Angular acceleration problem

  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. lightgrav

    lightgrav 1,231
    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?

    Second,
    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"
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook