# Moment of inertia and angular speed of skater

1. Jul 22, 2007

### Confused_07

1. The problem statement, all variables and given/known data
A 60kg skater begins a spin with an angular speed of 6 rad/s. By changing the position of her arms, the skater decreases her moment of intertia by 50%. What is the skater's final angular speed?

2. Relevant equations
I understand that I=m*(r^2), so if the radius decreases when she puts her arns in, then the initertia decreases.

3. The attempt at a solution

My attempt at this is that since it decreases by 50%, then the angular speed must increase by two to compensate. Is this correct and that her final speed is 9 rad/s?

2. Jul 22, 2007

### Dick

I=m*r^2 describes how to decrease moment of inertia - but doesn't tell you much about angular speed. You'll want to be thinking about angular momentum, which is conserved.

3. Jul 22, 2007

### SpitfireAce

"My attempt at this is that since it decreases by 50%, then the angular speed must increase by two to compensate"

you are correct but 6 x 2 does not equal 9

4. Jul 22, 2007

### Confused_07

Ah.....

Lf = Lo
[I(final) * w(final)] = [I(initial) * w(initial)]
w(final) = [I(initial) * w(initial)] / I(final)

*If I(initial) = 2, and, I(initial) * 50%= 1, then her final speed is 12 rad/s.

5. Jul 22, 2007

### Confused_07

Why I typed 9 I don't know... :D, I meant 12....