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

From a inertial frame of reference when we see the (man wheel) system, we can find that the wheel moving at a speed of $\omega$ and the man with $\Big(\omega+\frac{\text{V}}{\text{R}}\Big)$ after the man has started walking. $(\omega'=$ angular velocity after walking, $\omega=$ angular velocity of the wheel before walking$).$ Since $\sum\text{l}=0$ Extended torque = 0 Therefore, $(1+\text{MR}^2)\omega=\text{l}\omega'+\text{mR}^2\Big(\omega'+\frac{\text{V}}{\text{R}}\Big)$ $\Rightarrow(1+\text{MR}^2)\omega=\text{l}\omega'+\text{mR}^2\omega'+\text{mVR}$ $\Rightarrow\omega'=\omega-\frac{\text{mVR}}{(1+\text{mR}^2)}.$

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