Question

If the transition of spherical shell of liquid to solid occurs when the shell is at its mean position, Which of the following statements is CORRECT?

• A. The angular amplitude decreases, the oscillation energy decreases and the speed at the bottom decreases.
• B. The angular amplitude remains the same, the oscillation energy decreases and the speed at the bottom decreases.
• C. The angular amplitude remains same, the oscillation energy remains same and the speed at the bottom remains same.
• D. The angular amplitude decreases, the oscillation energy increases and the speed at the bottom decreases.

Answer 1

Answer: (D)

The angular amplitude decreases, the oscillation energy increases and the speed at the bottom decreases.

Answer 2

The sphere rolls without slipping at the bottom of a spherical bowl of radius R. The sphere has a radius r. The center of the sphere moves along a circular path of radius $(R-r)$. Let $\theta$ be the angular position of the center of the sphere with respect to the vertical. The height of the center of the sphere above the lowest point is $h = (R-r)(1-\cos\theta)$. The linear velocity of the center of the sphere is $v = (R-r)\dot{\theta}$. Due to the no-slipping condition, the angular velocity of the sphere about its center is $\omega = v/r = \frac{R-r}{r}\dot{\theta}$.

The sphere consists of a thin spherical shell of mass m and a substance of mass M inside. When the substance is an ideal liquid, it does not rotate with the shell. Its kinetic energy is purely translational, $\frac{1}{2}Mv^2$. The rotational kinetic energy is only due to the shell, $KE_{rot, shell} = \frac{1}{2}I_{shell}\omega^2$, where $I_{shell} = \frac{2}{3}mr^2$. The translational kinetic energy of the combined system is $\frac{1}{2}(m+M)v^2$. Total kinetic energy in the liquid state: $KE_{liquid} = \frac{1}{2}(m+M)v^2 + \frac{1}{2}I_{shell}\omega^2 = \frac{1}{2}(m+M)v^2 + \frac{1}{2}(\frac{2}{3}mr^2)(\frac{v}{r})^2 = \frac{1}{2}(m+M)v^2 + \frac{1}{3}mv^2 = (\frac{1}{2}m + \frac{1}{2}M + \frac{1}{3}m)v^2 = (\frac{5}{6}m + \frac{1}{2}M)v^2$. Potential energy: $PE = (m+M)gh = (m+M)g(R-r)(1-\cos\theta)$. Total energy in the liquid state: $E_{liquid} = (\frac{5}{6}m + \frac{1}{2}M)v^2 + (m+M)g(R-r)(1-\cos\theta)$. At the mean position ($\theta=0$, $PE=0$), the speed is maximum, $v_{0,liquid}$. $E_{liquid} = (\frac{5}{6}m + \frac{1}{2}M)v_{0,liquid}^2$. At the extreme position ($\theta=\theta_{max,liquid}$, $v=0$), the energy is $E_{liquid} = (m+M)g(R-r)(1-\cos\theta_{max,liquid})$. By conservation of energy, $(\frac{5}{6}m + \frac{1}{2}M)v_{0,liquid}^2 = (m+M)g(R-r)(1-\cos\theta_{max,liquid})$.

When the substance is solid, it forms a solid sphere of mass M and moment of inertia $I_{solid} = \frac{2}{5}Mr^2$ about its diameter. The total moment of inertia of the sphere about its center is $I_{total} = I_{shell} + I_{solid} = \frac{2}{3}mr^2 + \frac{2}{5}Mr^2 = (\frac{2}{3}m + \frac{2}{5}M)r^2$. Total kinetic energy in the solid state: $KE_{solid} = \frac{1}{2}(m+M)v^2 + \frac{1}{2}I_{total}\omega^2 = \frac{1}{2}(m+M)v^2 + \frac{1}{2}(\frac{2}{3}m + \frac{2}{5}M)r^2(\frac{v}{r})^2 = \frac{1}{2}(m+M)v^2 + (\frac{1}{3}m + \frac{1}{5}M)v^2 = (\frac{1}{2}m + \frac{1}{2}M + \frac{1}{3}m + \frac{1}{5}M)v^2 = (\frac{5}{6}m + \frac{7}{10}M)v^2$. Total energy in the solid state: $E_{solid} = (\frac{5}{6}m + \frac{7}{10}M)v^2 + (m+M)g(R-r)(1-\cos\theta)$. At the mean position ($\theta=0$, $PE=0$), the speed is maximum, $v_{0,solid}$. $E_{solid} = (\frac{5}{6}m + \frac{7}{10}M)v_{0,solid}^2$. At the extreme position ($\theta=\theta_{max,solid}$, $v=0$), the energy is $E_{solid} = (m+M)g(R-r)(1-\cos\theta_{max,solid})$. By conservation of energy, $(\frac{5}{6}m + \frac{7}{10}M)v_{0,solid}^2 = (m+M)g(R-r)(1-\cos\theta_{max,solid})$.

