A solid sphere of mass 4 kg and radius 28 c | vedclass

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Question

(D) For a solid sphere rolling down an inclined plane without sliding,the acceleration is given by $a_{roll} = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}

Vedclass · Accepted Answer

$4.9$

Vedclass · Answer

(D) For a solid sphere rolling down an inclined plane without sliding,the acceleration is given by $a_{roll} = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$.For a solid sphere,the moment of inertia $I = \frac{2}{5} MR^2$,so $a_{roll} = \frac{g \sin 	heta}{1 + 2/5} = \frac{5}{7} g \sin 	heta$.Given $a_{roll} = 3.5 \ m \ s^{-2}$,we have $3.5 = \frac{5}{7} g \sin 	heta$,which implies $g \sin 	heta = 3.5 	imes \frac{7}{5} = 4.9 \ m \ s^{-2}$.When the sphere slides down without rolling,there is no rotational motion,so the acceleration is simply $a_{slide} = g \sin 	heta$.Therefore,$a_{slide} = 4.9 \ m \ s^{-2}$.

$A$ solid sphere of mass $4 \ kg$ and radius $28 \ cm$ is on an inclined plane. If the acceleration of the sphere when it rolls down without sliding is $3.5 \ m \ s^{-2}$,then the acceleration of the sphere when it slides down without rolling is (in $m \ s^{-2}$)

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