1. Nov 16, 2005

jrd007

Problem: A person stands, hands at his side, on a platform that is rotating at a rate of 1.30 rev/s. If he raises his arms to a horizontial position the speed of rotation decreases to 0.80 rev/s. (a) Why? (b)By what factor has the moment of inertia changed? Answer: (b) 1.6

Okay, so I know part a which is because his rotational inertia increases.

2. Nov 16, 2005

Päällikkö

$$\vec{L} = I \vec{\omega}$$
L, angular momentum, is conserved. Can you take it from here?

Last edited: Nov 16, 2005
3. Nov 16, 2005

jrd007

But how do I find I ? I know w is 1.30 - 0.80....

4. Nov 17, 2005

Päällikkö

You don't.
You have Ii and If = kIi (where k is a factor by which Ii must be multiplied to get If)

I's should cancel out.

5. Nov 17, 2005

FredGarvin

You do realize what conservation of linear momentum imples...

$$\vec{L}_{initial} = \vec{L}_{final}$$

6. Nov 17, 2005

jrd007

I am going to have to ask my instructor. I have no idea where the 1.6 is coming from. sorry.

7. Nov 17, 2005

FredGarvin

Like I mentioned the previous time you brought this question up, the factor I get is 1.4. I think there is an error in the answer you were given.

EDIT: After looking at sig figs, I ended up with the factor of 1.6.

$${L}_1 = I_1 {\omega_1}$$ and $${L}_2 = I_2 {\omega_2}$$
Now Since $${L}_1 = {L}_2$$ then you can set the rest equal $$I_1 \omega_1 = I_2 \omega_2$$
To get the ratio you are looking for solve that for $$\frac{I_2}{I_1}$$