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Rotational kinematics problem

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data
    A hollow cylinder of length L and radius R has a weight W. Two cords are wrapped in the same direction around the cylinder, one near each end. The cords are fixed to the ceiling. The cylinder is held horizontally with the cords vertical. The cylinder is then released. Find the tension, T, in each of the cords.
    http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/oldexams/exam3/sp07/fig21.gif [Broken]

    2. Relevant equations

    [tex]\tau=rFsin\theta[/tex]

    [tex]\alpha=\frac{\tau_{net}}{I}[/tex]

    3. The attempt at a solution
    I drew the picture from the side, with the tension T pointing up on the side of the cylinder, and the weight W pointed down from the center of the cylinder. Using the parallel axis theorem, the moment of inertia would be MR^2 + Md^2 (with R = d) = 2MR^2 . I'm assuming weight W is in newtons, which means this needs to be rewritten as 2(W/g)R^2 to get it into known variables.

    I'm trying to solve it like any other rolling / rotational motion problem, but am getting confused. If the axis of rotation is at the end of the cylinder where the ropes are, they should be causing no torque at all on the cylinder, since the distance from the axis of rotation would be zero. Starting with a confusion like this, I don't know where to go. Any tips would be greatly appreciated!
     
    Last edited by a moderator: May 4, 2017
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  3. Dec 6, 2009 #2

    Redbelly98

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    What force, other than the rope tension, acts on the cylinder?
     
  4. Dec 6, 2009 #3
    The force of gravity. I already took this into account with W being pointed down from the center of the circle.
     
  5. Dec 7, 2009 #4

    Redbelly98

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    Okay, good. That will cause a torque about the pivot axis you have chosen, where the ropes meet the cylinder.
     
  6. Dec 7, 2009 #5
    see page 152 of this link that has solved physics problems

    [link deleted]
     
    Last edited by a moderator: Dec 8, 2009
  7. Dec 7, 2009 #6

    ideasrule

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    What's the purpose of doing a homework question if the OP just copies from an answer sheet?
     
  8. Dec 8, 2009 #7
    The moment of inertia about the axis (center in this case) is just MR^2, you don't need to use the parallel axis theorem.
    And torque is zero if the force goes through the axis of rotation. The Tensions are on the side (tangent) of the cylinder, not center.
     
  9. Dec 8, 2009 #8

    Redbelly98

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    You can choose any axis to solve the problem. If you choose the axis where the strings contact the cylinder, the parallel axis theorem is needed to find the M.O.I. about that axis. As ahmadmz says, the P.A.T. is not need if one chooses the center cylinder axis.

    Either axis will have a torque about it, and if you use the correct M.O.I. for that axis, you will get the same final solution.
     
  10. Dec 8, 2009 #9
    Answer to "Why get help from a solved problem"

    For the same reason your teacher does similar problems dummy...when you see how a similar problem is solved and learn the logic, you can then apply that to your problem. I apologize the link did not copy properly, for you could see the solved problem was not an exact match, it instead had masses attached. However, the solved problem involved the same tension logic. Sometimes, a little help goes a long way.
     
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