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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?

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?
 

Grade:11

2 Answers

Arun
25750 Points
6 years ago
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)
Yash Chourasiya
askIITians Faculty 256 Points
3 years ago
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

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