Question

a Hollow cylinder rolls down an inclined plane which has sufficient friction to prevent slipping. the ratio of translational to total kinetic energy is​

Answer 1

Given:

A hollow cylinder rolls down an inclined plane.

To find:

The ratio of translational energy to total kinetic energy.

Solution:

The translational kinetic energy = 1/2*mv²

Where m = mass of hollows cylinder,

v = sped of centre of mass of the hollow cylinder.

Rotational kinetic energy = 1/2*Iω²

I = mR²

ω= v/R

∴ Rotational kinetic energy = 1/2*mv²

Total kinetic energy = 1/2*mv²+ 1/2*mv² = mv²

Ratio of translational kinetic energy to total kinetic energy = (1/2*mv²)/ (mv²) = 1/2

The required ratio is 1:2.