remaining resistance value can be constructed using the forced current method. Inject $19I_{1}$ into node $B$, and each of the other nodes emits a current of 1. Then consider injecting $I$ into each of the other nodes while node $C$ emits $191$. Thus, the current flowing through the branch:
$$
I_{BC}=20I
$$
Voltage:
$$
U_{BC}=\frac{19Ir}{3}\cdot2
$$
Thus:
$$
U_{BC}=I_{BC}R_{BC}^{\prime}
$$
Resulting in:
$$
R_{BC}^{\prime}=\frac{19}{30}r
$$
Then, parallel it with $-2r$:
$$
\frac{1}{R_{BC}}=\frac{1}{R_{BC}^{\prime}}+\frac{1}{-2r}
$$
Finally obtained:
$$
R_{BC}=\frac{38}{41}r
$$ | $$
R_{BC} = \frac{38}{41}r
$$ |
318 | MECHANICS | Consider a small cylindrical object with radius $R$ and height $h$. Determine the expression for the force $F$ acting on the cylinder when a sound wave passes through it. The axial direction of the cylinder is the direction of wave propagation. In the sound wave, the displacement of a particle from its equilibrium position is $\psi = A\cos kx\cos 2\pi ft$, the ambient pressure is $P_0$, and the adiabatic index of air is $\gamma$. | The force on the cylinder will be
$$
F = -\pi R^{2}(p(y+h)-p(y)) = -\pi R^{2}h\frac{d p}{d y}
$$
For a traveling wave, substitute
$$
\Delta p(y,t) = -\gamma p_{0}A k\cos(k y-2\pi f t)
$$
$$
\frac{d p}{d y} = \gamma p_{0}A k^{2}\sin(k y-2\pi f t)
$$
Therefore, the force is
$$
F = -\pi R^2 h\gamma p_0 A k^2\sin(ky-2\pi ft)
$$
However, for a standing wave, we have
$$
\psi = A\cos{k x}\cos{2\pi}f t
$$
Thus,
$$
\Delta p = \gamma P_{0}k A\sin k x\cos2\pi f t
$$
Therefore,
$$
\frac{d p}{d x} = \gamma P_{0}k^{2}A\cos k x\cos2\pi f t
$$
Thus, the force is
$$
F = -\pi R^2 h\gamma P_0 k^2 A\cos kx \cos 2\pi ft
$$ | $$
F = -\pi R^2 h \gamma P_0 k^2 A \cos(kx) \cos(2\pi ft)
$$ |
62 | ADVANCED | In certain solids, ions have spin angular momentum and can be regarded as a three-dimensional real vector $\vec{S}$ with a fixed length under the semiclassical approximation, where the length $\vert\vec{S}\vert = S$ is a constant. The magnetic moment $\overrightarrow{M}$ of the ions is usually proportional to the spin vector, ${\overrightarrow{M}} = \gamma{\overrightarrow{S}}$, where $\gamma(>0)$ is a constant. The spin-spin interaction between ions $i$ and $j$ is usually the Heisenberg interaction, $-J_{i,j}\vec{S}_{i} \cdot \vec{S}_{j}$. Under a static uniform external magnetic field $\overrightarrow{B}$, the total energy (Hamiltonian) of the system is typically given by $\begin{array}{r}{H = -\sum_{(i,j)}J_{i,j}\vec{S}_{i}\cdot\vec{S}_{j} - }\end{array}$ $\begin{array}{r}{\gamma\overrightarrow{B}\cdot\sum_{i}\overrightarrow{S}_{i}}\end{array}$. Here, $(i,j)$ indicates a sum over all distinct pairs without order, meaning $(i,j)$ and $(j,i)$ represent the same pair and are not counted twice. The semiclassical time evolution of the spin vector satisfies the Heisenberg equation of motion (here $\times{}$ represents the vector cross product):
$$
\frac{\mathrm{d}}{\mathrm{d}t}\vec{S}_{i}=\frac{\partial H}{\partial \vec{S}_{i}}\times\vec{S}_{i}
$$
Consider two spins, $H = -J\vec{S}_{1}\cdot\vec{S}_{2} - \gamma\vec{B}\cdot\left(\vec{S}_{1}+\vec{S}_{2}\right)$, where $J>0$. Without loss of generality, let $\overrightarrow{B}$ be along the $\mathbf{Z}$ direction, $\vec{B}=B\hat{\bf z}$, $B \geq 0$. The ground state (lowest energy state) of the system is $\vec{S}_{1}=\vec{S}_{2}=S\hat{\mathbf{z}}$, and this state does not evolve over time. Consider states close to the ground state, $\vec{S}_{1}=S\big(x_{1},y_{1},\sqrt{1-{x_{1}}^{2}-{y_{1}}^{2}}\big)$, $\vec{S}_{2}=S\big(x_{2},y_{2},\sqrt{1-{x_{2}}^{2}-{y_{2}}^{2}}\big)$, where $\left|x_{1,2}\right|\ll1, \left|y_{1,2}\right|\ll1$. Find the product of the final normal mode frequencies. | Solution: The Heisenberg equation of motion is
$$
\begin{array}{r}{\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\vec{S}_{1}=-\left(J\vec{S}_{2}+\gamma\vec{B}\right)\times\vec{S}_{1}}\ {\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\vec{S}_{2}=-\left(J\vec{S}_{1}+\gamma\vec{B}\right)\times\vec{S}_{2}}\end{array}
$$
Note that $\vec{S}_{\pm}=\vec{S}_{1}\pm\vec{S}_{2}$ can be defined, then $\begin{array}{r}{H=\frac{J}{2}\big|\overrightarrow{S}_{+}\big|^{2}-J S^{2}-\gamma\overrightarrow{B}\cdot\overrightarrow{S}_{+}}\end{array}$,
$$
\frac{\mathrm{d}}{\mathrm{d}t}\vec{S}_{+}=-\gamma\vec{B}\times\vec{S}_{+}
$$
$$
\frac{\mathrm{d}}{\mathrm{d}t}\vec{S}_{-}=-\left(\vec{J S}_{+}+\gamma\vec{B}\right)\times\vec{S}_{-}
$$
The equations of motion for $x_{1,2},y_{1,2}$ in the lowest order approximation are
$$
\begin{array}{c}{{\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}(x_{1}+x_{2})=\gamma B\cdot(y_{1}+y_{2})}}\ {{\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}(y_{1}+y_{2})=-\gamma B\cdot(x_{1}+x_{2})}}\ {{\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}(x_{1}-x_{2})\approx(2J S+\gamma B)\cdot(y_{1}-y_{2})}}\ {{\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}(y_{1}-y_{2})\approx-(2J S+\gamma B)\cdot(x_{1}-x_{2})}}\end{array}
$$
The product of the characteristic frequencies is
$\gamma B(2J S+\gamma B)$ | $$\gamma B(2JS + \gamma B)$$ |
143 | MECHANICS | Consider an elastic soft rope with original length $a$, and elastic coefficient $k$. With one end fixed, the other end is attached to a particle with a mass of $m$. And the rope remains horizontal. The particle moves on a smooth horizontal surface. Initially, the rope is stretched to a length of $a+b$, and then the particle is released. Determine how much time it takes for the particle to return to the original release position. | First consider \( x > a \). At this time, the particle only experiences the elastic force \( -k(x-a) \) in the horizontal direction. According to Newton's second law, the equation of motion for the particle is
$$
m{\ddot{x}} = -k(x-a),
$$
which is the equation of simple harmonic motion. To solve equation (1), make the substitution \(\xi = x - a\), then equation (1) becomes
$$
m\ddot{\xi} = -k\xi.
$$
Further, with the transformation \(\ddot{\xi} = \frac{\mathrm{d}^{2}\xi}{\mathrm{d}t^{2}} = \frac{\mathrm{d}\xi}{\mathrm{d}t} \frac{\mathrm{d}}{\mathrm{d}\xi}\left(\frac{\mathrm{d}\xi}{\mathrm{d}t}\right) = \frac{1}{2} \frac{\mathrm{d}\dot{\xi}^{2}}{\mathrm{d}\xi}\), equation (2) becomes
$$
\mathrm{d}\dot{\xi}^{2} = -2\omega^{2}\xi\mathrm{d}\xi,
$$
where \(\omega = \sqrt{k/m}\). Integrating the above, and setting the initial conditions \(\xi = \xi_{0}, \dot{\xi} = v_{0} \) at \(t=0\), yields
$$
\dot{\xi}^{2} = v_{0}^{2} - \omega^{2}(\xi^{2} - \xi_{0}^{2}) = (v_{0}^{2} + \omega^{2}\xi_{0}^{2}) - \omega^{2}\xi^{2},
$$
which implies
$$
\frac{\mathrm{d}\xi}{\mathrm{d}t} = \pm\sqrt{(v_{0}^{2} + \omega^{2}\xi_{0}^{2}) - \omega^{2}\xi^{2}} = \pm\omega\sqrt{A^{2} - \xi^{2}},
$$
where \({\cal A} = \sqrt{\frac{v_{0}^{2}}{\omega^{2}} + \xi_{0}^{2}}\). Further integration and considering the initial conditions gives
$$
\xi = A\sin(\pm\omega t + \arcsin \frac{\xi_{0}}{A}) = A\cos(\omega t + \alpha),
$$
where \(\alpha = \pm\arcsin{\frac{\xi_{0}}{A}} \mp {\frac{\pi}{2}}\). Thus, the general solution of equation (1) is
$$
x = a + A\cos(\omega t + \alpha).
$$
According to the problem, at \(t=0\), \(x=a+b, \dot{x}=0\). Substituting into equation (3) determines the constants of integration to be \(A=b, \alpha=0\). Substituting back into equation (3), we obtain
$$
x = a + b\cos\omega t.
$$
This shows that the motion of the particle in the domain \(x > a\) is periodic. The time for the particle to move from \(x = a + b\) to \(x = a\) and back is the period, which is
$$
t_{0} = {\frac{1}{2}}\frac{2\pi}{\omega} = \pi\sqrt{\frac{m}{k}}.
$$
When \(x < -a\), the particle experiences an elastic force of \(-k(x+a)\) in the horizontal direction. Note that because \(x < -a\), \( -(x+a) > 0\), so the elastic force points in the positive \(x\) direction, which is consistent with the actual situation. The equation of motion is
$$
m{\ddot{x}} = -k(x+a).
$$
Using a similar method to the above, the general solution of the equation is
$$
x = -a + B\cos(\omega t + \beta),
$$
where \(B, \beta\) are constants of integration. According to analysis in \(\textcircled{2}\), \(\dot{x}|_{x=a} = \dot{x}|_{x=-a}\), we determine \(B=b\). Equation (5) indicates that the motion of the particle in the region \(x < -a\) is also simple harmonic, with the same frequency as in the region \(x > a\). The time for the particle to move from \(x=-a\) to \(x=-(a+b)\) and back is also \(t_{0}\).
In summary, the total time taken for the particle to move from \(x = a + b \) to \(x = a\) and from \(x = -a\) to \(x = -(a + b)\) and back is
$$
t_{1} = 2t_{0} = 2\pi\sqrt{\frac{m}{k}}.
$$
The previous discussion also shows that considering just \(x > a\) is sufficient to determine the characteristics of the particle's motion in the interval \(|x| > a\), especially concerning the motion's time.
\(\textcircled{2}\) When \(|x|<a\), i.e., \(-a<x<a\), the particle is not subject to any external force in the horizontal direction, and thus will move uniformly with the velocity it has at \(x = \pm a\). The speed of the particle at \(x = a\) is
$$
\dot{x}|_{x=a} = \dot{x}|_{t=t_{1}} = -b\omega = -b\sqrt{\frac{k}{m}}.
$$
The time for the particle to travel from \(x=a\) to \(x=-a\) and back to \(x=a\) is
$$
t_{2} = 4\frac{a}{|\dot{x}|_{x=a}|} = \frac{4a}{b}\sqrt{\frac{m}{k}}.
$$
The process for the particle to return to its release point is \(x = a+b \rightarrow x = a \rightarrow x = 0 \rightarrow x = -a \rightarrow x = -(a+b) \rightarrow x = -a \rightarrow x = 0 \rightarrow x = a \rightarrow x = a+b\), so the total time experienced is
$$
t = t_{1} + t_{2} = 2\left(\pi + \frac{2a}{b}\right)\sqrt{\frac{m}{k}}.
$$ | $$2\left(\pi+\frac{2a}{b}\right)\sqrt{\frac{m}{k}}$$ |
401 | MECHANICS | A certain planet has a smooth and elastic surface, perfect for playing bounce games. Let the planet have a radius $R$, and the first cosmic velocity at the surface is $v_{0}$. The ground's restitution coefficient is a fixed value $\varepsilon$. Imagine launching a bouncy ball from the surface at an angle $\theta$ with a speed of $k v_{0}$. The bouncy ball will rotate around the planet's center while continuously bouncing off the surface.
If we take $k<1$ and the launch angle $\theta$ to be a small amount and make reasonable approximations, find the total angle $\varphi$ that the bouncy ball rotates around the planet's center before it stops bouncing on the ground and rolls around the surface. | The two components of the initial throw speed are:
$$
v_{r}=k v_{0}\sin\theta\quad,\quad v_{\theta}=k v_{0}\cos\theta
$$
Based on the symmetry of the starting and landing points with respect to the major axis of the elliptical orbit, and combining the definition of the restitution coefficient under the conditions of a smooth ground. When the object undergoes the $(m+1)$-th segment of projectile motion, the two velocity components at the surface are generally:
$$
v_{m r}=\varepsilon^{m}k v_{0}\sin\theta\quad,\quad v_{m\theta}=k v_{0}\cos\theta
$$
Using the celestial body's angular momentum formula, we can obtain its semi-latus rectum $p$:
$$
L=m\sqrt{G M p}
$$
$$
L=m v_{m\theta}R
$$
Using the mechanical energy formula of the celestial body, we can obtain its orbital semi-major axis $A$:
$$
E=-{\frac{G M m}{2A}}
$$
$$
E=\frac{1}{2}m\left(v_{m r}^{2}+v_{m\theta}^{2}\right)-\frac{G M m}{R}
$$
The first cosmic velocity is:
$$
v_{0}={\sqrt{\frac{G M}{R}}}
$$
Finally, the orbital eccentricity is determined, and the standard equation is written as:
$$
r=\frac{p}{1-e\cos\varphi}\quad,\quad p=(1-e^{2})A
$$
The reason for choosing the denominator as negative is because we use the highest point of the orbit rather than the periapsis direction as the polar axis direction in polar coordinates. The calculation yields:
$$
r={\frac{k^{2}R\cos^{2}\theta}{1-{\sqrt{(1-k^{2}\cos^{2}\theta)^{2}+\varepsilon^{2m}k^{4}\sin^{2}\theta\cos^{2}\theta}}}}\cos\varphi
$$
To calculate the angle turned around the Earth's center during each projectile motion, take $r=R$, and obtain:
$$
\cos\varphi={\frac{1-k^{2}\cos^{2}\theta}{\sqrt{(1-k^{2}\cos^{2}\theta)^{2}+\varepsilon^{2m}k^{4}\sin^{2}\theta\cos^{2}\theta}}}
$$
Thus, the positive and negative acute solutions of $\varphi$ represent the angle turned. Additionally, converting the equation above into arctangent gives the rotation angle for this process:
$$
\Delta\varphi_{m}=2\arctan\frac{\varepsilon^{m}k^{2}\sin\theta\cos\theta}{1-k^{2}\cos^{2}\theta}
$$
Perform a small angle approximation for $\Delta\varphi_{m}$:
$$
\Delta\varphi_{m}\approx{\frac{2\varepsilon^{m}k^{2}\theta}{1-k^{2}}}
$$
Perform infinite summation for all angles:
$$
\varphi=\Delta\varphi_{0}+\Delta\varphi_{1}+\Delta\varphi_{2}+\cdot\cdot\cdot=\frac{2k^{2}\theta}{1-k^{2}}\sum_{m=0}^{\infty}\varepsilon^{m}
$$
Geometric series:
$$
\sum_{m=0}^{\infty}\varepsilon^{m}={\frac{1}{1-\varepsilon}}
$$
Thus:
$$
\boxed{\varphi=\frac{2k^{2}\theta}{(1-\varepsilon)(1-k^{2})}}
$$ | $$
\boxed{\varphi=\frac{2k^{2}\theta}{(1-\varepsilon)(1-k^{2})}}
$$ |
443 | ELECTRICITY | Surface plasmon polariton (SPP) is a research hotspot in micro-nano optics. It is not difficult to discover through calculations that the interaction between free electrons and photons in the metal-dielectric interface region can form specific electromagnetic modes. The free electron density of the metal we use is denoted as $n$, the electrons are modeled as quasiparticles with an equivalent mass $m$ and charge $-e$. Because under optical conditions, the light frequency $\omega$ is often smaller than the plasma frequency $\omega_{p}=\sqrt{n e^{2}/\varepsilon_{0}m}$, it can be found that the corresponding equivalent relative permittivity of the metal is a negative real number.
If a plane electromagnetic wave at this frequency, with a relative permittivity $\varepsilon_{1}>1$ for the dielectric, enters the metal (which at this frequency has an equivalent relative permittivity $\varepsilon_{2}<0$) with a wavevector $k_{x}=\alpha$, $k_{y}=-\mathrm{i}\beta$ ($\alpha$ and $\beta$ are real numbers), the electric field lies in the $xy$ plane, and the interface is defined as the $y=0$ plane, with the positive $y$-axis pointing from the dielectric towards the metal, there exists a specific horizontal wavevector $k_{x}$ such that no reflected light is generated. Instead, a unique optical mode satisfying Maxwell’s equations is formed solely by the incident and refracted light—this corresponds to the surface plasmon polariton mode.
However, in this problem, we need to consider the case where the metal actually has a relative complex permittivity $\widetilde{\varepsilon}_{2}=\varepsilon_{21}+\mathrm{i}\varepsilon_{22}$, where $\varepsilon_{21}<0$ and the imaginary part $\varepsilon_{22}$ is a small quantity. Under these conditions, the surface plasmon polariton mode not only propagates but also attenuates. Please express the wavelength $\lambda$ of the surface plasmon polariton mode in terms of the dielectric relative permittivity $\varepsilon_{1}$, the metal’s relative complex permittivity, and the vacuum wavelength $\lambda_{0}$. Approximate to the lowest order. | The magnetic field strength in the $z$ direction is tangentially continuous:
$$
H_{1z}=H_{2z}=H
$$
According to Maxwell's equations:
$$
\nabla\times\pmb{H}=\varepsilon\frac{\partial\pmb{E}}{\partial t}
$$
Furthermore, since the electric field is tangentially continuous:
$$
E_{1x}=E_{2x}
$$
The relationship obtained is:
$$
\frac{k_{1y}}{\varepsilon_{1}}=\frac{-\mathrm{i}\beta_{1}}{\varepsilon_{1}}=\frac{\mathrm{i}\beta_{2}}{\varepsilon_{2}}=\frac{k_{2y}}{\varepsilon_{2}}
$$
$$
\begin{array}{r}{k_{1}^{2}=\alpha^{2}-\beta_{1}^{2}=\varepsilon_{1}k_{0}^{2}>0}\\ {k_{2}^{2}=\alpha^{2}-\beta_{2}^{2}=\varepsilon_{2}k_{0}^{2}<0}\end{array}
$$
Solving for $k_{1x}=k_{2x}=k_{x}=\alpha$, we find:
$$
k_{x}=\frac{\omega}{c}\sqrt{\frac{\varepsilon_{1}\varepsilon_{2}}{\varepsilon_{1}+\varepsilon_{2}}}
$$
Squaring the above expression and rewriting, we get:
$$
(k_{x1}+\mathrm{i}k_{x2})^{2}=k_{0}^{2}\frac{\varepsilon_{1}(\varepsilon_{21}+\mathrm{i}\varepsilon_{22})}{\varepsilon_{1}+\varepsilon_{21}+\mathrm{i}\varepsilon_{22}}\approx k_{0}^{2}\frac{\varepsilon_{1}\varepsilon_{21}}{\varepsilon_{1}+\varepsilon_{21}}\left(1+\mathrm{i}\frac{\varepsilon_{22}}{\varepsilon_{21}}\right)\left(1-\mathrm{i}\frac{\varepsilon_{22}}{\varepsilon_{1}+\varepsilon_{21}}\right)
$$
By comparing the real and imaginary parts and making approximations:
$$
\begin{array}{c}{{k_{x1}^{2}=k_{0}^{2}\displaystyle\frac{\varepsilon_{1}\varepsilon_{21}}{\varepsilon_{1}+\varepsilon_{21}}}}\\ {{2k_{x1}k_{x2}=k_{0}^{2}\displaystyle\frac{\varepsilon_{1}\varepsilon_{21}}{\varepsilon_{1}+\varepsilon_{21}}\cdot\displaystyle\frac{\varepsilon_{1}\varepsilon_{22}}{\varepsilon_{21}\left(\varepsilon_{1}+\varepsilon_{21}\right)}}}\end{array}
$$
Substituting into:
$$
k_{0}=\frac{2\pi}{\lambda_{0}}\quad,\quad k_{x1}=\frac{2\pi}{\lambda}\quad,\quad k_{x2}=\frac{2\pi}{d}
$$
Finally, we obtain:
$$
\lambda=\sqrt{\frac{\varepsilon_{1}+\varepsilon_{21}}{\varepsilon_{1}\varepsilon_{21}}}\lambda_{0}
$$ | $$
\lambda=\sqrt{\frac{\varepsilon_{1}+\varepsilon_{21}}{\varepsilon_{1}\varepsilon_{21}}}\lambda_{0}
$$ |
115 | ADVANCED | Consider a charged rod undergoing relativistic rotation. Within an infinitely long cylinder with radius $R$, positive charges with number density $n$ and charge $q$ revolve around the axis with a uniform angular velocity of $\omega$ in a relativistic motion. The original reference frame is $S$, and now we switch to a reference frame $S^{\prime}$ that moves parallel to the axis with a velocity $\beta c$ relative to frame $S$. We consider the radiation from the accelerating charges. In the relativistic case, the instantaneous outgoing radiated electromagnetic power of a charge $q$ with arbitrary velocity $\pmb{v}$ and arbitrary acceleration $\pmb{a}$ follows the Larmor formula:
$$
P={\frac{\gamma^{6}q^{2}}{6\pi\varepsilon_{0}c^{3}}}\left[a^{2}-{\frac{(v\times a)^{2}}{c^{2}}}\right]
$$
where $\gamma=1/\sqrt{1-v^{2}/c^{2}}$. Assume all charge radiation is incoherent. Without using energy or power transformation formulas, calculate directly the radiation power per unit length of the rod in frame $S^{\prime}$ through the charge radiation method. The speed of light in a vacuum is $c$. | Charge and current distribution:
$$
\rho=n q
$$
$$
j_{\theta}=\rho v=\Im\Im q\omega\tau
$$
Note that the above equation is valid only for $r<R$, otherwise it is zero. There is an electric field inside and outside:
There is only a magnetic field inside:
$$
E_{\mathrm{r}}(r)=\int_{0}^{\operatorname*{min}\{r,R\}}\frac{\mathrm{d}\lambda}{2\pi\varepsilon_{0}\tau}\quad,\quad\mathrm{d}\lambda=\rho(r^{\prime})\cdot2\pi r^{\prime}\mathrm{d}\tau^{\prime}
$$
The integration yields the result:
$$
\mathcal{B}_{z}(r)=\int_{r}^{R}\mu_{0}\mathrm{d}i~~~,\mathrm{d}i=j_{\theta}(r^{\prime})\cdot\mathrm{d}r^{\prime}
$$
$$
E_{\mathrm{r}}(r)={\left\{\begin{array}{l l}{{\displaystyle{\frac{n q r}{2\varepsilon_{0}}}}}&{(r<R)}\ {{\displaystyle{\frac{n q R^{2}}{2\varepsilon_{0}r}}}}&{(r>R)}\end{array}\right.}
$$
$$
\mathcal{B}_{z}(r)=\left\{\begin{array}{l l}{\frac{\mu_{0}n q\omega(R^{2}-r^{2})}{2}}&{(r<R)}\ {0}&{(r>R)}\end{array}\right.
