Angular Momentum: Work Done & Frequency Change

[Dell]

1st of all i apologise if my terminology is incorrect, as i am translating into english..

on a spinning chair a man is sitting and holding 2 weights, of 10kg each, at a radius of 0.5m, the chair is turning at a frequency of 1Hz., the total moment of the man and the chair is I=2.5kgm². how will the frequency change is the man moves the weights to a radius of 0.2m? what is the work done by the man in this case?

what i did was, using conservation of momentum

I_weights=mr²=20*0.5²=5kgm²
I_man+chair=2.5kgm²
L=I$$\omega$$=(5+2.5)(2$$\Pi$$f)=15$$\Pi$$

conservation of momentum
I_weights=mr²=20*0.2²=0.8kgm²
I_man+chair=2.5kgm²

L=15$$\Pi$$=(0.8+2.5)(2$$\Pi$$f)

f=25/11Hz

is this correct,

now to find the work done, can i say- work done is the change of energy and using that, say,

E_i=E_k=0.5mv²
v_i=2$$\Pi$$f_i*R_i=$$\Pi$$
E_i=10$$\Pi$$²j

E_f=E_k=0.5mv²
v_f=2$$\Pi$$f_f*R_f=(10/11)$$\Pi$$
E_f=81.566j

W=$$\Delta$$E=0.5m(v_f²-v_i²)
=10*(8.157-9.87)=-17.13j

W=17.13j

 

[Dell]

the correct answer (in my book) for the work done comes out around 188.4j so I'm way off.- what do i need to change. there's no answer as far as the frequency goes

 

[LowlyPion]

The first part looks correct.

But for the second wouldn't you want to consider the change in rotational kinetic energy?

KE = ½Iω²

ΔKE = ½I₁ω₁² - ½I₂ω₂²

 

[Dell]

thanks a lot

 

[Dell]

could you take a please look at this post for me, also momentum, thanks

 

[Dell]

sorry forgot to post the link
https://www.physicsforums.com/showthread.php?t=282747

i think i got it, what i need to do is
mV=mu+Mu/2
then find u using energy, is this correct