Question

Show that the rotational kinetic energy of a ball rolling over a horizontal plane is (2/7th of its total kinetic energy.

Answer 1

If a solid sphere of mass M and radius R is rolling perfectly on a rough horizontal surface, what is the percentage of the rotational kinetic energy?

Te total kinetic energy (KE) of an object is given by:

KE=12mv2+12Iω2

where I is the moment of inertia of the object and ω its angular momentum. For a solid sphere, the moment of inertia is given by:

I=25mr2

angular velocity would be derived from its radius and linear velocity:

ω=vr

so the total equation would be:

KE=12mv2+1225mr2vr2

=12mv2+210mv2

To get the percentage attributed to rotational energy, we’d divide the rotational part of the energy by the total energy:

210mv212mv2+210mv2=210mv2710mv2=27