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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
25763 Points
3 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
11 months 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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