Question

I am stuck on this problem below. I think it may be 1/3 because of I, but I am unsure. Could someone explain if possible?

A solid uniform cylinder is rolling without slipping. What fraction of its kinetic energy is
rotational?

Answer 1

Here ,

let the velocity of the cycliner is v

radius of the cyclinder is R

for the angular speed , w = v/R

translational kinetic energy = 0.50 m * v^2

rotational kinetic energy = 0.50 * I * w^2

rotational kinetic energy = 0.50 * (0.50 *m * R^2) * (v/R)^2

rotational kinetic energy = 0.25 * m * R^2

fraction of energy which was rotational = (0.25 * m * v^2)/(0.50 * mv^2 + 0.25 mv^2)

fraction of energy which was rotational = 0.25/0.75

fraction of energy which was rotational = 1/3

the fraction of energy which was rotational is 1/3