Question

A disc of mass $2\text{kg}$ and radius $0.5\text{m}$ is doing translational and rolling motion on a surface as shown in the figure. If $v_{com}=4\text{m/s}$ and $\omega =1{\text{rad/s}}^2$ then, find the total kinetic energy of the disc.

• A. $15\text{J}$
• B. $16.125\text{J}$
• C. $14.252\text{J}$
• D. $15.753\text{J}$

Answer 1

The correct option is B $16.125\text{J}$


Given:
$m=2\text{kg},r=0.5\text{m},\omega =1{\text{rad/s}}^2$ and $v_{com}=4\text{m/s}$
Moment of inertia of disc about axis passing through COM and perpendicular to the plane of disc is $I=\frac{m{r}^2}{2}=\frac{2\times (0.5{)}^2}{2}=0.25{\text{kg-m}}^2$
We know that rolling is combination of translational as well as rotational motion.
Therefore, kinetic energy of the disc undergoing rolling
$=\frac{1}{2}m{v}_{com}^2+\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}\times 2\times 4^2+\frac{1}{2}\times 0.25\times 1^2=16.125\text{J}$