A solid cylinder of radius r and mass M rests on a curved path  of radius R as  shown figure. When displaced through a small angle θ  as shown and left to itself, it  executes simple harmonic motion. Then the time period of oscillation is (Assume  that the cylinder rolls without slipping)

Restoring torque acting on a cylinder after a small displacement θ , about an axis passing through point of contact O with the curved path, τ=−Mgrsinθ, for small angles   sinθ≈θTherefore,  τ=−Mgrθ………….(1)Angular acceleration of the cylinder is  α=d2ϕdt2  From the diagram, rϕ=(R−r)θ    ϕ=(R−r)θr    Consequently,  α=d2dt2(R−r)θr   α=(R−r)rd2θdt2……………………. (2)   Moment of inertia of the cylinder about a point of contact is I=32Mr2      Using the definition of torque  τ=Iα and substituting equation (2) in it,  τ=32Mr2(R−r)rd2θdt2      Substituting equation (1) in the above relation, −Mgrθ=32Mr2(R−r)rd2θdt2     −Mgθ=32M(R−r)d2θdt2   −gθ=32(R−r)d2θdt2  This equation can be written as  0=d2θdt2+23(R−r)gθ   Where  d2θdt2=α, Therefore  −23(R−r)gθ=α   Comparing this with the equation of motion for angular simple harmonic motion,  α=−ω2θ ,  ω2=2g3(R−r) Hence angular frequency is  ω=2g3(R−r)2πT=2g3(R−r), where   ω=2πT T=2π3(R−r)2g