Question

A rolling wheel of $12kg$ is on an inclined plane at position $P$ and connected to a mass of $3kg$ through a string of fixed length and pulley as shown in figure. Consider $PR$ as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom $Q$ of the inclined plane $PQ$ will be $\frac{1}{2}\sqrt{xgh}m/s$. The value of $x$ is ______

Answer 1

Answer: 3

Net loss in $PE=$ Gain in $KE$
$12gh-3gh=\frac{1}{2}3v^2+\frac{1}{2}12v^2+\frac{1}{2}\left[12r^2\right]{\left(\frac{v}{r}\right)}^2$
$9gh=\frac{1}{2}[3+12+12]v^2$
$v^2=\frac{2gh}{3}\Rightarrow v=\frac{1}{2}\sqrt{\frac{8}{3}gh}$
$x=\frac{8}{3}\simeq 3$