Rotation: Friction acts East (Right) at the bottom. This creates a torque about the center that rotates the sphere Counter-Clockwise (top moves Left).
Viewed from the South (facing North), this is Anticlockwise.

Speed: Friction accelerates the sphere East. The linear speed relative to ground increases.
Relative to the belt, the speed is initially \( v_0 \) and decreases to 0 (at pure rolling). Thus, speed relative to belt decreases.

Answer: (b) and (d)

Equations of Motion:

• Linear: \( v = a t = \mu g t \)
• Angular: \( \omega = \alpha t = \frac{5 \mu g}{2 R} t \)

Condition for Pure Rolling on Moving Belt: \( v_{bottom} = v_{belt} \).
Since rotation is CCW, bottom velocity is \( v + \omega R \). \[ v + \omega R = v_0 \] \[ \mu g t + \frac{5}{2} \mu g t = v_0 \] \[ \frac{7}{2} \mu g t = v_0 \implies t = \frac{2 v_0}{7 \mu g} \]

Velocity: \( v = \mu g t = \frac{2 v_0}{7} \)

Answer: (b) and (d)

Work relative to Ground (\( W_G \)): Equals change in total Kinetic Energy. \[ W_G = K_f = \frac{1}{2}m v^2 + \frac{1}{2}I \omega^2 \] Using \( v = \frac{2}{7}v_0 \) and \( \omega R = v_0 – v = \frac{5}{7}v_0 \): \[ W_G = \frac{1}{2}m \left( \frac{4}{49}v_0^2 \right) + \frac{1}{2} \left( \frac{2}{5}m \right) \left( \frac{25}{49}v_0^2 \right) = \frac{1}{7} m v_0^2 \]

Work relative to Belt (\( W_B \)): Change in KE in belt frame. \[ v_{rel, i} = -v_0, \quad v_{rel, f} = v – v_0 = -\frac{5}{7}v_0 \] \[ W_B = \Delta K_{belt} = [ \frac{1}{2}m (\frac{25}{49}v_0^2) + K_{rot} ] – [ \frac{1}{2}m v_0^2 + 0 ] \] \[ W_B = \frac{1}{2}m v_0^2 [ \frac{25}{49} + \frac{10}{49} – 1 ] = -\frac{1}{7} m v_0^2 \]

Answer: (a) and (d)