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Problem I must solve

  1. Dec 19, 2005 #1
    I'm studying for a highschool physics midterm and I need some help here. The question involves 2 masses and a pulley.

    A pulley of mas 3m and radius r is mounted on a frictionless bearing supported by a stand of mass 4m. I=(3/2)mr^2 Passing over the pulley is a massless cord supporting a block of mass m on the left and a block of mass 2m on the right. The cord does not slip.

    A) Write the following
    i. The equations of translational motion for each of the two blocks
    ii. The analogous equation for the rotational motion of the pulley

    Now, I have the answers, but I do not understand how collegeboard arrived at these answers, and I'm tired of pondering.

    T(1)=tension in string over mass "m"
    T(2)=tension in string over mass "2m"

    i T(1) - mg = ma
    2mg - T(2) = 2ma

    ii I(alpha) = (T(2)-T(1))r

    Can someone explain how these answers were arrived at?

    Also, when 2 objects, of different mass, are hanging like this, the tension in the string is uniform, how can T(2) be different from T(1)?
  2. jcsd
  3. Dec 19, 2005 #2
    Think of the "net tension" exerted on the wheel as a torque- it is what is causing the pulley to begin spinning.
  4. Dec 19, 2005 #3
    Alright, I'll try again.
  5. Dec 19, 2005 #4
    Why are then tensions on the 2 blocks different?
  6. Dec 21, 2005 #5
    because they have to rotate the wheel, so there must be a difference in the tensions
  7. Dec 21, 2005 #6
    Yes, I recently had a question on a final involve a pulley with 2 block on either side of it, where T1 was not equal to T2, for the reason of the pulley.

    The one thing that is constant between them is the acceleration of the two blocks, however. Therefore, the tangential acceleration of the pulley, given a radius you can find alpha. When you draw free body diagrams, include the pulley, then you can relate T1 and T2 to a.