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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?

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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.

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# Homework Help: Pivoting rod help

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