Question

A 1.2 m radius cylinder with a mass of 5.9 kg rolls without slipping down a hill which is 8.2 meters high. At the bottom of the hill, what fraction of its total kinetic energy is invested in rotational kinetic energy?

Answer 1

In this case, we use the conservation of energy:

At the initial position, we have the gravitational energy and at the bottom, we have kinetic energy (translational and rotational)

The translational kinetic energy is:

the rotational kinetic energy is:

remember that:

,

so the rotational kinetic energy is:

so the total kinetic energy is:

then, the ratio is: