SATHEE: Rotational Motion Question 169

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:

$L = m r^{2} ω = 2 π m r^{2} f$

$∵ W = 2 π f$

If frequency is halved then,

$L^{'} = m r^{2} \frac{ω}{2} = π m r^{2} f ∴ L^{'} = \frac{L}{2}$

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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:

Answer:

Solution:

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