* A particle performs uniform circular motion with the angular momentum L. If the frequency of the particle motion is double and its kinetic energy is half...what happens to its angular momentum??? *

We have,

Angular momentum = L = Iω

Where, I is the moment of inertia and ω is the angular velocity.

So frequency of revolution is, f = ω/2π

=> ω = 2πf

Kinetic energy, K = ½ Iω^{2} = ½ Lω = ½ (2πfL) = πfL ……………………..(1)

Now, frequency is doubled, so, let f^{/} = 2f = 2(ω/2π) = ω/π

The kinetic energy is halved, so, K^{/} = K/2 = ½ (πfL)

If L^{/} is the new angular momentum, then using (1) we can write,

K^{/} = πf^{/}L^{/}

=> ½ (πfL) = π(2f)L^{/}

=> L^{/} = L/4

So, angular momentum becomes one fourth of its original value.

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