Prove that result that the velocity v of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by `v^(2)=(2gh)/((1+(k^(2))/(R^(2))))` using dynamical consideration (i.e. by consideration of forces and torques). Note k is the radius of gyration of the body about its symmetry axis, and R is the radius of the body. The body starts from rest at the top of the plane.

Explanation

Suppose a body of mass M and radius R is rolling on inclined plane of height h, length l and angle of inclination `theta` is show as in figure. The forces acting on body are shown as in figure.

Suppose a is acceleration in downward direction. Equation of motion this body
`N-Mgcostheta=0` and
`F=Ma-f`
`therefore F=Mgsintheta-f....(1)`
Torque necessary against friction force f
`tau=Rxxf`
`Ialpha=Mk^(2)((a)/(R))` where `alpha=(a)/(R )`
`therefore Rf=(Mk^(2))/(R)a`
`therefore f=(Mk^(2))/(R^(2))a`
`therefore` From eqn. (1),
`Ma=Mgsintheta-(Mk^(2))/(R^(2)).a [F=Ma]`
`(M+(Mk^(2))/(R^(2)))a=Mgsintheta`
`therefore a=(Mgsintheta)/(M((R^(2)+k^(2))/(R^(2))))`
`therefore a=(gsintheta)/((R^(2)+k^(2))/(R^(2)))=(gsintheta)/(1+(k^(2))/(R^(2)))`
Let the height of slope is h and its length is l.
`therefore` From figure `(h)/(l)=sinthetaimpliesl=(h)/(sintheta)`
The velocity at the bottom of slope is v and at maximum height is `u=0`
`therefore` In eqn. `v^(2)-u^(2)=2al,u=0 and a=(gsintheta)/(1+(k^(2))/(R^(2)))`
`v^(2)-0=2xx(gsintheta)/(1+(k^(2))/(R^(2)))xx(h)/(sintheta)`
`therefore v^(2)=(2gh)/(1+(k^(2))/(R^(2)))` is proved.