Pure roll condition equivalence In this video: https://www.youtube.com/watch?v=rjz0XeK0gr0&t=315s at 5 minutes, Professor Walter Lewin says that considering the pure roll condition, having the disk rotating about its center of mass is the same as having it rotate about point $P$, so that from there we can use the torque to solve the problem. How do we prove that?
 A: I'll be presenting an intuitive argument,, which I'll try to make detailed enough that a proof can be sketched out of it.
Pure rolling is basically having translation and rotation simultaneously such that the point of contact does not slip relative to the surface on which it rolls (i.e., it is stationary in our lab frame because I assume that the surface is stationary in our frame, and hence anything stationary relative to the surface is also stationary relative to us).
Now the argument that Dr. Lewin made is independent of the surface's geometry ( for instance, whether it is curved or not doesn't matter). This can be easily understood by realizing that all we care about is what are the instantaneous velocities of other particles relative to the point of contact and about the center of mass at some instant in time. Note that the point of contact will be continuously changing as time evolves. So, if we can prove in general that the rate of rotation about the point of contact at some instant is equal to the rate of rotation about the center of mass at that instant, then we are essentially done with proving the problem. So for simplicity, we can assume that the particle is traveling along a plane path.
Some facts about rotation are that it involves a change in angle with time, and this change in angle is the same for all points (except the center). So, the velocities might differ for different points, but the angle traversed will be the same. More precisely, the velocities will increase as the distance from the point increases, and that increase is linearly (Simply $v=\omega r$, and this can be easily proven). Moreover, this velocity's direction is perpendicular to the radius vector's direction drawn to that point from the origin.
With all these simple facts, imagine a cylinder is rolling on a horizontal plane. Then, we can easily see that the center of mass is moving at some velocity $V$ at some instant, and the point of contact is stationary in our frame. Interestingly, you can intuitively feel that the point that is diametrically opposed to the point of contact must be moving with the greatest speed (in fact, it moves with double the speed of the center of mass). This can be realized by noting that the cylinder is assumed to be a rigid body, and what it can basically do is translate ( that is, the collective movement of the body as a whole) or rotate about the center of mass. You can intuitively realize this by noting that if the center of mass is stationary, the only possible movement is rotation because, for a rigid body, the particles in the body can't approach each other (because then it will no longer be rigid). Therefore the best they can do is to move perpendicular to the line joining them, and if you impose that on all particles, then the only thing they can do is to rotate ( if not translating i.e, if the center of mass is stationary).
So, let the rate of rotation (or angular velocity) be $\omega$. Then, we know that if we shift to the center of the mass frame, every particle will get a velocity $-V$ since the center of mass was moving with velocity $V$ in our frame, and now it is stationary in its own frame. So, we see that the point of contact (which was stationary in the lab frame) will now have a velocity $-V$. So, we get the relation, $V=\omega R$ where R is the radius of the rolling cylinder. By this relation, we realize that the diametrically opposite point has a velocity $V$ in the COM frame and hence a velocity $2V$ in our lab frame.
Now, looking from the point of view of the point of contact P, the center of mass is moving with velocity $V$, and we realize that all the other points are also moving perpendicular to the radius vector drawn from point P to them ( It can't be any otherway since it is a rigid body). Because the centre is infact at a distance R away from the point of contact, we might suspect that the angular velocity about P is also $\omega$ and this is what turns out to be. In fact, you can prove this for any other point too, which is not outside the circle of the cylinder. This can be proved easily, and is an interesting fact. Therefore, as it turns out to be true for any such point, it is also true for the point of contact, which is more "special" because it is stationary in the lab frame.
I realized the answer became a bit long but I hope that it makes some sense.
