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Rotational Motion/Pulley Problem

  1. Dec 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Two uniform disks with the same mass are connected by a light inextensible string supported by a massless pulley, on a frictionless axis. The string is attached to a point on the circumference of disk A. The string is wound around disk B so that the disk will rotate like a yo-yo when dropped. Describe the outcome when both disks hit the floor.

    2. Relevant equations

    3. The attempt at a solution
    Okay I know that the two tensions are equal because the pulley is massless. I wanted to prove it to myself with equations which is the right answer. I decided that the translation accelerations for each mass would be different because one has the string wound around it and the other is simply hanging by the string.
    I wasn't sure though what to make the direction of both accelerations, I just assumed negative for the calculations below:

    Disk A

    Disk B


    I substituted T=mA/2 into the second equation for disk B to get:

    -mA=mA/2 - mg
    mA=mg- mA/2
    A= 6.53

    Then I solved for T and just put the mass as 1kg
    T= 6.53/2=3.27

    Then I used T to solve for a for disk A
    -a= 3.27 - 9.8
    a= 6.53

    Does this seem right at all? They both accelerate the same in magnitude, but opposite in direction?
    I think I remember being my teacher doing something like a + A = 0, but I may have just imagined it.
    Confirmation or correction would be amazing! Thanks!
  2. jcsd
  3. Dec 2, 2008 #2
    How do you get different directions (both a and A have the same sign in your solution!)? If you ignore the torque on disk B and just look at the forces acting on the centers of mass, do you get different equations of motion for the two disks?

    Your solution is almost right, except that you obtain positive values for a and A. This means both disks are accelerating upwards?

    Hint: Look at your torque equation, and draw a diagram to relate the angular acceleration to the acceleration of the center of mass.
  4. Dec 2, 2008 #3
    I have no idea. I guess I'm not sure if the equations are the same because I don't know if they are both accelerating downwards. How do I know?

    Also when I calculated the torque, did I use the right equation or should it have been negative?
    Thanks for your help!
  5. Dec 3, 2008 #4
    Well, I think you are going about this the wrong way. If the equations are the same, the acceleration is the same. (Instead of saying the equations are the same if the acceleration is the same). This is because it is easy to write down the equations of motion. You have (correctly) written down the equations for the centers of mass. Are they the same?

    I can give you the answer to this, but that doesn't really serve the purpose. What you need to do is draw a picture, labelling the directions of the linear and angular accelerations. Then you can figure out for yourself if the acceleration should have been negative or otherwise. I can help you if you are stuck.
  6. Dec 3, 2008 #5
    Okay I think I still might be lost, but I'm thinking that the tension force on disk 2 will produce a torque that has the same direction as the angular acceleration, so even if it were negative they would both cancel out to give Tr=I(alpha). I dont really know where I am going with this, because I feel like I am kind of just doing circles around the problem. From the diagram I drew I would say that the angular acceleration is clockwise and the translational acceleration is downward.
    Am I even in the right realm of thought?
    Sorry that you have to keep answering questions, I really appreciate your help!
  7. Dec 3, 2008 #6
    Is it true that a1+a2+r(alpha)2=0 or pretty much that a1 + 2a2 = 0?
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