The transition from liquid to solid occurs at the mean position. The external influence causing the transition does not exert any force or torque. This means the velocity of the center of mass $v$ and the angular velocity $\omega$ of the sphere (shell) are instantaneously unchanged during the transition. Let the speed at the mean position just before the transition be $v_0$. Before transition (liquid): $E_{before} = (\frac{5}{6}m + \frac{1}{2}M)v_0^2$. This is the total energy of oscillation in the liquid state. After transition (solid): The speed at the mean position is still $v_0$. The total energy in the solid state is $E_{after} = (\frac{5}{6}m + \frac{7}{10}M)v_0^2$. Comparing the energies: $\frac{1}{2}M = \frac{5}{10}M$ and $\frac{7}{10}M$. Since $M>0$, $\frac{1}{2}M < \frac{7}{10}M$. So, $(\frac{5}{6}m + \frac{1}{2}M) < (\frac{5}{6}m + \frac{7}{10}M)$. Thus, $E_{before} < E_{after}$. The oscillation energy increases instantaneously at the mean position.

After the transition, the sphere is in the solid state and starts oscillating with the new energy $E_{after}$ and the new kinetic energy formula. The speed at the bottom immediately after the transition is $v_0$. This is the maximum speed for the subsequent oscillation in the solid state, so $v_{0,solid} = v_0$. The oscillation energy in the solid state is $E_{solid} = E_{after} = (\frac{5}{6}m + \frac{7}{10}M)v_0^2$. The oscillation energy in the liquid state was $E_{liquid} = E_{before} = (\frac{5}{6}m + \frac{1}{2}M)v_0^2$. Since $E_{after} > E_{before}$, the oscillation energy increases.

Now let's consider the oscillation amplitude. In the solid state, the maximum amplitude $\theta_{max,solid}$ is reached when the speed is zero. The energy is purely potential at this point. $E_{solid} = (m+M)g(R-r)(1-\cos\theta_{max,solid})$. Substituting $E_{solid} = (\frac{5}{6}m + \frac{7}{10}M)v_0^2$: $(\frac{5}{6}m + \frac{7}{10}M)v_0^2 = (m+M)g(R-r)(1-\cos\theta_{max,solid})$. Let's compare this with the liquid state oscillation where the initial energy was $E_{liquid} = (\frac{5}{6}m + \frac{1}{2}M)v_0^2$ (assuming the sphere was oscillating with amplitude $\theta_{max,liquid}$ such that its speed at the bottom was $v_0$). $E_{liquid} = (m+M)g(R-r)(1-\cos\theta_{max,liquid})$. Since $E_{solid} > E_{liquid}$ and the potential energy function is the same, the maximum height reached in the solid state will be greater than the maximum height reached in the liquid state, starting from the same speed $v_0$ at the bottom. $(m+M)g(R-r)(1-\cos\theta_{max,solid}) > (m+M)g(R-r)(1-\cos\theta_{max,liquid})$. $1-\cos\theta_{max,solid} > 1-\cos\theta_{max,liquid}$. $\cos\theta_{max,solid} < \cos\theta_{max,liquid}$. Since $\theta_{max}$ is typically a small angle in oscillations, $\cos\theta$ is a decreasing function for small positive angles. Therefore, $\theta_{max,solid} > \theta_{max,liquid}$. The angular amplitude increases.

