Simple Harmonic Motion

Total energy at an angle θ with the vertical

U = \frac{1}{2}mv^{2} + \frac{1}{2}Iw^{2} + mg(R - (R-r)cos\Theta )

Since total energy it conserved at any time, therefore
dUdt = 0

On differentiating and simplifying we get,

\frac{7}{5}va = -g(R-r)\sin \Theta \frac{d\Theta }{dt}

sinθ ≈ θ,
θ = xR-r
dθdt = vR-r

on substituting we get,

a = -\frac{5gx}{7(R-r)}

Therefore time period is 2Ï€√7(R-r)5g

2\pi \sqrt{\frac{7(R-r)}{5g}}