4th Putnam 1941

Problem B5

A wheel radius r is traveling along a road without slipping with angular velocity ω > √(g/r). A particle is thrown off the rim of the wheel. Show that it can reach a maximum height above the road of (rω + g/ω)2/(2g). [Ignore air resistance.]



Suppose the pebble leaves the wheel from a point on the rim which is at an angle θ to the vertical. Its point of departure is a distance r + r cos θ above the road. Its upward velocity is r ω sin θ, so it ascends a further (r ω sin θ)2/2g. Thus the total height is r + r2ω2/2g + r cos θ - r2ω2/2g cos2θ = r + g/2ω2 + r2ω2/2g - r2ω2/2g (cos θ - g/rω2) (*). We are given that g/rω2 < 1, so (*) has a maximum when cos θ = g/rω2 and the maximum value is 1/2g (r2ω2 + 2gr + g22) = (rω + g/ω)2/2g.



4th Putnam 1941

© John Scholes
5 Mar 2002