However, the question states "If the transition ... occurs when the shell is at its mean position". This implies that the oscillation before the transition had a certain amplitude, and we are looking at the effect of the transition on the oscillation after the transition. Let the angular amplitude before the transition be $\theta_{max,before}$. The speed at the mean position before the transition is $v_{0,before}$. The energy before the transition is $E_{before} = (\frac{5}{6}m + \frac{1}{2}M)v_{0,before}^2 = (m+M)g(R-r)(1-\cos\theta_{max,before})$. At the mean position, the transition occurs. The speed immediately after the transition is $v_{0,after} = v_{0,before}$. The energy immediately after the transition is $E_{after} = (\frac{5}{6}m + \frac{7}{10}M)v_{0,after}^2 = (\frac{5}{6}m + \frac{7}{10}M)v_{0,before}^2$. This energy $E_{after}$ is the total energy of the subsequent oscillation in the solid state. Let the maximum amplitude in the solid state be $\theta_{max,after}$. $E_{after} = (m+M)g(R-r)(1-\cos\theta_{max,after})$. $(\frac{5}{6}m + \frac{7}{10}M)v_{0,before}^2 = (m+M)g(R-r)(1-\cos\theta_{max,after})$. Comparing $E_{before}$ and $E_{after}$, we have $E_{after} > E_{before}$. So, $(m+M)g(R-r)(1-\cos\theta_{max,after}) > (m+M)g(R-r)(1-\cos\theta_{max,before})$. $1-\cos\theta_{max,after} > 1-\cos\theta_{max,before}$. $\cos\theta_{max,after} < \cos\theta_{max,before}$. This implies $\theta_{max,after} > \theta_{max,before}$. The angular amplitude increases.

Let's re-read the question options carefully. They all say "The angular amplitude decreases". This contradicts our finding that the amplitude increases. Let's check if I misinterpreted the problem or the physics.

The instantaneous increase in kinetic energy at the mean position leads to an increase in total energy, which should lead to a larger amplitude.

Let's examine the options again. All options state that the angular amplitude decreases. If this is true, then there must be a decrease in energy or a change in the relationship between energy and amplitude. The relationship between energy and amplitude $(m+M)g(R-r)(1-\cos\theta_{max})$ is the potential energy at the maximum displacement, which is correct. So, if amplitude decreases, energy must decrease.

Let's re-examine the kinetic energy calculation. When liquid is ideal, it does not participate in rotation. So $I_{liquid} = 0$. The total mass is $m+M$. $KE_{liquid} = \frac{1}{2}(m+M)v^2 + \frac{1}{2}I_{shell}\omega^2 = \frac{1}{2}(m+M)v^2 + \frac{1}{2}(\frac{2}{3}mr^2)(\frac{v}{r})^2 = (\frac{1}{2}(m+M) + \frac{1}{3}m)v^2 = (\frac{5}{6}m + \frac{1}{2}M)v^2$. This seems correct.

When solid, the moment of inertia is $I_{total} = \frac{2}{3}mr^2 + \frac{2}{5}Mr^2 = (\frac{2}{3}m + \frac{2}{5}M)r^2$. $KE_{solid} = \frac{1}{2}(m+M)v^2 + \frac{1}{2}I_{total}\omega^2 = \frac{1}{2}(m+M)v^2 + \frac{1}{2}(\frac{2}{3}m + \frac{2}{5}M)r^2(\frac{v}{r})^2 = (\frac{1}{2}m + \frac{1}{2}M + \frac{1}{3}m + \frac{1}{5}M)v^2 = (\frac{5}{6}m + \frac{7}{10}M)v^2$. This also seems correct.

The instantaneous transition at constant $v$ and $\omega$ (and thus constant $v$) results in an instantaneous increase in kinetic energy, and thus total energy. $E_{after} > E_{before}$. Since the potential energy at the maximum amplitude is the total energy, $E = (m+M)g(R-r)(1-\cos\theta_{max})$, a higher energy means a larger $1-\cos\theta_{max}$, which means a smaller $\cos\theta_{max}$, and thus a larger $\theta_{max}$. So the amplitude should increase.

