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Q.

A particle performing uniform circular motion has angular momentum L wrt centre. If its angular frequency is doubled and its kinetic energy is 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

K {.E}_{1} = \frac{1}{2} L_{1} ω_{1} \to (1)

$K {.E}_{2} = \frac{1}{2} L_{2} ω_{2} \to (2)$

$\frac{(2)}{(1)} = \frac{K .E/2}{K .E} = \frac{\frac{1}{2} L_{2} 2 ω}{\frac{L}{2} L_{1} ω}$

$L_{2} = \frac{1}{4} L$

$L_{2} = \frac{L}{4}$

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