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(D) For an object rolling down an inclined plane,the acceleration is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$.Since the moment of inert

Vedclass · Accepted Answer

$t_{	ext{sphere}} > t_{	ext{disc}} > t_{	ext{ring}}$

Vedclass · Answer

(D) For an object rolling down an inclined plane,the acceleration is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$.Since the moment of inertia $I$ for a ring is $MR^2$,for a disc is $\frac{1}{2}MR^2$,and for a solid sphere is $\frac{2}{5}MR^2$,we have $I_{	ext{ring}} > I_{	ext{disc}} > I_{	ext{sphere}}$.Consequently,the acceleration follows the order $a_{	ext{sphere}} > a_{	ext{disc}} > a_{	ext{ring}}$.The speed at the bottom is $v = \sqrt{2ah} / \sqrt{\sin 	heta} = \sqrt{\frac{2gh}{1 + \frac{I}{MR^2}}}$,so $v_{	ext{sphere}} > v_{	ext{disc}} > v_{	ext{ring}}$.The time of descent is $t = \sqrt{\frac{2s}{a}}$,which means $t \propto \frac{1}{\sqrt{a}}$. Therefore,$t_{	ext{sphere}} Comparing these with the given options,option $D$ is the wrong statement.

$A$ ring,sphere,and disc are rolling down an inclined plane from the same height. Find the wrong statement: (where $t$ is the time of descent,$a$ is the acceleration,and $v$ is the speed at the bottom).

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