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Angular Acceleration from Net Torque

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data
    A long straight rod has the following:
    mass = 0.016 kg
    length = 0.16 m

    It is being pulled on by a force on its right end. The force pulls down on the rod, making an angle of 22.5 degrees with the vertical.

    The magnitude of the force is 3.2E-2 N. Give the angular acceleration of the rod at the instant shown.


    2. Relevant equations



    3. The attempt at a solution
    I chose the center of mass to be the rotation axis, I hope that's okay.

    Then the torque due to this force is cos(22.5)(L/2)F = cos(22.5)(0.16 m/2)(3.2E-2 N) = 0.00237 Nm.

    Torque = I * angular acceleration, so angular acceleration = torque/I

    I = (1/12)mL2 = 3.41E-5 kgm2

    0.00237 Nm / 3.41E-5 kgm2 = 69.29 rad/s2

    But this is wrong, the answer should be 115 rad/s2

    Can someone help?
     
  2. jcsd
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