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A particle is moving in a circular path. If its kinetic energy is doubled, keeping its angular momentum unchanged, then new angular velocity of rotation is ______

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$\frac{ω}{4}$

$\frac{ω}{2}$

$2 ω$

$4 ω$

answer is B.

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Detailed Solution

$\begin{array}{l} L = m ω r^{2} a n d k = \frac{1}{2} m ω^{2} r^{2} = \frac{1}{2} L ω \\ ∴ \frac{1}{2} L ω^{'} = 2 k = 2 (\frac{1}{2} L ω) \Rightarrow ω^{'} = 2 ω . \end{array}$

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