- #1
etotheipi
Someone that I tutor asked a simple but pretty good question today which I thought I'd share the answer to. In a tidied up form: a disc with centre at the origin and central axis parallel to a unit vector in the plane rotates with a constant angular velocity about this axis. A point on the disc revolves around this axis in a time . However, the time it takes for to revolve once around the axis is also clearly , and not so what actually is and why does a point on the disk rotate about the -axis faster than ?
By symmetry, we can restrict our attention to the first quadrant of the motion. Consider that at time the point is at polar angle about the -axis. The first infinitesimal rotation is about the -axis, and takes to . This changes this polar angle to .
Now let the point be the the orthogonal projection of onto the -axis, and consider the triangle . The second infinitesimal rotation by about the -axis takes the triangle to a triangle . For convenience define . Since the arc is of length , if is the polar angle about the -axis (which satisfies ) then the amount by which the -coordinate increases is . This can easily be related to an increment in by the relation .
Thus the second infinitesimal rotation about the -axis actually contributes a further of rotation about the -axis. Defining the angle of tilt of the disk relative to the plane as , we have and thus the total increment in the polar angle after both infinitesimal rotations is but since and further since , this gives simply , as expected. The main thing to notice is that the actual rate of rotation about the -axis is greater than , precisely because the infinitesimal rotation about the -axis also constitutes a little bit of rotation about the -axis.
Hope I didn't make any mistakes, I've only briefly checked it over . Anyway hope it's useful to someone!
By symmetry, we can restrict our attention to the first quadrant of the motion. Consider that at time
Now let the point
Thus the second infinitesimal rotation about the
Hope I didn't make any mistakes, I've only briefly checked it over . Anyway hope it's useful to someone!