Angular momentum of a particle performing uniform circular motion is L . If kinetic energy of partic

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6.System of Particles and Rotational Motion

easy

The angular momentum of a particle performing uniform circular motion is $L$. If the kinetic energy of partical is doubled and frequency is halved, then angular momentum becomes

A

$\frac{L}{2}$

B

$2 L$

C

$\frac{L}{4}$

D

$4 L$

Solution

(d)

$L=\%$

$K=\frac{1}{2} l \omega^2$

$2 K=\frac{1}{2} I^{\prime}\left(\frac{\omega}{2}\right)^2$

$\frac{1}{2}=\frac{I}{I^{\prime}}(4)$

$I^{\prime}=8 I$

$L^{\prime}=(8 I)\left(\frac{\omega}{2}\right)=4 / \omega$

$\Rightarrow L^{\prime}=4 L$

Standard 11

Physics

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