A solid sphere `(radius = R)` rolls without slipping in a cylindrical throuh`(radius = 5R)`. Findthe time period of small oscillations.

Explanation

Let us locate solide cylinder when it is displaced from its stable equilibrium position by the small angle `theta` during its oscialltion ( figure). If `v_(c)` be the instantaneous speed of the `C.M. (C)` of the solid cylinder which is in pure rolling, then its angular velocity about its own centre `C` is
`omega = v_(c)/r...(1)`
Since `C` moves in a circle of radius `(R-r)`, the speed of `C` at the same moment can be written as
`v_(c)=dot(theta)(R-r) ..(2)`
Thus from Eqns `(1)` and `(2)`
`omega=dot(theta)((R-r))/(r) ...(3)`
As a mechanical energy of oscillation of the solide cylinder is conserved, `i.e. E=T+U=` constant
So, `(1)/(2)m v_(c)^(2)=(1)/(2)I_(c)omega^(2)+mg(R-r)(1-cos theta)=` constant
(Where `m` is the mass of solid cylinder and `I_(c)` is the moment of inertia of the solide cylinder about an axis passing through its `C.M. (C) ` and perpendicular to the plane of figure, of solide cylinder `)`
or, `(1)/(2) momega^(2)r^(2)+(1)/(2) (mr^(2))/(2)omega^(2)+mg(R_r)(1-cos theta)=` constant `(` using Eqn and `I_(c)= m r //)`
`(3)/(4)r^(2)(dot(theta))((R-r)^(2))/(r^(2))+g(R-r)(1-cos theta)=` constant, `(` using Eqn. 3 `)`
Differentiating w.r.t. time
`(3)/(4) (R-r)2 dot (theta)ddot(theta)+g sin theta dot(theta)=0`
So, `ddot (theta)=-(2g)/(3(R-r))theta`,( because for small `theta, sin theta~= theta)`
Thus `omega_(0)=sqrt((2g)/(3(R-r)))`
Hence the sought time period
`T=(2pi)/(omega_(0))=2pisqrt((3(R-r))/(2g))`