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(B) The acceleration of a body rolling down an inclined plane without slipping is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$, where $I$ i

Vedclass · Accepted Answer

$2.8$

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(B) The acceleration of a body rolling down an inclined plane without slipping is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$, where $I$ is the moment of inertia about the center of mass.For a solid sphere, $I_{sphere} = \frac{2}{5} MR^2$. Thus, $a_{sphere} = \frac{g \sin 	heta}{1 + 2/5} = \frac{g \sin 	heta}{7/5} = \frac{5}{7} g \sin 	heta = 3 \,ms^{-2}$.This implies $g \sin 	heta = \frac{3 	imes 7}{5} = 4.2 \,ms^{-2}$.For a thin uniform circular disc, $I_{disc} = \frac{1}{2} MR^2$. Thus, $a_{disc} = \frac{g \sin 	heta}{1 + 1/2} = \frac{g \sin 	heta}{3/2} = \frac{2}{3} g \sin 	heta$.Substituting the value of $g \sin 	heta$, we get $a_{disc} = \frac{2}{3} 	imes 4.2 = 2.8 \,ms^{-2}$.

$A$ solid sphere and a thin uniform circular disc of same radius are rolling down an inclined plane without slipping. If the acceleration of the sphere is $3 \,ms^{-2}$, then the acceleration of the disc is (in $\,ms^{-2}$)

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