Posted by **rosa** on Monday, November 11, 2013 at 11:22pm.

A figure skater stands on one spot on the ice(assumed frictionless) and spins around with her arms extended. When she pulls in her arms, her moment of inertia decreases to 3/4 of the initial value. Assume that her spin angular momentum is conserved. What is the ratio of her final rotational kinetic energy to her initial rotational kinetic energy?

- Physics Urgent please help -
**Anonymous**, Tuesday, November 12, 2013 at 5:11am
4/3

- Physics Urgent please help -
**Damon**, Tuesday, November 12, 2013 at 8:14am
I omega, angular momentum, does not change so:

I1 omega1 = I2 omega2

I1 omega1 = (3/4) I1 omega2

so

omega2 = omega 1 * 4/3

then

original ke = (1/2) I1 omega1&2

final Ke = (1/2) I2 omega2^2

so

final Ke=(1/2)(3/4)I1 *(4/3)^2 omega1^2

= 4/3 of original

(she had to do work to pull her arms in)

- Physics Urgent please help -
**rosa**, Tuesday, November 12, 2013 at 5:36pm
Thank you Damon

## Answer This Question

## Related Questions

- Physics - An ice skater spins with his arms outstretched. When he pulls his arms...
- physics - An ice skater spins about a vertical axis at an angular speed of 15 ...
- Physics - A figure skater is spinning with an initial angular speed of 5.6rad/s...
- Physics - A figure skater is spinning with an initial angular speed of 6.9rad/s...
- physics - an ice skater spins about a vertical axis at an angular speed of 15 ...
- Physics - . (a) Calculate the angular momentum of an ice skater spinning at 6.00...
- Physics - A figure skater rotates about a vertical axis through her center with ...
- ice skater world problem - A skater spins with an angular speed of 12.4 rad/s ...
- Physics - An ice skater spins with her arms extended and then pulls her arms in...
- physics - an ice skater starts her spin with her arms stretched straight out. ...

More Related Questions