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Question
Suppose the platform of the previous problem is brought to rest with the ball in the hand of the kid standing on the rim. The kid throws the ball horizontally to his friend in a direction tangential to the rim with a speed \[\nu\] as seen by his friend. Find the angular velocity with which the platform will start rotating.
Solution
Initial angular momentum of the system is zero (L1 = 0)
Let the platform starts rotating with an angular velocity ω.
Therefore, we have
Final angular momentum of the system = L2
\[= I_\omega + M R^2 \omega - mvR\]
(m = mass of the ball, v = velocity of the ball)
Total external torque = 0
Also,
Initial angular momentum = Final angular momentum
\[\Rightarrow I\omega + M R^2 \omega - mvR = 0\]
\[ \Rightarrow \omega = \frac{mvR}{\left( I + M R^2 \right)}\]
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