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For an Atwood's with a cylinder and a block, why do the objects have the same accel?

  1. Dec 1, 2011 #1

    wu7

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    1. The problem statement, all variables and given/known data

    One pulley, on one side we have a block with mass m.
    On the other side we have a cylinder with mass m.
    Cylinder (radius R) has unlimited string (massless, negligible thickness, no slippage)

    So you can imagine two blocks falling as more string unravels from the cylinder.

    I am told that these two objects have the same acceleration downwards. Why?

    2. Relevant equations

    F=ma
    ma=mg-T

    torque? T*R=I*[itex]\alpha[/itex]

    3. The attempt at a solution

    The problem made it sound like this was a quick, obvious argument.
    I proceeded to a longer argument, finding that the accelerations of both masses were (2/3)g downward:

    starting with the cylinder-
    TR=I*[itex]\alpha[/itex]
    T*R2 = (1/2)M*R2 * a
    Finding T= (1/2)ma, then plugging T back into the standard F=ma equations.
    Then I find the acceleration of the block to be (2/3)g as well.

    Is this even right? If it is, was there a way to show that the objects' accelerations were the same without finding the actual accelerations?

    The reason I don't think this is right is because I did the problem using conservation of energy using the assumption that the objects fell down at the same acceleration.
    With conservation of energy I got acceleration=(1/2)g (conserving KE, PE, angular KE)
     
  2. jcsd
  3. Dec 1, 2011 #2
    Re: For an Atwood's with a cylinder and a block, why do the objects have the same acc

    I think you don't need to use torque.
    By F= m * a(center of mass) you get that their accel will only depend on forces applied to them. In both cases block and cylinder you have F = T - mg.
    Then => m * aBlock = m * aCylinder => aBlock = aCylinder
     
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