A figure skater rotating on one spot with both arms and one leg extended has mom

Question

A figure skater rotating on one spot with both arms and one leg extended has moment of inertia Ii. She then pulls in her arms and the extended leg, reducing her moment of inertia to 0.75 Ii. What is the ratio of her final to initial kinetic energy?
1. 3/8
2. 2
3. 8/3
4. 1
5. 16/9
6. 3/4
7. 4/3
8. 1/2
9. 9/16

Consider the following statements for the figure skater:
I. Angular momentum was conserved.
II. Mechanical energy was conserved.
III. The kinetic energy changed because of energy dissipation due to friction.
IV. Her rotation rate changed in response to a torque exerted by pulling in her arms and leg.

Which is the correct combination of statements?
1. I and II
2. I, II, IV
3. I, II, III
4. II
5. I

Explanation / Answer

let initial moment of inertia = I1 and angular speed = 1

final = I2 and 2 ,, given I2= 0.75I1

Conservation of angular momentum we get

I11 = I22 = 0.75I12, 1/2 = 0.75

Kinetic energy inital = (1/2)I1(1)^2

Final = (1/2)I2(2)^2

Ratio of final to initial = (I2/I1)(2/1)^2 = 0.75*(1/0.75)^2 = 4/3 = 1.333

Option 2

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