Question

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

Answer 1

From a initial frame a reference when we see the (man wheel) system, we can find that the wheel moving at a speed of $\omega$ and the man with $(\omega +V/R)$ after the man has started walking.

($\omega =$ angular angular velocity after walking $\omega =$ angular velocity of the wheel before walking)

Since $\sum I=0$

Extended torque $=0$

Therefore $(1+M{R}^2)\omega =I{\omega }^{\prime }+m{R}^2({\omega }^{\prime }+V/R)$

$\Rightarrow (1+m{R}^2)\omega +I{\omega }^{\prime }+m{R}^2{\omega }^{\prime }+mVR$

$\Rightarrow {\omega }^{\prime }=\omega -\frac{mVR}{(1+m{R}^2)}$