Man holding weight, rotating and bringing them closer

[Karol]

Homework Statement

[/B]A man sits on a rotating chair. the moment of inertia of them both is I_m. 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.

Homework Equations

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

The Attempt at a Solution

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

 

[Orodruin]

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

 

[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.

 

[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

 

[Karol]

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

 

[haruspex]

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

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

 

[Karol]

Thank you Haruspex and Orodruin