A rolling wheel of 12 , kg is on an inclined | vedclass

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(D) Let $M = 12 \, kg$ be the mass of the wheel and $m = 3 \, kg$ be the hanging mass. When the wheel moves down by a vertical height $h$,the mass $m$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let $M = 12 \, kg$ be the mass of the wheel and $m = 3 \, kg$ be the hanging mass. When the wheel moves down by a vertical height $h$,the mass $m$ moves up by a vertical height $h$.Applying the principle of conservation of mechanical energy:Loss in potential energy of the wheel = Gain in potential energy of the mass $m$ + Gain in kinetic energy of the system.$Mgh = mgh + \frac{1}{2} Mv^2 + \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2$Assuming the wheel is a disc,$I = \frac{1}{2} Mr^2$ and $\omega = \frac{v}{r}$.$(M - m)gh = \frac{1}{2} Mv^2 + \frac{1}{2} (\frac{1}{2} Mr^2) (\frac{v}{r})^2 + \frac{1}{2} mv^2$$(12 - 3)gh = \frac{1}{2} (12)v^2 + \frac{1}{4} (12)v^2 + \frac{1}{2} (3)v^2$$9gh = 6v^2 + 3v^2 + 1.5v^2 = 10.5v^2$$v^2 = \frac{9gh}{10.5} = \frac{90gh}{105} = \frac{6}{7} gh$$v = \sqrt{\frac{6}{7} gh} = \frac{1}{2} \sqrt{\frac{24}{7} gh}$.Comparing with $\frac{1}{2} \sqrt{xgh}$,we get $x = \frac{24}{7} \approx 3.43$. Given the options,the closest integer value is $3$.

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