$$
In a new frame of reference, the number density scales directly:
$$
n^{\prime}=\frac{n}{\sqrt{1-\beta^{2}}}
$$
The radius remains unchanged:
$$
\boldsymbol{r}^{\prime}=\boldsymbol{r}
$$
But velocity acquires a new component:
$$
v_{z}^{\prime}=-\beta c
$$
The original components also change. Since one loop in the original reference frame corresponds to proper time, the angular velocity of the new helical motion becomes:
Thus, the new charge and current distribution is:
$$
v_{\theta}^{\prime}\cdot T^{\prime}=v_{\theta}T
$$
$$
T^{\prime}=\frac{T}{\sqrt{1-\beta^{2}}}
$$
$$
\rho^{\prime}=\pi^{\prime}q=\frac{n q}{\sqrt{1-\beta^{2}}}
$$
$$
j_{\theta}^{\prime}=\rho^{\prime}v_{\theta}^{\prime}=n q\omega r
$$
$$
j_{z}^{\prime}=\rho^{\prime}v_{z}^{\prime}=-\frac{n q\beta c}{\sqrt{1-\beta^{2}}}
$$
Thus, the original two electromagnetic fields are scaled proportionally as described above:
$$
E_{r}^{\prime}(r)=\left\{\begin{array}{l l}{\displaystyle\frac{n q r}{2\varepsilon_{0}\sqrt{1-\beta^{2}}}}&{(r<R)}\ {\displaystyle\frac{n q R^{2}}{2\varepsilon_{0}\sqrt{1-\beta^{2}}r}}&{(r>R)}\end{array}\right.
$$
$$
B_{\natural}^{\prime}(r)=\left\{{\begin{array}{l l}{{\frac{\mu_{0}n q\omega r^{2}}{2}}}&{(r<R)}\ {0}&{(r>R)}\end{array}}\right.
$$
However, an axial current is also produced, leading to a surrounding magnetic field, expressed as:
$$
B_{\theta}(r)=\int_{0}^{r}\frac{\mu_{0}\mathrm{d}I}{2\pi r}\quad,\quad\mathrm{d}I=j_{z}^{\prime}(r^{\prime})\cdot2\pi r^{\prime}\mathrm{d}r^{\prime}
$$
The integration yields the result:
$$
B_{\theta}^{\prime}(r)=\left\{\begin{array}{l l}{-\frac{\mu_{0}n q\beta c r}{2\sqrt{1-\beta^{2}}}}&{(r<R)}\\ {-\frac{\mu_{0}n q\beta c R^{2}}{2\sqrt{1-\beta^{2}}r}}&{(r>R)}\end{array}\right.
$$
(3)
Radiation power of a single particle in the original frame:
$$
P_{1}=\frac{q^{2}}{6\pi\varepsilon_{0}c^{3}}\cdot\frac{\omega^{4}r^{2}\left(1-\omega^{2}r^{2}/c^{2}\right)}{\left(1-\omega^{2}r^{2}/c^{2}\right)^{3}}
$$
Particle number per unit length:
$$
\mathrm{d}N=n\cdot2\pi\tau\mathrm{d}r=\pi n\mathrm{d}\left(r^{2}\right)
$$
Integration:
$$
P=\int_{0}^{R}P_{1}\mathrm{d}N
$$
Gives:
$$
P=\frac{n q^{2}c}{6\varepsilon_{0}}\left[\frac{\omega^{2}R^{2}/c^{2}}{1-\omega^{2}R^{2}/c^{2}}+\ln\left(1-\frac{\omega^{2}R^{2}}{c^{2}}\right)\right]
$$
Radiation power of a single particle in the new frame:
$$
P_{1}^{\prime}=\frac{q^{2}}{6\pi\varepsilon_{0}c^{3}}\cdot\frac{\left(v_{\theta}^{\prime2}/r\right)^{2}\cdot\left[1-(v_{\bar{s}}^{\prime2}+v_{\theta}^{\prime2})/c^{2}\right]}{[1-(v_{\bar{s}}^{\prime2}+v_{\bar{\varepsilon}}\cdot\bar{\mathbf{\xi}}}
$$
Reorganized as:
$$
\mathcal{P}_{1}^{\prime}=\mathcal{P}_{1}
$$
However, the number of particles per unit length increases by a factor of $1/\sqrt{1-\beta^{2}}$. Hence, the total power:
$$
P^{\prime}=\frac{P}{\sqrt{1-\beta^{2}}}
$$
Gives:
$$
P^{\prime}=\frac{n q^{2}c}{6\varepsilon_{0}\sqrt{1-\beta^{2}}}\left[\frac{\omega^{2}R^{2}/c^{2}}{1-\omega^{2}R^{2}/c^{2}}+\ln\left(1-\frac{\omega^{2}R^{2}}{c^{2}}\right)\right]
$$ | $$
P'=\frac{n q^2 c}{6 \varepsilon_0 \sqrt{1-\beta^2}} \left[\frac{\omega^2 R^2 / c^2}{1-\omega^2 R^2 / c^2}+\ln\left(1-\frac{\omega^2 R^2}{c^2}\right)\right]
$$ |
493 | ELECTRICITY | There is a uniform rigid current-carrying circular ring with mass $M$ and radius $R$, carrying a current of $I$. At a distance $h$ from the center of the ring, there is a magnetic dipole with a magnetic dipole moment of $m$, which is fixed and cannot rotate. The line connecting the magnetic dipole and the center of the ring is perpendicular to the plane of the ring. The ring can only rotate about a certain diameter axis passing through the center of the ring. Initially, the plane of the ring is perpendicular to the direction of the magnetic dipole moment. Now, the ring is slightly rotated by a small angle. Find the period $T_1$ of the simple harmonic vibration of the ring after release (assuming that all parameters satisfy the stable equilibrium condition, and in the initial state, the direction of the magnetic field produced by the current on the ring at the position of the magnetic dipole is the same as that of the magnetic dipole). Given the vacuum permeability is $\mu_0$. | When the ring is rotated by an angle \( \theta \), a cylindrical coordinate system is established, and the coordinates corresponding to the position of \( \vec{m} \) are
\[
\begin{cases}
\rho = h \sin \theta \\
z = h \cos \theta
\end{cases}
\]
The magnetic field component \( B_z(0, z) \) is given by
\[
B_z(0, z) = \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}}
\]
The expression for the energy of simple harmonic motion involves the second-order term of \( \theta \), so both \( B_z \) and \( B_\rho \) are retained up to the second-order term in \( \rho \) (where \( B_{\rho,0} \), \( B_{z,1} \), and \( B_{\rho,2} \) are all zero):
\[
\begin{cases}
B_{z,0} = B_z(0, z) = \frac{\mu_0 I R^2}{2(R^2 + z^2)^{3/2}} \\
B_{\rho,1} = -\frac{\rho}{2} \frac{\mathrm{d}B_z(0, z)}{\mathrm{d}z} = \frac{3 \mu_0 I R^2 z \rho}{4(R^2 + z^2)^{5/2}} \\
B_{z,2} = -\frac{\rho^2}{4} \frac{\mathrm{d}^2 B_z(0, z)}{\mathrm{d}z^2} = \frac{3 \mu_0 I R^2 (R^2 - 4z^2) \rho^2}{8(R^2 + z^2)^{7/2}}
\end{cases}
\]
Using the small-angle approximation \( \sin \theta \approx \theta \) and \( \cos \theta \approx 1 - \frac{1}{2} \theta^2 \):
\[
\begin{cases}
B_{z,0} \approx \frac{\mu_0 I R^2}{2(R^2 + h^2)^{3/2}} \left(1 + \frac{3}{2} \frac{h^2 \theta^2}{h^2 + R^2}\right) \\
B_{\rho,1} \approx \frac{3 \mu_0 I R^2 h^2 \theta}{4(R^2 + h^2)^{5/2}} \\
B_{z,2} \approx \frac{3 \mu_0 I R^2 h^2 \theta^2 (R^2 - 4h^2)}{8(R^2 + h^2)^{7/2}}
\end{cases}
\]
The potential energy at the angular displacement \( \theta \) is
\[
W = -\vec{m} \cdot \vec{B} = -m(B_z \cos \theta + B_\rho \sin \theta) = -m(B_{z,0} \cos \theta + B_{z,2} \cos \theta + B_{\rho,1} \sin \theta)
\]
\[
= -m \left[ \frac{\mu_0 I R^2}{2(R^2 + h^2)^{3/2}} \left(1 + \frac{3}{2} \frac{h^2 \theta^2}{h^2 + R^2}\right) \left(1 - \frac{\theta^2}{2}\right) + \frac{3 \mu_0 I R^2 h^2 \theta^2 (R^2 - 4h^2)}{8(R^2 + h^2)^{7/2}} \left(1 - \frac{\theta^2}{2}\right) + \frac{3 \mu_0 I R^2 h^2 \theta^2}{4(R^2 + h^2)^{5/2}} \right]
\]
\[
\approx -\frac{m \mu_0 I R^2}{2(R^2 + h^2)^{3/2}} + \frac{1}{2} \frac{\mu_0 m I R^2 (2h^4 + 2R^4 - 11R^2 h^2)}{4(h^2 + R^2)^{7/2}} \theta^2
\]
In the above derivation, all terms higher than the second order in \( \theta \) are neglected.
The rotational kinetic energy of the ring is
\[
K = \frac{1}{2}\left(\frac{1}{2}MR^{2}\right)\dot{\theta}^{2}
\]
According to the principle of determining simple harmonic motion by energy methods, the resonant period is
\[
T_{1}=2\pi\sqrt{\frac{\frac{1}{2}MR^{2}}{\frac{\mu_{0}mIR^{2}(2h^{4}+2R^{4}-11R^{2}h^{2})}{4(h^{2}+R^{2})^{7/2}}}} = 2\pi\sqrt{\frac{2M(h^{2}+R^{2})^{7/2}}{\mu_{0}mI(2h^{4}+2R^{4}-11R^{2}h^{2})}}
\] | \[
T_{1}= 2\pi\sqrt{\frac{2M(h^{2}+R^{2})^{7/2}}{\mu_{0}mI(2h^{4}+2R^{4}-11R^{2}h^{2})}}
\] |
69 | MECHANICS | In 2023, the team of astronomers led by Jerome Oros at San Diego State University in the United States released their latest research findings, discovering a new binary star system called "Kepler-49". Meanwhile, astronomical observations indicate that nearby planets are influenced by this binary star system, resulting in orbital precession. This problem aims to explain this phenomenon concisely.
"Kepler-49" is a binary star system mainly composed of two stars (Saka and Juipte) with masses $M_{S}$ and $M_{J}$ (both known). Under the mutual gravitational interaction, they perform stable circular motion around the center of mass. With an angular frequency $\Omega$, the respective motion radii $R_{S}$, $R_{J}$ (unknown quantities) and the distance between them $R_{0}$ (unknown quantity) can be determined. The gravitational constant $G$ is known.
The rotation center of the two stars is set as the origin of coordinates, and a polar coordinate system is established with the closest point as the radial distance to describe the movement of the planet. Experimental observation shows that the scale of the Saka and Juipte system is much smaller than the distance to the planet (i.e., $R_{0} \ll r$). Meanwhile, during one period of $\Omega$, the planet can be approximately considered as not moving. The modified potential energy $V_{\mathrm{eff}}$ (unknown quantity) of the planet can be determined, retaining the lowest order correction term $\delta V$.
It is known that the energy of the planet is $E < 0$, the orbital angular momentum is $\bar{L}$, and the mass is $m$. Only considering the zeroth-order term $V_{0}$ of the gravitational potential energy, the motion equation of the planet $r(\theta)$ (unknown quantity) can be determined.
Question: Consider now the form of potential energy $V = V_{0} + \delta V$ for the planet. In this case, $\vec{B}$ is no longer conserved. Solve for the orbital precession angle $\Delta{\alpha}$ (express the answer in terms of $M_{J}, M_{S}, G, \Omega, m, L$).
Hint 1: The Laplace-Runge-Lenz vector is introduced in the Kepler problem, and in this problem, it is defined as
$$
\vec{B} = \vec{v} \times \vec{L} - G(M_{J} + M_{S})m\hat{r}.
$$
Consider the planet moving only under the potential energy $V = V_{0}$, and in this case, let $\vec{B}_{0} = \vec{B}$. It can be proved that $\vec{B}_{0}$ is conserved (the proof method is unknown), and the expression for $|\vec{B}_{0}|$ can be determined (unknown quantity).
Hint 2: The precession angle is defined as the rotation angle of the closest point during a single period of the planet. Consider the LRL vector pointing to the closest point, retaining the lowest order term:
$$
\Delta\alpha = \left|\frac{\vec{B}_{0}}{|\vec{B}_{0}|} \times \left(\frac{\vec{B}}{|\vec{B}|} - \frac{\vec{B}_{0}}{|\vec{B}_{0}|}\right)\right| \approx \frac{|\vec{B}_{0} \times \Delta\vec{B}|}{|\vec{B}_{0}|^{2}}.
$$ | From
$$
\frac{G M_{J}M_{S}}{(R_{J}+R_{S})^{2}}=M_{J}R_{J}\Omega^{2}=M_{S}R_{S}\Omega^{2}
$$
we solve
$$
\begin{array}{c}{{R_{J}=\displaystyle\frac{{M}_{S}}{{ M}_{S}+{ M}_{J}}\left(\frac{{ G}({ M}_{S}+{ M}_{J})}{\Omega^{2}}\right)^{\frac{1}{3}}}}\ {{{}}}\ {{R_{S}=\displaystyle\frac{{ M}_{J}}{{ M}_{S}+{ M}_{J}}\left(\frac{{ G}({ M}_{S}+{ M}_{J})}{\Omega^{2}}\right)^{\frac{1}{3}}}}\end{array}
$$
Further
$$
R_{0}=R_{J}+R_{S}=\left({\frac{G(M_{S}+M_{J})}{\Omega^{2}}}\right)^{\frac{1}{3}}
$$
As stated in the problem, during a single period of rotation of the binary system, it is approximately assumed that the planet's position does not change; at time $t$, the potential energy is
$$
V=-{\frac{G M_{J}m}{\sqrt{r^{2}+R_{J}^{2}-2r R_{J}\cos\left(\Omega t-\varphi_{0}\right)}}}-{\frac{G M_{S}m}{\sqrt{r^{2}+R_{S}^{2}+2r R_{S}\cos\left(\Omega t-\varphi_{0}\right)}}}
$$
Expand the gravitational potential in a power series in $\scriptstyle{\frac{R_{0}}{r}}$:
$$
V=-{\frac{G M_{J}m}{r}}\left(1+\cos\left(\Omega t-\varphi_{0}\right){\frac{R_{J}}{r}}+{\frac{3\cos^{2}\left(\Omega t-\varphi_{0}\right)-1}{2}}{\frac{R_{J}^{2}}{r^{2}}}\right)\ldots
$$
$$
\ldots-{\frac{G M_{s}m}{r}}\left(1-\cos\left(\Omega t-\varphi_{0}\right){\frac{R_{s}}{r}}+{\frac{3\cos^{2}\left(\Omega t-\varphi_{0}\right)-1}{2}}{\frac{R_{s}^{2}}{r^{2}}}\right)
$$
Consider the time-averaged result and retain the lowest-order correction term, yielding $V_{\mathbf{eff}}$:
$$
V_{\mathrm{eff}}={\bar{V}}={\frac{\Omega}{2\pi}}\int_{0}^{\frac{2\pi}{n}}V\mathrm{d}t=V_{0}+\delta V=-{\frac{G(M_{J}+M_{S})m}{r}}-{\frac{G M_{J}M_{S}m}{4(M_{J}+M_{S})}}\cdot{\frac{R_{0}^{2}}{r^{3}}}
$$
When considering only the zero-order term of the potential energy
$$
V_{0}(r)=-\frac{G(M_{S}+M_{J})m}{r}
$$
For $\pmb{E}<\mathbf{0}$, the planet is on an elliptical orbit; provided without proof (to be memorized as fundamental knowledge):
$$
e=\sqrt{1+\frac{2E L^{2}}{G^{2}(M_{S}+M_{J})^{2}m^{3}}}
$$
$$
p={\frac{L^{2}}{G(M_{J}+M_{S})m^{2}}}
$$
As stated in the problem, taking the position near the center as the polar radius, we solve
$$
r(\theta)=\frac{\frac{L^{2}}{G(M_{J}+M_{S})m^{2}}}{1+\sqrt{1+\frac{2E L^{2}}{G^{2}(M_{S}+M_{J})^{2}m^{3}}}\cos\theta}
$$
(4)
Considering the form of the potential energy at this time
$$
V=V_{0}=-\frac{G(M_{J}+M_{S})m}{r}
$$
$\vec{B}$ conservation equivalently means it does not change over time. Consider the conservation of angular momentum
$$
{\frac{\mathrm{d}{\vec{B}}}{\mathrm{d}t}}={\frac{1}{m}}({\vec{F}}\times L)-G(M_{J}+M_{S})m{\dot{\theta}}{\hat{\theta}}={\frac{1}{m}}\cdot{\frac{G(M_{J}+M_{S})m}{r^{2}}}\cdot m r^{2}{\dot{\theta}}{\hat{\theta}}-G(M_{J}+M_{S})m{\dot{\theta}}{\hat{\theta}}=0
$$
For computational convenience, select the special position where the planet is near the center at $\theta=0$:
$$
\vec{B}_{0}=\frac{L^{2}(1+e)}{p}-G(M_{J}+M_{S})m\hat{r}=G(M_{J}+M_{S})m\sqrt{1+\frac{2E L^{2}}{G^{2}(M_{S}+M_{J})^{2}m^{3}}}\hat{r}
$$
where $\hat{r}$ is the direction of the planet near the center.
(5)
From the above derivation, we know that in the case where only the zero-order term of the potential energy is considered, $\vec{B}$ is conserved and its direction points to the near-center point. Since $\vec{B}$ reflects information about the orbit's shape and orientation, consider the calculation of $\Delta\vec{B}$ at this time
$$
\frac{\mathrm{d}\vec{B}}{\mathrm{d}t}=\frac{1}{m}(\vec{F}\times L)-G(M_{J}+M_{S})m\dot{\theta}\hat{\theta}=-\frac{1}{m}\frac{\partial(\delta V)}{\partial r}\hat{r}\times\vec{L}=\frac{3G M_{J}M_{S}m}{4(M_{J}+M_{S})}\cdot\frac{R_{0}^{2}}{r^{2}}\dot{\theta}\hat{\theta}
$$
Integrate over a single period
$$
\Delta\vec{B}=\int_{0}^{T}\frac{\mathrm{d}\vec{B}}{\mathrm{d}t}\cdot\mathrm{d}t=\int_{0}^{2\pi}\frac{3G M_{J}M_{S}m}{4(M_{J}+M_{S})}\cdot\frac{R_{0}^{2}}{r^{2}}\mathrm{d}\theta\hat{\theta}
$$
For the lowest-order correction term, use the elliptical orbit equation obtained in (3):
$$
r={\frac{p}{1+e\cos\theta}}
$$
$$
\Delta\vec{B}=\int_{0}^{2\pi}\frac{3G M_{J}M_{S}m R_{0}^{2}}{4(M_{J}+M_{S})}\cdot\frac{(1+e\cos\theta)^{2}}{p^{2}}(-\sin\theta\hat{x}+\cos\theta\hat{y})\mathrm{d}\theta=\frac{3G M_{J}M_{S}m R_{0}^{2}}{2(M_{J}+M_{S})}\cdot\frac{e\pi}{p^{2}}\hat{y}
$$
Using the hint in the problem
$$
\Delta\alpha={\frac{|\vec{B_{0}}\times\Delta\vec{B}|}{|\vec{B_{0}}|^{2}}}={\frac{3\pi}{2}}\cdot{\frac{G^{\frac{8}{3}}M_{J}M_{S}(M_{S}+M_{J})^{\frac{2}{3}}m^{4}}{\Omega^{\frac{4}{3}}L^{4}}}
$$ | $$\Delta\alpha = \frac{3\pi}{2} \cdot \frac{G^{8/3} M_J M_S (M_S + M_J)^{2/3} m^4}{\Omega^{4/3} L^4}$$ |
65 | ELECTRICITY | In the upper half-space, there is a uniform magnetic field with magnitude \( B \) directed vertically upwards, and the ground is sufficiently rough.
