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DEEPTO CHAKRABARTY: Now that we've seen how to describe a rotating vector, we can use that to analyze the motion of our gyroscope. So again, I'll draw a side view of my pivot, my rod. Here's the wheel. I'll call this point S. That's a distance d. And we'll assume that the angular velocity is in that direction, so that the spin angle velocity vector is pointing outwards in the plus r hat direction. That's the k hat direction and theta hat is into the screen. Now, again, the weight is acting downwards at the center of mass of the wheel. There's a normal force acting upwards. So the spin angular momentum with respect to point S is just equal to the moment of inertia of the disk about its center of mass times the angular speed of the spin, and that's directed in the plus r hat direction. And this is the angular momentum with respect to point S. Now, the torque with respect to point S, again, it's R cross F relative to point S. That'...

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...