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A particle performing uniform circular motion has angular momentum L. If its angular frequency is halved and kinetic energy is doubled then new angular momentum is

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$\frac{L}{2}$

$\frac{L}{4}$

$2 L$

$4 L$

answer is C.

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Detailed Solution

$K = \frac{1}{2} I ω^{2} = \frac{1}{2} L ω \Rightarrow L \propto \frac{K}{ω}$

$\Rightarrow L_{2} = \frac{k_{2}}{k_{1}} \frac{ω_{1}}{ω_{2}} L_{1}$

$L_{2} = \frac{2 k}{k} . \frac{ω}{ω} 2 L = 4 L$

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