Homework Help: Why does work have the wrong units?

1. Feb 26, 2017

Vitani11

1. The problem statement, all variables and given/known data
A particle of mass m is moving on a frictionless horizontal table and is attached to a massless string that passes through a tiny hole of negligible radius in the table, and I am holding the other end of the string underneath the table. Initially the particle is moving in a circle of radius r0 with angular velocity ω0, but I now pull the string down until the radius reaches r. Assuming that I pull the string so slowly that we can approximate the particle’s path by a circle of slowly shrinking radius, calculate the work I did by pulling the string, and compare it to your answer in (5.1).

2. Relevant equations
W = ∫F⋅dr

3. The attempt at a solution
W = ∫F⋅dr = ∫mv2/r dr= ∫mω2rdr = (mω2/2)(r2-ro2)

From conservation of angular momentum I get that ω = (ωoro2)/r. replacing this with the result gives (mωo2ro4)/2[((ro/r)2-1)] and the units don't check out. Shouldn't this be the same as KE?

Last edited: Feb 26, 2017
2. Feb 26, 2017

Staff: Mentor

This equation is not dimensionally consistent and thus cannot be true.

3. Feb 26, 2017

Vitani11

Oh that is v not omega. Crap, stupid mistakes. I got (mωo2ro4)/2r2[(1-(ro/r)2)]

4. Feb 26, 2017

Vitani11

To check this answer I set it equal to KE to see if it turned to 0 but instead it gave me that r=-ro which isn't true. Is this answer still correct or is this not a good way to check your answer?

5. Feb 26, 2017

Arman777

The problem is in here is $F$ is not constant as you can see $F=\frac {mv^2} {r}$ but here $v$ depends on $r$ (angular momentum conservation).So you have to find first $F(r)$

6. Feb 26, 2017

haruspex

Please post your steps there. How do you handle the fact that r and ω are both varying?

7. Feb 26, 2017

Vitani11

Typo. It should have been W = ∫F⋅dr = ∫mv2/r dr= ∫mω2rdr = (mω2/2)(r2-ro2)

8. Feb 26, 2017

haruspex

But that is wrong. In the integration you treated ω as constant.

9. Feb 26, 2017

Vitani11

Got it, thanks