# Angular momentum

1. Dec 27, 2008

### 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

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

conservation of momentum
Iweights=mr2=20*0.22=0.8kgm2
Iman+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,

Ei=Ek=0.5mv2
vi=2$$\Pi$$fi*Ri=$$\Pi$$
Ei=10$$\Pi$$2j

Ef=Ek=0.5mv2
vf=2$$\Pi$$ff*Rf=(10/11)$$\Pi$$
Ef=81.566j

W=$$\Delta$$E=0.5m(vf2-vi2)
=10*(8.157-9.87)=-17.13j

W=17.13j

2. Dec 27, 2008

### 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. theres no answer as far as the frequency goes

3. Dec 27, 2008

### 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

4. Dec 27, 2008

thanks a lot

5. Jan 5, 2009

### Dell

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

6. Jan 5, 2009