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(D) The angular momentum of a particle in circular motion is given by $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular ve

Vedclass · Accepted Answer

$\frac{L}{4}$

Vedclass · Answer

(D) The angular momentum of a particle in circular motion is given by $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular velocity.The kinetic energy is given by $K = \frac{1}{2}I\omega^2$.We can express kinetic energy in terms of angular momentum as $K = \frac{1}{2}L\omega$.From this, the angular momentum is $L = \frac{2K}{\omega}$.Since the frequency $f$ is doubled, the angular velocity $\omega = 2\pi f$ is also doubled, so $\omega_2 = 2\omega_1$.The kinetic energy is halved, so $K_2 = \frac{K_1}{2}$.Now, calculating the new angular momentum $L_2$:$L_2 = \frac{2K_2}{\omega_2} = \frac{2(K_1/2)}{2\omega_1} = \frac{K_1}{2\omega_1} = \frac{1}{4} \left( \frac{2K_1}{\omega_1} \right) = \frac{L_1}{4}$.Therefore, the new angular momentum is $\frac{L}{4}$.

$A$ particle performs uniform circular motion with an angular momentum $L$. If the frequency of the particle's motion is doubled and its kinetic energy is halved, then its angular momentum becomes

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