A solid sphere of mass 2M and a thin hollow s | vedclass

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Question

(A) The time taken to reach the bottom of an inclined plane is given by $t = \frac{1}{\sin 	heta} \sqrt{\frac{2h}{g} \left( 1 + \frac{K^2}{R^2} igh

Vedclass · Accepted Answer

The solid sphere will reach the bottom first.

Vedclass · Answer

(A) The time taken to reach the bottom of an inclined plane is given by $t = \frac{1}{\sin 	heta} \sqrt{\frac{2h}{g} \left( 1 + \frac{K^2}{R^2} ight)}$.For a solid sphere,the radius of gyration factor is $\frac{K^2}{R^2} = \frac{2}{5} = 0.4$.For a hollow spherical shell,the radius of gyration factor is $\frac{K^2}{R^2} = \frac{2}{3} \approx 0.67$.Since the time $t$ is directly proportional to $\sqrt{1 + \frac{K^2}{R^2}}$,the object with the smaller value of $\frac{K^2}{R^2}$ will take less time to reach the bottom.Because $\left( \frac{K^2}{R^2} ight)_{	ext{solid}} < \left( \frac{K^2}{R^2} ight)_{	ext{hollow}}$,the solid sphere reaches the bottom first.

$A$ solid sphere of mass $2M$ and a thin hollow spherical shell of mass $M$ of the same radius roll down an inclined plane simultaneously. Then,

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