Finding Kinetic Energy of Person Spinning on Chair w/ Weights

1. dvdqnoc

23
How do you find the Kinetic Energy of a person spinning on a chair with 2 equal weights on each arm?

A specific problem asks what the change in KE is if he moves his originally extended arms inward.

I tried doing (1/2)(I)(wf)^2 - (1/2)(I)(wi)^2 where wi is initial angular velocity and wf is final angular velocity, but its wrong.

Any help?

2. learningphysics

4,124
What numbers/variables do they give?

3. dvdqnoc

23
Ahh, sorry, I suppose I should have included this information in the main thread. My bad.

Well I is the same for initial and final, which is 8. Therefore,

Initial I = 8
Final I = 8
Initial w = .7
Final w = 1.15

So I did:

(1/2)(I)(wf)^2 - (1/2)(I)(wi)^2
= (1/2)(8)(1.15)^2 - (1/2)(8)(.7)^2
= 5.31118 - 1.96
= 3.35118116 j

...but apparently it's wrong.

4. learningphysics

4,124
I can't be the same initial and final... if he brings his arms in the moment of inertia changes. post the question exacty as it is stated...

5. dvdqnoc

23
Alright. But just fyi, I know it cant be the same I because the r changes, but the problem says to just assume its the same. I used the same I for initial and final to get the wf, and it was correct, so i know I is same initial and final. Anyways, heres the problem:

A student sits on a rotating stool holding two
3 kg objects. When his arms are extended
horizontally, the objects are 1 m from the axis
of rotation, and he rotates with angular speed
of 0.7 rad/sec. The moment of inertia of the
student plus the stool is 8 kgm^2 and is assumed
to be constant. The student then pulls
the objects horizontally to a radius 0.29 m
from the rotation axis.

a) Calculate the final angular speed of the