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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
