First slide

Rotational motion

Question

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

Moderate

A

$4 L$
B

$2 L$
C

$\frac{L}{2}$
D

$\frac{L}{4}$

Solution

$L = Iω = \frac{\frac{1}{2} {Iω}^{2}}{\frac{1}{2} ω} = \frac{2 K}{ω}, where K = \frac{1}{2} {Iω}^{2}$ is the kinetic energy. If $ω$ is doubled and K is halved, the value of L becomes one-fourth. Hence the correct choice is (4).

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