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Pivoting Rod

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data
    A rod with mass M and length L is mounted on a central pivot. A ball is attached at the end of the rod. The rod initially makes an angle [tex]q = 27^{\circ}[/tex] with the horizontal. The rod is released from rest. What is the angular acceleration immediately after the rod is released?


    2. The attempt at a solution
    M = 1.4 kg
    m = 0.7 kg
    L = 1.2 m

    [tex]\sum\tau = I\alpha[/tex]

    [tex]\alpha = \frac{\tau}{I}[/tex]

    [tex]\tau = F_{g}rcos(q) = (mg)(\frac{L}{2})cos(27^{\circ})[/tex]

    [tex]I = (\frac{1}{12})ML^{2} + (M(\frac{L}{2})^{2})[/tex]

    [tex]\alpha = \frac{(mg)(\frac{L}{2})cos(27^{\circ})}{(\frac{1}{12})ML^{2} + (M(\frac{L}{2})^{2})}[/tex]

    [tex]\alpha = \frac{(0.7kg * 9.81 m/s^{2})(\frac{1.2m}{2})cos(27^{\circ})}{(\frac{1}{12})(1.4 kg)(1.2 m)^{2} + (1.4kg(\frac{1.2 m}{2})^{2})} = \frac{3.67113 \frac{kgm^{2}}{s^{2}}}{0.672 kgm^{2}} = 5.46 \frac{rad}{s^{2}}[/tex]

    I believe I did everything correctly, however, the computer is saying I'm not. Would any of you please point out what's wrong with my solution?

    Thanks!
     
    Last edited: Nov 14, 2008
  2. jcsd
  3. Nov 14, 2008 #2

    Hootenanny

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    Welcome to Physics Forums,

    You may want to recheck your moment of inertia, particularly the masses...
     
  4. Nov 15, 2008 #3
    Thanks for the reply. Upon further inspection of the MOI, I noticed that I used the incorrect mass for the ball (M instead of m). I've corrected the issue and got the correct answer.
     
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