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Torque with pulley question.

  1. Nov 21, 2012 #1
    1. The problem statement, all variables and given/known data

    A 2.85 kg bucket is attached to a disk-shaped pulley of radius .121 m and mass .742 kg. If the bucket is allowed to fall, what is the linear acceleration, angular acceleration, and how far does it drop in 1.5 seconds.

    I really only need help with the first part since the other two are pretty easy and depend on the first answer.

    2. Relevant equations

    t - mg = ma
    alpha = a / r

    3. The attempt at a solution

    I'm a bit confused on which signs I should be using. This was my first try and I'd love some input to see where I went wrong.

    T - mg = ma
    Since we don't know or a we need to use the torque produced by the pulley.
    torque = TR = I alpha
    T= .5Mr^2 alpha / R
    alpha = a / R
    So then T = .5Ma ----> we plug this into T in the original equation

    .5Ma - mg = ma,

    .5(.742)(a) - (2.85)(9.8) = 2.85a
    .371a - 27.93 = 2.85a

    -27.93 = 2.85a - .371a
    = -27.93 = 2.479a
    a = - 11.26

    Now I'm pretty sure I messed up on the signs somewhere. Wouldn't it all depend on how you set up the diagrams? So in this case the tension is pointing upwards so it's positive, and MG is - so it's negative. Also, how would you know if the pulley is going clockwise or counter clockwise?

    Would love any clarification of how to set up the signs. Thanks guys!
     
  2. jcsd
  3. Nov 21, 2012 #2

    haruspex

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    It shouldn't matter what you assume about the directions things will move in provided you're consistent.
    According to your first equation, T - mg = ma, you're measuring acceleration as upwards. The equation alpha = a / R then implies alpha is in the direction of reeling in the bucket. But the string effectively reverses the direction of T; if it's upwards on the bucket then it's downwards on the pulley. So for the third equation you would have T= -.5MR2 alpha / R.
     
  4. Nov 21, 2012 #3

    tiny-tim

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    hi dolpho! :smile:
    that's fine down to the last line, which should be .5Ma - mg = -ma

    you assumed that α was positive in the direction of the string (ie downwards), so T had the same sign as α

    but if α is positive downwards, then the bucket is falling, so a (of the bucket) must also be positive downwards :wink:

    (of course, you could have assumed α was positive upwards, then a would be also, and they would both come out as having negative values!)

    [btw, there's a simple way of checking your result (but which i don't think would be approved of in the exam) …

    use the "rolling mass", I/r2 (= m/2 for a cylinder), so the total weight is Mg, and the total effective mass is M + m/2 :wink:]​
     
  5. Nov 21, 2012 #4
    Great, thanks for the help. I'll definitely try that check :D
     
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