Work on an object moving in a circle.

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The discussion revolves around calculating the total work done on two individuals, A and B, riding a ferris wheel, as they move from the bottom to the top and vice versa. The key point is that the work done on both individuals is zero due to constant velocity and the perpendicular direction of motion to the force exerted by the ferris wheel. The Work-Energy Theorem is referenced, indicating that any change in kinetic energy is zero since their speeds remain constant. Additionally, the gravitational potential energy changes as they move vertically, but this does not contribute to the total work done by the ferris wheel itself. The conversation emphasizes understanding the relationship between work, energy, and the forces acting on objects in circular motion.
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[SOLVED] Work on an object moving in a circle.

Homework Statement


I have a question for a test review that has two people, a with a certain mass, let's say m_A and m_B riding on a spinning ferris wheel with a certain radius, let's 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.


Homework Equations


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

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?
 
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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?
 
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 ?
 
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
 
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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 \Delta U+\Delta K=0[/tex] What does that say about W?
 
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?
 
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?
 
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The above was right..thanks
 
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