Problem Overview

A wheel of mass $m$, radius $r$, and radius of gyration $k$ is placed on a conveyor belt moving with velocity $u$. Friction acts between the wheel and the belt until slipping ceases (pure rolling is established). We calculate the additional energy delivered by the motor.

Key Concept: The work done by the motor equals the work done by the friction force over the displacement of the belt.

$$ W_{motor} = \int \vec{f} \cdot \vec{u} \, dt = u \int f \, dt = u \cdot J $$

Where $J = \int f dt$ is the linear impulse provided by friction.

Part (a): Axle Held Stationary

The wheel is pressed down with the axle fixed. The wheel only rotates.

1. Impulse-Momentum Equation:
   Angular impulse changes angular momentum:
   $$ \int \tau \, dt = \Delta L \implies \int (f r) \, dt = I \omega_f $$ $$ r \int f \, dt = I \omega_f \implies r J = I \omega_f $$
2. Final State (No Slipping):
   The rim velocity must match the belt velocity $u$.
   $$ \omega_f r = u \implies \omega_f = \frac{u}{r} $$
3. Calculating Impulse $J$:
   $$ J = \frac{I \omega_f}{r} = \frac{I (u/r)}{r} = \frac{I u}{r^2} $$ Using $I = mk^2$: $$ J = \frac{mk^2 u}{r^2} $$
4. Energy Delivered:
   $$ W_{motor} = u \cdot J = u \left( \frac{mk^2 u}{r^2} \right) = \frac{m k^2 u^2}{r^2} $$

Answer (a): $\displaystyle \frac{m k^2 u^2}{r^2}$

Part (b): Wheel Gently Released

The wheel is free to move. Friction accelerates the center of mass ($v$) and causes rotation ($\omega$).

1. Impulse Equations:
   Linear: $J = m v_f \implies v_f = J/m$
   Angular: $J r = I \omega_f \implies \omega_f = \frac{J r}{mk^2}$
2. No Slip Condition:
   The velocity of the point of contact must match the belt velocity $u$.
   $$ v_f + \omega_f r = u $$ Substituting expressions in terms of $J$: $$ \frac{J}{m} + \left( \frac{J r}{mk^2} \right) r = u $$ $$ J \left( \frac{1}{m} + \frac{r^2}{mk^2} \right) = u $$ $$ J \left( \frac{k^2 + r^2}{mk^2} \right) = u $$ $$ J = \frac{m u k^2}{k^2 + r^2} $$
3. Energy Delivered:
   $$ W_{motor} = u \cdot J = \frac{m u^2 k^2}{k^2 + r^2} $$

Answer (b): $\displaystyle \frac{m u^2 k^2}{k^2 + r^2}$