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Uniform disk tension problem

  1. Jan 22, 2005 #1
    I have a diagram for this at http://snipurl.com/diagr

    There is a uniform disk with radius R, mass M, it has a string wrapped around it and is attached to a fixed rod.
    Part 1: Show the tension in the string is 1/3 the weight of the disk

    what i have done for this is

    [tex] Torque=Fd [/tex]
    and setting the force equal to Tension
    [tex] Torque=TR [/tex]
    then [tex] \sum Torque=I\frac{a}{R}[/tex] and i used [tex]I=\frac{1}{2}MR^2[/tex]for the moment of inertia

    Then [tex] TR=\frac{1}{2}MRa[/tex]
    [tex]T=\frac{1}{2}Ma[/tex]
    and then acceleration is just g, so i get the tension equals 1/2 the weight. where have i gone wrong????
     
  2. jcsd
  3. Jan 23, 2005 #2

    ehild

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    The acceleration of the centre of mass is not just g, but it is determined by the resultant of all forces acting on the pulley.

    ehild
     
  4. Jan 23, 2005 #3

    Andrew Mason

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    Your method is correct except for equating a to g. The forces are Mg and T, so Mg-T=ma. You have found that T = Ma/2 so T = Mg/3. Here is another way of looking at it:

    The weight is Mg. The tension is T. The net force provides the acceleration of the wheel:

    (1) Mg - T = Ma

    To find the acceleration of the center of mass:
    The torque of the center of mass about the point of rotation (point of contact between wheel and string) is MgR. That torque divided by the moment of inertia of the wheel about that contact point provides the angular accleration of the centre of mass = a/R.

    So:

    (2)[tex]MgR/I = a/R [/tex]

    The moment of inertia about that point is NOT the same as the moment of inertia about the center. Use the parallel axis theorem: [itex]I = I_{centre} + MR^2 = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2[/itex]

    Plugging that into (2) results in a = 2g/3

    Then it is just a matter of plugging that into (1) to get:

    T = Mg(1 - 2/3) = Mg/3

    AM
     
    Last edited: Jan 23, 2005
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