Let's check if I misunderstood the behavior of an ideal liquid inside a rotating sphere. An ideal liquid has no viscosity. When the shell rotates, the liquid does not get dragged along by viscosity. If the sphere is rolling, its center is moving and it is rotating. The liquid inside is subject to gravity and inertial forces (centrifugal and Coriolis in the rotating frame of the shell). However, if we consider the energy, the rotational kinetic energy of the liquid relative to the center of the sphere is zero if it's ideal and not rotating. Its translational kinetic energy is the same as the sphere's center of mass, $\frac{1}{2}Mv^2$. So, the total kinetic energy is the sum of the translational kinetic energy of the total mass and the rotational kinetic energy of the shell. This is what I used.

Perhaps the question implies that the initial oscillation had a certain amplitude, and the transition happens when it passes through the mean position. Then the subsequent oscillation will have a different amplitude. Let the initial amplitude be $\theta_{max,1}$. The energy is $E_1 = (m+M)g(R-r)(1-\cos\theta_{max,1})$. At the mean position, the speed is $v_{0,1}$ such that $E_1 = (\frac{5}{6}m + \frac{1}{2}M)v_{0,1}^2$. At the mean position, the transition occurs. The speed remains $v_{0,1}$. The energy of the subsequent oscillation is $E_2 = (\frac{5}{6}m + \frac{7}{10}M)v_{0,1}^2$. Since $(\frac{5}{6}m + \frac{7}{10}M) > (\frac{5}{6}m + \frac{1}{2}M)$, we have $E_2 > E_1$. The amplitude of the subsequent oscillation $\theta_{max,2}$ is given by $E_2 = (m+M)g(R-r)(1-\cos\theta_{max,2})$. Since $E_2 > E_1$, $(m+M)g(R-r)(1-\cos\theta_{max,2}) > (m+M)g(R-r)(1-\cos\theta_{max,1})$. $1-\cos\theta_{max,2} > 1-\cos\theta_{max,1}$, which means $\cos\theta_{max,2} < \cos\theta_{max,1}$. For angles between 0 and $\pi/2$, this implies $\theta_{max,2} > \theta_{max,1}$. The angular amplitude increases.

This conclusion is consistently reached, but it contradicts all the given options. Let me check if there is any other interpretation.

Could the "oscillation energy" refer to something else? In simple harmonic motion, energy is proportional to the square of the amplitude. Here it's not SHM, but oscillation. The total mechanical energy is the oscillation energy.

Could the speed at the bottom decrease? The speed at the bottom of the subsequent oscillation is the maximum speed, $v_{0,2}$. The energy is $E_2 = (\frac{5}{6}m + \frac{7}{10}M)v_{0,2}^2$. We know $E_2 = (\frac{5}{6}m + \frac{7}{10}M)v_{0,1}^2$. So $v_{0,2} = v_{0,1}$. The speed at the bottom immediately after the transition is the same as immediately before the transition. So the maximum speed of the subsequent oscillation is the same as the maximum speed of the previous oscillation. So, the speed at the bottom remains the same as it was at the instant of transition.

This means the speed at the bottom remains the same. This contradicts all options which state that the speed at the bottom decreases or remains the same in one option that is incorrect about amplitude and energy.

Let's re-examine the initial state. The sphere is rolling back and forth. It has a certain oscillation. Let's assume the initial oscillation had amplitude $\theta_{max,1}$ and speed at the bottom $v_{0,1}$. The transition occurs when it is at the mean position, with speed $v_{0,1}$. After the transition, the substance is solid. The new oscillation starts from the mean position with speed $v_{0,1}$. The energy of this new oscillation is $E_2 = (\frac{5}{6}m + \frac{7}{10}M)v_{0,1}^2$. The maximum amplitude of this new oscillation is $\theta_{max,2}$ such that $E_2 = (m+M)g(R-r)(1-\cos\theta_{max,2})$. The new speed at the bottom is $v_{0,2}$ such that $E_2 = (\frac{5}{6}m + \frac{7}{10}M)v_{0,2}^2$. From the transition, we have $v_{0,2} = v_{0,1}$. So the speed at the bottom remains the same.

If the speed at the bottom remains the same, and the oscillation energy increases, then the amplitude must increase. $E_{liquid} = (\frac{5}{6}m + \frac{1}{2}M)v_{bottom}^2$. $E_{solid} = (\frac{5}{6}m + \frac{7}{10}M)v_{bottom}^2$. If $v_{bottom}$ is the same, then $E_{solid} > E_{liquid}$. $E = (m+M)g(R-r)(1-\cos\theta_{max})$. If E increases, $\theta_{max}$ increases.

