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Acceleration of a mass connected to 2 different disks

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data
    The attachment gives the setup, but I can do the first part (a). The second question (not in the attachment) is "Now consider the same system as part (a) except M = .250 kg. After the system is released from rest, what will the acceleration of the .250kg mass be?


    2. Relevant equations
    sum of torques = T1*(r1) - T2*(r2) = I*angular acceleration
    sum of forces on M= Mg-T1=Ma
    sum of forces on m = T2-mg=ma


    3. The attempt at a solution
    I multiplied the two force equation by the r of the disk that each mass is attached to, then added all three equations together. The tensions cancel, giving:
    RMg-rmg = RMa + rma + I*ang acc

    I don't know what to do from here though. Maybe I = I big disk + I small disk. I don't know what ang acc is though. I would usually set it to a/r, but there are two different r's in this case. Any help is greatly appreciated!
     

    Attached Files:

  2. jcsd
  3. Dec 9, 2008 #2

    Doc Al

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    Staff: Mentor

    Good.
    Careful! The masses have different accelerations. How do the accelerations relate?

    Yep!

    Good--there are two different accelerations, also. :wink:
     
  4. Dec 9, 2008 #3
    Let's see.

    Is it Ma = MR(ang acc)
    ma = mr(ang acc)?

    If so, does the final equation, after adding all three, look like:

    g(RM + rm) = ang acc (R(^2)M + r(^2)m + .5MR(^2) + .5mr(^2) ?

    If so, you can calculate ang acceleration. Then do you just multiply that by R to get a of block 1?

    edit: actually the final equation should be (I think)
    g(RM - rm) = ang acc (R(^2)M + r(^2)m + .5MR(^2) + .5mr(^2)

    oh boy, edit again. the M and m next to .5 should be masses of the disks
     
    Last edited: Dec 9, 2008
  5. Dec 9, 2008 #4

    Doc Al

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    Staff: Mentor

    That looks OK. (Except for using the same letters to represent different masses!)
    Yep.
     
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