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 Iω² = 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.