A child with mass m is standing at the edge of a playground merry-go-round (A large uniform disc which rotates in horizontal plane about a fixed vertical axis in parks) with moment of inertia Vs I, radius R, and initial angular velocity w as shown in the figure. The child jumps off the edge of the merry-go-round with a velocity v with respect to the ground in direction tangent to periphery of the disc as shown. The new angular veloity of the merry-go-round is:

Question

A child with mass m is standing at the edge of a playground merry-go-round (A large uniform disc which rotates in horizontal plane about a fixed vertical axis in parks) with moment of inertia Vs I, radius R, and initial angular velocity w as shown in the figure. The child jumps off the edge of the merry-go-round with a velocity v with respect to the ground in direction tangent to periphery of the disc as shown. The new angular veloity of the merry-go-round is:, $\sqrt{\frac{I \omega^2-m v^2}{I}}$, $\frac{\left(\mathrm{I}+\mathrm{mR}^2\right) \omega-\mathrm{mvR}}{\mathrm{I}}$, $\sqrt{\frac{\left(I+m R^2\right) \omega^2-m v^2}{I}}$, $\frac{\mathrm{I} \omega-\mathrm{mvR}}{\mathrm{I}}$

MARKS App · Accepted Answer

Let the angular velocity of the merry-go-round be w'. As there is no external torque, thus $\mathrm{L}$ is conserved. Initial angular momentum, $\mathrm{L}_{\mathrm{i}}=\mathrm{Iw}+\mathrm{mR}^2 \mathrm{w}$ Final angular momentum, $\mathrm{L}_{\mathrm{f}}=\mathrm{I} w^{\prime}+\mathrm{mvR}$ $\begin{aligned} & \mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{f}} \Rightarrow \mathrm{Iw}+\mathrm{mwR}^2=\mathrm{Iw}^{\prime}+\mathrm{mvR} \ & \Rightarrow \mathrm{w}^{\prime}=\frac{\left(\mathrm{I}+\mathrm{mR}^2\right) \mathrm{w}-\mathrm{mvR}}{\mathrm{I}} \end{aligned}$

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