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 P R as friction free surface. Th

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 P R 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 1 2 x g h m s - 1 . The value of x (rounded off to the nearest integer) is _____.

TestHub · Accepted Answer

If we take the wheel as a ring then its moment of inertia will be, $I = M r^{2}$ .

Using mechanical energy conservation,

$M g h - m g h = \frac{1}{2} m v^{2} + \frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2} \Rightarrow 12 g h - 3 g h = \frac{1}{2} 3 v^{2} + \frac{1}{2} 12 v^{2} + \frac{1}{2} 12 r^{2} \frac{v^{2}}{r^{2}}$

$\Rightarrow 9 g h = \frac{15 v^{2} + 12 v^{2}}{2}$

$\Rightarrow v = \sqrt{\frac{2}{3} g h} = \frac{1}{2} \sqrt{\frac{8}{3} g h}$

Therefore, $x = \frac{8}{3} ≃ 3$

Physics - W.E.P. Question with Solution | TestHub

Doubts & Discussion