Work on an object moving in a circle.

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Homework Help Overview

The discussion revolves around a physics problem involving two individuals on a ferris wheel, focusing on the work done on each person as they move from the bottom to the top and vice versa. The subject area includes concepts of work, energy, and forces in circular motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between work and energy, questioning how to apply the work-energy theorem and the concept of gravitational work in the context of constant velocity.

Discussion Status

The discussion is active, with participants sharing thoughts on the implications of constant velocity and vertical displacement on work done. Some guidance has been offered regarding the work-energy theorem and gravitational work, but no consensus has been reached on the calculations or interpretations.

Contextual Notes

Participants are navigating the problem without specific numerical values and are considering the implications of constant mass and velocity on the work done. There is an emphasis on understanding the setup and definitions involved in the problem.

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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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