A cylinder is rolling down on an inclined pla | vedclass

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Question

(D) For a body rolling down an inclined plane without slipping,the acceleration $a$ is given by the formula:$a = \frac{g \sin 	heta}{1 + \frac{I_{cm}

Vedclass · Accepted Answer

$10$

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(D) For a body rolling down an inclined plane without slipping,the acceleration $a$ is given by the formula:$a = \frac{g \sin 	heta}{1 + \frac{I_{cm}}{MR^2}}$For a solid cylinder,the moment of inertia about its central axis is $I_{cm} = \frac{1}{2} MR^2$.Substituting this into the formula:$a = \frac{g \sin 	heta}{1 + \frac{\frac{1}{2} MR^2}{MR^2}} = \frac{g \sin 	heta}{1 + \frac{1}{2}} = \frac{g \sin 	heta}{\frac{3}{2}} = \frac{2}{3} g \sin 	heta$Given $g = 10 \ m/s^2$ and $	heta = 60^{\circ}$:$a = \frac{2}{3} 	imes 10 	imes \sin(60^{\circ}) = \frac{20}{3} 	imes \frac{\sqrt{3}}{2} = \frac{10 \sqrt{3}}{3} = \frac{10}{\sqrt{3}} \ m/s^2$Comparing this with the given expression $\frac{x}{\sqrt{3}} \ m/s^2$,we get $x = 10$.

$A$ cylinder is rolling down on an inclined plane of inclination $60^{\circ}$. Its acceleration during rolling down will be $\frac{x}{\sqrt{3}} \ m/s^2$,where $x=$ . . . . . . . (use $g=10 \ m/s^2$)

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