Now, there is a solid insulating sphere with uniformly distributed mass \( m \) and radius \( R \) . On its surface at the equator, there are 3 mutually orthogonal wires with negligible mass and thickness and a unit length resistance of \( \lambda \). These wires are connected at the intersection point. The sphere is placed on the ground so that the circular surface formed by one of the wires is perpendicular to the direction of the magnetic field. It is known that the sphere's center has acquired a velocity \( v_0 \) in the horizontal plane, and its initial angular velocity is such that it is in a pure rolling state. What is the subsequent change in velocity? | By noticing that the cutting of the magnetic field by the wire induces an electromotive force, which in turn generates Ampere force and torque, the motion state of the sphere changes. Now, due to the completeness of the circuit, the magnitude of the Ampere force is zero, and we only need to consider torque. Since the magnetic field here is uniform, the problem is to determine the magnetic moment $\mu$.
We use a fundamental yet concise method to solve this problem: we move to the rest frame of the sphere to analyze the problem, and establish the coordinate system as shown in the figure. In this frame, the magnetic field appears to rotate. It should be noted that the focus here is on solving the current distribution. This change of frame is purely mathematical and does not involve relativistic effects.
Below, we only consider the variation of the magnetic field in the $\pmb{x}$ direction, denoted as $\dot{B_{x}}$, while the variations in the $y/z$ directions are analyzed analogously. Since the relationships between electromotive force, current, and magnetic moment are linear, superposition can be performed in the end! Here, due to the symmetry introduced by the circuit, the electromotive forces and resistances for the eight mesh loops in the front and back are exactly the same, meaning all the loop equations for each mesh are identical. This results in no current through the cross-shaped wires in the center.
Only the circular arcs perpendicular to the ${\pmb x}$ direction have current, with the electromotive force given by:
$$
\varepsilon=-\pi R^{2}\dot{B_{x}}
$$
Thus, the current is:
$$
I={\frac{\varepsilon}{2\pi R\lambda}}
$$
The $\pmb{\mathscr{x}}$ component of the total magnetic moment is:
$$
\mu_{x}=I S=-\frac{\pi R^{3}}{2\lambda}\dot{B_{x}}
$$
Therefore, the total magnetic moment is:
$$
\vec{\mu}=-\frac{\pi R^{3}}{2\lambda}\dot{\vec{B}}
$$
In the rest frame, the magnetic field rotates at an angular velocity of $-\alpha$ (referring to the angular velocity of the sphere's rotation). Hence, $\dot{\vec{B}}=-\vec{\omega}\times\vec{B}$. In the ground frame, we arrive at a remarkably elegant result:
$$
\vec{\mu}=\frac{\pi\mathcal{R}^{3}}{2\lambda}\vec{\omega}\times\vec{B}
$$
This result is remarkably coincidental and symmetric, allowing us to utilize a simple vector method for mechanical analysis. Following the coordinate system established in the first question:
The theorem of the motion of the center of mass is:
$$
m{\frac{d{\vec{v}}}{d t}}={\vec{f}}
$$
Rotational motion law:
$$
\frac{2}{5}m R^{2}\frac{d\vec{\omega}}{d t}=R\vec{n}\times\vec{f}+\vec{\mu}\times\vec{B}
$$
Pure rolling constraint:
$$
\vec{v}+R\vec{\omega}\times\vec{n}=0
$$
By taking the cross product with $\vec{n}$ on both sides of the rotational motion law, we get:
$$
\frac{2}{5}m R^{2}\frac{d\vec{n}\times\vec{\omega}}{d t}=-R\vec{f}+\frac{\pi R^{3}}{2\lambda}\vec{n}\times\left[\left(\vec{B}\cdot\vec{\omega}\right)\vec{B}-B^{2}\vec{\omega}\right]
$$
Noting that $\vec{n}\times\vec{B}=0$, the second term on the right-hand side of the above equation is omitted. Substituting the pure rolling constraint yields:
$$
\frac{2}{5}m\frac{d\overrightarrow{v}}{d t}=-\overrightarrow{f}-\frac{\pi R B^{2}}{2\lambda}\overrightarrow{v}
$$
Combining the theorem of the motion of the center of mass:
Direct integration gives:
$$
\frac{7m}{5}\frac{d\overrightarrow{v}}{d t}=-\frac{\pi R B^{2}}{2\lambda}\overrightarrow{v}
$$
$$
\vec{v}=\vec{v_{0}}e^{-\frac{5\pi R B^{2}}{14m\lambda}t}
$$
This result indicates that while the direction of velocity of the sphere does not change, its magnitude continually decreases. | $$
\vec{v} = \vec{v_0} e^{-\frac{5\pi R B^2}{14m\lambda} t}
$$ |
131 | ELECTRICITY | Place a small magnetic needle with a magnetic moment \( m \) directly above a superconducting sphere with a radius \( R \) at a distance \( a \). It is known that the magnetic field outside the superconducting sphere can be regarded as the superposition of the magnetic field of \( m \) and a magnetic field produced in space by an internal magnetic moment \( m' \) inside the superconducting sphere. Find the magnitude of \( m' \). (It is known that the magnetic moment \( m \) has a magnetic induction field distribution in space as: \( \mathbf{B}={ \frac{ \mu_{0}m}{4 \pi r^{3}}}{ \left(2 \cos \theta \mathbf{e}_{r}+ \sin \theta \mathbf{e}_{ \theta} \right)} \)) | Based on symmetry, the form is definitely on the line connecting with the center of the sphere. As shown in the figure, assume \(m' \) is at a height \(r' \) above point O, in the downward direction \( \boldsymbol{r^{ \prime}} \) (if the direction is upward, the calculated value of \(m \) would be negative). The normal component of the magnetic induction intensity on the surface of a superconductor is zero, allowing us to calculate the normal component of the magnetic field at points P and Q. At point P: $$ B_{p}= \frac{ \mu_{0}}{4 \pi} \frac{2m}{ \left(a-R \right)^{3}}- \frac{ \mu_{0}2m}{4 \pi \left(R-b \right)^{3}}=0 $$ At point Q: $$ B_{ \varrho}= \frac{ \mu_{0}}{4 \pi} \frac{2m}{ \left(a+R \right)^{3}}- \frac{ \mu_{0}2m^{ \prime}}{4 \pi(R+b)^{3}}=0 $$ Solving equations (1) and (2) simultaneously, we get $$ \begin{array}{c} \ {{m^{ \prime}= \displaystyle \frac{R^{3}}{a^{3}}m}} \end{array} $$ | $$m' = \frac{R^3}{a^3} m$$ |
709 | MECHANICS | In a zero-gravity space, there is a ring made of a certain material, with a radius of $R$, density $\rho$, Young's modulus $E$, and a circular cross-section with a radius of $r$ ($r \ll R$). The center of the ring is stationary in a rotating reference frame with an angular velocity $\Omega$, and the normal of the ring is in the same direction as $\Omega$. $\Omega$ is much smaller than the natural frequency of the ring's vibration. A perturbation is applied, causing vibrations on the ring. Considering a standing wave solution of order $m$, it is known that the antinodes of the standing wave will precess. Find the precession angular velocity $\delta\omega$. The positive direction is defined to be the same as $\Omega$. Ignore motion perpendicular to the plane of the ring; the order $m$ indicates that the displacement has the form of $\cos(m\theta)$ or $\sin(m\theta)$.
| Assume the radial and angular displacements of a particle on the ring are $w, v$, and the kinetic energy, potential energy line density are given as $$ T = \frac{1}{2}\rho S\left((\dot v + \Omega (R + w))^2 + (\dot w - \Omega v)^2\right) $$ $$ V = \frac{E I}{2 R^4}\left(w + \partial_\theta^2w\right)^2 $$ The incompressibility constraint $$ \partial_\theta v + w = 0 $$ Thus, the Lagrangian density is $$ \mathcal{L} = \frac{1}{2}\rho S\left((\dot v + \Omega (R + w))^2 + (\dot w - \Omega v)^2\right) - \frac{E I}{2 R^4}\left(w + \partial_\theta^2w\right)^2 - \lambda (\partial_\theta v + w) $$ Substituting into the Lagrange's equations, we get $$ \begin{cases} \ddot v + 2 \Omega \dot w + \partial_\theta \lambda = 0, \\ \ddot w - 2 \Omega \dot v + \frac{EI}{\rho S R}\left(\partial_\theta^4 + 2\partial_\theta^2 + 1\right)w - \lambda = 0, \\ \partial_\theta v + w = 0 \end{cases} $$ Eliminate variables to obtain $$ \partial_t^2\partial_\theta^2 w - \partial_t^2 w + 4 \Omega \partial_t \partial_\theta w + \frac{EI}{\rho S R}(\partial_\theta^6 + 2\partial_\theta^4 + \partial_\theta^2)w = 0 $$ Substituting the $m$-th order traveling wave solution $w = e^{i(\omega t + m\theta)}$ gives $$ (m^2+1)\omega^2-4m\Omega\omega-\frac{EI}{\rho S R}(m^6-2m^4+m^2)=0 $$ Solving gives $$ \omega = \frac{4m\Omega\pm\sqrt{16m^2\Omega^2+\frac{4EI}{\rho S R}(m^6-2m^4+m^2)}}{2(m^2+1)}=\omega_1 \pm \omega_{0} $$ The precession standing wave solution is $$ w = \cos((\omega_1 + \omega_{0})t+m\theta)+\cos((\omega_1 - \omega_{0})t+m\theta) = 2 \cos(\omega_1 t + m\theta)\cos(\omega_0 t) $$ It can be observed that the wave crest will retreat with a precession angular velocity, according to our sign convention, $$ \delta\omega = - \frac{2}{m^2+1}\Omega $$ Of course, it can also be solved using Newtonian mechanics methods, but it is a bit more tedious.
| $$
\delta \omega = -\frac{2}{m^2+1} \Omega
$$ |
423 | THERMODYNAMICS | One mole of a substance is a simple $p,v,T$ system. The coefficient of body expansion in any case is
$$
\alpha=3/T
$$
The adiabatic equation in any case is
$$
p v^{2}=C
$$
The isobaric heat capacity of the system in the arbitrary case is exactly: $$
$$
c_{p}\propto \frac{p v}{T}
$$
The scale factor is a state-independent dimensionless constant, and
To determine the entropy change $$S_D-S_A$$ of the D state compared to the A state, two states $$A,D$$ on the $$p-v$$ graph are chosen, knowing that $$p_{A}$$,$$v_{A}$$,$$T_{A}$$,$$p_{D}$$,$$v_{D}$$ and that the entropy change of the D state compared to the A state is $$S_D-S_A$$.
| First of all
$$
\alpha={\frac{1}{v}}\left({\frac{\partial v}{\partial T}}\right)_{p}={\frac{3}{T}}
$$
The equation of state can be obtained
$$
v=T^{3}h(p)
$$
Write the full differential of entropy and utilize Maxwell's relation:
$$
d S={\frac{\partial S}{\partial T}}d T+{\frac{\partial S}{\partial p}}d p
$$
$$
\frac{\partial S}{\partial p}=-\frac{\partial v}{\partial T}=-3T^{2}h
$$
During adiabatic process $$ d S=0$$
$$
{\frac{d p}{d T}}={\frac{\frac{\partial S}{\partial T}}{3T^{2}h}}}
$$
And the adiabatic process is
$$
p v^{2}=T^{6}p h^{2}=C
$$
The logarithmic differentiation yields
$$
\frac{d p}{d T}=-\frac{6p h}{T(h+2p h^{\prime})}
$$
Comparing the two derivatives gives
$$
\frac{\partial S}{\partial T}=-\frac{18T p h^{2}}{h+2p h^{\prime}}}
$$
Due to the properties of full differentiation, the two partial derivatives continue to conform to
$$
\frac{\partial}{\partial p}\left(-\frac{18T p h^{2}}{h+2p h^{\prime}}\right)=\frac{\partial}{\partial T}\left(-3T^{2}h\right)
$$
Get the differential equation
$$
\left({\frac{3p h^{2}}{h+2p h^{\prime}}\right)^{\prime}=h
$$
Simplify it like this
$$
h=\left[\frac{3p h^{3}}{(p h^{2})^{\prime}}\right]^{\prime}=-3p h^{3}\cdot\frac{(p h^{2})^{\prime\prime}}{[(p h^{2})^{\prime}]^{2}}}+3\ frac{(p h^{2})^{\prime}h+p h^{2}h^{\prime}}{(p h^{2})^{\prime}}
$$
$$
{\frac{(p h^{2})^{\prime}}{p h^{3}}}\cdot h={\frac{(p h^{2})^{\prime}}{p h^{2}}}=-3{\frac{(p h^{2})^{\prime\prime}}{(p h^{2})^{\prime}}}}}+ 3{\frac{(p h^{2})^{\prime}}{p h^{2}}}}+3{\frac{h^{\prime}}{h}}}
$$
The solution is
$$
(p h^{2})^{-2/3}(p h^{2})^{\prime}=C h=C p^{-1/2}\cdot(p h^{2})^{1/2}
$$
Further moving the right-hand side $(p h^{2})^{1/2}$ to the left-hand side and integrating over $P$ finally yields
$$
h={\frac{1}{C_{1}{\sqrt{p}}({\sqrt{p}}+C_{2})^{3}}}}
$$
And thus the final equation of state
$$
T=C_{1}^{\frac{1}{3}}p^{\frac{1}{6}}(\sqrt{p}+C_{2})v^{\frac{1}{3}}
$$
Since the isobaric heat capacity is
$$
c_{P}=T{\frac{\partial S}{\partial T}}
$$
Substituting gives the result
$$
c_{P}=-{\frac{18T^{2}p h^{2}}{h+2p h^{\prime}}}}={\frac{6T^{2}}{C_{1}({\sqrt{p}}+C_{2})^{2}}}}
$$
Since
$$
v=T^{3}\cdot{\frac{1}{C_{1}{\sqrt{p}}({\sqrt{p}}+C_{2})^{3}}}}
$$
Substituting the relationship yields
$$
\lambda=6\left(1+{\frac{C_{2}}{\sqrt{p}}}\right)
$$
In order to make $\lambda$ constant we deduce that
$$
C_{2}=0
$$
Thus, only
Since $$ C_{2}=0 $$ the previous conclusion reduces to
$$
\lambda=6
$$
Since $C_{2}=0$ simplifies to
$$
h={\frac{1}{C_{1}p^{2}}}
$$
$$
T^{3}=C_{1}p^{2}v
$$
$$
d S=\frac{6T}{C_{1}p}d T-\frac{3T^{2}}{C_{1}p^{2}}d p
$$
Thus, according to the equation of state
$$
C_{1}=\frac{3T_A^{3}}{p_{A}v_{A}^2}
$$
$$
\left({\frac{T_{D}}{T_{A}}}\right)^{3}=\left({\frac{p_{D}}{p_{A}}}\right)^{2}\left({\frac{v_{D}}}{v_{A}}}\right)
$$
Get
$$
T_{D}=T_{A}\left(\frac{p_D^2v_{D}}{p_A^2v_{A}}\right)^{\frac{1}{3}}
$$
The entropy integral gives
$$
S={\frac{3T^{2}}{C_{1}p}}+C^{\prime}=3{\sqrt{\frac{T v}{C_{1}}}}+C^{\prime}
$$
Since the third law of thermodynamics
$$
C^{\prime}=0\quad,\quad S=\frac{3T^{2}}{C_{1}p}
$$
Thus entropy changes
$$
S_{D}-S_{A}={\frac{3T_{D}^{2}}{C_{1}p_{D}}}-{\frac{3T_{A}^{2}}{C_{1}p_{A}}}=\frac{3p_Av_A}{T_A}\left(\left(\frac{p_Dv_D^2}{p_Av_A^ 2}\right)^{\frac{1}{3}}-1\right)
$$
| $$
S_{D}-S_{A}=\frac{3p_Av_A}{T_A}\left(\left(\frac{p_Dv_D^2}{p_Av_A^2}\right)^{\frac{1}{3}}-1\right)
$$
|
36 | MECHANICS | Assume that the Earth is initially a solid sphere with a uniform density of $\rho$ and a volume of $V={\frac{4}{3}}\pi R^{3}$, and is incompressible. The gravitational constant $G$ is known.
First, consider the effect of Earth's rotation. Earth rotates around its polar axis with a constant angular velocity $\omega$, forming an oblate spheroid under the action of inertial centrifugal forces.
