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Finding the right integral

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A small mass m is pulled to the top of a frictionless halfcylinder (of radius R) by a cord that passes over the top of the cylinder. (a) If the mass moves at a constant speed, show that [tex]F=mg cos(\theta)[/tex]. The angle is between the horizontal and the radius drawn to the mass.

    (b) By directly integrating
    [tex]\int{Fds}[/tex]
    find the work done in moving the mass at constant speed from the bottom to the top of the half-cylinder. Here ds represents an incremental displacement of the small mass.

    2. Relevant equations

    3. The attempt at a solution

    The a-part was easy when I drew a diagram. The b-part is the one I'm struggling with. With the Work-Energy theorem I get that the work done by F is mgR. But what integral should I compute and why? Have no idea whatsoever..:redface:
     
  2. jcsd
  3. Feb 13, 2009 #2

    tiny-tim

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    Hi Ailo! :smile:

    (I'm not sure what you mean by mgR)

    The integral is given to you … ∫F ds, where s is the displacement.

    (Remember, the string is always tangent to the cylinder.)
     
  4. Feb 13, 2009 #3
    Hi! Thx for the answer, but the problem is how to set it up. I should get an expression, integrate it, and end up with the answer mg*R. I've got the force as a function of the angle, and I don't understand how to integrate it over a distance.

    Maybe I didn't explain the situation good enough. The half cylinder lies on the ground, and we pull the mass up along the quartercircle. Does anybody understand? =)
     
  5. Feb 13, 2009 #4

    tiny-tim

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    Hi Ailo! :smile:

    Just decide what ds is (in terms of θ), and then integrate mgcosθ ds, and you should get mgR. :wink:
     
  6. Feb 13, 2009 #5
    That's the problem. I've never done a problem like this before...
     
  7. Feb 13, 2009 #6

    tiny-tim

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    ok …

    what is ds in terms of θ?

    in other words, if you increase the angle by dθ, how much do you increase the length (s) of the string by? :smile:
     
  8. Feb 13, 2009 #7
    My best guess is to make a triangle with sides R, R and ds. Will that work?
     
  9. Feb 13, 2009 #8
    Ohh! Now I get it. It's (theta)*R, right?

    *palmslap
     
  10. Feb 13, 2009 #9

    tiny-tim

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    Yup! :biggrin:

    ds = Rdθ :smile:
     
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