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:


14156403203232.jpg


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, ET = KERotational + KETranslational


ET = 1/2 Iω 2 + 1/2 mv2 …(i)


We know that ,


I = mK2


ω = v/R


Using in equation (i)


ET = 1/2 mK2(v/R) 2 + 1/2 mv2


ET = 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 = ET


mgh = 1/2 mv2 (1 + K2/R2)


⇒ v = 2gh/(1 + K2/R2)


Hence proved.


27