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.]

**Solution**

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 + r^{2}ω^{2}/2g + r cos θ - r^{2}ω^{2}/2g cos^{2}θ = r + g/2ω^{2} + r^{2}ω^{2}/2g - r^{2}ω^{2}/2g (cos θ - g/rω^{2}) (*). We are given that g/rω^{2} < 1, so (*) has a maximum when cos θ = g/rω^{2} and the maximum value is 1/2g (r^{2}ω^{2} + 2gr + g^{2}/ω^{2}) = (rω + g/ω)^{2}/2g.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002