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Mechanics Problem (Tension)

  1. Nov 17, 2013 #1
    Recently stumbled on this problem:

    A uniform disk (I =½ MR2
    ) of mass 8.0 kg can rotate without
    friction on a fixed axis. A string is wrapped around its
    circumference and is attached to a 6.0 kg mass. The string does
    not slip. What is the tension in the cord while the mass is
    falling?

    The answer is 24 Newtons, but i would like to know why.
    If anyone can provide a solution/explanation, that would be greatly appreciated.
    I am having trouble figuring out the effect the disk's inertia has on the system's acceleration.
     
  2. jcsd
  3. Nov 17, 2013 #2
    post it in the Homework section
     
  4. Jan 6, 2014 #3
    Hello! To do that problem, you have to, as you know, understand how the rotation of the pulley affects the tension in the cord. Because Tension T is pulling down on the point A where the string is perpendicular to the center of the pulley, and pulling up further down the rope to counteract the mg of the block, we have that:
    1) T - mg = ma
    Since we know what mg is, we have to find a in terms of T, and solve for T
    We know that the Torque about the pulley is Torque = r x F, and since Mg passes through the pulley's center, we don't have to consider it. We only consider T.
    2) Torque = r x F = TR
    Now, remember that Torque = I(alpha), and alpha = a/R?
    TR = I(alpha) [Let's do some substitution]
    TR = MR^2/2 * a/R
    So, a = 2T/M.
    Now let's solve for T!
    Go back to 1) and rearrange it. T + ma = mg. Now plug in a.
    So, we have that T = mg/(1+2m/M). T = 23.52N which is 24.
     
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