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Homework Help: Classical Physics Pulley Problem

  1. Aug 28, 2007 #1
    A massless string is placed over a massless pulley, and each end is wound around and fastened to a vertical hoop. The hoops have masses M1 and M2 and radii R1 and R2. The apparatus is placed in a uniform gravitation field g and released with each end of the string aligned along the field.

    I have to show that the tension is T = gM1M2/(M1+M2)

    I can sort of solve the problem by just saying that (M1M2)/(M1+M2) is the reduced mass. Then all we have to do is say the tension is balanced with the weight, so that T = Mg = gM1M2/(M1+M2). But then doesn't this imply that the hoop isn't moving? I think I'm missing something here...
  2. jcsd
  3. Aug 28, 2007 #2


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    You don't know anything about whether they are moving or not. What you do know is that the acceleration of each hoop is equal (one up and one down). Write down a free body diagram, F=ma for each hoop and eliminate the acceleration to solve for the tension.
  4. Sep 2, 2009 #3
    I realize that this is an old post, but I have the same exact problem.

    If we assume that the rings are point masses, we see that

    F1_net = T - (m1)g = (m1)a
    F2_net = T - (m2)g = -(m2)a

    where the right hand side must be of different sign (one mass accelerates upward, the other downward). Rearranging the equations so that each is equal to "a" and then equating gives

    T = 2g(m1)(m2)/(m1 + m2)

    but this is not the answer--it is off by a factor of 2. Of course, my work assumes point particles, so what is the distinction between point particles and rings to obtain this factor difference?
    Last edited: Sep 2, 2009
  5. Sep 2, 2009 #4


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    I was assuming the problem was to be read that the rings were in rotational acceleration as well as linear acceleration. The moment of inertia should make some difference. Try doing it that way. If you don't get it let me know and I'll travel two years back in time...
    Last edited: Sep 2, 2009
  6. Sep 3, 2009 #5
    No need for time travel. Your right about the rotational acceleration. Thanks. Sometimes understanding what the problem is itself can be the hardest part of solving a problem.
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