A solid cylinder of mass $' M'$ and radius $' R'$ is rotating along its axis with angular velocity $ω$ without friction. A particle of mass $' m'$ moving with velocity $v$ collides against the cylinder and sticks to its rim. After the impact calculate angular velocity of cylinder.

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$ω' = \frac{I ω + m v R}{I + m R^{2}}$

$ω' = \frac{I ω + m v R}{m R^{2}}$

$ω' = \frac{m v R}{I + m R^{2}}$

$ω' = \frac{I ω - m v R}{I + m R^{2}}$

answer is D.

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Detailed Solution

Initial angular momentum of cylinder $= I ω$

Initial angular momentum of particle $= m v R$

Before collision the total angular momentum $L_{1} = I ω + m v R$

After collision the total angular momentum $L_{2} = (I + m R^{2}) ω'$

$L_{1} = L_{2} \Rightarrow (I + m R^{2}) ω' = I ω + m v R$

New angular velocity $ω' = \frac{I ω + m v R}{I + m R^{2}}$

Note: Initial kinetic energy of the system $= \frac{1}{2} I ω^{2} + \frac{1}{2} m v^{2}$

Final kinetic energy of the system $= \frac{1}{2} (I + m R^{2}) ω'^{2}$

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