Question

A rolling wheel of $12\, kg$ is on an inclined plane at position $P$ and connected to a mass of $3\, kg$ 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 $P Q$ will be $\frac{1}{2} \sqrt{ xgh } m / s$. The value of $x$ is ______

Answer 1

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