Competitive exam Question: Work done to shorten string

1. Oct 10, 2009

Q) A particle of mass 'm' attached to a string rotates with velocity Vo when the length of the string is Ro. How much work is done in shortening the string to R?

One way I thought about doing this was:
W= {(m*r*w^2) * r}dr and integrate this from R to Ro
But I am not sure if that is correct. So if someone can help me with this question I will really appreciate it.

Last edited: Oct 10, 2009
2. Oct 10, 2009

Delphi51

Sounds good.
What happens to the velocity as r decreases?
I suppose angular momentum will be conserved.

3. Oct 10, 2009

You think my solution is correct? Can someone else also please confirm this.

v=rw, w is constant. r decreases so v decreases as well. Correct me if I am wrong.

Angular momentum I believe is conserved.

4. Oct 10, 2009

Delphi51

I don't think w is constant. The constant is k = mrv so v = k/(mr)
and w = v/r = k/(mr^2).

v increases as r decreases.

5. Oct 10, 2009

I guess you are right. I was taking the wrong assumption. If angular momentum is conserved then velocity increases if radius decreases.
Now when you have corrected me I think my solution to the problem was not correct either. Can you please verify that for me as well?
Thanks

6. Oct 10, 2009

Delphi51

I haven't seen a solution yet. I just agreed that integrating Force*dr would be the way to do it. And you have to be careful to express v or w in terms of k and r because v and w are not a constants.

7. Oct 10, 2009

Solution:

dW= F.dr
dW= {(m*v^2)/r}dr (v=k/mr)
dW= {(k^2/m)*(1*r^3)}dr
Integrate this with limits Ro to R

W = [(k^2)/(2m)]*[(1/Ro^2) - (1/R^2)] <----Answer

Please check and let me know if this seems correct to you?
Thanks

8. Oct 10, 2009

Delphi51

That is precisely what I got!
Of course I am not infallible! I'm just a retired high school teacher missing the good feeling of helping my students.

9. Oct 10, 2009