Question

show that Kinetic energy of a rolling body is more than the kinetic energy of a sliding body.​

Answer 1

Energy of a Rolling Object

INTRODUCTION

In this experiment, we will apply the Law of Conservation of Energy to objects rolling down a ramp. As an object rolls down the incline, its gravitational potential energy is converted into both translational and rotational kinetic energy. The translational kinetic energy is

( 1 )

KEtrans = (1/2)mv2

whereas the rotational kinetic energy is

( 2 )

KErot = (1/2)Iω2

In this last equation ω is the angular velocity in radians/sec, and I is the object's moment of inertia. For objects with simple circular symmetry (e.g. spheres and cylinders) about the rotational axis, I may be written in the form:

( 3 )

I = kmr2

where m is the mass of the object and r is its radius. The geometric factor k is a constant which depends on the shape of the object:

k = 2/5 = 0.4 for a uniform solid sphere,

k = 1/2 = 0.5 for a uniform disk or solid cylinder,

k = 1 for a hoop or hollow cylinder.

If the object rolls without slipping, then the object's linear velocity and angular speed are related by

v = rω.

Substituting equation 3 and the expression for v into equation 2, we obtain:

( 4 )

KErot = (1/2)kmv2

Figure 1

Figure 1

Consider a round object rolling down a ramp as in the illustration above. Assuming no loss of energy we may write the conservation of energy equation as:

total energy at top of ramp = total energy at bottom of ramp,