# Work on an object moving in a circle.

1. Feb 24, 2008

### jj8890

[SOLVED] Work on an object moving in a circle.

1. The problem statement, all variables and given/known data
I have a question for a test review that has two people, a with a certain mass, lets say m_A and m_B riding on a spinning ferris wheel with a certain radius, lets say R, in carts opposite to one another. One (A) is originally at the bottom of the ferris wheel while the other (B) is at the top of the ferris wheel. As the wheel turns, B comes to the bottom while A arrives at the top. Neglect air resistance. I need to find the magnitude of the total work done on A and B moving from the bottom to top and top to bottom respectively.
I don't want to give the numbers because I want to work it out myself. I just need help figuring out how to set up the problem. I would appreciate any help.

2. Relevant equations
The total work is the sum of the work done by all of the forces on the body, W total = F_net · ds.

3. The attempt at a solution
I was thinking that the W_total on student A from bottom to top would be found by 2(Ma-Mb)gR but I am not sure that this looks right.

Wouldn't the Work done on student B by Ferris wheel is be 0 because the direction of motion is always perpendicular to force?

2. Feb 24, 2008

### jj8890

I have been trying to work it out and now I think that it is switched, the total work done on student A=0 because it says that the velocity is constant, and the masses certainly aren't changing and the change in kinetic energy equals change in total work so since there is not change in mass or velocity W_total must be zero. In order to clarify, it asks for the total work done on A (which would be zero) but only the work done on B, is there a difference? Would the work done on B also be 0, how would I go about this?

3. Feb 24, 2008

Have you by chance studied the Work-Energy Theorem yet? $$W=\Delta K$$ or Conservation of Mechanical Energy? Or have you only learned that W= F*displacement ?

4. Feb 24, 2008

### jj8890

yes we have used W=deltaK.... W=K_2-K_1 but how would I use this to find the work done on B? I know that K=1/2mv^2 but since the speed is not changing how would I work this out? That is why I was wondering if it would be the same...zero

Last edited: Feb 24, 2008
5. Feb 24, 2008

I wish I was not so tired right now, else I could think this through. Regardless of constant velocity, if an object moves through a vertical displacement, there is work done by gravity (I'm fairly confident that is a true statement).

Now. If you know that by W-E theorem $W=\Delta K$ and by conservation of mechanical energy [itex]\Delta U+\Delta K=0[/tex] What does that say about W?

6. Feb 24, 2008

### jj8890

Well, obviously deltaK=-deltaU so W=-DeltaU or -U_2 +U_1 but where would i go from there? Or am I totally on the wrong path?

7. Feb 24, 2008

### jj8890

Well, if the radius was R and the mass of the person was m_B, would the work done on B by the ferris wheel be 2m_B*g*R because person B moves 2 times the radius...or would I have to include the mass of person A in there somewhere?

Last edited: Feb 24, 2008
8. Feb 24, 2008

### jj8890

The above was right..thanks