Picture is attached. A rod with mass M = 1.4 kg and length L = 1.2 m is mounted on a central pivot . A metal ball of mass m = .7 kg is attached to one end of the rod. You may treat the metal ball as a point mass. The system is oriented in the vertical plane and gravity is acting. The rod initially makes an angle theta = 27 degrees with respect to the horizontal. The rod is released from rest. What is the angular acceleration of the rod immediately after it is released? ----------------------------------- Ok, first I converted 27 degrees to .471 radians. Now, I know that Net torque =I*alpha or net torque = [I(rod)+I(point mass)]*alpha so I got it set up like this r-r*m2*g*sin .471=[I(rod)+I(point mass)]*alpha where r is 1.2m/2=.6m m2=mass of ball=.7kg m1=1.4kg I(rod)=1/12*mL^2=(1/12)*(1.4kg)*(1.2m)^2 I(ball)=mr^2=.7kg*(.6m)^2 So yea, does that look like I set it up right? I'm not to sure if I set up my net torque right, but I think I got the other half of the equation correct.