Also consider the tidal forces exerted by Earth's satellite—the Moon. For simplicity, we establish the following simplified model: consider the Moon as a homogeneous ring with mass line density $\textstyle{\lambda={\frac{m}{2\pi r}}}$, located in the Earth's equatorial plane and centered at the Earth's center with a radius of $r$ ($r\gg R$). Find the eccentricity of the Earth when stable $\begin{array}{r}{e=\frac{\sqrt{a^{2}-b^{2}}}{a}}\end{array}$, where $a$ is the semi-major axis and $b$ is the semi-minor axis. | The incompressibility condition is given as
$$
{\frac{4\pi a^{2}b}{3}}={\frac{4\pi R^{3}}{3}}
$$
Thus,
$$
a={\frac{R}{(1-e^{2})^{\frac{1}{6}}}}
$$
A uniformly dense ellipsoid can be divided into many thin ellipsoid shells with the same eccentricity. It can be proven that the surface mass density distribution of each ellipsoid shell is identical to the surface charge density distribution of ellipsoid conductors in electrostatics. The gravitational potential inside an ellipsoid shell is zero, so the self-energy of each ellipsoid shell can be calculated and summed. The total gravitational self-energy is
$$
V_{g}=\int_{0}^{a}-{\frac{4\pi}{3}}{\sqrt{1-e^{2}}}a^{3}{\frac{4\pi G\rho a{\sqrt{1-e^{2}}}}{e}}\mathrm{sin}^{-1}e d a
$$
After integration,
$$
V_{g}=-{\frac{16\pi^{2}}{15}}G\rho^{2}R^{5}{\frac{(1-e^{2})^{\frac{1}{6}}}{e}}\sin^{-1}e
$$
The total centrifugal potential energy is
$$
V_{i}=\iiint-{\frac{1}{2}}\rho\omega^{2}(x^{2}+y^{2})d V
$$
After integration,
$$
V_{g}=-\frac{4\pi}{15}\rho\omega^{2}R^{5}\frac{1}{(1-e^{2})^{\frac{1}{3}}}
$$
The tidal potential energy per unit mass is
$$
v_{t}=\int_{0}^{2\pi}-\frac{G\lambda r d\varphi}{[(x-r\cos\varphi)^{2}+(y-r\sin\varphi)^{2}+z^{2}]^{\frac{1}{2}}}
$$
After approximation,
$$
v_{t}=-{\frac{G m}{r}}+{\frac{G m}{4r^{3}}}(2z^{2}-x^{2}-y^{2})
$$
The total tidal potential energy is
$$
V_{t}=\iiint[-{\frac{G m\rho}{r}}+{\frac{G m\rho}{4r^{3}}}(2z^{2}-x^{2}-y^{2})]d V=-{\frac{4\pi G m\rho R^{3}}{3}}-{\frac{2\pi G m\rho R^{5}}{15}}{\frac{e^{2}}{r^{3}(1-e^2)^{1/6}}}
$$
The total potential function is the sum of gravitational self-energy, centrifugal potential energy, and tidal potential energy. After approximation,
$$
V=-\frac{16\pi^{2}}{15}G\rho^{2}R^{5}-\frac{4\pi}{15}\rho\omega^{2}R^{5}-\frac{4\pi}{3}\frac{G m\rho R^{3}}{r}-\frac{4\pi}{45}\rho\omega^{2}R^{5}e^{2}-\frac{2\pi}{15}\frac{G m\rho R^{5}}{r^{3}}e^{2}+\frac{16\pi^{2}}{675}G\rho^{2}R^{5}e^{4}
$$
When stable, the potential function reaches a minimum,
$$
e={\sqrt{\frac{30\omega^{2}+45{\frac{G m}{r^{3}}}}{16\pi\rho G}}}
$$
**[Solution Method 2] (Alternative Method: Surface Equipotential Approach)**
The surface equation for a rotational ellipsoid with a short axis in spherical coordinates is
$$
r^{\prime}(\theta)^{2}=\frac{a^{2}b^{2}}{a^{2}\cos\theta^{2}+b^{2}\sin\theta^{2}}
$$
From equation $\textcircled{2}$,
$$
r^{\prime}(\theta)\approx R+\frac{R}{6}(1-3\cos\theta^{2})
$$
Using the multipole expansion method in electrostatics and expanding to the quadrupole moment of a rotational ellipsoid, the gravitational potential per unit mass is
$$
v_{g}=-{\frac{4\pi G\rho R^{3}}{3r^{\prime}}}-{\frac{2\pi G\rho R^{5}}{15{r^{\prime}}^{5}}}{\frac{e^{2}}{(1-e^{2})^{\frac{1}{3}}}}{({x^{\prime}}^{2}+{y^{\prime}}^{2}-2{z^{\prime}}^{2})}
$$
The centrifugal potential per unit mass is
$$
v_{i}=-\frac{1}{2}\omega^{2}({x^{\prime}}^{2}+{y^{\prime}}^{2})
$$
The tidal potential per unit mass is
$$
v_{t}=-{\frac{G m}{r}}+{\frac{G m}{4r^{3}}}(2{z^{\prime}}^{2}-{x^{\prime}}^{2}-{y^{\prime}}^{2})
$$
The total surface potential per unit mass, which is the sum of the three potentials, is approximated as
$$
v(\theta)=-\frac{4\pi G\rho R^{2}e^{2}}{45}(3\cos\theta^{2}-1)-\frac{1}{2}\omega^{2}R^{2}\sin\theta^{2}-\frac{G m}{r}+\frac{G m R^{2}}{4r^{3}}(3\cos\theta^{2}-1)
$$
The surface equipotential condition requires $v$ to be independent of $\theta$, yielding
$$
e={\sqrt{\frac{30\omega^{2}+45{\frac{G m}{r^{3}}}}{16\pi\rho G}}}
$$ | $$
\sqrt{\frac{30\omega^2 + \frac{45Gm}{r^3}}{16\pi\rho G}}
$$ |
333 | ELECTRICITY | A dielectric cylinder with a radius $R$, mass $m$, and uniform length $L \gg R$ is permanently polarized, where the magnitude of the polarization intensity vector at any point a distance $r$ from the center is $P_{0}e^{k r}$, directed radially outward. Initially, the cylinder is at rest, and a constant external torque $M$ is applied to the cylinder. Find the angular acceleration $\beta$ of the cylinder. The final result is expressed in terms of $M, m, k, R, L, P_0, \mu_0$. | When the angular velocity of the cylinder is $\omega$, the surface current density is
$$
i=\sigma\omega R=P_{0}\omega R e^{k R}
$$
The volume current density in region C is
$$
j_{2}=\rho r\omega=-(1+k r)P_{0}\omega e^{k r}
$$
From Ampère's circuital theorem:
$$
\begin{array}{l}{{\displaystyle{\cal B}=\int_{r}^{R}\mu_{\scriptscriptstyle0}j d r+\mu_{\scriptscriptstyle0}i}}\ {{\mathrm{}}}\ {{\displaystyle{\cal B}=\mu_{\scriptscriptstyle0}P_{0}\omega r e^{k r}}}\end{array}
$$
Magnetic flux
$$
\Phi=\int_{0}^{r}\boldsymbol{B}\cdot2\pi r d r
$$
$$
\Phi=2\pi\mu_{\scriptscriptstyle0}\omega P_{\scriptscriptstyle0}(\frac{r^{\prime}{}^{2}}k e^{k r}-\frac r{k^{2}}e^{k r}+\frac2{k^{3}}e^{k r})
$$
Electromagnetic induction
$$
\begin{array}{l}{E=-\displaystyle\frac{1}{2\pi r}\frac{\partial}{\partial t}\Phi}\ {E=-\displaystyle\mu_{0}\beta P_{0}e^{\displaystyle k r}(\frac{r}{k}{-}\frac{2}{k^{2}}{+}\frac{2}{k^{3}r})}\end{array}
$$
Electromagnetic torque
$$
M_{c}=\int_{0}^{R}E\cdot\rho L\cdot2\pi r^{2}d r+E R\sigma\cdot2\pi R L_{\mathrm{~/~}}
$$
After integrating
From the momentum theorem:
$$
M+M_{e}=\frac{1}{2}m R^{2}\beta\stackrel{\_}{\_}.
$$
Solving for $\beta$
$$
\beta = \frac{2M}{mR^2+4\mu_0P_0^2\pi L[\frac 3{8k^4}+(\frac{R^3}{2k}-\frac{3R^2}{4k^2}+\frac{3R}{4k^3}-\frac 3 {8k^4})e ^{2kR}}]
$$ | $$
\beta = \frac{2M}{mR^2 + 4\mu_0 P_0^2 \pi L \left(\frac{3}{8k^4} + \left(\frac{R^3}{2k} - \frac{3R^2}{4k^2} + \frac{3R}{4k^3} - \frac{3}{8k^4}\right)e^{2kR}\right)}
$$ |
50 | THERMODYNAMICS | A cylindrical container with a cross-sectional area $S$ is fixed on the ground. The container walls are thermally insulated, and the top is sealed with a thermally insulated piston of mass $p_{0}S/g$ (where $g$ is the gravitational acceleration). The lower surface of the piston is parallel to the bottom surface of the container. The exterior of the container is a vacuum, and the interior contains a diatomic ideal gas with the mass of a single molecule being $m$ and the total number of molecules $N$. The temperature of the gas throughout the container is uniform at $T_{0}$.
Slowly sprinkle a total mass of powder $(p_{1}-p_{0})S/g$ onto the piston. Neglect all friction, and consider the effect of gravity on the distribution of gas molecules. Find the final temperature of the gas inside the container, $T_{1}$. | In equilibrium in a gravitational field, the molecular number density at temperature $T$ follows the Boltzmann distribution:
$$
n(z) = n_{0} \exp\left(-\frac{m g z}{k T}\right)
$$
In the formula, $n_{0}$ is the molecular number density at $z=0$ (the bottom of the container), which is an unknown to be determined; $k$ is the Boltzmann constant. The total number of molecules in the container remains constant, always equal to $N$:
$$
N = \int_{0}^{H} S n(z) \operatorname{d}z = S n_{0} \int_{0}^{H} \exp\left(-\frac{m g z}{k T}\right) \operatorname{d}z = \frac{S n_{0} k T}{m g} \left[1 - \exp\left(-\frac{m g H}{k T}\right)\right]
$$
Since the system is adiabatic and the powder is spread slowly, it can be assumed that the gas inside the container undergoes a quasi-static adiabatic process. At any moment during the process, let the temperature of the gas inside the container be $T$, the pressure at the lower surface of the piston inside the container be $p$, and the height from the lower surface of the piston to the bottom of the container be $H$. According to the first law of thermodynamics, for the gas inside the container, we have
$$
-p S \mathrm{d}H = \frac{5}{2}N k \mathrm{d}T + \mathrm{d}E_{p}
$$
In this formula, $\frac{5}{2}N k \mathrm{d}T$ represents the increase in internal energy of a diatomic ideal gas, and $\mathrm{d}E_{p}$ is the increase in gravitational potential energy of the gas in the container. Taking $z=0$ as the zero point of gravitational potential energy, we have
$$
\begin{array}{l}{E_{p} = \int_{0}^{H} m g z S n(z) \mathrm{d}z = m g S n_{0} \int_{0}^{H} z \exp\left(-\frac{m g z}{k T}\right) \mathrm{d}z } \\ { = -k T S H n_{0} \exp\left(-\frac{m g H}{k T}\right) + k T S n_{0} \cdot \frac{k T}{m g} - k T S n_{0} \exp\left(-\frac{m g H}{k T}\right) \cdot \frac{k T}{m g}} \end{array}
$$
At the lower surface of the piston, according to the ideal gas pressure formula, we have
$$
p = n(H) \cdot k T = n_{0} \exp\left(-\frac{m g H}{k T}\right) k T
$$
Substituting into equation $\textcircled{4}$ gives
$$
E_{p} = -p S H + (p S + N m g) \frac{k T}{m g} - p S \frac{k T}{m g} = -p S H + N k T
$$
Here, $p + \frac{N m g}{S} = n_{0} k T$ is the pressure at the bottom of the container. Taking the differential of the above equation gives $\mathrm{d}E_{p} = -p S \mathrm{d}H - S H \mathrm{d}p + N k \mathrm{d}T$, substituting into equation $\textcircled{3}$ gives
$$
S H \mathrm{d}p = \frac{7}{2} N k \mathrm{d}T
$$
Combining equations $\textcircled{2}$ and $\textcircled{5}$ gives
$$
H = \frac{k T}{m g} \ln\frac{p S + N m g}{p S}
$$
Substituting into equation $\textcircled{7}$ gives
$$
\begin{array}{c}{{\displaystyle{\frac{S}{m g}}\ln\frac{p S + N m g}{p S} \mathrm{d}p = \frac{7}{2} N \frac{\mathrm{d}T}{T}}} \\ {\mathrm{}} \\ {\displaystyle{\ln(p S + N m g) \mathrm{d}(p S) - \ln(p S) \mathrm{d}(p S) = \frac{7}{2} N m g \frac{\mathrm{d}T}{T}}} \end{array}
$$
According to piston equilibrium, initial state $T = T_{0}$, $p = p_{0}$, final state $T = T_{1}$, $p = p_{1}$, integrating the above equation gives
$$
\frac{p_{1} S}{N m g} \ln\left(1 + \frac{N m g}{p_{1} S}\right) - \frac{p_{0} S}{N m g} \ln\left(1 + \frac{N m g}{p_{0} S}\right) + \ln\frac{p_{1} S + N m g}{p_{0} S + N m g} = \frac{7}{2} \ln\frac{T_{1}}{T_{0}}
$$
Therefore, the temperature we seek is
$$
T_{1} = T_{0} \frac{\left(1 + \frac{N m g}{p_{1} S}\right)^{\frac{2 p_{1} S}{7 N m g}}}{\left(1 + \frac{N m g}{p_{0} S}\right)^{\frac{2 p_{0} S}{7 N m g}}} \left(\frac{p_{1} S + N m g}{p_{0} S + N m g}\right)^{\frac{2}{7}}
$$ | $$
T_0 \frac{\left(1+\frac{Nmg}{p_1 S}\right)^{\frac{2p_1 S}{7Nmg}}}{\left(1+\frac{Nmg}{p_0 S}\right)^{\frac{2p_0 S}{7Nmg}}} \left(\frac{p_1 S+Nmg}{p_0 S+Nmg}\right)^{\frac{2}{7}}
$$ |
221 | MODERN | A three-dimensional relativistic oscillator moves in a space filled with uniform "dust." During motion, "dust" continuously adheres to the oscillator, which is assumed to increase the rest mass of the sphere without altering its size. The collision is adiabatic, and the "dust" quickly replenishes the region the sphere just passed. The original length of the spring is vanishing, and the potential energy $V$ can be expressed as $V=\frac{1}{2}k x^2$, where $k$ is a known constant. The cross-sectional area of the sphere is $A$, the density of the "dust" is $\rho$, the initial rest mass of the sphere is $m_{0}$, and the speed of light in vacuum is $c$. Initially, the sphere performs uniform circular motion with a radius $R_{0}$. Find the time $t$ required when the circular motion radius changes to $R$. Assume $R^2 \gg A$ and $m_{0} \gg \rho A R$. | From the problem statement, it can be assumed that the sphere is constantly maintaining uniform circular motion
$$
v=\omega r
$$
$$
\omega={\sqrt{\frac{k}{\gamma m}}}
$$
Also, angular momentum conservation
$$
\gamma m v r=c o n s t
$$
Where $\gamma = \begin{array}{r}{\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}}\end{array}$
$$
\gamma m r^{4}=c o n s t=B
$$
Next, solve for the initial $B$. From
$$
{\frac{1}{\gamma^{2}}}=1-{\frac{v^{2}}{c^{2}}}=1-{\frac{k r^{2}}{\gamma m c^{2}}}
$$
Simplify to obtain
$$
\gamma^{2}-{\frac{k r^{2}}{m c^{2}}}\gamma-1=0
$$
Therefore
$$
\gamma=\frac{\frac{k r^{2}}{m c^{2}}+\sqrt{\left(\frac{k r^{2}}{m c^{2}}\right)^{2}+4}}{2}
$$
Substitute initial conditions, we have
$$
\gamma_{0}=\frac{\frac{k R_{0}^{2}}{m_{0}c^{2}}+\sqrt{\left(\frac{k R_{0}^{2}}{m_{0}c^{2}}\right)^{2}+4}}{2}
$$
$$
B=\frac{\frac{k R_{0}^{2}}{m_{0}c^{2}}+\sqrt{\left(\frac{k R_{0}^{2}}{m_{0}c^{2}}\right)^{2}+4}}{2}m R_{0}^{4}
$$
Adiabatic process implies total energy conservation, and mass changes with time
$$
{\frac{d}{d t}}\gamma m=\rho A v-{\frac{k r}{c^{2}}}{\frac{d r}{d t}}
$$
Substitute angular momentum conservation
$$
d(\frac{B}{r^{4}})=\rho A\sqrt{\frac{k}{B}}r^{3}d t-d(\frac{k r^{2}}{2c^{2}})
$$
Resulting in
$$
\Delta(-\frac{k}{c^{2}r}+\frac{4B}{7r^{7}})=\rho A\sqrt{\frac{k}{B}}t
$$
Simplify to get
$$
t=\frac{1}{\rho A}\sqrt{\frac{B}{k}}(\frac{k}{c^{2}}(-\frac{1}{R}+\frac{1}{R_{0}})+\frac{4}{7}B(\frac{1}{R^{7}}-\frac{1}{R_{0}^{7}}))
$$ | $$
t=\frac{1}{\rho A}\sqrt{\frac{\left(\frac{k R_{0}^{2}}{m_{0}c^{2}}+\sqrt{\left(\frac{k R_{0}^{2}}{m_{0}c^{2}}\right)^{2}+4}\right)m R_{0}^{4}}{2k}}\left(\frac{k}{c^{2}}\left(-\frac{1}{R}+\frac{1}{R_{0}}\right)+\frac{4}{7}\frac{\left(\frac{k R_{0}^{2}}{m_{0}c^{2}}+\sqrt{\left(\frac{k R_{0}^{2}}{m_{0}c^{2}}\right)^{2}+4}\right)m R_{0}^{4}}{2}\left(\frac{1}{R^{7}}-\frac{1}{R_{0}^{7}}\right)\right)
$$ |
120 | THERMODYNAMICS | Consider an ideal gas with a known fixed constant $\gamma$ undergoing a rectangular cycle on a $p-V$ diagram, which consists of two isochoric processes and two isobaric processes forming a positive cycle. It is known that the area of this cycle is fixed as $W$ (where $W$ is sufficiently small, negating the possibility of negative pressure or volume). The rectangle must contain the point $(p_0,V_0)$. Find the maximum efficiency of this cycle. | If the center shifts to $\mathcal{p}^{\prime}, V^{\prime}$, then the heat absorbed is:
$$
Q={\frac{W}{2}}+{\frac{\Delta p V^{\prime}+\gamma p^{\prime}\Delta V}{\gamma-1}}
$$
Let us first consider the translation of a fixed rectangular shape with ${\Delta p,\Delta V=W/\Delta p}$ in the $\mathcal{P}V$ diagram. Clearly, minimizing the heat absorbed requires minimizing the values of the center's $p,V$.Thus, it becomes optimal only when the rectangle is translated so that the $(p_0, V_0)$ point is at the upper-right corner of the rectangle:
$$
p^{\prime}=p_0-{\frac{\Delta p}{2}}\quad,\quad V^{\prime}=V_0-{\frac{\Delta V}{2}}
$$
Substituting back, the heat absorbed becomes:
$$
Q={\frac{\Delta p V+\gamma p\Delta V}{\gamma-1}}-{\frac{W}{\gamma-1}}
$$
From the same inequality, the maximum efficiency is:
$$
\eta=\frac{\gamma-1}{2\sqrt{\gamma p_0 V_0/W}-1}
$$ | $$
\eta=\frac{\gamma-1}{2\sqrt{\gamma p_0 V_0/W}-1}
$$ |
121 | MODERN | Two spacecraft are traveling in a space medium that is flowing uniformly at a constant velocity $u$ with respect to an inertial frame $S$. The spacecraft are moving through the medium with equal relative velocities $v$ with respect to the medium. Neither of the velocities is known. Ultimately, the velocities of the two spacecraft with respect to the inertial frame $S$ are $v_{1}$ and $v_{2}$, and the angle between the directions of these velocities is an acute angle $\alpha$. These three quantities are given. The speed of light is $c$. Considering relativistic effects, determine the minimum possible value of $u$. | First, let's derive a byproduct of velocity transformation. The velocity transformation is:
$$
v_{x}^{\prime}={\frac{v_{x}-u}{1-u v_{x}/c^{2}}}
$$
$$
v_{y,z}^{\prime}=v_{y,z}\cdot\frac{\sqrt{1-u^{2}/c^{2}}}{1-u v_{x}/c^{2}}
$$
Hence, we can calculate:
$$
\begin{array}{l}{\gamma=\cfrac{1}{\sqrt{1-(v_{x}^{2}+v_{y}^{2}+v_{z}^{2})/c^{2}}}}\ {=\cfrac{1}{\sqrt{1-(v_{x}^{\prime2}+v_{y}^{\prime2}+v_{z}^{\prime2})/c^{2}}}}\end{array}
$$
Obtain the transformation:
$$
\gamma^{\prime}=\gamma\cdot\frac{1-{u v_{x}}/{c^{2}}}{\sqrt{1-{u^{2}}/{c^{2}}}}
$$
The minimum value of $v$ can be calculated by first determining the relative velocity of two planes:
$$
\gamma_{12}=\frac{1-\beta_{1}\beta_{2}\cos\alpha}{\sqrt{(1-\beta_{1}^{2})(1-\beta_{2}^{2})}}
$$
Consider a system where two spaceships have equal but opposite velocities, this velocity $v=\beta c$ is what we seek. At this moment, the relative velocity of the two planes can be obtained:
Thus, jointly obtain:
$$
\gamma_{12}=\frac{1+\beta^{2}}{\sqrt{(1-\beta^{2})(1-\beta^{2})}}=\frac{1+\beta^{2}}{1-\beta^{2}}
$$
$$
v=c{\sqrt{\frac{c^{2}-v_{1}v_{2}\cos\alpha-{\sqrt{(c^{2}-v_{1}^{2})(c^{2}-v_{2}^{2})}}}{c^{2}-v_{1}v_{2}\cos\alpha+{\sqrt{(c^{2}-v_{1}^{2})(c^{2}-v_{2}^{2})}}}}}
$$
The minimum value of $u=\beta_{\mathsf{u}}c$ can be set such that its magnitude and direction in the original reference system are between angles $\varphi_{1},\varphi_{2}$ of both velocities. Therefore, after transformation, shared velocity:
$$
=\gamma_{i}\frac{1-\beta_{u}\beta_{\mathrm{{i}c o s\varphi_{i}}}}{\sqrt{1-\beta_{u}^{2}}}
$$
$$
\gamma_{1}^{\prime}=\gamma_{2}^{\prime}
$$
Obtain velocity:
$$
\beta_{u}=\frac{\gamma_{1}-\gamma_{2}}{\gamma_{1}\beta_{1}\cos\varphi_{1}-\gamma_{2}\beta_{2}\cos\varphi_{2}}
$$
The denominator can be expanded using the auxiliary angle formula to find extremes:
$$
\cdots=(\gamma_{1}\beta_{1}-\gamma_{2}\beta_{2}\cos\alpha)\cos\varphi_{1}-\gamma_{2}\beta_{2}\sin\alpha\sin\varphi_{1}\leq{\sqrt{(\gamma_{1}\beta_{1}-\gamma_{2}\beta_{2}\cos\alpha)^{2}+(\gamma_{2}\beta_{2}\sin\alpha)^{2}}}
$$
Substitute to find the extreme value:
$$
u=c^{2}{\frac{\left|{\sqrt{c^{2}-v_{1}^{2}}}-{\sqrt{c^{2}-v_{1}^{2}}}\right|}{{\sqrt{c^{2}\left(v_{1}^{2}+v_{2}^{2}\right)-2v_{1}^{2}v_{2}^{2}-2v_{1}v_{2}\cos\alpha{\sqrt{(c^{2}-v_{1}^{2})(c^{2}-v_{2}^{2})}}}}}}
$$ | $$
u = c^2 \frac{\left| \sqrt{c^2 - v_1^2} - \sqrt{c^2 - v_1^2} \right|}{\sqrt{c^2 (v_1^2 + v_2^2) - 2 v_1^2 v_2^2 - 2 v_1 v_2 \cos \alpha \sqrt{(c^2 - v_1^2)(c^2 - v_2^2)}}}
$$ |
648 | MECHANICS | In a vacuum, there are concentric spherical shells made of insulating material with radii $R$ and $2R$. The inner shell is uniformly charged with a total charge of $Q(Q>0)$, while the outer shell is uncharged. There is also a small insulating sphere with a charge of $+q(q>0)$ and a mass $m$, which can be launched from a point on the inner shell with a certain velocity. The direction can be adjusted, and its motion is confined between the inner and outer shells. Collisions with the inner and outer shells are perfectly elastic, and the charges do not change during collisions. It is known that the sphere can return to its starting point after making one full orbit around the center. If the sphere is launched from point $A$ on the inner shell and it collides with the inner shell a total of $n$ times (counting the final return to the starting point as one collision), determine the minimum launch speed $v$, ignoring gravity.
