A ball of mass `m` and radius `r` rolls along a circular path of radius `R` its speed at the bottom `(theta=0^(@))` of the path is `v_(0)` find the force of the path on the ball as a function of `theta`.

Explanation

`h=(R-r)(1-costheta)`
`K=(7)/(5)((1)/(2)mv_(0)^(2))-mgh`
`because` In case of pure rolling
`K_(T)=(5)/(7)K`
`therefore(1)/(2)mv^(2)=(1)/(2)mv_(0)^(2)-(5)/(7)mgh`
`thereforev^(2)=v_(0)^(2)-(10)/(7)gh` ...(ii)
Equation of motion at angle `theta` is
`N-mgcostheta=(mv^(2))/((R-r))`
`thereforeN=mgcostheta=(m)/((R-r))(v_(0)^(2)-(10)/(7)gh)`
Subsituting value of `h` from eq. (i)
`N=mgcostheta+((m)/(R-r))`
`{v_(0)^(2)-(10)/(7)g(R-r)(1-costheta)}`
`=(mg)/(7)(17costheta-10)+(mv_(0)^(2))/(R-r))`
Force of friction,
`f=(mgsintheta)/(1+(mr^(2))/(I))` (for pure rolling to take place)
`=(mgsintheta)/(1+(5)/(2))` `(I=(2)/(5)mr^(2))`
`=(2)/(7)mgsintheta`