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DEEPTO CHAKRABARTY: A pivoted rod held horizontal, parallel to the ground, and released from rest will simply fall, rotating about the pivot point. In particular, suppose we have a mass attached to one end of a pivoted rod. So here is my pivot. Here's my pivoted rod, which we'll assume is massless. And I have-- and it has a length d, and I attach a mass m to one end. If I let this go from rest, it will simply fall, rotating about the pivot point, which I'll call s. And it's easy to understand that in terms of the action of gravity which is acting downward and the resulting torque about point s. If I replace this point mass with a wheel of the same mass, so this disk is a wheel with, let's say, a radius r and the same mass m attached to a pivoted rod. The rod is massless and has length d. If I hold this horizontal, parallel to the ground, and release it from rest, it will still just fall to the ground. Not a surprising result. What is surprising is if I...

Consider a rod of mass m, and suppose I apply a force to this rod. Let's say a force like that. So we know that as a result of that applied force, the center of mass of the rod, which we can imagine is right at the center the rod, will translate with some acceleration, such that the vector f is equal to the mass of the rod times the acceleration of the center of mass. Now recall that for a rigid body, this equation will be true regardless of where on the rod I apply the force. So for example, if I draw the rod again over here, and I apply the same force, vector f, but I apply it, let's say, on the right hand side, I'll still have the same f equals ma. All I specify with f is its magnitude and direction. But for a rigid body, no matter where on the body I apply that force, the acceleration of the center of mass will be the same. Now, we know from experience, however, that the motion of the rod is different if I push it at the center at one end or at the other e...