| Let the distance from the center of the sphere to the position of the ball at the extreme point be $r_{1}$, and the speed be $ u_{\mathrm{_I}}$. According to the conservation of angular momentum and energy: $$ m u_{0}R\sin\theta = m u_{1}r_{1} = L $$ $$ {\frac{1}{2}}{m{ u_{0}}^{2}}+{\frac{k Q q}{R}} = {\frac{1}{2}}{m{ u_{1}}^{2}}+{\frac{k Q q}{r_{1}}} = E = {\frac{k Q q}{2a}} $$ In an appropriate polar coordinate system, we have: $$ p = \frac{L^{2}}{kQq m}, \quad e = \sqrt{1+\frac{2E}{m}\frac{L^{2}}{k^{2}{Q}^{2}q^{2}}} $$ When $Q>0$, the charged ball undergoes hyperbolic motion. Each time it collides with the inner sphere shell, it also collides with the outer sphere shell, and returns to point $A$. The angular displacement around the center during inner and outer sphere collisions is: $$ \alpha = {\frac{2\pi}{2n}} = {\frac{\pi}{n}} $$ The curve $AB$ is its trajectory, with point $O$ as one focus of the hyperbola, and $C$ as the other focus. According to conic section knowledge, the trajectory of the hyperbola to focus $O$ is: $$ r = \frac{p}{e\cos\theta^{\prime}-1} $$ where $\theta^{\prime}$ is the angle between the trajectory of the ball and its polar axis. Considering that: $$ 2a = {\overline{{A O}}} - {\overline{{A C}}} = R - {\overline{{A C}}} = {\overline{{B O}}} - {\overline{{B C}}} = 2R - {\overline{{B C}}} $$ $$ {\overline{B C}} - {\overline{{A C}}} = R $$ Thus the possible trajectory of point $C$ is a hyperbola, and since $r$ continuously increases from $A$ to $\boldsymbol{B}$, the polar axis of the hyperbola must not pass through the trajectory $AB$. To minimize the launch speed $ u_{0}$, and thus minimize energy $E$, $2a$ should be maximized, making $AC$ minimal, satisfying the condition when focus $C$ is at the extended line $OA$ and intersects the asymptote of the hyperbola at point $D$. At this point: $$ \theta = \frac{\pi}{2} $$ The speed is minimal, and according to geometric relations: $$ a+c = R $$ Additionally, according to previous equations: $$ \theta^{\prime} = 0,\quad r = {\frac{p}{e\cos0-1}} = R $$ $$ \begin{array}{c c}{{\theta^{\prime}=\alpha,}}&{{r=\displaystyle\frac{p}{e\cos{\alpha}-1}}=2R} {{}}&{{}} \end{array} $$ $$ {{\displaystyle a=\frac{2\cos{\alpha}-1}{2\cos{\alpha}}R}} $$ Substituting into previous equations: $$ u_{0} = \sqrt{\frac{2k Q q}{m R}\frac{1-\cos\frac{\pi}{n}}{2\cos\frac{\pi}{n}-1}} $$
| $$
\sqrt{\frac{2kQq}{mR} \cdot \frac{1-\cos\left(\frac{\pi}{n}\right)}{2\cos\left(\frac{\pi}{n}\right)-1}}
$$ |
72 | THERMODYNAMICS | At time $t=0$, an adiabatic cylinder with a cross-sectional area $S$ is divided into two equal parts with a volume of $V_{0}$ by an adiabatic, thin and lightweight movable piston. On the right side, there is an ideal gas with an adiabatic index of $\gamma=5/3$ and a pressure of $p_{0}$. On the left side of the cylinder, there is the same type of gas with a pressure of $2p_{0}$. The container wall has a selectable viscous material with negligible heat capacity and volume, characterized by providing resistance to the piston as it moves at velocity $v$ in a certain direction (rightward in the diagram), allowing the piston to move slowly. Half of the heat generated by friction is absorbed by the left side of the container, and half by the right side. The piston is released from rest; calculate the displacement $l$ after the piston stops moving (express the result as a coefficient with two significant figures), shown in the form:
$$
l=x \frac{V_{0}}{S} (x \text{ unknown, to two decimal places})
$$
This problem assumes that all processes are quasistatic processes. | From the proportional relationship between work done and heat absorbed:
$$
\frac{1}{2}\left(p_{1}\mathrm{d}V_{1}+p_{2}\mathrm{d}V_{2}\right)=T_{1}\mathrm{d}S_{1}=T_{2}\mathrm{d}S_{2}
$$
From the first law of thermodynamics:
$$
\mathrm{d}U_{i}=T_{i}\mathrm{d}S_{i}-p_{i}\mathrm{d}V_{i}
$$
And the equation of state:
$$
U_{i}=\frac{3}{2}\nu_{i}R T_{i}=\frac{3}{2}p_{i}V_{i}
$$
We obtain:
$$
\mathrm{d}U_{1}={\frac{3}{2}}\mathrm{d}\left(p_{1}V_{1}\right)=-{\frac{1}{2}}p_{1}\mathrm{d}V_{1}+{\frac{1}{2}}p_{2}\mathrm{d}V_{2}
$$
$$
\mathrm{d}U_{2}={\frac{3}{2}}\mathrm{d}\left(p_{2}V_{2}\right)=-{\frac{1}{2}}p_{2}\mathrm{d}V_{2}+{\frac{1}{2}}p_{1}\mathrm{d}V_{1}
$$
To solve these two differential equations, we note that due to energy conservation, the total internal energy remains unchanged:
$$
U_{1}+U_{2}={\frac{3}{2}}\left(p_{1}V_{1}+p_{2}V_{2}\right)={\frac{9}{2}}p_{0}V_{0}
$$
Thus, for the above two equations, we define:
$$
V_{1}=(1+x)V_{0}\quad,\quad V_{2}=(1-x)V_{0}
$$
Simplifying the first equation:
$$
\frac{4}{3}p_{1}V_{2}\mathrm{d}V_{1}+\frac{1}{3}p_{1}V_{1}\mathrm{d}V_{2}+V_{1}V_{2}\mathrm{d}p_{1}=p_{0}V_{0}\mathrm{d}V_{2}
$$
Multiplying both sides by $V_{1}^{1/3}V_{2}^{-2/3}$, the left side conveniently forms a total differential:
$$
\mathrm{d}\left[p_{1}(1+x)^{\frac{1}{3}}(1-x)^{\frac{1}{3}}\right]=-p_{0}{\frac{(1+x)^{1/3}}{(1-x)^{2/3}}}\mathrm{d}x
$$
Due to conservation of internal energy, the pressure at final equilibrium is equal on both sides:
$$
p\cdot2V_{0}=3p_{0}V_{0}\quad,\quad p={\frac{3}{2}}p_{0}
$$
Thus, we finally obtain the integral equation:
$$
2-\frac{3}{2}(1+y)^{\frac{1}{3}}(1-y)^{\frac{1}{3}}=\int_{0}^{y}\frac{(1+x)^{1/3}}{(1-x)^{2/3}}\mathrm{d}x
$$
Solution by numerical methods:
$$
y=0.20
$$
That is:
$$
l=0.20\frac{V_{0}}{S}
$$ | $$
l=0.20\frac{V_0}{S}
$$ |
603 | ELECTRICITY | The characteristics of a medium in an electric field are described as follows: $D=\varepsilon E=\varepsilon_0 E+P$ where $E$ and $D$ are the electric field intensity and electric displacement, respectively, $\varepsilon$ is the dielectric constant of the medium, $P$ is the polarization intensity (electric dipole moment per unit volume), and $\varepsilon_0$ is the permittivity of free space. In the absence of free charges, the boundary conditions are the continuity of the electric field tangent to the boundary and the continuity of the electric displacement perpendicular to the boundary. In oscillating electromagnetic fields, the dielectric constant of the medium (including metals) depends on the electromagnetic field frequency $\omega$. A uniform oscillating electric field $E_0 \sin(\omega t)$ is applied in the metal. Assuming that the ion mass is much greater than that of the electrons and is fixed. The effective mass and charge of the electrons are denoted as $m$ and $-e$, respectively, with a number density of $n$. Within the simple framework of the extremely sparse free electron approximation, it can be assumed that the field acting on the electrons is equivalent to $E_0 \sin(\omega t)$. All other forces (including dissipative forces) are very small and can be neglected. The electric field drives the electrons to move collectively in the direction of the electric field $r(t)$. Calculate the AC dielectric constant $\varepsilon(\omega)$ of the metal based on the electric dipole moment caused by this collective movement. | Write the Newton's second law for the motion of the electron: \[ m\frac{d^2x}{dt^2}=eE_0 \sin(\omega t) \] We obtain \[ r=\frac{eE_0}{m\omega^2}\sin(\omega t) \] Since the motion of ions is not considered, they do not contribute to the total polarizability. The electric dipole moment of the electron and the total polarizability are \[ p=ex \] \[ P=np=-\frac{ne^2E_0}{m\omega^2}\sin(\omega t) \] The formula for calculating the dielectric constant is \[ \varepsilon \varepsilon_0 E=\varepsilon_0 E+P \] We obtain the dielectric constant \[ \varepsilon =1-\frac{ne^2}{\varepsilon_0 m\omega^2} \] | $$\varepsilon = 1 - \frac{ne^2}{\varepsilon_0 m \omega^2}$$ |
670 | ELECTRICITY | Consider a plasma system composed of protons and electrons, where the equilibrium number density of positive and negative charges is \( n_0 \). A point charge \( q \) is placed in this plasma. Due to the Coulomb interaction, the point charge will attract opposite charges and repel like charges, causing a change in the charge distribution within the plasma: within a certain spatial range around the point charge, the densities of positive and negative charges are no longer equal, thereby weakening the electric field of the point charge. This phenomenon is known as Debye shielding. It is known that in spherical coordinates, the Poisson equation for the electrostatic field is: \[ \nabla^2 \varphi = \frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{\mathrm{d}\varphi}{\mathrm{d}r} \right) = -\frac{\rho(r)}{\varepsilon_0}, \] where \( \varphi(r) \) is the electric potential distribution. Hint: try assuming a trial solution of the form \( \varphi(r) = \frac{u(r)}{r} \). Assume the plasma satisfies the following conditions: At equilibrium, the number density \( n(\varphi) \) of protons and electrons follows the Boltzmann distribution; The plasma is overall electrically neutral; \( e\varphi \ll kT \) (\( e \) is the elementary charge, \( k \) is the Boltzmann constant, \( T \) is the plasma temperature). The Debye radius \( \lambda_D \) is defined as the distance at which the potential decays to \( \frac{1}{e} \) of the vacuum point charge potential. Derive the expression for \( \lambda_D \). | The number density of positive and negative ions are respectively: $$ n_{i}(\varphi)=n_{i0}e^{-\frac{e\varphi}{k T}} $$ $$ n_{e}(\varphi)=n_{e0}e^{\frac{e\varphi}{k T}} $$ Charge density: $$ \rho=n_{i}q+n_{e}(-q)=n_{i0}e\cdot e^{-{\frac{e\varphi}{k T}}}-n_{e0}e\cdot e^{\frac{e\varphi}{k T}} $$ The plasma is electrically neutral, i.e., $$ n_{i0}=n_{e0}=n_{0} $$ Thus, we can get: $$ \rho=-2\frac{e^{2}\varphi}{k T}n_{0} $$ Then the Poisson's equation becomes: $$ \frac{1}{r^{2}}\frac{\mathrm{d}}{\mathrm{d}r}(r^{2}\frac{\mathrm{d}\varphi}{\mathrm{d}r})=\frac{2e^{2}\varphi n_{0}}{k T\varepsilon_{0}} $$ Let $\varphi={\frac{u}{r}}$, substitute and simplify to obtain: $$ \frac{\mathrm{d}^{2}u}{\mathrm{d}r^{2}}=\frac{2e^{2}u n_{0}}{k T\varepsilon_{0}} $$ $\lambda_{D}=\sqrt{\frac{k T\varepsilon_{0}}{2e^{2}n_{0}}}$ From the original equation, it is easy to get: $$ u=C_{1}e^{\frac{r}{\lambda_{D}}}+C_{2}e^{-\frac{r}{\lambda_{D}}} $$ Considering the actual situation, the plasma is electrically neutral at infinity, so the potential must converge, thus $C_{1}=0$, and the potential tends towards that of a point charge at infinitesimal distances, hence $$ \varphi(r)={\frac{u}{r}}={\frac{q}{4\pi\varepsilon_{0}r}}e^{-{\frac{r}{\lambda_{D}}}} $$ According to the problem statement, the characteristic scale of the Debye shielding is given directly as: $$ \lambda_{_D}=\sqrt{\frac{k T\varepsilon_{0}}{2e^{2}n_{0}}} $$ | $$
\lambda_D = \sqrt{\frac{k T \varepsilon_0}{2 e^2 n_0}}
$$ |
140 | MECHANICS | Three identical homogeneous balls are placed on a smooth horizontal surface, touching each other and are close enough to each other. A rope is wrapped around the spheres at the height of their centers, tying them together. A fourth identical sphere is placed on top of the three spheres. Find the tension $T$ in the rope. It is given that the weight of each sphere is $P$. | In the text, $P$ represents the gravity on sphere $O$.
Since the spheres are completely identical, the string forms part of an equilateral triangle, so the tension $T$ along the direction $CG$ for sphere $C$ should equal the component of $N$ in that direction, i.e.,
$$
2T\cos{\frac{\pi}{6}} = N\cos\theta.
$$
From equation (1) and equation (2), we have
$$
T = {\frac{P}{3{\sqrt{3}}}}\cot\theta.
$$
Since $G$ is the center of the equilateral triangle $ABC$ formed by the centers of spheres $A$, $B$, and $C$ (Figure 1.11(c)), from geometric relationships, we know
$$
CG = \frac{2}{3}(2R)\cos{\frac{\pi}{6}} = \frac{2R}{\sqrt{3}},
$$
thus,
$$
\cos\theta = {\frac{CG}{CO}} = {\frac{1}{\sqrt{3}}}, \quad \cot\theta = {\frac{1}{\sqrt{2}}}.
$$
Substituting equation (4) into equation (3), we get
$$
T = {\frac{\sqrt{6}}{18}}P.
$$ | $$\frac{\sqrt{6}}{18}P$$ |
653 | ELECTRICITY | A massless insulating string of length $\alpha$ is fixed at one end, with a conductor ring of mass $m$ and radius $r$ ($r \ll \alpha$) suspended on the other end, forming a simple pendulum. Initially, the pendulum string is vertical, and the ring is just inside a sector of a uniform magnetic field. The center of the sector is at the suspension point, with its edges just tangent to the ring. The magnetic field is perpendicular to the plane of the ring and directed inward, with a magnetic flux density of $B$. If the ring is given an initial horizontal velocity (perpendicular to the magnetic field) with a magnitude of $v$, what is the minimum initial velocity $v_{0}$ required for the pendulum string to reach the horizontal direction? The gravitational acceleration is $g$, the magnetic field is sufficiently strong, and the torque due to the Ampere force is much greater than the torque due to gravity when the ring moves within the magnetic field. The ring is a superconducting ring with a self-inductance coefficient $L$. | If the ring is a superconducting ring, when it moves in a magnetic field, due to the properties of the superconductor, the magnetic flux in the ring remains unchanged, i.e.: $$ L I + B S = B \cdot \pi r^2 $$ Thus, the current in the ring is: $$ \begin{array}{l}{I = \displaystyle\frac{B}{L}(\pi r^{2} - S)}\ = \displaystyle\frac{B}{L}\left\{r^{2}\arcsin\left[\frac{a(\theta_{0}-\theta)}{r}\right] - \sqrt{2r a\theta-a^{2}\theta^{2}} \cdot a(\theta_{0}-\theta)\right\}\end{array} $$ $$ \phi = \operatorname{arccos}\left[\frac{a(\theta_{0}-\theta)}{r}\right] $$ We have: $$ \cos\phi = \frac{a(\theta_{0}-\theta)}{r}, \sin\phi = \sqrt{1-\cos^{2}\phi} = \sqrt{1-\frac{a^{2}(\theta_{0}-\theta)^{2}}{r^{2}}} $$ Thus: $$ I = \frac{B}{L}(r^{2}\phi - r^{2}\sin\phi\cos\phi), l = 2r\sin\phi $$ The torque about the suspension point due to the Ampère force acting on the part of the ring within the magnetic field is approximately: $$ \tau = I B l a = \frac{2B^{2}r a}{L}(r^{2}\phi - r^{2}\sin\phi\cos\phi)\sin\phi $$ The dynamic equation of the ring is: $$ m a^{2}\frac{d\omega}{dt} = -\frac{2B^{2}r a}{L}(r^{2}\phi - r^{2}\sin\phi\cos\phi)\sin\phi $$ Since $\frac{d\omega}{dt} = \omega\frac{d\omega}{d\theta}$, we have: $$ d\theta = \frac{r}{a} \sin (\phi) d\phi $$ Thus: Integrating both sides, given that initially when $\theta = 0$, $\phi = 0$, and $\omega \approx \frac{v_{0}}{a}$, we have: $$ m a^{2}\int_{\frac{v_{0}}{a}}^{\omega}\omega d\omega = -\frac{2B^{2}r^{4}}{L}\int_{0}^{\phi}\left(\phi - \sin\phi\cos\phi\right)\sin^{2}\phi d\phi $$ We obtain: $$ \frac{1}{2}m a^{2}\left(\omega^{2}-\frac{v_{0}^{2}}{a^{2}}\right) = -\frac{B^{2}r^{4}}{2L}\left[\phi^{2} - \phi\sin2\phi - \frac{1}{2}(\cos2\phi - 1) - \sin^{4}\phi\right] $$ Thus: $$ \omega^{2} = \frac{v_{0}^{2}}{a^{2}} - \frac{B^{2}r^{4}}{m a^{2}L}\left[\phi^{2} - \phi\sin2\phi - \frac{1}{2}(\cos2\phi - 1) - \sin^{4}\phi\right] $$ When the ring has just completely left the magnetic field area, $\theta = 2\theta_{0}$, corresponding to $\phi = \pi$, substituting into the above equation, we get: $$ \omega_{2}^{2} = \frac{v_{0}^{2}}{a^{2}} - \frac{\pi^{2}B^{2}r^{4}}{m a^{2}L} $$ To make the pendulum reach the horizontal direction, it should satisfy: $$ v_{0} = \sqrt{2g a + \frac{\pi^{2}B^{2}r^{4}}{m L}} $$ | $$\sqrt{2g a+\frac{\pi^2 B^2 r^4}{m L}}$$ |
719 | THERMODYNAMICS | The martian atmosphere can be considered as composed only of very thin $CO_{2}$. The molar mass of this atmosphere is denoted by $\mu$, and the atmosphere at the same height can be considered as an ideal gas in equilibrium. The mass of Mars is $M_{m}$ (far greater than the total mass of the martian atmosphere), and its radius is $R_{m}$. Assume the relationship between the atmospheric mass density and the height $h$ above the surface of Mars is given by $$ \alpha(h)=\rho_{\mathrm{o}}(1+\frac{h}{R_{m}})^{1-n} $$ where $\rho_{\mathrm{0}}$ is a constant, and $\textit{n}(n>4)$ is also a constant. Derive the expression for the temperature $T(h)$ of the martian atmosphere as a function of the height $h$. Express the result using the following physical quantities: the mass of Mars $M_{m}$, radius $R_{m}$, molar mass of the atmosphere $\mu$, constant $n$, gravitational constant $G$, and ideal gas constant $R$. | (1) Basic Assumptions and Ideal Gas Law In the atmosphere at a certain height \( h \) on Mars, a thin horizontally placed box-shaped region is defined. The volume of this region is \( V \), the mass of the atmosphere inside is \( M \), the pressure is \( P(h) \), and the temperature is \( T(h) \). Because the box is very thin, the density, pressure, and temperature of the gas can be considered constant. According to the ideal gas law: $$ P(h)V = n R T(h) $$ The volume of the atmosphere inside the box is: $$ V = \frac{M}{\rho(h)} $$ The number of moles of the Mars atmosphere inside the box is: $$ n = \frac{M}{\mu} $$ By combining equations (1), (2), and (3), we get: $$ T(h) = \frac{P(h) \mu}{\rho(h) R} $$ (2) Determine the Pressure \( P(h) \) Consider a thin spherical cap-shaped atmospheric layer from height \( x \) to \( x + dx \), subtending a solid angle \( \Delta \Omega \). Let \( \Delta \Omega \) be very small, so that the radial lines on the side of the spherical cap can be considered parallel to each other. From the force analysis, we know: $$ [P(x + dx) - P(x)] \cdot \Delta \Omega (R_m + x)^2 + \rho(x) g(x) \cdot \Delta \Omega (R_m + x)^2 dx = 0 $$ That is: $$ \frac{dP}{dx} = - \rho(x) g(x) $$ Where the gravitational acceleration from the surface of Mars to height \( x \) is: $$ g(x) = \frac{G M_m}{(R_m + x)^2} $$ Substituting into equation (6) and integrating, we obtain: $$ P(h) = G M_m \int_h^{\infty} \frac{\rho(x)}{(R_m + x)^2} dx $$ (3) Substitute and Determine the Temperature Expression Substitute equation (7′) into equation (4), and use the density distribution expression provided in the problem: $$ \rho(x) = \rho_0 \left(1 + \frac{x}{R_m}\right)^{1 - n} $$ Finally, we get: $$ T(h) = \frac{\mu G M_m}{n R (R_m + h)} $$ Final Result (Temperature Distribution of Martian Atmosphere) $$ \boxed{T(h) = \frac{\mu G M_m}{n R (R_m + h)}} $$ This result indicates that the temperature of Mars' atmosphere decreases inversely with height \( h \). | $$T(h) = \frac{\mu G M_m}{n R (R_m + h)}$$ |
211 | MODERN | In the ground frame S, two trains with an intrinsic length of $L_{0}$ are stationary along the $x$ axis, with the left side of train A located at $x=0$. The gap between the left side of train B and the right side of train A is $D$. At time $t=0$ in the ground frame S, both trains A and B synchronize their carriage times to $t_{A}=t_{B}=0$, and the clocks are located on the left side of each train. At this moment, the two trains start moving with constant accelerations relative to themselves, $a_{A}$ and $a_{B}$.
