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Angular acceleration problem

  1. Nov 6, 2005 #1
    A rod of length 57.0 cm and mass 1.90 kg is suspended by two strings which are 41.0 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.

    I tried using torque= I* alpha
    torque= L x f= (.57)(18.62)=10.6
    I got from Newton's 2nd law, (9.8)(1.90)
    so, 10.6 = I* alpha
    I= (1/12)(mL^2)= (1/12)(1.90 * (.57)^2= .051
    so, 10.6 = .051 alpha
    alpha = 207.8 rad/s^2
    alpha= a/L
    207.8 = a / .57
    a= 118.4 m/s ^2

    This isn't right... can someone please help me?
  2. jcsd
  3. Nov 6, 2005 #2


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    Homework Helper

    First, the place that it is rotating around is at one end of the rod.
    How far from this is the rod's center-of-mass?

    Newton's 2nd law is "Sum of Forces = ma"
    Maybe you mean Newton's 4th Law "Force by gravity = m g = m GM/r^2"
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