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Acceleration of falling mass and tension in the cord

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data

    A 25 kg wheel has a radius of 40cm and turns freely on a horizontal axis. The radius of gyration is 30cm. A 1.2 kg mass hangs at the end of a cord that is wound around the rim of the wheel. This mass falls and causes the wheel to rotate. Find the acceleration of the falling mass and the tension in the cord

    2. Relevant equations

    I= Mk^2
    a= [itex]\alpha[/itex] r


    3. The attempt at a solution

    I figured out the inertia, with that being I=25*.3^2 to get 2.26.

    The downward force on the object would be mg-Ft
    The upward force is just Ft, I believe.

    But somewhere in there has to go acceleration, which is a=[itex]\alpha[/itex]*.4

    I think I need another equation to set everything to, but I don't know what to do...
     
  2. jcsd
  3. Dec 4, 2011 #2
    eqn 1.............F(net) = ma for the falling mass

    eqn 2 ...........torque = Iα for the rotating disk

    a = Rα
     
  4. Dec 4, 2011 #3
    Why do I need torque?
     
  5. Dec 4, 2011 #4
    Without the torque equation you will end up with too many unknown quantities.
     
  6. Dec 4, 2011 #5
    But torque is another unknown.

    I have

    Ft= 25*a

    and

    Torque = 2.25 * angular acceleration

    And I have a=.4*angular acceleration

    So if I solve for either of them, I still wind up with two unknowns in the equation.
     
  7. Dec 4, 2011 #6
    What is this equation?
     
  8. Dec 4, 2011 #7
    Force of the tension = mass times acceleration.

    Should it be mass of the object?

    so Ft= 1.2a

    But either way, that doesn't solve my too many variables problem
     
  9. Dec 4, 2011 #8
    F(net) = ma...............for falling mass

    But F(net) downwards = mg - Tension

    therefore

    mg - Tension = ma

    1.2 x 9.8 - tension = 1.2a

    In above we have still 2 unknows

    But we also have torque = I x ang acc and linear acc a - R x ang acc
     
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