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Moment of Inertia/Acceleration of a Swinging Rod

  1. Apr 7, 2005 #1
    I've been at this problem for almost an hour and I know that I'm going in the right direction but I think that I'm using something wrong within the calculations and what not. But well...here's the problem:

    A rod of length 61.0 cm and mass 1.10 kg is suspended by two strings which are 44.0 cm long, one at each end of the rod. The string on side B is cut. Find the magnitude of the initial acceleration of end B.

    http://homework.inst.physics.ucsb.edu/msuphysicslib/Graphics/Gtype20/prob13a_1016full.gif

    What I've done so far is that I found the downward FORCE for the rod which is (1.10 kg)(9.81 m/s^2) = 10.80N. I then assumed that once the end of B is cut the TORQUE would then be T=L x F ==> T= 0.61m * 10.80N = 6.59 Nm. The moment of Inertia of this rod is I= (1/3)ML^2 and that equals 0.136 kgm^2. I then plugged the MOMENT OF INERTIA and TORQUE into the equation T = I x a(alpha) and solved for the angular acceleration which I got (alpha) = 6.59/0.136 = 48.46 rad/s^2. Since I know alpha = a/r I used the R value of .44m and tried to solve for acceleration but it is wrong.

    Help is much appreciated!
     
    Last edited: Apr 7, 2005
  2. jcsd
  3. Apr 8, 2005 #2

    Galileo

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    You can consider the gravity to act in the COM of the rod, which is the middle, so the torque is not LxF, but 1/2LxF.
     
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