So, angular amplitude increases, oscillation energy increases, and speed at the bottom remains the same. This combination is not among the options.

There must be a subtlety I am missing. Let's check the problem statement again. "If the transition of spherical shell of liquid to solid occurs when the shell is at its mean position". This is clear. "Which of the following statements is CORRECT?".

Let's consider the possibility that the speed at the bottom refers to the speed at the bottom in the steady oscillation state, before and after the transition. Suppose the initial oscillation in the liquid state had a certain amplitude $\theta_{max,1}$. The speed at the bottom was $v_{0,1}$ such that $(\frac{5}{6}m + \frac{1}{2}M)v_{0,1}^2 = (m+M)g(R-r)(1-\cos\theta_{max,1})$. The transition occurs when the sphere is at the mean position, with speed $v_{0,1}$. Immediately after the transition, the substance is solid, and the speed at the mean position is still $v_{0,1}$. The energy of the subsequent oscillation is $E_2 = (\frac{5}{6}m + \frac{7}{10}M)v_{0,1}^2$. The speed at the bottom of this new oscillation is $v_{0,2}$ such that $E_2 = (\frac{5}{6}m + \frac{7}{10}M)v_{0,2}^2$. So $v_{0,2} = v_{0,1}$. The speed at the bottom of the oscillation remains the same.

Let's consider the possibility that the question is flawed or there is a standard interpretation I am unaware of. However, based on the physics principles, the speed at the bottom should remain unchanged immediately after the transition, and this speed is the maximum speed for the subsequent oscillation.

Let's look at the options again. All options say amplitude decreases. This strongly suggests that the amplitude should decrease. If amplitude decreases, energy must decrease. If energy decreases, and the kinetic energy at the bottom is $E = (\frac{5}{6}m + \frac{7}{10}M)v_{bottom}^2$ in the solid state, then $v_{bottom}$ must decrease. So, amplitude decreases, energy decreases, speed at the bottom decreases. This is the first option.

If the first option is correct, then the oscillation energy must decrease during the transition at the mean position. This would mean $E_{after} < E_{before}$. $E_{after} = (\frac{5}{6}m + \frac{7}{10}M)v_0^2$ and $E_{before} = (\frac{5}{6}m + \frac{1}{2}M)v_0^2$. For $E_{after} < E_{before}$ to be true, we would need $(\frac{5}{6}m + \frac{7}{10}M) < (\frac{5}{6}m + \frac{1}{2}M)$, which means $\frac{7}{10}M < \frac{1}{2}M = \frac{5}{10}M$. This is false since $M>0$. So the oscillation energy increases during the transition.

Perhaps the liquid was not ideal in the sense that it was also rotating with the shell due to some mechanism not described, and upon solidification, the moment of inertia changes in a way that reduces the kinetic energy at constant speed. But the problem explicitly says "usually it behaves like an ideal liquid (i.e. without any viscosity)". This implies no rotation of the liquid relative to the center.

Let's assume there is a mistake in the question or options, and proceed with our derived result: angular amplitude increases, oscillation energy increases, and speed at the bottom remains the same.

However, since I am required to choose from the given options, let me reconsider the possibility that the speed at the bottom decreases. If the speed at the bottom decreases, and the kinetic energy at the bottom is $E = (\frac{5}{6}m + \frac{7}{10}M)v_{bottom}^2$, then the energy of oscillation decreases. If the energy of oscillation decreases, then the amplitude decreases. So, if speed at the bottom decreases, then energy decreases, and amplitude decreases. This matches the first option.

For the speed at the bottom to decrease during the instantaneous transition at the mean position, the kinetic energy must decrease. $KE_{after} < KE_{before}$ means $(\frac{5}{6}m + \frac{7}{10}M)v_{0,after}^2 < (\frac{5}{6}m + \frac{1}{2}M)v_{0,before}^2$. If $v_{0,after} = v_{0,before}$, then $KE_{after} > KE_{before}$, which contradicts the need for energy to decrease. So, it must be that $v_{0,after} < v_{0,before}$.

But why would the speed decrease instantaneously without external force or torque? Let's consider the angular momentum about the point of contact. However, the point of contact is not fixed.

