Math Puzzles

Solution to The February 2008 PoW

Consider a spinning disk with radius 10 cm. A quarter is rolled along a diameter at a rate of 10cm/sec. If the disk is spinning clockwise at the rate of w = p radians per second, find the path the quarter takes relative to the disk. Assume the quarter moves along the disk along a diameter, and is not affected by the spin of the disk.

Also find the path for w = 2p, 3p, 4p, p/2, 1 and 4 (radians per second).

SOLUTION:
Picture the disk rotating clockwise. The dashed curve is the path the quarter will travel on the disk. The solid curve is the path the quarter has already traveled.
quarter roll

For w = p (pictured above), let's fix a particular t with 0 < t < 1. The quarter will be at the point (x, y) = (10-10t, 0), which is 10-10t units from the origin, (0,0). Unwinding the disk where this point came from, we see it would be on the circle with radius 10-10t, center (0,0), and an angle of q = wt = pt (counter-clockwise) from the positive x-axis. This is the point (x, y) where
x(t) = (10-10*t)*cos(p*t), y(t) = (10-10*t)*sin(p*t); and this will be good for all points such that 0 < t < 1.

Now suppose 1 < t < 2. The quarter will be at the point (x, y) = (10-10t, 0), which is 10t-10 units from the origin, (0,0). Unwinding the disk where this point came from, we see it would be on the circle with radius 10-10t, center (0,0), and an angle of q = wt = pt (counter-clockwise) from the negative x-axis, which will be at an angle of q = wt + p = pt + p (counter-clockwise) from the positive x-axis. This is the point (x, y) where
x(t) = (10*t-10)*cos(p*t + p), y(t) = (10*t-10)*sin(p*t + p); and this will be good for all points such that 1 < t < 2.

Since cos(q + p) = -cos(q) and sin(q + p) = -sin(q), this will simplify to x(t) = (10-10*t)*cos(p*t), y(t) = (10-10*t)*sin(p*t); and so this will be good for all points such that 0 < t < 2.

Similarly for other values of w:
x(t) = (10-10*t)*cos(w*t),
y(t) = (10-10*t)*sin(w*t);
0 < t < 2.

Click on a picture to see an animation: