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) $\frac{L}{2}$

B) L

C) $\frac{L}{3}$

D) $\frac{L}{4}$

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Answer:

Correct Answer: A

Solution:

[a] Angular momentum of particle is given by:

$L=mr^{2}\omega =2\pi mr^{2}f$

$\because W=2\pi f$

If frequency is halved then,

$L’=mr^{2}\frac{\omega }{2}=\pi mr^{2}f\therefore L’=\frac{L}{2}$



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