# Man holding weight, rotating and bringing them closer

1. Sep 3, 2016

### Karol

1. The problem statement, all variables and given/known data
A man sits on a rotating chair. the moment of inertia of them both is Im. he holds two weights m, each in a spread out hand, and rotates at frequency f1. the distance each mass from the chair's axis is r1. he then pulls his hands closer, each to r2.
What's the new f2 and the work done.

2. Relevant equations
Conservation of momentum: $m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$
Kinetic energy of a solid body: $E_k=\frac{1}{2}I\omega^2=\frac{1}{2}I4\pi^2 f^2$

3. The attempt at a solution
Conservation of momentum:
$$(I_m+2mr_2^2)f_2^2=(I_m+2mr_1^2)f_1^2~~\rightarrow~~f_2^2=\frac{I_m+2mr_1^2}{I_m+2mr_2^2}f_1^2$$
$$W=\Delta E=\frac{1}{2}[I_2\omega_2^2-I_1\omega_1^2]=2\pi m(f_2^2r_2^2-f_1^1r_1^2)$$

2. Sep 4, 2016

### Orodruin

Staff Emeritus
You are using the conservation of angular momentum, not the conservation of momentum. Apart from that, what is your question?

3. Sep 4, 2016

### haruspex

For the work done, you dropped a pi, making a slight mess.
Since f2 is not a given, I think you should eliminate it from the expression for work done.

4. Sep 4, 2016

### Karol

$$W=\Delta E=\frac{4\pi^2 m}{2}(I_2f_2^2-I_1f_1^2)=...=\frac{[(1-2m)I_m+2mr_2](r_2^2-r_1^2)}{I_m+2mr_2^2}2\pi^2 mf_1^2$$
It looks bad since the dimensions aren't consistent, i will check again

5. Sep 4, 2016

### Karol

$$W=\Delta E=\frac{4\pi^2 m}{2}(I_2f_2^2-I_1f_1^2)=...=2\pi^2 (r_1^2-r_2^2)\frac{mI_m}{I_m+2mr_2^2}f_1^2$$

6. Sep 4, 2016

### haruspex

I think you mean (r22-r12). Other than that, looks right.

7. Sep 4, 2016

### Karol

Thank you Haruspex and Orodruin