Let's assume that the total kinetic energy decreases during the transition. This would happen if, for the same speed v, the effective inertia term $(\frac{5}{6}m + \frac{1}{2}M)$ decreased to $(\frac{5}{6}m + \frac{7}{10}M)$. But the coefficient of $M$ increased from $1/2$ to $7/10$. So the effective inertia term increased. $I_{eff,liquid} = (\frac{5}{6}m + \frac{1}{2}M) \frac{r^2}{(R-r)^2} + (m+M)$. $E_{liquid} = \frac{1}{2}M_{eff,liquid} v^2 + PE$, where $M_{eff,liquid} = (\frac{5}{6}m + \frac{1}{2}M) \frac{(R-r)^2}{r^2} + (m+M)$. No, this is not correct. The kinetic energy is $KE = \frac{1}{2}M_{total} v_{CM}^2 + \frac{1}{2}I_{CM} \omega^2$. We have $v_{CM} = v$ and $\omega = v/r$. $KE_{liquid} = \frac{1}{2}(m+M)v^2 + \frac{1}{2}I_{shell}(v/r)^2 = \frac{1}{2}(m+M)v^2 + \frac{1}{3}mv^2 = (\frac{1}{2}m + \frac{1}{2}M + \frac{1}{3}m)v^2 = (\frac{5}{6}m + \frac{1}{2}M)v^2$. $KE_{solid} = \frac{1}{2}(m+M)v^2 + \frac{1}{2}I_{total}(v/r)^2 = \frac{1}{2}(m+M)v^2 + \frac{1}{2}(\frac{2}{3}m + \frac{2}{5}M)r^2(v/r)^2 = (\frac{1}{2}m + \frac{1}{2}M + \frac{1}{3}m + \frac{1}{5}M)v^2 = (\frac{5}{6}m + \frac{7}{10}M)v^2$.

The kinetic energy at the mean position is $KE_{mean} = (\frac{5}{6}m + \frac{1}{2}M)v_0^2$ before transition and $KE_{mean} = (\frac{5}{6}m + \frac{7}{10}M)v_0^2$ after transition, where $v_0$ is the speed at the mean position. Since the transition is instantaneous and without external force/torque, the speed $v_0$ does not change instantaneously. Thus, the kinetic energy increases, and so does the total energy. This leads to an increase in amplitude.

Given the options, it is highly probable that there is an error in the problem statement or the options provided. However, if forced to choose the "most likely correct" option based on a potential misunderstanding of the physics, let's consider what might cause a decrease in amplitude. A decrease in amplitude implies a decrease in energy. A decrease in energy at the mean position (where PE=0) means a decrease in kinetic energy. If the speed at the mean position remains the same, then the coefficient of $v^2$ in the kinetic energy expression must decrease. But it increases from $(\frac{5}{6}m + \frac{1}{2}M)$ to $(\frac{5}{6}m + \frac{7}{10}M)$.

Let's reconsider the problem from a different angle. Maybe the liquid was not ideal, but was rotating with the shell due to viscosity, and upon solidification, the internal friction ceases, or the moment of inertia changes in a way that reduces the energy. But the problem states "ideal liquid (i.e. without any viscosity)". This implies no rotation of the liquid relative to the center.

Let's assume there is a mistake in the problem and the kinetic energy at the bottom decreases. If $v_{bottom}$ decreases, then the oscillation energy $E = (\frac{5}{6}m + \frac{7}{10}M)v_{bottom}^2$ decreases. If energy decreases, the amplitude decreases. This aligns with the first option.

For the speed at the bottom to decrease during the instantaneous transition at the mean position, the kinetic energy must decrease. $KE_{after} < KE_{before}$ means $(\frac{5}{6}m + \frac{7}{10}M)v_{0,after}^2 < (\frac{5}{6}m + \frac{1}{2}M)v_{0,before}^2$. If $v_{0,after} = v_{0,before}$, then $KE_{after} > KE_{before}$, which contradicts the need for energy to decrease. So, it must be that $v_{0,after} < v_{0,before}$.

Let's assume there is an error in the question or options. If forced to choose, the first option presents a consistent set of decreases.

The final answer is $\boxed{The angular amplitude decreases, the oscillation energy decreases and the speed at the bottom decreases.}$.