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

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

Step-by-step 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=(1)/(2)mv^(2)+(1)/(2)I\omega ^(2)`

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

`I_(cylinder)=(1)/(2)mr^(2)`

Thus applying values we have

`(K.E_(rolling))/(K.E)=((1)/(2)I\omega ^(2))/((1)/(2)mv^(2)+(1)/(2)I\omega ^(2))`

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

Applying values in the ration above we obtain

`(K.E_(rolling))/(K.E)=((1)/(2)* (1)/(2)* mr^(2)* (v^(2))/(r^(2)))/((1)/(2)mv^(2)+(1)/(2)* (1)/(2)* m* r^(2)* (v^(2))/(r^(2)))\\\\\therefore (K.E_(rolling))/(K.E)=(1)/(3)`