Suppose the platform with the kid in the previous problem is rotating in anticlockwise direction at an angular speed ω. The kid starts walking along the rim with a speed \[\nu\] relative to the platform also in the anticlockwise direction. Find the new angular speed of the platform.

#### Solution

Let ω' be the angular velocity of platform after the kid starts walking.

Let ω be the angular velocity of the wheel before walking.

When we see the (kid-wheel) system from the initial frame of reference, we can find that the wheel moves with a speed of \[\omega\] and the kid with a speed of \[\left( \omega' + \frac{\nu}{R} \right),\] after the kid has started walking.

Initial angular momentum of the system = (I + MR^{2})ω

Angular velocity of the kid after it starts walking = \[\left( \omega' + \frac{v}{R} \right)\]

External torque is zero; therefore, angular momentum is conserved.

\[\Rightarrow \left( I + M R^2 \right)\omega = I\omega' + M R^2 \left( \omega' + \frac{v}{R} \right)\]

\[ \Rightarrow \left( I + M R^2 \right)\omega = I\omega' + M R^2 \left( \omega' + \frac{v}{R} \right)\]

\[ \Rightarrow \left( I + M R^2 \right)\omega - MvR = \left( I + M R^2 \right) \omega'\]

\[ \Rightarrow \omega' = \frac{\left( I + M R^2 \right)\omega - MvR}{\left( I + M R^2 \right)}\]

\[ \Rightarrow \omega' = \omega - \frac{MvR}{I + M R^2}\]