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Acceleration of objects related to moment of inertia

  1. Jan 4, 2012 #1
    1. The problem statement, all variables and given/known data
    A system is given below. Pulley A has radius R and mass 2M and pulley B has radius 1/2 R and mass M. If the pulley is assumed as solid cylinder, find the acceleration of each object
    3.jpg

    2. Relevant equations
    τ = I.α
    F = m.a
    a = α.R

    3. The attempt at a solution
    Equation of motion for block M:
    ƩF = m.a
    T1 - M.g = M.a1

    Equation of motion for second block:
    ƩF = m.a
    3/2 M.g - T2 = 3/2 M . a2

    Equation relating torque:
    τ = I.α

    I'm not sure how to find the moment of inertia of the system, is it Isystem = I1 + I2 = 1/2 (2M) (R)2 + 1/2 (M) (1/2 R) ?
     
  2. jcsd
  3. Jan 5, 2012 #2

    Simon Bridge

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    Note - the two pulleys will have the same angular acceleration; torque is force time radius.
    [edit] hit the wrong button ... sorry: yes, you add the I's.
    You get the I for the pulley from the I for a point mass by integrating - "integrate" is a fancy way of adding things up.
     
    Last edited: Jan 5, 2012
  4. Jan 5, 2012 #3
    Well i guess you can add the moments to get net moment, after all, thats what we do in integration.

    But aren't the pulleys given as 2 (or separate?) so will they have same acc. Its not specified anywhere !
     
  5. Jan 5, 2012 #4

    Simon Bridge

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    The question is not very clear on that point is it?
    When it tells you "the pulley is a solid cylinder", I take that to mean the composite pulley. Otherwise the question does not make a lot of sense.

    If you like OP can do it either way - pulleys locked together or pulleys not connected at all. (No mention of a friction coefficient for limited slip between them so it is one or the other.)
     
  6. Jan 5, 2012 #5
    I think there are two concentric pulleys.

    Next, after finding the Isystem = I1 + I2 = 1/2 (2M) (R)2 + 1/2 (M) (1/2 R)2, I proceed as follow:

    Equation of motion for block M:
    ƩF = m.a
    T1 - M.g = M.a1
    T1 - M.g = M.α.R1......(1)

    Equation of motion for second block:
    ƩF = m.a
    3/2 M.g - T2 = 3/2 M . a2
    3/2 M.g - T2 = 3/2 M.α.R2......(2)

    τ = I.α
    T2.1/2 R - T1.R = 1/2 (2M) (R)2 + 1/2 (M) (1/2 R)2.α...(3)

    By using elimination, solve for α. Am I right?

    I don't get the meaning. Can there be two cases? Can they have different α?

    Thanks
     
  7. Jan 5, 2012 #6
    Yes Your method is correct !!

    but Well when i read the question again, i guess you can ignore my statement, the method is fine ...
     
  8. Jan 5, 2012 #7
    Thank you and Simon for the help :)
     
  9. Jan 5, 2012 #8

    Simon Bridge

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    If the pulleys are not joined together, then there would be two cases to solve for.
    You are not explicitly told the two pulleys are joined but it is kinda implied in the question ... interpreting problems scientifically is something you have to learn how to do. A variation would be to have the two pulleys separate but leave the second mass unknown ... you get told that the two masses hit the ground at the same time and asked to calculate the unknown mass. I think this one wants you to notice that the heavier mass does not dominate the motion.

    But it's all good fun - enjoy :)
     
  10. Jan 10, 2012 #9
    OK now let say that the pulleys are not joined together. Honestly, I can't imagine how the system works.

    If the pulleys are joined, then they will have same angular acceleration and one will move upwards while the other downwards.

    I can't even imagine the figure if the pulleys are not joined. In my mind, there are two separate pulleys and they are independent to each other, one will move clockwise and the other moves anti-clockwise. Both masses will move downwards and the equation. We will get equations from each pulley that are not related to each other, unless there is additional information given from the question (such as the masses reach the ground at the same time).

    Am I correct this far? Thanks
     
  11. Jan 10, 2012 #10
    Yes you are correct. These type of questions are not mostly asked because of their simplicity while solving ...
     
  12. Jan 10, 2012 #11

    Simon Bridge

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    That's correct.

    Exploit the third dimension ... imagine the big pulley is completely separate, separate axle and everything, and it is several feet behind the small one, but the axles line up. That would give you the same picture viewed end-on.

    If you've seen block and tackle, you get two pulleys mounted side-by-side which can rotate separately.
     
  13. Jan 10, 2012 #12
    OK thanks again :)
     
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