Question

A ring of radius $2\,m$ weights $100\,kg$ . It is having pure rolling motion on a horizontal floor so that its centre of mass has a speed of $20\,cms^{- 1}$ . If work done to stop it is $xJ$ . Then $x$ will be

Answer 1

$W=\Delta K$
$=\frac{1}{2}\left(mv\right)^{2}\left(1 + \frac{K^{2}}{R^{2}}\right)$
$=\frac{1}{2}\left(mv\right)^{2}\left(\right.1+1\left.\right)$
$=100(0.2$
$=4\,J$