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Rotational motion-why are torques not in opposite directions

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data
    A cylinder of mass m is suspended through 2 strings wrapped around it at its ends, connected to the ceiling. Both strings have equal tension, and the cylinder rolls without slipping. r is the distance between the CM of the cylinder and each end. I is the moment of inertia of the cylinder.
    Find the tension T in the string and the speed of the cylinder as it falls through a distance h.

    2. Relevant equations
    Equations for linear motion in vertical direction for center of mass of the cylinder, and that of rotational motion about the CM.

    3. The attempt at a solution
    Linear:
    mg - 2T = ma

    Rotational:
    The solution says 2Tr = I*alpha (where alpha is angular acceleration).
    After this we simplify using angular acceleration= a/r, and solve the equations.

    I understand that part, but I don't get why 2Tr = Ia...both tensions are acting on the top of the cylinder, so the torques should be in opposite directions and cancel each other out, right? When I use torque = r x T, the directions come out of the page and into the page. But the equations say that both tensions contribute an equal torque in one direction. Why?
     
    Last edited: Apr 28, 2015
  2. jcsd
  3. Apr 28, 2015 #2

    SteamKing

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    Is their a picture or other diagram which is furnished describing how the cylinder is suspended initially?
     
  4. Apr 28, 2015 #3
    image.jpg
     
  5. Apr 28, 2015 #4

    SteamKing

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    Based on this sketch, it appears that the cylinder is initially suspended with both strings coming off one side.

    You should sketch a free body diagram of the cylinder and the strings to determine the total forces acting on the cylinder as it starts to move.
     
  6. Apr 28, 2015 #5
    I did, and that's how I got the linear equations, but the torques appear to be in opposite directions, unlike the equation that the solution suggests (2Tr = I*alpha). This equation suggests that the torques are in the same direction.
     
  7. Apr 28, 2015 #6

    haruspex

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    That's because you are looking at torque around the wrong axis. The diagram is confusing you. Consider what it looks like from one end of the cylinder. You should see a circle (cylinder end) with a string rising almost vertically from one side. The other string is directly behind it. Each string exerts the same torque about the cylinder's axis.
    Note that it's really a bit more complicated. The strings won't quite be vertical, but you can't take that into account because it depends on string length.
     
  8. Apr 28, 2015 #7
    Ohh, yes that makes sense. Then both torques would be in the same direction. Thanks :)
     
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