Prove the 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 torque). 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

When a body rolls down an inclined plane of height h, we use the law of conservation of energy of the rolling body, i.e. there is no loss of energy due to friction. The loss of potential energy (mgh) is equal to gain in kinetic energy `((1)/(2)mv^(2))`
`mgh=(1)/(2)mv^(2)+(1)/(2)I omega^(2)`
Since `I=mk^(2) and omega=(v)/(R)`
where K= radius gyration
`=omega` angular velocity
v= linear velocity

So `mgh=(1)/(2) mv^(2)+(1)/(2) mK^(2)(v^(2))/(R^(2))`
`mgh=(1)/(2) mv^(2)(1+(K^(2))/(R^(2)))`
`v^(2)=(2gh)/((1+(K^(2))/(R^(2)))`