Rotational Motion Question 7

Question: A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is

[AIEEE 2003]

Options:

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

B) 2L

C) 4L

D) $ \frac{L}{2} $

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Answer:

Correct Answer: A

Solution:

  • [a] Angular momentum $ L=l\omega $ (i)

    Kinetic energy $ K=\frac{1}{2}l{{\omega }^{2}}=\frac{1}{2}L\omega $ [from Eq.(i)]

    $ \therefore $ $ L=\frac{2K}{\omega } $

    Now, the new angular momentum

    $ L’=\frac{2( \frac{K}{2} )}{2\omega } $

    $ ( \because K’=\frac{K}{2}and\omega ‘=2\omega ) $

    $ \Rightarrow $ $ L’=\frac{L}{4} $



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