Question

A particle performing uniform circular motion. Its angular frequency is doubled and its kinetic energy halved, the new angular momentum is
A. $\dfrac{L}{4}$
B. $2L$
C. $4L$
D. $\dfrac{L}{2}$

Answer 1

Hint: Find the dependency of angular momentum on kinetic energy and angular frequency. Assume the angular momentum in each case and find the relative change by taking the ratio.

Formula Used:
Angular momentum of the particle is given by,
$L=I\omega$

The kinetic energy of a rotating particle is given by,
$K=\dfrac{1}{2}I{{\omega }^{2}}$

Where,
$I$ is the moment of inertia of the particle
$\omega$ is the angular frequency

Complete step by step answer:

Let’s assume that the moment of inertia of the particle is I.
Hence, the kinetic energy of the particle can be given by,
$K=\dfrac{1}{2}I{{\omega }^{2}}$

So, we can write,
$I=\dfrac{2K}{{{\omega }^{2}}}$.....................(1)

Let’s assume that the initial kinetic energy of the particle was $K$ and the final kinetic energy is ${{K}_{1}}$.

Given that,
${{K}_{1}}=\dfrac{K}{2}$

Also, let the initial angular frequency was $\omega$ and the final angular frequency is ${{\omega }_{1}}$.

Let the final angular momentum is ${{L}_{1}}$

Angular momentum of the particle is given by,
$L=I\omega$ ............(2)

We can use the expression found in equation (1) in equation (2).
$L=(\dfrac{2K}{{{\omega }^{2}}})\omega$
$\Rightarrow L=\dfrac{2K}{\omega }$.
So, we can write,
$L=\dfrac{2K}{\omega }$

Hence, we can write the following two equations:
$L=\dfrac{2K}{\omega }$
&
${{L}_{1}}=\dfrac{2{{K}_{1}}}{{{\omega }_{1}}}$
We can take the ratio of the above two equations and find,
$\dfrac{{{L}_{1}}}{L}=\dfrac{{{K}_{1}}}{K}\times \dfrac{\omega }{{{\omega }_{1}}}$
$\Rightarrow \dfrac{{{L}_{1}}}{L}=\dfrac{1}{2}\times \dfrac{1}{2}$
$\Rightarrow {{L}_{1}}=\dfrac{L}{4}$

Hence, the new angular momentum is,
$\dfrac{L}{4}$

So, the correct answer is (A).

Note: Angular momentum depends on the axis and reference. When nothing is mentioned we consider the axis of rotation to be the reference axis. Angular momentum is a vector quantity. So, it always has a direction. In this question, the direction of the angular momentum vector is either out of the page or into the page, depending on the motion of the particle.