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Torque with Pulley and Hanging Masses

  1. Jan 27, 2008 #1
    1. The problem statement, all variables and given/known data

    We've just started Torque and I hate it, just to let everyone know. Can't wrap my head around anything remotely three dimensional.

    Anyyyyway.

    A 15 kg object and a 10 kg object are suspended, joined by a cord that passes over a pulley with a radius of 10 cm and a mass of 3kg. The cord has a negligible mass and does not slip on the pulley, which rotates on its axis without friction. The objects start from rest 3 m apart. treat the pulley as a uniform disk, and determine the speeds of the two objects as they pass each other as prescribed, below.
    a) Calculate the acceleration of the two masses using Newton's laws in order to determine their speed when the meet.
    b) Calculate speed directly using conservation of energy.

    2. Relevant equations

    T = r x F
    Summation T = Ia (angular acceleration)
    I for Disk = 1/2Mr^2

    3. The attempt at a solution

    I really have very little idea what to do, but this is what Ive done so far for (a)

    Isolate mass One
    Fnet = -ma
    Ft - mg = -ma

    Isolate pulley
    Summation Torque = Ia
    I = 1/2Mr^2
    = 1/2(3)(0.1)^2
    And then I honestly don't know what to do :(
     
  2. jcsd
  3. Jan 27, 2008 #2

    jambaugh

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    Begin with the forces that the weight of the masses places on the pully. Use the right hand rule to get the signs correct: (Make a fist with R-hand and with thumb extended. A force twisting around your thumb in the direction of your fingers creates a torque in the direction of the thumb. So if you draw the pulley on paper laying flat on your desk and with up being positive the the left mass will pull the pulley CCW which is a + torque and the right mass pulls the pulley CW which is a negative torque.

    Now you'll need to write down a bunch of equations but remember both masses accelerate at the same rate (but in opposite directions) and that will also relate to the disk. Remember the angular acceleration times the radius gives the tangential acceleration. (Again using a right hand rule to get the signs right or [itex] \vec{a} = \vec{\alpha}\times\vec{r}[/itex]).

    You should be able to relate all three accelerations using this and then as well the forces and torques. With both sets of equations you'll be able to solve for the magnitude of a.
     
    Last edited: Jan 27, 2008
  4. Jan 29, 2008 #3
    Okay... well I got 89 m/s ^2 for ang. acceleration, which doesn't seem logical to me, but it makes the tangential about 8.9, which is okay, right? So maybe I did it correctly. Thanks for all your help
     
  5. Jan 30, 2008 #4

    jambaugh

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    Here is a trick to help solve the problem. The pulley will have an effective mass equivalent to the point mass you need to place on the edge to give it the same moment of inertia:

    Note that [itex] T = r\times F = I\alpha[/itex] so
    [tex] r\times m_{eff} a = I \alpha[/tex]
    or
    [tex] m_{eff} r\times(r\times \alpha) = I \alpha[/tex]
    or with angular acceleration orthogonal to r:
    [tex] m_{eff} = I/r^2[/tex]
    In this case the effective mass of the pulley is then half its actual mass.

    Then you can treat the problem as if you had a massless pulley and the effective mass attached to the cord (but never acted on by gravity).
     
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