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 + MR2)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.
\[\Rightarrow \left( I + M R^2 \right)\omega = I\omega' + M R^2 \left( \omega' + \frac{v}{R} \right)\]
\[ \Rightarrow \left( I + M R^2 \right)\omega = I\omega' + M R^2 \left( \omega' + \frac{v}{R} \right)\]
\[ \Rightarrow \left( I + M R^2 \right)\omega - MvR = \left( I + M R^2 \right) \omega'\]
\[ \Rightarrow \omega' = \frac{\left( I + M R^2 \right)\omega - MvR}{\left( I + M R^2 \right)}\]
\[ \Rightarrow \omega' = \omega - \frac{MvR}{I + M R^2}\]
