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Rigid Body Kinetics

  1. Oct 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A long slender rod has a mass of 0.66 kg and a length of 648.0 mm. The friction torque at the hinge is 0.41 Nm.

    What is the angular acceleration of the rod when it is 38 degrees below the horizontal?


    2. Relevant equations



    3. The attempt at a solution

    Torque = moment of inertia x angular acceleration
    Moment of inertia for a rod = (1/12)*m*(l^2)
    = (1/12)*0.66*(0.648^2) = 0.0231 Kgm^2
    Torque = Radius * Force
    = (0.648*cos(38))*(9.81*0.66) = 3.306 Nm

    Angular acceleration = Torque/Moment of inertia
    = 3.306/0.0231 = 143.12

    Which is wrong.
     

    Attached Files:

  2. jcsd
  3. Oct 12, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That's the moment of inertia about what axis?

    And when you find the net torque, be sure to include the torque due to friction.
     
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