Question

A particle performs uniform circular motion with an angular momentum L. If the frequency of particle’s motion is doubled and its kinetic energy is halved, the angular momentum becomes
(A). $2L$
(B). $4L$
(C). $\dfrac{L}{2}$
(D). $\dfrac{L}{4}$

Answer 1

Hint: The angular momentum is the product of moment of inertia and angular velocity of the particle. The frequency is directly proportional to the angular velocity. So as the frequency increases angular velocity will also increase. The change in kinetic energy will cause a change in moment of inertia. Using new values of angular velocity and moment of inertia, we can calculate new angular momentum and compare it with the old value.
Formulas used:
$L=I\omega $
$\omega =2\pi f$
$K=\dfrac{1}{2}I{{\omega }^{2}}$

Complete answer:
Give, a particle is moving in a circular motion. Let its angular velocity be $\omega $ and its moment of inertia be $I$, then its angular momentum will be
$L=I\omega $ - (1)
We know that,
$\omega =2\pi f$
Here, $f$ is the frequency of the particle
Given that its frequency is doubled, so $f'=2f$, the new angular velocity will be
$\begin{align}
  & \dfrac{f'}{f}=\dfrac{\omega '}{\omega } \\
 & \Rightarrow \dfrac{2f}{f}=\dfrac{\omega '}{\omega } \\
 & \therefore \omega '=2\omega \\
\end{align}$
Therefore the new angular velocity is two times the old angular velocity
The initial kinetic energy of the particle is
$K=\dfrac{1}{2}I{{\omega }^{2}}$ - (2)
Here, $K$ is the kinetic energy of the particle
The new kinetic energy will be-
$K'=\dfrac{1}{2}I'\omega {{'}^{2}}$
$\Rightarrow \dfrac{K}{2}=\dfrac{1}{2}I'{{(2\omega )}^{2}}$ - (3)
Dividing eq (2) and eq (3), we get,
$\begin{align}
  & \dfrac{K}{\dfrac{K}{2}}=\dfrac{\dfrac{1}{2}I{{\omega }^{2}}}{\dfrac{1}{2}I'4{{\omega }^{2}}} \\
 & \Rightarrow 2=\dfrac{I}{I'4} \\
 & \therefore I'=\dfrac{I}{8} \\
\end{align}$
Therefore, the new moment of inertia is $\dfrac{I}{8}$ and the new angular velocity is $2\omega $. The new angular momentum will be-
$\begin{align}
  & L'=\dfrac{I}{8}2\omega \\
 & \Rightarrow L'=\dfrac{I\omega }{4} \\
 & \therefore L'=\dfrac{L}{4} \\
\end{align}$
Therefore, the new angular momentum is $\dfrac{L}{4}$.

Hence, the correct option is (D).

Note:
The angular velocity, angular momentum, moment of inertia are analogous to velocity, momentum and mass in motion in a straight line. The centripetal force is responsible for the circular motion of a particle. The frequency is the number of rotations per second. It is the reciprocal of time period.