66Consider the following diagram.
Let the radius of the bigger circle be R while that of the smaller circle be r. Its easily seen that ADB is a right triangle and so is AEC. Hence, we get
AD = 2r cos θ and AE = 2R cos θ.
So that
DE = AE - AD = 2(R-r) cos θ
On the other hand the acceleration along the segment DE is g cosθ. So the time taken is obtained by
12 (g cosθ) t2 = DE = 2(R-r)cosθ
from where we obtain the time
t = 2√(R-r)/g
which is independent of θ and accordingly constant for a given pair of circles.
