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Classical mechanics problem

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data

    A disc of mass M, which may be considered to be a point mass, is placed on a frictionless horizontal table. A massless string is fastened to the disc and is passed through a small hole at the centre of the table. The lower end of the string is tied to the end of a flexible rope of mass m per unit length which lies on the floor just under the hole in the table. Initially the lower end of the massless string is held firmly just at floor level and the disc is made to move in a circle, radius r0 about the hole, with a velocity of constant magnitude v0.

    At a subsequent time, the end of the massless string is released from the floor, allowing the radius r of the circular motion of the disc to vary and the rope to be lifted off the floor. You may assume that the contact point between the rope and the floor remains vertically below the hole in the table and may neglect the effects of any horizontal motion of the rope along the floor.

    The subsequent velocity of the disc has a radial component vr and an azimuthal component v. Show that the azimuthal component of the velocity is given by v = v0r0/r.


    2. Relevant equations

    3. The attempt at a solution

    Need a quantity that is conserved under the change of situation. Is it angular momentum?
     

    Attached Files:

    Last edited: Apr 19, 2012
  2. jcsd
  3. Apr 19, 2012 #2
    remember the rope with mass gains P.E. while moving upwards, what form of energy does the mass on the table have? some of that is converted to P.E., and so the (insert answer to my question here) should remain constant.
     
  4. Apr 19, 2012 #3
    The rope with mass gains P.E. while moving upwards; the mass on the table has kinetic energy. Some of that is converted to P.E., and so the total energy should remain constant.

    Thank you so much!

    But i'm not sure how the fact that the energy is conserved leads to the formula. Any hints?
     
  5. Apr 19, 2012 #4
    i doubt it will lead to that, but yes, angular momentum is still conserved to answer your first question. To get the formula conservation of angular momentum would still work, though it would seem like a one step (so short it seems like a trick) solution imho.
     
  6. Apr 19, 2012 #5
    L = r X p. Initally, the disc has only no radial component, so the angular momentum is mass*v0*r0. Later, the disc has a radial component and an azimuthal component, but the cross product of radial component with the the position gives zero, so that the angular momentum is mass*v*r. Equating the two gives the above formula. What do you think?
     
  7. Apr 22, 2012 #6
    Would anyone care to say yes or no?
     
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