Question

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 υ² = [ 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

Answer 1

The total energy of a solid body at the top of a plane is,

( TE ) 0 =mgh

Here, m is the mass of the solid body, g is the gravitational acceleration and h is the vertical height of the inclined plane.

The equation for the moment of inertia of the solid body is,

I=m k 2

Here, k is the radius of gyration of the body about its symmetrical axis.

The equation for the linear velocity of the solid body at the bottom of the plane is,

v=Rω

Here, ω is the angular speed of the solid body and R is the radius of the solid body.

Solving for ω the equation is rearranged as,.

ω= v R

The equation for the total kinetic energy of the solid body at the bottom of the plane is,

( TE ) f = 1 2 m v 2 + 1 2 I ω 2

Substituting the values of I and ω in the above equation, we get:

( TE ) f = 1 2 m v 2 + 1 2 ( m k 2 ) ( v R ) 2 = 1 2 m v 2 + 1 2 m( k 2 R 2 ) v 2 = 1 2 m v 2 ( 1+ k 2 R 2 )

The expression for the law of conservation of energy is,

( TE ) 0 = ( TE ) f

Substituting the calculated values in the above equation we get:

mgh= 1 2 m v 2 ( 1+ k 2 R 2 ) gh= 1 2 v 2 ( 1+ k 2 R 2 ) v 2 = 2gh ( 1+ k 2 R 2 )

Thus, the velocity of translation of a rolling body at the bottom of the inclined plane, v is given by the equation:

v 2 = 2gh ( 1+ k 2 / R 2 ) .