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 PQ will be $\frac{1}{2} \sqrt{x g h} \mathrm{~m} / \mathrm{s}$The value of x is ______
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Solution
Net loss in PE = Gain in KE
$$
\begin{aligned}
& 12 g h-3 g h=\frac{1}{2} 3 v^2+\frac{1}{2} 12 v^2+\frac{1}{2}\left[12 r^2\right]\left(\frac{v}{r}\right)^2 \\
& 9 g h=\frac{1}{2}[3+12+12] v^2 \\
& v^2=\frac{2 g h}{3} \Rightarrow v=\frac{1}{2} \sqrt{\frac{8}{3} g h} \\
& x=\frac{8}{3} \approx 3
\end{aligned}
$$
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