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. 2 L
• B. 4 L
• C. L / 2
• D. L / 4

Answer 1

The correct option is D L / 4

angular momentum $\left(L\right)=I\omega$,

where $I$ is moment of inertia and $\omega$ is angular velocity,

Frequency $f=\frac{\omega }{2\pi }$ $\Rightarrow \omega =2\pi f$ .....1

Kinetic energy $K=\frac{1}{2}I{\omega }^2$ $\to I=\frac{2K}{{\omega }^2}$ .......2

Put equation 1 and 2 in angular momentum formula,

$L=\frac{k}{f\pi }$

using the given condition in question, angular momentum becomes $\frac{L}{4}$