Question:

A circular disk of mass $M$ and radius $R$ is initially at rest on a frictionless horizontal surface. A small block of mass $m$ is gently placed on the edge of the disk, and the block starts moving in a circular path with constant speed. At this moment, the block is $r$ distance away from the center of the disk. The disk and the block are connected by a string, with the string passing over a pulley of negligible mass. The string is initially taut and stationary.

a) Explain, in terms of angular momentum, why the disk will start rotating in the opposite direction when the block is placed on it.

b) Calculate the angular speed of the disk when the block is at a distance $r$ away from the center of the disk.

c) If the block slides off the disk, explain whether the rotational speed of the disk will increase, decrease, or remain the same. Justify your answer.

Assume that there are no external torques acting on the system.

Answer:

a) When the block is placed on the disk, it starts moving in a circular path with constant speed. According to the conservation of angular momentum, the angular momentum of the block-disk system must remain constant since there are no external torques acting on the system. Initially, the disk is at rest, so its angular momentum is zero. Therefore, when the block starts rotating in a circular path, it possesses an angular momentum in the clockwise direction. To maintain the conservation of angular momentum, the disk must acquire an equal magnitude of angular momentum but in the opposite direction (counterclockwise). This causes the disk to start rotating in the opposite direction.

b) The angular momentum of the block-disk system can be expressed as $L = I_{\text{disk}}\omega_{\text{disk}} + I_{\text{block}}\omega_{\text{block}}$, where $I_{\text{disk}}$ is the moment of inertia of the disk, $I_{\text{block}}$ is the moment of inertia of the block (treated as a point mass), $\omega_{\text{disk}}$ is the angular speed of the disk, and $\omega_{\text{block}}$ is the angular speed of the block.

Since the block is at a distance $r$ away from the center of the disk, its linear speed can be given as $v_{\text{block}} = \omega_{\text{block}}r$, where $v_{\text{block}}$ is the linear speed of the block.

Assuming the string between the block and disk remains taut and stationary, the linear speed of the disk at the point of contact with the block is equal to the linear speed of the block, i.e., $v_{\text{disk}} = v_{\text{block}}$.

For the disk, its moment of inertia can be calculated as $I_{\text{disk}} = \frac{1}{2}MR^2$, and for the block, its moment of inertia (treated as a point mass) is $I_{\text{block}} = mr^2$.

Since both the block and the disk move at a constant linear speed, their angular speeds can be written as $\omega_{\text{disk}} = \frac{v_{\text{disk}}}{R}$ and $\omega_{\text{block}} = \frac{v_{\text{block}}}{r}$, respectively.

Substituting these values into the expression for angular momentum:

$L = I_{\text{disk}}\omega_{\text{disk}} + I_{\text{block}}\omega_{\text{block}}$

$L = \frac{1}{2}MR^2\left(\frac{v_{\text{disk}}}{R}\right) + mr^2\left(\frac{v_{\text{block}}}{r}\right)$

$L = \frac{1}{2}Mv_{\text{disk}} + mv_{\text{block}}$

The conservation of angular momentum tells us that this expression remains constant. Therefore, the angular speed of the disk when the block is at a distance $r$ away from the center of the disk is given by rearranging the above equation:

$\omega_{\text{disk}} = \frac{L - 2mvr}{MR^2}$

c) When the block slides off the disk, its angular speed changes since the angular momentum is no longer conserved due to the external torque exerted by the block on the disk as it slides off. The rotational speed of the disk will decrease as a result.

This can also be derived by considering the conservation of mechanical energy. When the block slides off, energy is being transferred from the rotational kinetic energy of the disk to the translational kinetic energy of the block. The total mechanical energy of the system decreases, consistent with a decrease in rotational speed.