A rod with mass M = 1.2 kg and length L = 1.17m is mounted on a central pivot. A

Question

A rod with mass M = 1.2 kg and length L = 1.17m is mounted on a central pivot. A metal ball of mass m =0.70 kg is attached to one end of the rod. You may treat the metalball as a point mass. The system is oriented in the vertical planeand gravity is acting. The rod initially makes an angle = 31° with respect to the horizontal. The rodis released from rest. What is the angular acceleration of the rodimmediately after it is released?
= 1rad/s2    

Explanation / Answer

We know that torque acting on the rod is                        = I and also we have                      = m g x      ( where x isperpendicular distance between line of action and weight ) ==> we have                     m g x = I         ==> = [ m g x / I ]                   = [ m g ( L cos / 2)   /    ( (1 / 3) m L2)                  = [ 1.5 g cos / L ]                   = [ 1.5 * 9.8 * cos 31 / 1.17 ]                   = 10.77 rad /s2

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