Question

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

• A. $K+M v_{CM}^{2} / 2$
• B. $K-M v_{C M}^{2} / 2$
• C. $K + Mv _{ CM }^{2}$
• 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 $\left(\frac{1}{2} M v_{c m}^{2}\right)$ and kinetic energy of rotation $\left( k =\frac{1}{2} I \omega^{2}\right)$

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