Find the trajectory of a particle which moves from rest in a vertical plane under (constant) gravity and a force kv perpendicular to its velocity v.
Take the x-axis horizontal and the y-axis vertically downwards. Then the equations of motion are: y'' = g - kx', x'' = ky'. If k = 0, then we have the usual constant acceleration downwards (x = 0, y = 1/2 gt2). So suppose k ≠ 0.
By inspection, x = A (cos kt - 1) + B (sin kt - kt) (we have chosen the terms - A and -kBt to give x(0) = x'(0) = 0). Hence (integrating x'' = ky') we have y = - A sin kt + B(cos kt - 1). But y'(0) = 0, so A = 0. Finally, to satisfy y'' = g - kx', we B = -g/k2. So x = gt/k - g/k2 sin kt, y = g/k2 (1 - cos kt). This is the equation of a cycloid. Note that the particle is never able to fall more than a finite distance, it keeps being returned to its starting level.
16th Putnam 1956
© John Scholes
4 Dec 1999