Question

A cylinder of mass 10.0 kg rolls without slipping on a horizontal surface. At a certain instant, its center of mass has a speed of 10.0 m/s. Determine (a) the translational kinetic energy of its center of mass, (b) the rotational kinetic energy about its center of mass, and (c) its total energy.

Short answer

(a) The translational energy is 500 J.

(b) The rotational energy is 250 J.

(c) The total energy is 750 J.

Step-by-step solution

Step: Definition of Energy.

Energy is defined as the capacity to do work. It can be of various types like- potential energy, kinetic energy, thermal energy, etc.

Calculate translational kinetic energy.

(a)

Given value:

$\begin{array}{rcl}m& =& 10 \mathrm{kg}\\ \mathrm{v}& =& 10 \mathrm{m}/\mathrm{s}\end{array}$

The translational energy,

$\begin{array}{c}{K}_{\mathrm{tran}}=\frac{1}{2}m{v}^2\\ K_{\mathrm{tran}}=\frac{1}{2}\left(10\right)(10{)}^2\\ =500 \mathrm{J}\end{array}$

Calculate rotational energy.

(b)

The moment of inertia of the cylinder is,

$I=\frac{1}{2}m{r}^2$

The angular velocity and relative velocity are related as :

$\omega =\frac{v}{r}$

The rotational energy,

$K_{\mathrm{rot}}=\frac{1}{2}I{\omega }^2$

Substitute all the values in the above equation.

$\begin{array}{rcl}{K}_{\mathrm{rot}}& =& \frac{1}{2}\left(\frac{1}{2}m{r}^2\right){\left(\frac{v}{r}\right)}^2\\ & =& \frac{1}{4}m{v}^2\end{array}$

Substitute the numerical values, and we get

$\begin{array}{c}{K}_{\mathrm{rot}}=\frac{1}{4}\left(10\right)(10{)}^2\\ =250 \mathrm{J}\end{array}$

Calculate total energy.

(c)

The total energy can be calculated as:

$K=K_{\mathrm{tran}}+K_{\mathrm{rot}}$

Substitute the numerical values, and we get

$\begin{array}{c}K=500+250\\ =750 \mathrm{J}\end{array}$

So, the total energy is 750 J.