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Rotational mechanics problem

  1. Feb 2, 2012 #1
    This is an example problem I am studying from a classical mechanics textbook.

    Wrap a light, flexible cable around a solid cylinder with mass M and radius R. The cylinder rotates with negligible friction about a stationary horizontal axis. Tie the free end of the cable to a block of mass m and release the object with no initial velocity at a distance h above the floor. As the block falls, the cable unwinds without stretching or slipping, turning the cylinder. Find the speed of the falling block and the angular speed of the cylinder just as the block strikes the floor.



    The solution starts as follows:

    The cable doesn't slip and friction does no work. The cable does no net work; at its upper end the force and displacement are in the same direction, and at its lower end they are in opposite directions. Thus the total work done by the two ends of the cable is zero. Hence, only gravity does work, and so mechanical energy is conserved. ... ... ...



    I am having trouble with the underlined part. The solution considers only the tension forces exerted by the upper and lower ends of the cable. Is this because the internal tension forces exerted by the inner parts of the cable cancel each other out (i.e. the work done by those forces equals zero)?
     
    Last edited: Feb 2, 2012
  2. jcsd
  3. Feb 2, 2012 #2

    Simon Bridge

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    That's what they are saying - work=force x displacement in the direction of the force.

    The cable is also "light" so they are probably not worried about the energy in the cable falling. All (almost all) energy goes into the falling mass and the rotating cylinder. This method can be used to measure the moment of inertia of structures that are not simple but turn smoothly.
     
  4. Feb 2, 2012 #3

    Ken G

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    Yes. This is typical of massless cables-- you only ever analyze the forces at both ends, as the internal tension is internal to the cable and we only care about how the cable is affecting other things. The one thing you do use that internal tension for is to be able to equate the magnitudes of the forces at both ends, but if you already know that property, you don't need to consider the internal forces. Note also that if the cable has mass, then its acceleration must be accounted for, and the forces at the ends will no longer be of the same magnitude.
     
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