Why does work have the wrong units?

[Vitani11]

Homework Statement

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).

Homework Equations

W = ∫F⋅dr
Answer for KE is [(mω_o²r_o²)/2]((r_o/r)²-1)

The Attempt at a Solution

W = ∫F⋅dr = ∫mv²/r dr= ∫mω²rdr = (mω²/2)(r²-r_o²)

From conservation of angular momentum I get that ω = (ω^or_o²)/r. replacing this with the result gives (mω_o²r_o⁴)/2[((r_o/r)²-1)] and the units don't check out. Shouldn't this be the same as KE?

 

[Doc Al]

From conservation of angular momentum I get that ω = (ω^or_o²)/r.

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

 

[Vitani11]

Oh that is v not omega. Crap, stupid mistakes. I got (mω_o²r_o⁴)/2r²[(1-(r_o/r)2)]

 

[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=-r_o which isn't true. Is this answer still correct or is this not a good way to check your answer?

 

[Arman777]

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

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 $F(r)$

 

[haruspex]

∫mω²rdr = (mω²/2)(rr₀²)

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

 

[Vitani11]

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

Typo. It should have been W = ∫F⋅dr = ∫mv²/r dr= ∫mω²rdr = (mω²/2)(r²-r_o²)

 

[haruspex]

Typo. It should have been W = ∫F⋅dr = ∫mv²/r dr= ∫mω²rdr = (mω²/2)(r²-r_o²)

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

 

[Vitani11]

Got it, thanks