A spherical ball of mass m and radius r rolls without slipping on a rough concave surface of large radius R. It makes small oscillations about the lowest point. Find the time period.

Explanation

Let at any instant, the boby is at angular position `theta` with respect to the vertical line drow from the center of mirror , if `phi` is the angular displacement of the ball about its senter, then

Restoring tarque acting on the ball `tau = - mg sin theta xxr`
For small `theta, sin theta` `:. tau = - mg sin theta xxr`
or `tau = - mg [(r)/(R - r)] phi r = mg (r^(2)/(R - r)) (- phi)`
Angular acceleration `alpha = (mg)/(l) ((r^(2)))/(R - r)) (- phi)`
Now comparing above equation with atandard eqation of SHM, `alpha = - omega ^(2) phi, we get omega = sqrt((mg)/(l) ((r^(2))/(R - r))`
Here `l` is moment of inertia of the rolling ball about the point of constant which is `l = 7//5 m r^(2)`
`sqrt((mg)/((7)/(5) m r^(2) (r^(2)/(R - r))) = sqrt((5)/(7) (g)/((R - r))))`
and `T = (2 pi)/(omega) = 2 pi sqrt((7(R - r))/(5 g))`
Alternate method: We can that when the `CM` of the boby is shifted slightly by an angle `theta` relative to the center of curvature of the surface, the magnitude of acceleration of the `CM` of the boby can be given as
`a = (g sin theta)/(1 + (k^(2))/(r^(2))`
(because the body is accelerating down the inclined plane of instancous angle `theta`)
Subastituting `sin theta = theta = (x)/(R - r)` , we have
`a = (- gx)/((R - r) (1 + (k^(2))/(r^(2)))`
(negative sign is given because `bara` and `barx` are oppositely directed)
Comparing the above equation with `a = - omega^(2) x`, we have
`omega = sqrt(( g)/((R - r) (1 + (k^(2))/(r^(2)))`
For solid sphere `k^(2) = 2//5 r^(2)`.
Hence `omega = sqrt((5)/(7) (g)/((R - r)))`