Angular acceleration and Torque when rod reaches vertical position?

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sweetpete28
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Angular acceleration and Torque when rod reaches vertical position??

A rod (mass 1.30 kg, length .670 m) attached to a frictionless axis is released from rest in the horizontal position. What is the magnitude of the torque and angular acceleration when the rod has reached the vertical position?

I know at horizontal position (immediately after release) Torque = FR = mg(L/2)= 4.27n-m and torque = I∂ giving ∂ = 21.9 rad/s^2...but how do I solve for when it hits vertical position...? Please help!
 
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Is it 0 b/c it is through the axis of rotation?
 


Both would be 0 then, right?