Let the distance between train carriages AB in frame S be $D(t)$. Express the value of $a_{B}$ when $D(t)\equiv D$ in terms of $L_{0}$, $a_{A}$, and the speed of light $c$. | Since the left end of the train is aligned with the ground clock, the left end of the train can be regarded as a point mass, and thus we can study uniform accelerated motion.
Let the velocity of the left end at time $t$ in the ground frame be $v(t)$, at time $t+d t$, first perform a Lorentz transformation to the rest frame with velocity $v(t)$. In the rest frame, obtain an infinitesimal velocity $adτ$:
Using the proper time formula:
$$
v+dv={\frac{v+adτ}{1+{\frac{v\times adτ}{c^{2}}}}}\rightarrow dv=adτ\left(1-{\frac{v^{2}}{c^{2}}}\right)
$$
$$
dτ={\sqrt{1-{\frac{v^{2}}{c^{2}}}}}dt \to dv=a{\sqrt{1-{\frac{v^{2}}{c^{2}}}^{3}}}dt
$$
Separate variables on both sides, yielding:
$$
at=\int_{0}^{v}{\frac{dV}{1-{V}^{2}/{c}^{2}}}={\frac{v}{1-{v}^{2}/{c}^{2}}}
$$
Solving for the velocity function:
$$
v(t)={\frac{at}{\sqrt{1+\left({\frac{at}{c}}\right)^{2}}}}\to x(t)=x_{0}+\int_{0}^{t}v(t)dt=x_{0}+{\frac{c^{2}}{a}}\left({\sqrt{1+\left({\frac{at}{c}}\right)^{2}}}-1\right)
$$
Select special cases A, B to write the displacement function:
$$
\left\{\begin{array}{l}{\displaystyle x_{l A}=\frac{c^{2}}{a_{A}}\left(\sqrt{1+\left(\frac{a_{A}t}{c}\right)^{2}}-1\right)}\ {\displaystyle x_{l B}=L_{0}+D+\frac{c^{2}}{a_{B}}\left(\sqrt{1+\left(\frac{a_{B}t}{c}\right)^{2}}-1\right)}\end{array}\right.
$$
Using the motion on the left side, we can calculate the proper time:
$$
τ=\int_{0}^{t}{\sqrt{1-\left({\frac{v}{c}}\right)^{2}}}dt=\int_{0}^{t}{\frac{dt}{\sqrt{1+\left({\frac{at}{c}}\right)^{2}}}}={\frac{c}{a}}\mathrm{arcsinh}{\frac{at}{c}}
$$
Apply to special cases A, B:
$$
\left\{\begin{array}{l l}{{\displaystyleτ_{A}=\frac{c}{a_{A}}\mathrm{arcsinh}\frac{a_{A}t}{c}}}\ {{\displaystyleτ_{B}=\frac{c}{a_{B}}\mathrm{arcsinh}\frac{a_{B}t}{c}}}\end{array}\right.
$$
Taking train A as an example, given the proper time $τ_{A}$, the speed of A at this time is:
$$
v_{A}(τ)={\frac{a_{A}t}{\sqrt{1+\left({\frac{a_{A}t}{c}}\right)^{2}}}}=c\operatorname{tanh}{\frac{a_{A}τ}{c}}
$$
The left side of the train system $(0,τ)$ and the ground frame $(x_{l A},t)$ are aligned, allowing for easy writing of the Lorentz transformation for the right end time and coordinates in the ground frame:
$$
t_{r,A}=t+{\frac{v_{A}L_{0}/c^{2}}{1-v_{A}^{2}/c^{2}}}=t+\cosh{\frac{a_{A}τ}{c}}\left({\frac{L_{0}}{c}}\operatorname{tanh}{\frac{a_{A}τ}{c}}\right)=t+{\frac{L_{0}}{c}}\sinh{\frac{a_{A}τ}{c}}=\left(1+{\frac{a_{A}L_{0}}{c^{2}}}\right)t
$$
$$
x_{r,A}=x_{l,A}+\frac{L_{0}}{1-v_{A}^{2}/c^{2}}=\left(\frac{c^{2}}{a_{A}}+L_{0}\right)\sqrt{1+\left(\frac{a_{A}t}{c}\right)^{2}}-\frac{c^{2}}{a_{A}}
$$
Since $x_{r,A}$ occurs at event S frame $t=t_{r,A}$, the correct $x_{r,A}(t)$ function should replace $t$:
$$
x_{r,A}={\frac{c^{2}}{{\frac{a_{A}}{1+{\frac{a_{A}L_{0}}{c^{2}}}}}}}{\sqrt{1+\left({\frac{a_{A}t/c}{1+{\frac{a_{A}L_{0}}{c^{2}}}}}\right)^{2}}}-{\frac{c^{2}}{a_{A}}
$$
At this moment, the distance between the right side of A and the left side of B is:
$$
=x_{l,B}-x_{r,A}=L_{0}+D+{\frac{c^{2}}{a_{B}}}\left(\sqrt{1+\left({\frac{a_{B}t}{c}}\right)^{2}}-1\right)+{\frac{c^{2}}{a_{A}}}-{\frac{c^{2}}{{\frac{a_{A}}{1+{\frac{a_{A}L_{0}}{c^{2}}}}}}}\sqrt{1+\left({\frac{a_{A}t/c}{1+{\frac{a_{A}L_{0}}{c^{2}}}}}\right)^{2}}
$$
If the distance is constant, then naturally:
$$
a_{B}={\frac{a_{A}}{1+{\frac{a_{A}L_{0}}{c^{2}}}}}
$$ | $$
\frac{a_A}{1+\frac{a_A L_0}{c^2}}
$$ |
139 | MECHANICS | Two smooth cylinders, each with a radius of $r_{1}$ but with weights $P_{1}$ and $P_{2}$ respectively, are placed in a smooth cylindrical groove with a radius of $R$. Find the tangent value of the angle $\varphi$ when the two cylinders are in equilibrium. The angle $\varphi$ is defined as the angle between the line connecting the center of the higher cylinder and the center of the cylindrical groove, and the horizontal line. Assume $\cos\beta = \frac{r}{R - r}$. Express the result in terms of $\beta$. | Under the action of these forces, cylinders 1 and 2 reach equilibrium. Triangle \( O O_{1}O_{2} \) is an isosceles triangle. Let \(\beta=\angle O O_{1}O_{2},\) then
\[
\cos \beta = \frac{r}{R-r}.
\]
For cylinder 1, the condition for force equilibrium is
\[
\begin{array}{r l}
&{P_{1}+f\sin(\beta-\varphi)-N_{1}\sin(2\beta-\varphi)=0,}\\
&{f\cos(\beta-\varphi)-N_{1}\cos(2\beta-\varphi)=0.}
\end{array}
\]
By eliminating \( N_{1} \) from the two equations in expression (2), we obtain
\[
P_{1}\cos(2\beta-\varphi)-f\sin\beta=0.
\]
Eliminating \( f \), we obtain
\[
P_{1}\cos(\beta-\varphi)-N_{1}\sin\beta=0.
\]
For cylinder 2, at equilibrium we have
\[
\begin{array}{r l}
&{P_{2}-f\sin(\beta-\varphi)-N_{2}\sin\varphi=0,}\\
&{f\cos(\beta-\varphi)-N_{2}\cos\varphi=0.}
\end{array}
\]
By eliminating \( N_{2} \) from the two equations in expression (5), we obtain
\[
P_{2}\cos\varphi-f\sin\beta=0.
\]
Eliminating \( f \), we obtain
\[
P_{2}\cos(\beta-\varphi)-N_{2}\sin\beta=0.
\]
By eliminating \( f \) from equations (3) and (6), we obtain
\[
\tan\varphi = \frac{P_{2}-P_{1}\cos(2\beta)}{P_{1}\sin(2\beta)}.
\]
This is the relationship that \(\varphi\) satisfies when the system is at equilibrium. | $$\varphi = \arctan \frac{P_2 - P_1 \cos(2\beta)}{P_1 \sin(2\beta)}$$ |
106 | MECHANICS | Two thin rods, each with mass $m$ and length $l$, are centrally connected by a thin string (with negligible mass). The upper ends are also connected by a very short flexible string (with negligible length). The other end of each rod is free to slide across a horizontal table without friction. The plane formed by the rods and strings is perpendicular to the tabletop, with gravitational acceleration $g$. Initially, the rods are stationary, forming an angle $\theta_{0}$ with the tabletop. When the thin string at the center is cut, the rods fall vertically until they hit the table. What is the velocity $v$ of the rods just before the ends connected at the table touch the tabletop? | Solution: Using the conservation of energy:
Initial mechanical energy: $\frac{1}{2}mgl\sin\theta_{0}$
Mechanical energy while in motion (angular velocity is $\dot{\theta}$): translational kinetic energy $^{+}$ rotational kinetic energy $^+$ gravitational potential energy:
$$
{\frac{1}{2}}m{\dot{\theta}}^{2}{\frac{l^{2}}{4}}+{\frac{1}{2}}{\frac{1}{12}}m l^{2}{\dot{\theta}}^{2}+{\frac{1}{2}}m g l \sin\theta={\frac{1}{6}}m l^{2}{\dot{\theta}}^{2}+{\frac{1}{2}}m g l \sin\theta
$$
We obtain the angular velocity $\dot{\theta}: \dot{\theta} = \sqrt{3g/l(\sin\theta_{0} - \sin\theta)}$. The terminal velocity is: $v=-\sqrt{3gl\sin\theta_{0}}$ | $$v = -\sqrt{3 g l \sin \theta_0}$$ |
58 | ELECTRICITY | An arbitrary multi-terminal resistor network structure, for example with $N+1$ terminals, sets one specific terminal (0 terminal) as the reference for electric potential, setting its potential at 0. The potentials for other terminals (numbered 1, 2, ..., N) are $v_{1}, v_{2}, \ldots, v_{N}$, and the current flowing into the network through these corresponding terminals (negative when flowing out) are $I_{1}, I_2, \ldots, I_n$. Therefore, we have:
$$
\left\{
\begin{array}{l}
V_{1} = a_{11} I_{1} + a_{12} I_{2} + \cdots + a_{1N} I_{N} \\
V_{2} = a_{21} I_{1} + a_{22} I_{2} + \cdots + a_{2N} I_{N} \\
\vdots \\
V_{i} = a_{i1} I_{1} + a_{i2} I_{2} + \cdots + a_{iN} I_{N} \\
\vdots \\
V_{N} = a_{N1} I_{1} + a_{N2} I_{2} + \cdots + a_{NN} I_{N}
\end{array}
\right.
$$
Due to symmetry, we know that $a_{ij} = a_{ji}$.
This means that the properties of the $N+1$ terminal network can be determined by $\frac{N(N+1)}{2}$ independent parameters: $a_{11}, a_{12}, \ldots , a_{ij}, \ldots , a_{NN} (i<j)$. These parameters are only determined by the resistor network itself and are unrelated to the potential at the terminals or the magnitude of the current flowing in.
Consider a square uniform conductive thin plate as a four-terminal resistor network. The four vertices are sequentially numbered clockwise as 0, 1, 2, 3. If the equivalent resistance between adjacent vertices is ${R}_{1}$ and the equivalent resistance between opposite vertices is ${{R}}_{2}$, use the method of finding the coefficients $a_{i j} (i,j=1,2,3)$ to solve: what is the equivalent resistance between the other two vertices if a pair of adjacent vertices is short-circuited? | For this situation, label the 4 endpoints sequentially as 0, 1, 2, 3. If current flows into endpoint 1 with current I (flowing out from endpoint 0), then:
$$
V_{1}=a_{11}I=I R_{1}
$$
Thus, we obtain: $a_{11}=R_{1}$
Similarly, if current flows into endpoints 2 or 3 with current I, we can respectively obtain: $a_{22}=R_{2}$ $a_{33}=R_{1}$. If current flows into endpoint 1 with current I and flows out from endpoint 2:
$$
V_{1}^{\prime}=a_{11}I-a_{12}I,V_{2}^{\prime}=a_{21}I-a_{22}I
$$
At this moment, the potential difference between endpoints 1 and 2 is:
$$
V_{1}^{\prime}-V_{2}^{\prime}=(a_{11}-2a_{12}+a_{22})I
$$
Due to symmetry, at this moment, the potential difference between endpoints 1 and 2 should be equal to the potential at endpoint 1 (between 1 and 0) when current flows only into endpoint 1 (out from endpoint 0), thus:
$$
(a_{11}-2a_{12}+a_{22})I=a_{11}I
$$
We obtain:
$$
a_{12}=\frac{1}{2}a_{22}=\frac{1}{2}R_{2}
$$
[This can also be directly obtained by the condition of current input into endpoint 2, where the potential at endpoint 1 is half of that at endpoint 2 due to symmetry.] Similarly, we obtain:
$$
a_{23}=\frac{1}{2}R_{2}
$$
Returning again to the situation with current flowing only into endpoint 1 (flowing out from endpoint 0), at this moment, the potentials at endpoints 2 and 3 are respectively:
$$
V_{2}=a_{21}I,V_{3}=a_{31}I
$$
Due to symmetry, the potential difference between endpoints 1 and 2 should be equal to the potential difference between endpoints 3 and 0, thus:
$$
V_{1}-V_{2}=V_{3}
$$
Therefore:
$$
a_{11}I-a_{21}I=a_{31}I
$$
We obtain:
$$
a_{13}=a_{31}=a_{11}-a_{12}=R_{1}-{\frac{1}{2}}R_{2}
$$
Now returning to the required situation of the problem, without loss of generality, we can assume endpoints 3 and 0 are short-circuited and calculate the equivalent resistance between endpoints 1 and 2. In this case, if current flows into endpoint 1 with current I and out from endpoint 2, since endpoints 3 and 0 are short-circuited, there must be current flowing from 0 through the short-circuit wire into endpoint 3. Let the current flowing through the short-circuit wire be $I^{\prime}$, then it can be considered as the original four-terminal situation with current flowing into endpoint 1, out from endpoint 2, and into endpoint 3 with $I^{\prime}$. At this moment, the potential at endpoint 3 is:
$$
V_{3}=a_{31}I-a_{32}I+a_{33}I^{\prime}
$$
Because 3 and 0 are practically short-circuited, $V_{3}$ should be 0. Thus we can compute:
$$
I^{\prime}=\frac{a_{32}-a_{31}}{a_{33}}I=\frac{R_{2}-R_{1}}{R_{1}}I
$$
We can then compute the potentials at endpoints 1 and 2:
The potential difference between endpoints 1 and 2:
$$
\begin{array}{c}{{V_{1}=a_{11}I-a_{12}I+a_{13}I^{\prime}=(R_{1}-\displaystyle{\frac{1}{2}}R_{2})\displaystyle{\frac{R_{2}}{R_{1}}}I}}\ {{V_{2}=a_{21}I-a_{22}I+a_{23}I^{\prime}=\displaystyle{\frac{R_{2}}{2}}\displaystyle{\frac{R_{2}-2R_{1}}{R_{1}}}I}}\end{array}
$$
$$
V_{1}-V_{2}=\frac{(2R_{1}-R_{2})R_{2}}{2R_{1}}I
$$
Thus, the equivalent resistance between endpoints 1 and 2 is:
$$
R={\frac{V_{1}-V_{2}}{I}}={\frac{(2R_{1}-R_{2})R_{2}}{2R_{1}}}=(2-{\frac{R_{2}}{R_{1}}})R_{2}
$$ | $$\left(2-\frac{R_2}{R_1}\right)R_2$$ |
412 | ELECTRICITY | Consider a semi-infinite solenoid with radius $R$, turns per unit length $n$, connected to a constant current source with current $I$. To facilitate subsequent calculations, we establish a cylindrical coordinate system such that the $z$-axis coincides with the symmetry axis of the solenoid. The solenoid extends from the origin along the negative $z$-axis to infinity, and the current is viewed as rotating counterclockwise when observed from above the $z$-axis, meaning the direction of the current is in the positive $\hat{\theta}$ direction. Now, a superconducting metal ring is placed such that some constraint keeps the symmetry axis of the current loop always coinciding with the $z$-axis, allowing it to move freely up and down in this state. It is known that at the instant the metal ring transitions from normal to superconducting state, a current $I_{0}$ flows within the ring, and there is no other external magnetic field at a sufficiently far distance from the solenoid. $I_{0}$ flows in the positive $\hat{\theta}$ direction. Subsequently, the metal ring remains in the superconducting state. Assume the gravitational acceleration is $\overrightarrow{g}=-g\widehat{z}$, the mass of the metal ring is $m$, and it is of uniform density. Let $z_{0}>0$ be known. If the equilibrium is stable, find the vibration period $T$ near the equilibrium position (expressed in terms of $n$, $I_{0}$, $r$, $R$, $L$, $m$), and the answer may include $z_{0}$. It is known that the radius of the metal ring is $r \ll R$, and the self-inductance is $L$. | The equilibrium condition is
$$
F(z)=m g
$$
$$
\frac{\pi}{2}\mu_{0}n I R^{2}r^{2}\frac{1}{(z^{2}+R^{2})^{3/2}}(\frac{\mu_{0}n I\pi r^{2}}{2L}(1-\frac{z}{\sqrt{z^{2}+R^{2}}})-I_{0})=m g
$$
Consider the stability of the equilibrium. If $\frac{d F}{d z}$ is negative, then the equilibrium is stable; otherwise, it is unstable.
$$
\left.{\frac{d F}{d z}}\right|_{z_{0}}={\frac{\pi}{2}}\mu_{0}n I R^{2}r^{2}\cdot[({\frac{\mu_{0}n I\pi r^{2}}{2L}}-I_{0}){\frac{-3z_{0}}{(z_{0}^{2}+R^{2})^{5/2}}}+{\frac{\mu_{0}n I\pi r^{2}}{2L}}{\frac{3z_{0}^{2}-R^{2}}{(z_{0}^{2}+R^{2})^{3}}}]
$$
Therefore, the requirement for stability is
$$
\frac{3z_{0}^{2}-R^{2}}{3z_{0}\sqrt{z_{0}^{2}+R^{2}}}<1-\frac{2I_{0}L}{\mu_{0}n I\pi r^{2}}
$$
If the equilibrium is stable, then
$$
T=2\pi\sqrt{\frac{m}{\left.\displaystyle-\frac{d F}{d z}\right|_{z_{0}}}}
$$
$$
T=2\pi\sqrt{\frac{m}{\frac{\pi}{2}\mu_{0}n I R^{2}r^{2}\cdot[(\frac{\mu_{0}n I\pi r^{2}}{2L}-I_{0})\frac{3z_{0}}{(z_{0}^{2}+R^{2})^{5/2}}-\frac{\mu_{0}n I\pi r^{2}}{2L}\frac{3z_{0}^{2}-R^{2}}{(z_{0}^{2}+R^{2})^{3}}]}}
$$ | $$
T=2\pi\sqrt{\frac{\frac{1}{(z_0^{2}+R^{2})^{3/2}}(\frac{\mu_{0}n I\pi r^{2}}{2L}(1-\frac{z_0}{\sqrt{z_0^{2}+R^{2}}})-I_{0})}{g\left[(\frac{\mu_{0}n I\pi r^{2}}{2L}-I_{0})\frac{3z_{0}}{(z_{0}^{2}+R^{2})^{5/2}}-\frac{\mu_{0}n I\pi r^{2}}{2L}\frac{3z_{0}^{2}-R^{2}}{(z_{0}^{2}+R^{2})^{3}}\right]}}
$$
|
43 | ELECTRICITY | Two dielectric materials with base areas both equal to $\pmb{A}$, thicknesses of $h_{1}$ and $h_{2}$ respectively, dielectric constants of $\varepsilon_{1}$ and $\varepsilon_{2}$ respectively, and conductivities of $\sigma_{1}$ and $\sigma_{2}$ respectively, are placed tightly between two highly conductive flat plates with the same area $\pmb{A}$ and a separation distance of $(h_{1}+h_{2})$, forming a capacitor. The system is connected into a circuit , with a power source of electromotive force $U_{0}$. Initially, the capacitor is uncharged, and at $\scriptstyle t=0$, the switch is turned on. Determine the time-dependent relationship of the current $I(t)$ flowing through the switch for $t > 0$. | Let the magnitudes of the electric field intensity in the two media be $E_{1}, E_{2}$ respectively, and according to the relationship between the electric potentials, we have
$$
E_{1}h_{1} + E_{2}h_{2} = U_{0}
$$
The magnitudes of the current density in the two media are
$$
\begin{aligned}
j_1 &= \sigma_1 E_1, \\
j_2 &= \sigma_2 E_2.
