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Homework Help: Acceleration / Torque Problem

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data

    A very thin uniform rod, 2.40 m long and of weight 135 N, has a frictionless hinge at its lower end. It starts out vertically from resting on a wall and falls, pivoting about the hinge. Just as it has rotated through an angle of 55 degrees, the acceleration of the end farthest from the hinge is what?

    2. Relevant equations

    [tex]\tau = I[/tex][tex]\alpha[/tex]
    [tex]\tau[/tex] = R X F = |R| |F| sin(angle between)

    I for thin rod,
    I = (1/3)MR2

    3. The attempt at a solution

    To find R X F I tried to find the angle between the force of gravity and the pivot. If you place both vectors at the pivot, R will be 35 degrees above the horizontal (after falling 55 degrees), and gravity will be directly down, so I believe the angle between them is 90+35 = 125 degrees? This is one thing I'm not entirely sure about.

    [tex]\tau[/tex] = 2.4M * 135N * sin(125)

    Then finding I,
    I = (1/3)(135N/9.8)*2.42

    And then doing T/I to find the angular acceleration,
    (135N * sin(125)) / ( (1/3)(135N/9.8)*2.4) ~= 9.69

    Finally, multiply that by the radius of 2.4M to get the linear acceleration. This didn't give me right answer though.
  2. jcsd
  3. Nov 18, 2009 #2


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    The interior angle between the force and rod is not correct. Draw a sketch and use some geometry. Also, where along the length of the rod should you apply the weight force acting on the rod?
  4. Nov 18, 2009 #3


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    Weight acts at the center of the rod.
    So the torque = F*R/2*sinθ
  5. Nov 18, 2009 #4
    I think the angle is actually correct, I just didn't know the weight acted on the center of mass.

    Torque = 135N * 1.2M * sin(125)

    Dividing that by the moment of inertia and them multiplying by the radius gave me the correct answer.

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  6. Nov 18, 2009 #5


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    To be technically and graphically correct in calculating the torque of the weight force about the pivot, it's T=r X F = 1.2*135*sin(55) clockwise, where r is the position vector from the cm to the pivot, F is the weight vector applied at the cm, and theta is the angle between the position vector and the force vector at the point of application of the force. You get the same answer, but be careful in drawing the sketch. The weight acts down at the cm, not at the pivot. Or its T= (perpendicuar distance from the pivot to the line of action of the force) *(force) = 1.2*sin(55)*135, clockwise, again, same result.
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