Q. The angular momentum of a particle describing uniform circular motion in $L$. If its kinetic energy is halved and angular velocity doubled, its new angular momentum is
A) $4L$
B) $\frac{L}{4}$
C) $\frac{L}{2}$
D) $2L$
Explanation
(b) We know
$L = I\omega$ …(i)
$L^{2} = 2KI$
From Eq. (i)
$L^{2} = 2K\frac{L}{\omega}$
$L = \frac{2K}{\omega}$
$L^{'} = \frac{2\left( \frac{K}{2} \right)}{2\omega} = \frac{L}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.