a particle performs a uniform circular motion with an angular momentum L. if the frequency of prticle’s motion is doubled and it’s kinetic energy is halved, what happens to its angular momentum?

Dear Shimi
 
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.

Regards
Arun (askIITians forum expert)

Dear Student

We Know That
Angular momentum (L) = Iω
Where, I is the moment of inertia and ω is the angular velocity.
So frequency of revolution (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.

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya