Question

$KE$ of rolling motion is $K\to$ Rotational $KE$ of system about $CM$ of the system

• A. $K+M{v}_{CM}^2/2$
• B. $K-M{v}_{CM}^2/2$
• C. $K+M{v}_{CM}^2$
• D. $K-M{v}_{CM}^2$

Answer 1

Answer: (A)

The kinetic energy of a rolling body is the sum of kinetic energy of translation $\left(\frac{1}{2}M{v}_{cm}^2\right)$ and kinetic energy of rotation $\left(k=\frac{1}{2}I{\omega }^2\right)$

So, total $KE$ of rolling motion $=\frac{1}{2}I{\omega }^2+\frac{1}{2}M{v}_{CM}^2$
$=K+\frac{1}{2}M{v}_{CM}^2$