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Homework Help: Disc pulled by a string

  1. Feb 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A solid uniform disk of mass 21.0 and radius 85.0 is at rest flat on a frictionless surface. The figure shows a view from above. A string is wrapped around the rim of the disk and a constant force of 35.0 is applied to the string. The string does not slip on the rim.
    (i) When the disk has moved a distance of 5.0 m, determine how fast it is moving.
    (ii) How fast it is spinning (in radians per second)
    (iii) How much string has unwrapped from around the rim.

    2. Relevant equations

    Work = tau * theta
    rotational energy = 0.5I (omega)^2
    translational KE = 0.5Mv^2

    3. The attempt at a solution
    I have tried using energy conservation but for some reason the answer isn't right. tau = 35*0.85. But I'm having trouble finding theta. I used theta = 5 / 0.85 but that isnt working since that would mean that the answer to question (iii) is 5 which it isn't. The work done equals the translational and kinetic energy. But without a correct theta, I can't work out any of the problem. The problem is from Physics for Scientists and Engineers by Giancoli Chapter 10, Problem 108

    Attached Files:

  2. jcsd
  3. Feb 10, 2010 #2
    Text of the problem says that the disk is at rest on a flat frictionless surface. It means that the disk is not rolling. String acts on the disk with the force equal to the force applied to the string and it's the only relevant force. It means that the disk is accelerating translationally because of that force (2. Newton's law). There is also a rotation present. It is caused by the torque of the force acting on the disk.

    Using [tex]F=ma[/tex] we can get the translational acceleration and also the speed of the disk.

    Then, using [tex]\tau=I\alpha=rF[/tex] we can also get angular acceleration and from there the angular velocity.

    Since the body is rotating, string unwrapps from the rim. Length of the unwrapped part is equal to [tex]l=r\varphi[/tex] where [tex]\varphi[/tex] is the angle traversed because of the rotation ([tex]\varphi[/tex] can be calculated using angular acceleration).
  4. Feb 10, 2010 #3
    Thank you very much. I have been at this problem since lunch and didn't manage to consider doing the translational and rotational motion individually. You are a life saver! Thanks again
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