A particle performs uniform circular motion with an angular momentum L. If the frequency of particle motion is doubled, its kinetic energy is halved, the new angular momentum becomes

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2 L

4 L

L/4

L/2

answer is D.

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Detailed Solution

$We know that L = {mr}^{2} ω K.E. = \frac{1}{2} {mv}^{2} = \frac{1}{2} m (rω)^{2} = \frac{1}{2} {mr}^{2} ω^{2} According to the given problem m {(r^{'})}^{2} (2 ω) = \frac{1}{2} [\frac{1}{2} {mr}^{2} ω^{2}] {(r^{'})}^{2} = r^{2} / 8 New angular momentum L^{'} = m {(r^{'})}^{2} (2 ω) ∴ L^{'} = m (r^{2} / 8) 2 ω = L / 4$

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