----- Original Message ---- From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Thursday, August 7, 2008 5:05:05 PM Subject: [math-fun] Rolling a ball (without slipping) along a smooth curve The problem of rolling a ball (without slipping) along a smooth curve seems quite interesting in itself. PUZZLE: Suppose a unit sphere rolls on the plane, once around the unit circle about the origin. What net spatial rotation (as a 3x3 matrix) does this impart to the sphere? --Dan --------------------------------------------------------------------------------- Let a ball of unit radius roll on a circle in the xy-plane of radius r in the positive (CCW) direction starting from the point (r, 0, 0) on the positive x-axis. Approximate the circle by N segments, and let t = 2 pi/N. When rolling on the k-th segment (0 <= k <= N-1), the ball rotates about the axis (cos kt, sin kt, 0) by angle rt in the negative sense. Along the 0-th segment, the rotation matrix is R[0] = Rx(-rt), a rotation about the x-axis. To determine the rotation matrix for the k-th segment, rotate the system by -kt about the z-axis, then apply R[0], and finally rotate back by kt about z. With rotation matrices applied right-to-left, we have R[k] = Rz(kt) Rx(-rt) Rz(-kt). The overall rotation for the complete roll around the circle is R = R[N-1] R[N-2] ... R[1] R[0] = Rz[(N-1)t] Rx(-rt) Rz[-(N-1)t] Rz[(N-2)t] Rx(-rt) Rz[-(N-2)t] ... Rz(t) Rx(-rt) Rz(-t) Rz(0) Rx(-rt) Rz(-0) = (Rz(-t) Rx(-rt))^N, using (N-1)t = -t mod 2 pi. For a unit vector n and any vector v, let Kn be the matrix such that Kn v = n X v (the cross product). Then the rotation by angle p about n is given by the matrix Rn(p) = exp(p Kn). We have R = (Rz(-2 pi/N) Rx(-2 pi r/N))^N = ((1 - (2 pi/N) Kz + O(N^-2)) (1 - (2 pi r/N) Kx +O(N^-2)))^N = (1 - (2 pi/N) (Kz + r Kx) + O(N^-2))^N ---> exp(- 2 pi (Kz + r Kx)). Thus the rotation angle is 2 pi sqrt(1 + r^2) in the negative sense about the axis (1, 0, r). When r --> 0, so that the ball is much larger than the circle, the ball rocks a little bit going around the circle, and the net rotation is nearly the identity. When r = 1, the ball and circle having equal radii, the rotation angle is 2 sqrt(2) pi. Suppose the ball is rolled, not along the complete circle, but only along the arc from a to b. Then, it is clear from the above derivation, that we have R(a, b) = Rz(b) exp(- (b - a) (Kz + r Kx)) Rz(-a). If we write this as R(a, b) = Rz(b - a) Rz(a) exp(- (b - a) (Kz + r Kx)) Rz(-a) = Rz(b - a) exp(- (b - a) (Kz + r Kx')), where x' is x rotated by angle a, it is clear that the x-axis plays no special role. If the rotation instead begins at (0, r, 0) on the y-axis, then the formula is the same as that obtained by replacing Kx in the original formula by Ky. Also note that for r --> 0, R(a,b) --> Rz(b - a) Rz(- (b - a)) = 1, and so, despite the appearance of 2 pi in the formula above, the rotation never deviates much from the identity. Before ending, I will mention a nice fact that is important in the theory of rigid bodies. Every rotation about a space fixed axis commutes with every rotation about a body fixed axis. Try it for yourself to see that this is true. It is a simple puzzle to reconcile this observation with the noncommutativity of the rotation group. Gene