\end{aligned}
$$
Let the free charge surface density at the contact of the two media be $\sigma_{\bullet}$, then
From this, we obtain
$$
\left\{\begin{array}{l l}{\displaystyle{\frac{\mathrm{d}\sigma_{\circ}}{\mathrm{d}t}}=j_{1}-j_{2}}\ {\sigma_{\circ}=\varepsilon_{2}E_{2}-\varepsilon_{1}E_{1}}\end{array}\right.
$$
$$
U_{0} = \frac{\mathrm{d}\sigma_{0}}{\mathrm{d}t}\cdot\left(\frac{h_{1}\varepsilon_{2} + h_{2}\varepsilon_{1}}{\sigma_{1}\varepsilon_{2} - \sigma_{2}\varepsilon_{1}}\right) + \sigma_{0}\cdot\left(\frac{\sigma_{1}h_{2} + \sigma_{2}h_{1}}{\sigma_{1}\varepsilon_{2} - \sigma_{2}\varepsilon_{1}}\right)
$$
That is
$$
\sigma_{0} = \frac{U_{0}\left(\sigma_{1}\varepsilon_{2} - \sigma_{2}\varepsilon_{1}\right)}{\sigma_{1}h_{2} + \sigma_{2}h_{1}}\cdot\left(1-\mathrm{e}^{\frac{t}{\tau}}\right)
$$
Where the time constant is
$$
\tau = \frac{h_{1}\varepsilon_{2} + h_{2}\varepsilon_{1}}{h_{1}\sigma_{2} + h_{2}\sigma_{1}}
$$
The current in the circuit can be obtained as
$$
I = A{\left(\sigma_{1}E_{1} + \varepsilon_{1}{\frac{\mathrm{d}E_{1}}{\mathrm{d}t}}\right)} = A{\left(\sigma_{2}E_{2} + \varepsilon_{2}{\frac{\mathrm{d}E_{2}}{\mathrm{d}t}}\right)}
$$
Substitute in to get
$$
I = A U_{0}\cdot\left[\frac{\sigma_{1}\sigma_{2}}{\sigma_{1}h_{2}+\sigma_{2}h_{1}} + \mathrm{e}^{\frac{\sigma_{1}h_{2}+\sigma_{2}h_{1}}{\varepsilon_{1}h_{2}+\varepsilon_{2}h_{1}}t}\cdot\left(\frac{\varepsilon_{2}\sigma_{1}+\varepsilon_{1}\sigma_{2}}{\varepsilon_{1}h_{2}+\varepsilon_{2}h_{1}} - \frac{\sigma_{1}\sigma_{2}}{\sigma_{1}h_{2}+\sigma_{2}h_{1}} - \frac{\varepsilon_{1}\varepsilon_{2}\left(\sigma_{1}h_{2}+\sigma_{2}h_{1}\right)}{\left(\varepsilon_{1}h_{2}+\varepsilon_{2}h_{1}\right)^{2}}\right)\right]
$$ | $$
I = A U_0 \left[\frac{\sigma_1 \sigma_2}{\sigma_1 h_2 + \sigma_2 h_1} + e^{-\frac{\sigma_1 h_2 + \sigma_2 h_1}{\varepsilon_1 h_2 + \varepsilon_2 h_1} t} \left(\frac{\varepsilon_2 \sigma_1 + \varepsilon_1 \sigma_2}{\varepsilon_1 h_2 + \varepsilon_2 h_1} - \frac{\sigma_1 \sigma_2}{\sigma_1 h_2 + \sigma_2 h_1} - \frac{\varepsilon_1 \varepsilon_2 (\sigma_1 h_2 + \sigma_2 h_1)}{(\varepsilon_1 h_2 + \varepsilon_2 h_1)^2} \right)\right]
$$ |
117 | MECHANICS | There is an available cycloidal valley, which can be described using a cycloid (downward along the $y$-axis):
$$
\left\{{\begin{array}{l}{x=R(\theta-\sin\theta)}\\ {y=R(1-\cos\theta)}\end{array}}\right.
$$
Someone wants to release a cube with a mass of $M$ and with equal length, width, and height of $r$ at rest from one end of the valley at $\theta = 0$. It is released facing the slope. A layer of snow with a surface density of $\sigma$ is attached to the valley. The snow is in contact with the ground but does not interact, remaining stationary until it touches the object, at which point it adheres to its surface without altering the object's shape. Assuming gravitational acceleration $g$, there are the approximate conditions $\sigma r R \ll M, r \ll R$ (results are retained up to the first order of $r$). Determine the highest position the cube can reach at the other end (considering no rolling occurs).
| In this sub-question, energy is not conserved, but analyzing its motion still presents certain difficulties. Considering that the problem only requires accuracy to the first-order term, the loss of ascent quotient can be divided into two parts for correction, with zero-order motion considered during the correction process. For the collision process between the square block and the snow, consider the conservation of momentum: $$ M v = (M + \sigma\gamma u{\mathrm{d}}t)(v - {\mathrm{d}}v) $$ This results in: $$ \mathrm{d}\upsilon = -\frac{\sigma r v^{2}}{M}\mathrm{d}t $$ The energy loss during the process is: $$ d E = \frac{1}{2}(M + \mathrm{d}M)(\upsilon + \mathrm{d}\upsilon)^{2} - \frac{1}{2}(M)\upsilon^{2} = -\frac{1}{2}\sigma r\upsilon^{2}\upsilon\mathrm{d}t = -\frac{1}{2}\sigma r\upsilon^{2}\mathrm{d}s $$ In this case, it can be treated as linear motion, and the equation is not influenced by acceleration. Therefore, it can be corrected independently, with the zero-order velocity expression as: $$ \upsilon_{0} = \sqrt{2g R(1-\cos\theta)} $$ Substituting into the integral: $$ \delta W = \int f\mathrm{d}s = \int_{0}^{2\pi} -g R(1-\cos\theta)\sigma\boldsymbol{r}\cdot2R\sin\frac{\theta}{2}\mathrm{d}\theta = -\frac{32}{3}g R^{2}r\sigma $$ This gives the energy loss caused by collisions, which we convert into a height correction, then combine it with the correction for the gravitational component of the snow from the first question, resulting in the total reached height: $$ \begin{array}{l} {{\delta y_{f} = \displaystyle\frac{32}{3}\displaystyle\frac{\sigma R^{2}r}{M}}}{{}}\\ {{\delta y_{g} = \displaystyle\frac{32}{3}\displaystyle\frac{\sigma R^{2}r}{M}}}{{}}\\ {{\delta y = \displaystyle\frac{64}{3}\displaystyle\frac{\sigma R^{2}r}{M}}} \end{array} $$ The two combined terms are both first-order and do not interfere with each other.
| $$
\delta y=\frac{64}{3}\frac{\sigma R^2 r}{M}
$$ |
42 | OPTICS | We consider a special rotationally symmetric refractive index distribution $n=n(r)$, such that the light trajectory is $r=a\cos q\theta$ (although it is not very realistic for the refractive index to diverge at $r=0$). In the space where $r<a$, find the refractive index $n(r)$ (since proportionally changing the refractive index does not affect the light's path, your answer should contain a dimensionless coefficient $n_{0}$). | Conservation of Angular Momentum
$$
L=m r^{2}{\dot{\theta}}
$$
Therefore
$$
{\frac{d}{d t}}={\frac{L}{m r^{2}}}{\frac{d}{d\theta}}
$$
Conservation of Mechanical Energy
$$
E=V+{\frac{1}{2}}m\left({\dot{r}}^{2}+r^{2}{\dot{\theta}}^{2}\right)=V+{\frac{L^{2}}{2m r^{4}}}\left(r^{2}+a^{2}q^{2}\sin^{2}q\theta\right)
$$
Thus
$$
V={\frac{L^{2}}{2m}}\left({\frac{q^{2}-1}{r^{2}}}-{\frac{a^{2}q^{2}}{r^{4}}}\right)
$$
And
$$
E=0
$$
$$
A={\frac{L^{2}}{2m}}\left(q^{2}-1\right)
$$
Using $B$ as a known quantity, we have
Since
Given the magnitude of the velocity
$$
v={\sqrt{{\frac{2}{m}}{\frac{B}{l^{2}}}\left({\frac{1}{l^{2}}}-{\frac{q^{2}-1}{a^{2}q^{2}}}\right)}}
$$
And because
$$
B=\frac{L^{2}}{2m}a^{2}q^{2}
$$
Let the angle between the velocity direction and the polar axis be $\varphi$, substituting in
$$
L=m v l\sin\varphi
$$
Solving
$$
\varphi=\pm\arcsin{\frac{1}{\sqrt{\left({\frac{a^{2}}{l^{2}}}-1\right)q^{2}+1}}}
$$
It's obvious that we also require
$$
l\leq a
$$
(2) The condition $E=0$ was previously mentioned, so in the analogy of light, there is no need to match the refractive index to satisfy $E=0$, simply let
$$
V=-{\frac{1}{2}}n^{2}
$$
This is feasible, such that
$$
n(r)={\sqrt{{\frac{L^{2}}{m}}\left({\frac{a^{2}q^{2}}{r^{4}}}-{\frac{q^{2}-1}{r^{2}}}\right)}}
$$
Taking into account the problem of dimensional analysis, proportional changes, and range, finally we get:
$$
n(r)=n_{0}a{\sqrt{\left({\frac{a^{2}q^{2}}{r^{4}}}-{\frac{q^{2}-1}{r^{2}}}\right)}}
$$ | $$
n_0 a \sqrt{\frac{a^2 q^2}{r^4} - \frac{q^2 - 1}{r^2}}
$$ |
45 | THERMODYNAMICS | Calculate the wave speed of longitudinal waves in a one-dimensional gas. Consider a section of the gas column with cross-sectional area $S$, assume its displacement along the $x$ direction is $\xi = \xi(x)$, the gas pressure distribution is $p = p(x)$, and the density of the gas when uncompressed is $\rho_{0}$. Consider the case of small amplitude and long wavelength, i.e. $\xi \ll x, \frac{\partial \xi}{\partial x}\ll 1, (p-p_0)\ll p_0, (\rho-\rho_0)\ll\rho_0$. Calculate the wave speed of longitudinal waves in a van der Waals gas. The state equation of one mole of the gas is given as:
$$
\left( p + \frac{a}{V^{2}} \right)(V - b) = RT\
$$
The molar heat capacity at constant volume is $c_{v}$, and the molar mass of the gas is $\mu$. Solve for the wave speed under adiabatic conditions, expressed in terms of $p, \rho, a, b,\mu, \gamma = \frac{c_{v} + R}{c_{v}}$.
Hint: You can use the formulas:
$$dU = c_{v}dT + \left( \frac{\partial U}{\partial V} \right)_{T}dV$$
and
$$\left( \frac{\partial U}{\partial V} \right)_{T} = T\left( \frac{\partial p}{\partial T} \right)_{V} - p$$
to calculate the adiabatic equation. | Consider the air column from $x \rightarrow x + dx$. From the dynamics equation, we obtain:
$$
\rho_{0}Sdx\frac{\partial^{2}\xi}{\partial t^{2}} = - \frac{dp}{dS}(1)
$$
After simplification, we get
$$
\rho_{0}\frac{\partial^{2}\xi}{\partial t^{2}} = - \frac{\partial p}{\partial x}(2)
$$
Consider the air column originally located at $x\rightarrow x+\mathrm d x$, now located at $x+\xi\rightarrow x+dx+\xi +d\xi$. By mass conservation:
$$
\rho(d\xi + dx) = \rho_{0}dx(3)
$$
Hence, we get
$$
\rho = \rho_{0}\left( 1 - \frac{\partial\xi}{\partial x} \right)(4)
$$
Using
$$
\frac{\partial p}{\partial x} = \left( \frac{\partial p}{\partial\rho} \right)_{c} \cdot \frac{\partial\rho}{\partial x}(5)
$$
we obtain
$$
\frac{\partial^{2}\xi}{\partial t^{2}} = \left( \frac{\partial p}{\partial\rho} \right)_{c}\frac{\partial^{2}\xi}{\partial x^{2}}(6)
$$
The wave speed is
$$
v = \sqrt{\frac{\partial p}{\partial \rho}}
$$
Using the formula from the hint, we obtain
$$
\left( \frac{\partial U}{\partial V} \right)_{T} = T\left( \frac{\partial p}{\partial T} \right)_{V} - p = \frac{a}{V^{2}}(7)
$$
First Law of Thermodynamics
$$
dQ = dU + dW(8)
$$
$$
0 = c_{v}dT + \frac{a}{V^{2}}dV + pdV = c_{v}dT + \frac{RT}{V - b}dV(9)
$$
Integrating and substituting into the state equation, we obtain
$$
\left( p + \frac{a}{V^{2}} \right)(V - b)^{\frac{c_{v} + R}{C_{v}}} = C(10)
$$
Using
$$
V = \frac{\mu}{\rho}(11)
$$
Taking the logarithm of equation (10) and then differentiating, we obtain:
$$
v = \sqrt{\left. \left(\frac{\partial p}{\partial\rho} \right. \right)_{s}} = \sqrt{\frac{\gamma\mu}{\mu\rho - b\rho^{2}}\left( p + \frac{a}{\mu^{2}}\rho^{2} \right) - \frac{2a\rho}{\mu^{2}}}(12)
$$ | $$\sqrt{\frac{\gamma\mu}{\mu\rho - b\rho^2}\left(p + \frac{a}{\mu^2}\rho^2\right) - \frac{2a\rho}{\mu^2}}$$ |
37 | MECHANICS | The curling can be approximated as a solid homogeneous cylinder with mass $m$, bottom surface radius $r$, and height $h$. Only the outer edge of the bottom is in contact with the ice surface, which has a friction coefficient $\mu$. The support force of the ice surface on the curling stone is $N=\int_{0}^{2\pi}n(\theta)\mathrm{d}\theta$,
where $n(\theta)$ is the distribution of the support force according to the polar coordinate $\theta$. When the curling is stationary on the horizontal ice surface, $n(\theta)=mg/2\pi$. Assume that the deformation of the outer edge of the curling is very small, and the elastic force satisfies Hooke's law. Therefore, it can be proven that the support force distribution has the form $n(\theta)=n_0+n_1\cos(\theta-\theta_0)$ (where $n_0$, $n_1$, and $\theta_0$ are unknown quantities). If the curling has a velocity $v$ in the positive direction along the x-axis, and also has an angular velocity $\omega$ in the clockwise direction (viewed from above), the curling will generate a lateral acceleration $a$ in the y-direction. It is through this that the athlete can achieve the curling's turn. Find the expression for $a_y$ in the limit case where $v\gg\omega r$. | The deformation at a particular point $\pmb{d}$ is a linear function of its projection distance $\boldsymbol{\mathbf{\rho}}_{x}$ in the direction of maximum deformation:
$$
d = d_{0} + x \tan\alpha
$$
Assuming the direction of maximum deformation is $\theta_{0}$, we have:
$$
x = r \cos{\left(\theta - \theta_{0}\right)}
$$
According to Hooke's law:
$$
n = k d = k d_{0} + k r \tan\alpha \cos\left(\theta - \theta_{0}\right) = n_{0} + n_{1} \cos\left(\theta - \theta_{0}\right)
$$
The curling stone is in vertical force equilibrium, so:
$$
N = \int_{0}^{2\pi} n\big(\theta\big)d\theta = m g
$$
Solving, we obtain $n_{0} = m g / 2\pi$.
The friction force acting on the curling stone is proportional to the normal pressure:
$$
f(\theta) = \mu n(\theta)
$$
Its direction is along the negative $\mathbf{x}$ axis.
During the motion of the curling stone, the overall torque is balanced:
$$
M_{x} = \int_{0}^{2\pi}\boldsymbol{n}(\theta) r \sin\theta d\theta + \frac{1}{2} \int_{0}^{2\pi} f_{y}\left(\theta\right) h d\theta = 0
$$
$$
M_{y} = -\int_{0}^{2\pi} \eta \left(\theta\right) r \cos\theta d\theta - \frac{1}{2} \int_{0}^{2\pi} f_{x}\left(\theta\right) h d\theta = 0
$$
When $\nu \gg \omega r$, the components of the friction force retain only the first-order approximations:
$$
f_{x} = -\frac{\nu + \omega r \sin\theta}{\sqrt{\left(\nu + \omega r \sin\theta\right)^{2} + \omega^{2} r^{2} \cos^{2}\theta}} \mu n(\theta) \approx -\mu n(\theta)
$$
$$
f_{y} = \frac{\omega r \cos{\theta}}{\sqrt{\left(\nu + \omega r \sin{\theta}\right)^{2} + \omega^{2} r^{2} \cos^{2}{\theta}}} \mu n(\theta) \approx \frac{\omega r \cos{\theta}}{\nu} \mu n(\theta)
$$
For convenience in solving, let $n\left(\theta\right) = n_{0} + a \cos\theta + b \sin\theta$. Substituting into equations (6) and (7) and solving simultaneously, we obtain:
$$
n\left(\theta\right) = \frac{m g}{2\pi} \left(1 + \frac{\mu h}{r} \cos\theta - \frac{\mu^{2} h^{2} \omega}{2 \nu r} \sin\theta\right)
$$
The final transverse acceleration is:
$$
a_{y} = \frac{1}{m} \int_{0}^{2\pi} f_{y} d\theta = \frac{\mu^{2} \omega h g}{2 \nu}
$$ | $$\frac{\mu^2\omega hg}{2v}$$ |
654 | MECHANICS | The forty-second batch of Trisolarian immigrants arrived on Earth aboard a spaceship propelled by photons. The spaceship has a rest mass of $M$, and it uses the method of annihilating matter and antimatter to produce photons, which are then emitted directly backward to provide thrust. In the reference frame of the spaceship, the frequency of the photons emitted backward is $ u _{0}$. For simplicity, we assume that the \"windward side\" of the Trisolarian spaceship is a flat surface with an area $S$. During flight, dust floating in the universe may collide with it. We assume that the cosmic dust is composed of particles with rest mass $m$. For an observer on Earth: the spaceship moves at a constant speed of $v _1$, the particles in the cosmic dust have a speed of $v_2$ and move in the same direction as the spaceship, and the particles in the dust are uniformly distributed with a density of $n$. If, according to the Trisolarian beings on the spaceship, the cosmic dust undergoes a \"perfectly elastic collision\" upon impact with the spaceship, what is the number $N^{'}$ of photons that need to be emitted backward per unit time in the reference frame of the spaceship to maintain its constant speed? Given are Planck's constant $h$, the speed of light in vacuum $c$, and relativistic effects need to be considered.
| There are two reference frames; let's set up the agreements first. The reference frame of the Earth observer is called the $\Sigma$ frame, and the reference frame of the spaceship is called the $\Sigma$ frame. We agree that: $$\beta_{1}=\frac{v_{1}}{c},\gamma_{1}=\frac{1}{\sqrt{1-\beta_{1}^{2}}},\beta_{2}=\frac{v_{2}}{c},\gamma_{2}=\frac{1}{\sqrt{1-\beta_{2}^{2}}},\beta_{21}=\frac{v_{21}}{c},\gamma_{21}=\frac{1}{\sqrt{1-\beta_{21}^{2}}}$$ $$\beta_{2}^{'}=\frac{v_{2}^{'}}{c},\gamma_{2}^{'}=\frac{1}{\sqrt{1-\beta_{2}^{'2}}},\beta_{21}^{'}=\frac{v_{21}^{'}}{c},\gamma_{21}^{'}=\frac{1}{\sqrt{1-\beta_{21}^{'2}}}$$ The resistance of the spaceship is equal to the impulse given to the spaceship by cosmic dust particles per unit time. In the reference frame of the spaceship, we have: $$ F^{'}=n^{'}S\left|v_{2}^{'}\right|\times\left|p_{2}^{'}-p_{21}^{'}\right|=n^{'}S\left|v_{2}^{'}\right|\times\left|2p_{2}^{'}\right|=2n^{'}S m \gamma_{2}^{'}v_{2}^{'2} $$ The calculation of $n'$ needs to take into account the length contraction effect. If the particle density is $n_{0}$ in the rest frame of the cosmic dust itself, we have: $$ n=n_{0}\gamma_{2}, n^{'}=n_{0}\gamma_{2}^{'} $$ Therefore, we can obtain $n^{\prime}=\frac{\gamma_{2}^{'}}{\gamma_{2}}n_{0}$. Since $S$ is perpendicular to the direction of velocity, it is identical in both reference frames. By relativistic velocity transformation: $$ v_{2}^{'}=\frac{v_{2}-v_{1}}{1-\frac{v_{2}v_{1}}{c^{2}}} $$ Substituting into the expression for $F$ yields: $$ F^{'}=2\frac{\gamma_{2}^{'}}{\gamma_{2}}n S m\gamma_{2}^{'} v_{2}^{'2}=2n S m\gamma_{1}^{2}\gamma_{2}(v_{1}-v_{2})^{2} $$ From momentum conservation in the spaceship reference frame, we can obtain: $$ \boldsymbol{F}^{'}=\frac{N^{'} h u_{0}}{c} $$ $$ N^{'}=\frac{2n S m c\gamma_{1}^{2}\gamma_{2}(v_{1}-v_{2})^{2}}{h u_{0}} $$
| $$
\frac{2nSmc(v_{1}-v_{2})^{2}}{(1-\left(\frac{v_1}{c}\right)^2)\sqrt{1-\left(\frac{v_2}{c}\right)^2}h\nu_{0}}
$$ |
244 | THERMODYNAMICS | Consider a non-ideal gas system. Without worrying about the internal composition and interaction details, we find that its macroscopic equation of state is:
$$
p = k V^{-4/5}T^{3/2}
$$
where $k$ is a constant coefficient.
