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

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.

Given:

A body rolling on an inclined plane of height h, is shown in the following figure:

Mass of the body = m

Radius of the body = R

Radius of gyration of the body = K

Translational velocity of the body = v

Height of the inclined plane = h

Acceleration due to gravity = g = 9.8 ms^-2

Total energy at the top of the plane, E₁ = mgh

Total energy at the bottom of the plane, E_T = KE_Rotational + KE_{Translational}

E_T = 1/2 Iω ² + 1/2 mv² …(i)

We know that ,

I = mK²

ω = v/R

Using in equation (i)

E_T = 1/2 mK²(v/R) ² + 1/2 mv²

E_T = 1/2 mv² (1 + K²/R²)

We know that total energy at the top will be equal to total energy at the bottom, by law of conservation of energy, so we can write,

E₁ = E_T

mgh = 1/2 mv² (1 + K²/R²)

⇒ v = 2gh/(1 + K²/R²)

Hence proved.