Question

A particle performs uniform circular motion with an angular momentum $L$. If the frequency of particle motion is doubled and its KE is halved the angular momentum becomes

• A. 2 L
• B. 4L
• C. $\frac{L}{2}$
• D. $\frac{L}{4}$

Answer 1

The Correct Option is D

$L=mvr=mr^2 \omega$ Also kinetic energy K =$\frac{1}{2}mv^2$ or $K=-\frac{1}{2}m(r \omega )^2$ $=\frac{1}{2}mr^2 \omega^2$ $K=\frac{1}{2}mr^2 \omega^2$ $K=\frac{1}{2}\frac{L}{\omega}\omega^2=\frac{L\omega}{2}$ $\Rightarrow L=\frac{2K}{\omega}$ hence $\omega '=2 \omega $ or $K'=\frac{1}{2}K$ hence $L'=\frac{2K'}{\omega'}=\frac{2\left(\frac{1}{2}K\right)}{2 \omega}=\frac{L}{4}$