# Work-Energy Thm

1. Jan 25, 2009

### roeb

1. The problem statement, all variables and given/known data
A particle of mass m is moving on a frictionless horizontal table, attached to a massless string, the other end of which passes through a hole in the table. It was rotating with angular velocity $$\omega_0$$, at a distance $$r_0$$ from the hole. Assuming that I pull the string so slowly that we can always approximate the path of the particle at any time by a circle of slowly shrinking radius, calculate the work I did while pulling the string. Show that the work-energy theorem is satisfied in this case.

2. Relevant equations
$$W = \int F . dr$$
[/tex]\delta W = KE_f - KE_i[/tex]

3. The attempt at a solution
Here is how I initially attempted the problem:
$$KE_i = \frac{1}{2}*m*(r_0 \omega_0)^2$$
$$KE_f = \frac{1}{2}*m*(r \omega_f)^2$$
And by taking the difference I was hoping that I would get the work done. Unfortunately it turns out that I need to do something a bit more complicated.

I know $$W = \int F . dr$$ but I'm not really sure what to do in order to apply it to this situation.

Anyone have any hints?

2. Jan 25, 2009

### Staff: Mentor

Find the force required as a function of radius. Hint: What's conserved? Apply Newton's 2nd law.

3. Jan 25, 2009

### roeb

For this situation, I believe energy conservation is important to consider (or perhaps conservation of angular momentum?)

I'm not sure if this is what you were refering to, but centripetal force comes to mind when trying to find a force as a function of radius.
$$F(r) = \frac{ mv^2 } {r} = \frac{m \omega_0^2 r^2}{r}$$
$$\int_{r_0}^{r} F(r) . dr = 1/2 m \omega_0^2 (r^2 - r_0^2)$$
Unfortunately this isn't correct, hmm. In this case I assumed that $$\omega_0$$ was a constant for both radii. Physically, this seems like a bad assumption, so I will have to try to think of something else.

Am I correct that I should be using centripetal force?

Last edited: Jan 25, 2009
4. Jan 25, 2009

### Staff: Mentor

Yes, you need to use centripetal force. (That tells you how hard you must pull on the string.)

No, ω is definitely not constant as the radius changes. But another quantity is constant, which will allow you to figure out the force as a function of radius. (And it's not energy!)

5. Jan 25, 2009

### roeb

Thank you Doc Al, I was able to figure it out.