Why does a rotating person slow down when raising their arms?

[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.

Someone please help explain where 1.6 is even coming from... I have yet to get a soild answer. :(

 

[Päällikkö]

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

 

[jrd007]

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

 

[Päällikkö]

You don't.
You have I_i and I_f = kI_i (where k is a factor by which I_i must be multiplied to get I_f)

I's should cancel out.

 

[FredGarvin]

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

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

 

[jrd007]

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

 

[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.

Since you're not grasping the conservation idea, how about this:

$${L}_1 = I_1 {\omega_1}$$ 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 $$\frac{I_2}{I_1}$$