where α, β, γ are arbitrary constants. Substituting in (23) we find

( 1 + M0a2) ẋ = a(βz − γy),   ( 1 + M0a2) ẏ = a(γx − αz),   ( 1 + M0a2) ż = a(αy − βx).
C cC cC c

(25)

Hence αẋ + βẏ + γż = 0, or

αx + βy + γz = const.;

(26)

which shows that the centre of the rolling sphere describes a circle. If the axis of z be taken normal to the plane of this circle we have α = 0, β = 0, and

( 1 + M0a2) ẋ = −γ ay,   ( 1 + M0a2) ẏ = γ ax.
C cC c

(27)

The solution of these equations is of the type