Work Done without using the Conservation of Energy

  • #1

Homework Statement



A small mass m is pulled over the top of a frictionless half cylinder of radius R by a massless cord passing over the top. Without using the conservation of energy, find the work done in moving the mass from the bottom to the top of the cylinder at a constant speed, in terms of m, g, and R.

Homework Equations



?


The Attempt at a Solution



Really not sure how to approach this... Can I somehow set up an integral?
 

Answers and Replies

  • #2
Doc Al
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Yes, you'll need to set up an integral of F*ds along the cylinder. Figure out the force that must be applied as a function of position along the cylinder. Hint: Measure position in terms of angle, from 0 to ∏/2 radians.
 
  • #3
I understand the concept, but I am still having trouble setting up the function. I guess I can disregard normal force, so the only force doing work is the tension.

So the work of tension is Force of T * displacement * cos angle. Since the tension is always in the same direction as the displacement, the angle will always be cos 0, or 1, so I can leave that out. The displacement will be R * angle. I don't know how to figure out the force of the tension at any given point since the weight is sometimes pulling down (like at 0 degrees) and sometimes at a right angle, where I think it should then be doing no work.
 
  • #4
Doc Al
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I understand the concept, but I am still having trouble setting up the function. I guess I can disregard normal force, so the only force doing work is the tension.
Yes. We only care about the work done by the tension force.
So the work of tension is Force of T * displacement * cos angle. Since the tension is always in the same direction as the displacement, the angle will always be cos 0, or 1, so I can leave that out. The displacement will be R * angle.
At any given position, the infinitesimal displacement will be Rdθ.
I don't know how to figure out the force of the tension at any given point since the weight is sometimes pulling down (like at 0 degrees) and sometimes at a right angle, where I think it should then be doing no work.
Since the speed is constant, the tension must be just enough to balance out the tangential component of the weight. Find that as a function of θ.
 
  • #5
Not sure I know how to do that...

I think the integral should be from 0 to pi/ 2 of Rdtheta * wcos theta?
 
  • #6
Doc Al
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Exactly!

Now just do that integral.
 
  • #7
Also: check your answer. you should be able to use conservation of energy to check it at least!

I did a similar problem (frictionless half-pipe) with some AP teachers in outreach last month.
 
  • #8
So the work done is just wR [sin theta]0 to pi/2 =

wR sin(pi/2) - wR sin(0) = wR ?
 
  • #9
Also, how can I check using the conservation of energy if I don't know the speed?
 
  • #10
Doc Al
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So the work done is just wR [sin theta]0 to pi/2 =

wR sin(pi/2) - wR sin(0) = wR ?
Yep.
 
  • #11
Doc Al
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Also, how can I check using the conservation of energy if I don't know the speed?
The speed doesn't change.
 
  • #12
OK. I get it. Thanks to all of you for your help!
 

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