Consider a very special case. If for this non-ideal gas, its specific heat ratio:
$$
\gamma = \frac{C_{p}}{C_{V}}
$$
is exactly a constant that is independent of temperature and volume. Calculate the value of $\gamma$. | Combine the formula for the difference in heat capacities with the heat capacity relation:
$$
C_{p}=\gamma C_{V}
$$
This directly solves for the heat capacity:
$$
C_{V}=\frac{45}{16}\frac{k V^{1/5}T^{1/2}}{\gamma-1}
$$
And at this point, it must satisfy
$$
\frac{\partial C_{V}}{\partial V}=g=\frac{3}{4}k V^{-4/5}T^{1/2}
$$
Substituting gives the only possible value for $\gamma$:
$$
\gamma=\frac{7}{4}
$$ | $$
\gamma = \frac{7}{4}
$$ |
25 | ELECTRICITY | On an infinitely large horizontal plane, establish polar coordinates ($r, \theta$) .On the plane there is a stationary magnetic field of magnitude $\frac{mv}{q\sqrt{ar}}$ directed vertically upwards. A particle with a charge-to-mass ratio of $\frac{3q}{2\sqrt{2}m}$ is projected along the polar axis direction($\theta=0$) from the origin with an initial velocity of $v_0$. Find the polar coordinate trajectory equation $r(\theta)$ for this particle. | Solution:
The dynamic equations are
$$
\left\{{\begin{array}{l}{m({\ddot{r}}-r{\dot{\theta}^{2}})=q B r{\dot{\theta}}}\\ {m{\frac{d\left(r^{2}{\dot{\theta}}\right)}{r d t}}=-q B{\dot{r}}}\end{array}}\right.
$$
Rearranging and integrating equation (21) yields:
$$
\begin{array}{l}{{d(r^{2}\dot{\theta})={\frac{3v_{0}\sqrt{r}}{2\sqrt{2a}}}d r}}\\ {{d(r^{2}\dot{\theta})=-d{\frac{v_{0}r^{\frac{3}{2}}}{\sqrt{2a}}}}}\\ {{~\therefore~r^{2}\dot{\theta}+{\frac{v_{0}r^{\frac{3}{2}}}{\sqrt{2a}}}=0}}\\ {{\dot{\theta}=-{\frac{v_{0}}{\sqrt{2a r}}}}}\end{array}
$$
Lorentz force does no work, energy conservation:
$$
\begin{array}{r}{\therefore\dot{r}=\sqrt{v_{0}^{2}-\frac{v_{0}^{2}r}{2a}}}\\ {\frac{d r}{d\theta}=-\sqrt{2a r-r^{2}}}\end{array}
$$
$$
\therefore r=a(1-\cos\theta)
$$ | $$
a(1-\cos\theta)
$$ |
100 | MECHANICS | Consider a metal beam with a width of $a$, thickness of $h$, and Young's modulus of $E$. The beam is placed on supports spaced at a distance $l$, and it is assumed that the length of the metal beam is slightly greater than $l$. A weight with a mass of $m$ is placed in the middle of the beam. Assume the angle between the beam and the horizontal line is very small, neglect friction and the weight of the beam, and take the gravitational acceleration as $g$.
Hint: Since the beam has thickness, it develops a situation where the inner side is compressed and the outer side is stretched when in equilibrium. Although the net force at the interface is zero, a torque is generated. This also causes every part of the beam to experience compression/stretch, thereby storing potential energy.
After the weight reaches equilibrium, apply a small horizontal perturbation along the rod’s direction to the weight and determine the vibration angular frequency of the system. | (1) Using the point where the heavy object is suspended as the origin, and the horizontal direction as the $\pmb{\mathscr{x}}$ axis, and the vertical direction as the $\textit{\textbf{y}}$ axis, establish a coordinate system.
First, consider a small curved metal beam segment with a central angle of $\theta$ and radius of $\scriptstyle{\mathcal{R}}$, then the strain at $z$ is:
$$
\epsilon = \frac{(R+z)\theta - R\theta}{R\theta} = \frac{z}{R}
$$
Resulting moment:
$$
M = \int_{-\frac{h}{2}}^{\frac{h}{2}}\frac{E a}{R}z^{2}d z = \frac{E a h^{3}}{12R}
$$
Given the approximation condition ${y}^{\prime}\ll1$:
$$
R = {\frac{(1+y^{\prime2})^{\frac{3}{2}}}{|y^{\prime\prime}|}} \approx {\frac{1}{|y^{\prime\prime}|}}
$$
Substitute into the previous equation:
$$
M = {\frac{E a h^{3}|y^{\prime\prime}|}{12}}
$$
For the part of the metal beam with $\mathfrak{x}(\mathfrak{x}>0)$ and above, set up the moment equilibrium at the $\pmb{x}$ position:
$$
M = \frac{m g}{2}(\frac{l}{2}-x)
$$
Substitute into the previous equation:
$$
y^{\prime\prime} = \frac{6m g}{E a h^{3}}(\frac{l}{2}-x)
$$
Initial conditions:
$$
\begin{array}{r}{y(0)=0}\ {y^{\prime}(0)=0}\end{array}
$$
Integrating yields:
$$
y = {\frac{3m g l}{2E a h^{3}}}x^{2} - {\frac{m g}{E a h^{3}}}x^{3}
$$
Substitute $\begin{array}{r}{x=\frac{1}{2}}\end{array}$ to obtain:
$$
\lambda = \frac{m g l^{3}}{4E a h^{3}}
$$
By the results of question (1), the metal beam can be equated to a spring with a stiffness coefficient of:
$$
k = \frac{4E a h^{3}}{l^{3}}
$$
This gives the angular frequency of vertical vibration:
$$
\omega_{1} = {\sqrt{\frac{k}{m}}} = {\sqrt{\frac{4E a h^{3}}{m l^{3}}}}
$$
As shown in the diagram, let the heavy object deviate from the center by $\delta l$, denote $\begin{array}{r}{\varepsilon=\frac{\delta\ell}{\ l}}\end{array}$
From mechanical equilibrium:
$$
N_{1} = \frac{m g}{2}(1-2\varepsilon)
$$
$$
N_{2} = \frac{m g}{2}(1+2\varepsilon)
$$
For the part above the horizontal axis, establish the moment equilibrium equation at (2 >0):
$$
M_{1} = {{N}_{2}}\left[\frac{l}{2}\left(1-2\varepsilon\right)-x\right]
$$
Similarly, for ${\mathfrak{x}}<{\mathfrak{0}}$:
$$
M_{2} = N_{1}\left[\frac{l}{2}\left(1+2\varepsilon\right)+x\right]
$$
Initial conditions:
$$
\begin{array}{c}{y\left(0\right)=0}\ {y^{\prime}\left(0\right)=k_{0}}\end{array}
$$
Substitute into equation (2) and solve using initial conditions:
Substitute boundary conditions:
$$
\begin{array}{c}{{y_{1}=k_{0}x+\displaystyle\frac{3m g l}{2E a h^{3}}\left(1-4\varepsilon^{2}\right)x^{2}-\displaystyle\frac{m g}{E a h^{3}}\left(1+2\varepsilon\right)x^{3}(x>0)}} \\ {{y_{2}=k_{0}x+\displaystyle\frac{3m g l}{2E a h^{3}}\left(1-4\varepsilon^{2}\right)x^{2}+\displaystyle\frac{m g}{E a h^{3}}\left(1-2\varepsilon\right)x^{3}(x<0)}}\end{array}
$$
$$
y_{1}\left(\frac{l}{2}\left(1-2\varepsilon\right)\right) = y_{2}\left(-\frac{l}{2}\left(1+2\varepsilon\right)\right) = \lambda^{\prime}
$$
Solve to obtain:
$$
\begin{array}{l}{{k_{0}=\displaystyle\frac{2m g l^{2}}{E a h^{3}}\left(1-4\varepsilon^{2}\right)\varepsilon}} \\ {{\lambda^{\prime}=\displaystyle\frac{m g l^{3}}{4E a h^{3}}\left(1-4\varepsilon^{2}\right)^{2}}}\end{array}
$$
When $\varepsilon\ll1$, we have:
$$
\lambda^{\prime} \approx \frac{m g l^{3}}{4E a h^{3}}\left(1-8\varepsilon^{2}\right)
$$
Therefore, the gravitational potential energy of the heavy object is:
$$
E_{p1} = \frac{2m^{2}g^{2}l^{3}}{E a h^{3}}\varepsilon^{2}
$$
Consider a small segment of the metal beam and its elastic potential energy:
$$
d E_{p2} = \int{\frac{1}{2}}E\epsilon^{2}d V = \int_{-{\frac{h}{2}}}^{{\frac{h}{2}}}{\frac{{\cal E}a y^{\prime\prime2}d x}{2}}z^{2}d z = {\frac{{\cal E}a h^{3}}{24}}y^{\prime\prime2}d x
$$
Integrating yields:
$$
E_{p2} = \int_{-\frac{1}{2}(1+2\varepsilon)}^{0}\frac{E a h^{3}}{24}y_{2}^{\prime\prime}d x + \int_{0}^{\frac{1}{2}(1-2\varepsilon)}\frac{E a h^{3}}{24}{y_{1}^{\prime\prime}}^{2}d x = \frac{m^{2}g^{2}l^{3}(1-4\varepsilon^{2})^{2}}{8E a h^{3}} \approx \frac{m^{2}g^{2}l^{3}(1-8\varepsilon^{2})}{8E a h^{3}}
$$
The kinetic energy of the ball is:
$$
E_{k} = {\frac{1}{2}}m l^{2}{\dot{\varepsilon}}^{2}
$$
Total energy of the system:
$$
E = E_{k} + E_{p1} + E_{p2} = {\frac{1}{2}}m l^{2}{\dot{\varepsilon}}^{2} + {\frac{m^{2}g^{2}l^{3}}{E a h^{3}}}\varepsilon^{2} + E_{0} = {\frac{1}{2}}M{\dot{\varepsilon}}^{2} + {\frac{1}{2}}K\varepsilon^{2} + E_{0}
$$
Giving the angular frequency of horizontal vibration:
$$
\omega_{2} = {\sqrt{\frac{K}{M}}} = {\sqrt{\frac{2m g^{2}l}{E a h^{3}}}}
$$ | $$
\sqrt{\frac{2m g^2 l}{E a h^3}}
$$ |
89 | ELECTRICITY | A typhoon is a vortex-like weather system, and as it passes, it is often accompanied by lightning. When the electric field strength in the air reaches a certain value, the air breaks down, generating lightning. The breakdown field strength of air is denoted as $E_{m}$. The typhoon system is modeled as an isothermal atmospheric ideal gas with a temperature denoted as ${T}_{0}$. Under strong ionization conditions, it can be considered as consisting of neutral monoatomic gas (with mass $m_{tot}=m_{N}+m_{e}$) and free electrons (with mass $m_{e}$) released through ionization. It is known that each atom can ionize to release at most one electron, with the total number of monoatomic gas particles being $N$.
The rotation speed of the typhoon is ${\overrightarrow{\omega}}=\omega{\hat{z}}$. In this problem, we consider a cylindrical typhoon system with radius ${R}$ and height $H(H \gg R)$. Both the electrons and the positive ions are confined within this cylindrical volume and do not escape. Interactions between free electrons and free electrons, free electrons and positive ions, and positive ions and positive ions are ignored. Additionally, the effects of internal potential distributions within the typhoon on the distribution of free electrons and positive ions are not considered (calculated results can be simplified using $k=\frac{\omega^2}{2k_BT_0}$). The permittivity of free space is $\varepsilon_0$.
After the typhoon has stabilized, researchers need to deploy unmanned aerial vehicles (UAVs) to gather data within the typhoon system. The UAV has a volume of $V(V^{\frac{1}{3}} \ll R)$ and mass ${m}$. To ensure stable operation during data collection, the UAV must remain at a radius ${r_{0}}$ while orbiting the typhoon center with the same angular velocity $\omega$. It is known that the UAV can only exert thrust in the vertical direction. At time $t=0$, the UAV in stable operation experiences a small radial perturbation and acquires an initial radial velocity $\overrightarrow{v_{0}}$ with $v_{0} \ll \omega r_{0}$. Observations show that the UAV exhibits periodic precession. Solve for its precession angular frequency $\Omega$ (the answer can be expressed in terms of $q_{0}$ and $r_{0}$). | According to the problem statement, a single ideal gas under the isothermal model can satisfy the Boltzmann distribution in any continuous potential energy field distribution. By entering the rotating reference frame of the typhoon, centrifugal potential energy is introduced:
$$
V_{\text{eff}} = -\frac{1}{2}m\omega^2r^2
$$
Thus, the mass density of free electrons and positive ions satisfies:
$$
\rho_{-}(r) = \rho_{-0}e^{m_{\text{e}}kr^2}, \rho_{+}(r) = \rho_{+0}e^{m_{N}kr^2}
$$
Consider that the total number of electrons and positive ions satisfies the normalization condition:
Solving yields:
$$
N = \frac{\int_{0}^{R}2\pi r H\times\rho_{-}(r)dr}{m_e}, N = \frac{\int_{0}^{R}2\pi r H\times\rho_{+}(r)dr}{m_N}
$$
$$
\rho_{-}(r) = \rho_{-0}e^{-m_{e}kr^2} = \frac{k m_{e}^2N}{\pi H(e^{m_{e}k R^2}-1)}e^{m_{e}kr^2}
$$
$$
\rho_{+}(r) = \rho_{+0}e^{-m_{N}kr^2} = \frac{k m_{N}^2N}{\pi H(e^{m_{N}k R^2}-1)}e^{m_{N}kr^2}
$$
From
$$
PV = nRT
$$
It follows that
$$
P(r) = \left(\frac{\rho_{+}}{m_{N}} + \frac{\rho_{-}}{m_{e}}\right)k_{B}T_{0}
$$
$$
P(r) = \left(\frac{k m_{e}N}{\pi H(e^{m_{e}k R^2}-1)}e^{m_{e}kr^2} + \frac{k m_{N}N}{\pi H(e^{m_{N}k R^2}-1)}e^{m_{N}kr^2}\right)k_{B}T_{0}
$$
We prioritize considering the electric field intensity at the maximum electric field, analyzing symmetry $\vec{E} = E(r)\hat{r}$
$$
E(r) = \frac{\int_{0}^{r}(e\rho_{+} - e\rho_{-})(2\pi r \, dr)}{2\pi r\varepsilon_{0}H} = \frac{N e}{2\pi r H\varepsilon_{0}}\left[\frac{e^{m_{N}kr^2}-1}{e^{m_{N}k R^2}-1} - \frac{e^{m_{e}kr^2}-1}{e^{m_{e}k R^2}-1}\right]
$$
Considering when $\pmb{E}$ reaches maximum value, it satisfies:
$$
\frac{dE(r)}{dr} = 0
$$
Simplifying, the $r_{\text{max}}$ at maximum electric field satisfies the equation when $r = r_{\text{max}}$:
$$
\left[\frac{e^{m_{N}kr^2}-1}{e^{m_{N}k R^2}-1} - \frac{e^{m_{e}kr^2}-1}{e^{m_{e}k R^2}-1}\right] - 2kr^2\left[m_{N}\frac{e^{m_{N}kr^2}}{e^{m_{N}k R^2}-1} - m_{e}\frac{e^{m_{e}kr^2}}{e^{m_{e}k R^2}-1}\right] = 0
$$
Substituting the data, solving yields:
$$
r_{\text{max}} = 719.96 \, \mathrm{m}
$$
At this time:
$$
E_{\text{max}} = \frac{N e}{2\pi r_{\text{max}}H\varepsilon_{0}}\left[\frac{e^{m_{N}kr_{\text{max}}^2}-1}{e^{m_{N}k R^2}-1} - \frac{e^{m_{e}kr_{\text{max}}^2}-1}{e^{m_{e}k R^2}-1}\right] = -9.83 \times 10^{18} \, \mathrm{V/m}
$$
Since:
$$
\left|E_{\text{max}}\right| > E_m
$$
It can break down air.
For analyzing the forces on the drone in the ground reference frame, the buoyancy it experiences satisfies:
$$
\vec{F}' = (\rho_{-}(r_{0}) + \rho_{+}(r_{0}))V g\hat{z} + (\rho_{-}(r_{0}) + \rho_{+}(r_{0}))V\omega^2r_{0}(-\hat{r})
$$
In the $\hat{r}$ direction, list the force balance equation:
$$
E(r_{0})q_{0} - (\rho_{-}(r_{0}) + \rho_{+}(r_{0}))V\omega^2r_{0} = -m\omega^2r_{0}
$$
Substituting specific values, solving gives:
$$
q_{0} = \frac{-m\omega^2r_{0} + \left(\frac{k m_{\circ}^2N}{\pi H(\circ^{m_{\circ}kr_{0}^2}-1)}e^{m_{\circ}kr_{0}^2} + \frac{k m_{N}^2N}{\pi H(\circ^{m_{N}kr_{0}^2}-1)}e^{m_{N}kr_{0}^2}\right)V\omega^2r_{\circ}}{\frac{N e}{2\pi r_{0}H\varepsilon_{0}}\left[\frac{e^{m_{N}kr_{0}^2}-1}{e^{m_{N}kr_{0}^2}-1} - \frac{e^{m_{e}kr_{0}^2}-1}{e^{m_{\circ}kr_{0}^2}-1}\right]}
$$
Further consider the force balance equation in the $\hat{z}$ direction:
$$
F_{0} + (\rho_{-}(r_{0}) + \rho_{+}(r_{0}))V g - m g = 0
$$
Substitute specific values, solving yields:
$$
F_{0} = -(\rho_{-}(r_{0}) + \rho_{+}(r_{0}))V g + m g
$$
$$
F_{0} = m g - \left(\frac{k m_{e}^2N}{\pi H(e^{m_{e}k R^2}-1)}e^{m_{e}kr_{0}^2} + \frac{k m_{N}^2N}{\pi H(e^{m_{N}k R^2}-1)}e^{m_{N}kr_{0}^2}\right)V g
$$
With conservation of angular momentum:
$$
L_{0} = m r_{0}^2\omega
$$
Assume a relative stable position $\boldsymbol{r}_{0}$ with a perturbation distance $\delta r$, and write out the dynamic equation:
$$
m\ddot{\delta r} = \frac{L_{0}^2}{m(r_{0}+\delta r)^3} + E(r_{0}+\delta r)q_{0} - [\rho_{-}(r_{0}+\delta r) + \rho_{+}(r_{0}+\delta r)]V\omega^2(r_{0}+\delta r)
$$
Analyze and calculate the results in each term:
Effective centrifugal force term:
$$
\frac{L_{0}^2}{m(r+\delta r)^3} = \frac{L_{0}^2}{m r_{0}^3} + \frac{\partial\frac{L_{0}^2}{m r^3}}{\partial r}\delta r = \frac{L_{0}^2}{m r_{0}^3} - 3m\omega^2\delta r
$$
Radial electric field term:
$$
E(r_{0}+\delta r)q_{0} = E(r_{0})q + \frac{N e q_{0}}{2\pi H\varepsilon_{0}r_{0}^2}\left[\frac{(2k m_{N}r_{0}^2 - 1)e^{m_{N}kr_{0}^2} + 1}{e^{m_{N}k R^2} - 1} - \frac{(2k m_{e}r_{0}^2 - 1)e^{m_{e}kr_{0}^2} + 1}{e^{m_{e}k R^2} - 1}\right]\delta r
$$
Radial component of buoyancy term:
$$
\begin{array}{rl}
-[\rho_{-}(r_{0}+\delta r) + \rho_{+}(r_{0}+\delta r)]V\omega^2(r_{0}+\delta r) =& -[\rho_{-}(r_{0}) + \rho_{+}(r_{0})]V\omega^2r_{0} - \ldots
\end{array}
$$
$$
\ldots \frac{k N V\omega^2}{\pi H}\left[\frac{(2k m_{N}r_{0}^2 + 1)e^{m_{N}kr_{0}^2}}{e^{m_{N}k R^2} - 1}m_{N}^2 + \frac{(2k m_{e}r_{0}^2 + 1)e^{m_{e}kr_{0}^2}}{e^{m_{e}k R^2} - 1}m_{e}^2\right]\delta r
$$
Since
$$
\frac{L_{0}^2}{m r_{0}^3} + E(r_{0})q_{0} - [\rho_{-}(r_{0}) + \rho_{+}(r_{0})]V\omega^2r_{0} = 0
$$
Zero-order term sum is zero, thus only considering the first-order terms, we obtain the precession angular frequency:
$$
\boxed{\Omega^2 = 3\omega^2 - \frac{N e q_{0}}{2\pi m H\varepsilon_{0}r_{0}^2}\left[\frac{(2k m_{N}r_{0}^2 - 1)e^{m_{N}kr_{0}^2} + 1}{e^{m_{N}k R^2} - 1} - \frac{(2k m_{\text{e}}r_{0}^2 - 1)e^{m_{\text{e}}kr_{0}^2} + 1}{e^{m_{\text{e}}k R^2} - 1}\right] + \frac{k N V\omega^2}{\pi m H}\left[\frac{(2k m_{N}r_{0}^2 + 1)e^{m_{N}kr_{0}^2}}{e^{m_{N}k R^2} - 1}m_{N}^2 + \frac{(2k m_{e}r_{0}^2 + 1)e^{m_{e}kr_{0}^2}}{e^{m_{e}k R^2} - 1}m_{e}^2\right]}
$$ | $$3\omega^2 - \frac{Ne q_0}{2\pi m H \varepsilon_0 r_0^2} \left( \frac{(2k m_N r_0^2 - 1)e^{m_N k r_0^2} + 1}{e^{m_N k R^2} - 1} - \frac{(2k m_e r_0^2 - 1)e^{m_e k r_0^2} + 1}{e^{m_e k R^2} - 1} \right) + \frac{k N V \omega^2}{\pi m H} \left( \frac{(2k m_N r_0^2 + 1)e^{m_N k r_0^2}}{e^{m_N k R^2} - 1} m_N^2 + \frac{(2k m_c r_0^2 + 1)e^{m_c k r_0^2}}{e^{m_c k R^2} - 1} m_c^2 \right)$$ |
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