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Q.

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

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a

$\frac{L}{4}$

b

2L

c

4L

d

$\frac{L}{2}$

answer is A.

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Detailed Solution

$L = Iω and K = \frac{1}{2} {Iω}^{2} \Rightarrow L = \frac{2 K}{ω}$

$\Rightarrow \frac{L_{f}}{L_{i}} = \frac{K_{f}}{K_{i}} \times \frac{ω_{i}}{ω_{f}} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$\Rightarrow L_{f} = \frac{L_{i}}{4} = \frac{L}{4}$

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