Question

A particle performs uniform circular motion with an angular momentum L. If the angular frequency of the particle is doubled and kinetic energy is halved, its angular momentum becomes:
a. $2L$
b. $4L$
c. $\dfrac{L}{2}$
d. $\dfrac{L}{4}$

Answer 1

Angular momentum is momentum of a body in rotational motion. Kilometer square per second is the SI units for angular momentum. Formula for calculating the angular momentum is $L = mvr$ where L is angular momentum, m is mass, v is velocity and r is radius. Torque is the rate of change of angular momentum. Kinetic energy is transferred between bodies and transformed into other forms of energies.
Formula for kinetic energy is $K.E = \dfrac{1}{2}m{v^2}$ where m is mass of the object and v is the velocity of the object.
Kinetic energy is a scalar quantity. Kinetic energy describes magnitude.
Formula Used:
$L = I\omega$

Complete step by step answer:
Given: Angular frequency of the particle is doubled
Kinetic energy of the particle is halved
Angular momentum, $L = I\omega$
Here L is angular momentum, I is moment of inertia, $\omega$ is angular velocity.
Frequency f =$\dfrac{\omega }{{2\pi }}$
Bring $2\pi$ to LHS
$\omega = f$ $2\pi$……..equation$1$
Kinetic energy (KE), K = $\dfrac{1}{2}I{\omega ^2}$
To find I bring values in RHS to LHS
$I = \dfrac{{2K}}{{{\omega ^2}}}$ ……..equation$2$
Put equation $1$ and equation$2$ in Angular momentum, $L = I\omega$
$L = \dfrac{{2K}}{{{{(f2\pi )}^2}}} \times f2\pi$
Cancelling the values and simplifying
$L = \dfrac{K}{{f\pi }}$
By applying the conditions given in question Angular frequency of the particle is doubled, Kinetic energy of the particle is halved
$K = \dfrac{1}{2}$
$f = 2$
$L = \dfrac{K}{{f\pi }}$
By applying the K and f values in above equation
$L = \dfrac{1}{{4\pi }}$
Therefore correct answer is $\dfrac{L}{4}$

Hence, the correct answer is option (D).

Note: Angular momentum is a vector quantity. It represents the object rotational of a product of a rotational velocity and inertia about a particular axis. Angular momentum $L\;$ is proportional to angular speed $\omega$ and moment of inertia $I$. It is measured in radians per second.
If there are no net external forces linear momentum is conserved, when the net torque is zero angular momentum is constant or conserved.