Rotational Motion Question 169
Question: A particle moving in a circular path has an angular momentum of L. If the frequency of rotation is halved, then its angular momentum becomes
Options:
A) $\frac{L}{2}$
L
C) $\frac{L}{3}$
D) $\frac{L}{4}$
Show Answer
Answer:
Correct Answer: A
Solution:
Angular momentum of a particle is given by:
$L=mr^{2}\omega =2\pi mr^{2}f$
$\because \omega=2\pi f$
If frequency is halved, then,
$L’=mr^{2}\frac{\omega }{2}=\pi mr^{2}f\therefore L’=\frac{L}{2}$
BETA
