Question

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

• A. $L/4$
• B. $L/2$
• C. $2L$
• D. $4L$

Answer 1

The correct option is A $L/4$

Given,
Final angular frequency, $\omega {}^{\prime }=2\omega$ and Final kinetic energy, $E^{\prime }=\frac{1}{2}E$
According to formula of angular momentum $L=I\omega$
Kinetic energy, $E=\frac{1}{2}I{\omega }^2$
$\frac{1}{2}I{\omega }^2=2\left(\frac{1}{2}{I}^{\prime }{{\omega }^{\prime }}^2\right)$
$I=8I^{\prime }$
Now new angular momentum $L^{\prime }=I^{\prime }{\omega }^{\prime }$
$L^{\prime }=\frac{I}{8}\left(2\omega \right)$
$L^{\prime }=\left(I\omega \right)/4$
$L^{\prime }=L/4$