Question

A solid cylinder rolls without slipping along a horizontal surface, at constant speed v. Calculate the fraction of the total kinetic energy of the cylinder that arises from rotational motion.

Answer 1

Answer:

The rotational kinetic energy is 1/3 of the total kinetic energy of the cylinder.

Explanation:

Since the kinetic energy of the cylinder will be composed of 2 parts

1) Kinetic energy due to translational motion.

2) Kinetic energy due to rotational motion of the cylinder.

Thus the total kinetic energy of the cylinder equals

$K.E=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega ^{2}$

We know that for solid cylinder the moment of Inertia (I) is given by

$I_{cylinder}=\frac{1}{2}mr^{2}$

Thus applying values we have

$\frac{K.E_{rolling}}{K.E}=\frac{\frac{1}{2}I\omega ^{2}}{\frac{1}{2}mv^{2}+\frac{1}{2}I\omega ^{2}}$

Now in rolling we know that $v=r\omega$

Applying values in the ration above we obtain

$\frac{K.E_{rolling}}{K.E}=\frac{\frac{1}{2}\times \frac{1}{2}\times mr^{2}\times \frac{v^{2}}{r^{2}}}{\frac{1}{2}mv^{2}+\frac{1}{2}\times \frac{1}{2}\times m\times r^{2}\times \frac{v^{2}}{r^{2}}}\\\\\therefore \frac{K.E_{rolling}}{K.E}=\frac{1}{3}$