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Homework Help: Pivoting rod help

  1. Dec 1, 2003 #1
    Picture is attached.

    A rod with mass M = 1.4 kg and length L = 1.2 m is mounted on a central pivot . A metal ball of mass m = .7 kg is attached to one end of the rod. You may treat the metal ball as a point mass. The system is oriented in the vertical plane and gravity is acting. The rod initially makes an angle theta = 27 degrees with respect to the horizontal. The rod is released from rest. What is the angular acceleration of the rod immediately after it is released?
    -----------------------------------
    Ok, first I converted 27 degrees to .471 radians.

    Now, I know that Net torque =I*alpha

    or net torque = [I(rod)+I(point mass)]*alpha

    so I got it set up like this

    r-r*m2*g*sin .471=[I(rod)+I(point mass)]*alpha

    where r is 1.2m/2=.6m
    m2=mass of ball=.7kg
    m1=1.4kg
    I(rod)=1/12*mL^2=(1/12)*(1.4kg)*(1.2m)^2
    I(ball)=mr^2=.7kg*(.6m)^2


    So yea, does that look like I set it up right? I'm not to sure if I set up my net torque right, but I think I got the other half of the equation correct.
     

    Attached Files:

    Last edited: Dec 1, 2003
  2. jcsd
  3. Dec 1, 2003 #2

    Doc Al

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    Staff: Mentor

    Looks good except for the left hand side of the above, which is the torque. I think you're mixing up the angles.
     
  4. Dec 1, 2003 #3
    Ok. I re-tried it with 207 degrees, or 3.61 radians, but that new angle still doesn't give me the right answer. I get an alpha of 5.85 /sec^2.
     
  5. Dec 1, 2003 #4

    Doc Al

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    Staff: Mentor

    Not sure what you are doing. The torque is r*F*sinα, where α is the angle between the r and F vectors. But in this case: torque = r*mg*sinα = r*mg*cosθ, by the definition of θ in your diagram. Make sense?
     
  6. Dec 1, 2003 #5
    Yea, now it does. Should have said that earlier, but I'm greatfully grateful for your help. I've spent at least an hour on this, but now I'm done. :)
     
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