A slender rod of length l and mass m rotates with constant angular verlocity omega about an axis which is at an angle alpha to the rod and pases through its centre. Obtain an expression for the torque acting on the rod and get eth equation of motion of the rod once it is released from its constraint at the centre. T = (1/24)ml^2*omega^2*sin(2*alpha) Here is what I have: Just replace theta by alpha. I chose the coordinate axes to be the principal axes so the cross terms in the moment of inertia vanish. I get T = (Iyy-Izz)*omega^2*sin(2*alpha) as you can see from the attachment. Can someone help me to figure out the values of Izz and Iyy? I would really find it vey useful if I could solve this question. Thanks, James.