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A tape,pulley,disk and undisclosed mass.

  1. May 16, 2007 #1
    the question is:
    A disk of mass M and radius R unwinds from a tape wrapped around it. the tape passes over a frictionless pulley, and a mass m is suspended from the other end. assume that the disk drops vertically.
    1. relate the acclerations of m and the disk, a and A, respectively to the angualr accelration of the disk.
    (the answer clue reveals that the naswer is: if A=2a, then alpha=3A/R.
    my answer is that the accleration of the disk equals -a+[tex]\alpha[/tex]R=A and then i get that if A=2a then alpha equals 3a/R, but it's the opposite it should be 3A/R, which i don't see how to arrive at this.

    anyway, in the attached file there's a pic of this, the above left pic.
    thanks in advance.
     

    Attached Files:

  2. jcsd
  3. May 16, 2007 #2
    I dunno if I'm doing this wrong, but this is what I get:

    Let the tension in the string be T, acceleration of the disk be A and of the mass be a and R be the radius of the disk.

    For the small mass m:

    T-mg=ma ---1

    For the disk:

    Mg-T=MA

    [tex]TR=I\alpha[/tex]

    [tex]I=\frac{MR^2}{2}[/tex]

    [tex]A=R\alpha[/tex]

    Solving these, A=2g/3, T=g/3, [tex]a=\frac{g(M-3m)}{3m}[/tex]. What did I do wrong?
     
  4. May 16, 2007 #3
    the problem is that it's not given to you that the disk rolls without slippering, if it were so, then obviously we would have A=R*(alpha).
     
  5. May 18, 2007 #4
    But if the rope is wound tightly across the disk, then it must roll without slipping as there is no other option. It cant slip (across what?). Therefore a=r(alpha) must hold.
     
  6. May 18, 2007 #5
    In any case, the velocity of the string along the tension must be the same at all the points on the string.
     
  7. May 21, 2007 #6
    the problem is that's not a string but a tape!
     
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