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(A) The time $t$ taken by a body rolling down an inclined plane of length $l$ and inclination $	heta$ is given by $t = \sqrt{\frac{2l(1 + K^2/R^2)}{g

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Solid cylinder

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(A) The time $t$ taken by a body rolling down an inclined plane of length $l$ and inclination $	heta$ is given by $t = \sqrt{\frac{2l(1 + K^2/R^2)}{g \sin 	heta}}$,where $K$ is the radius of gyration and $R$ is the radius of the body.For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $K^2 = \frac{1}{2}R^2$,which means $K^2/R^2 = 0.5$.For a hollow cylinder,the moment of inertia $I = MR^2$,so $K^2 = R^2$,which means $K^2/R^2 = 1$.Since the time $t$ is directly proportional to $\sqrt{1 + K^2/R^2}$,the body with the smaller $K^2/R^2$ ratio will take less time to reach the bottom.Comparing the two,the solid cylinder has a smaller $K^2/R^2$ ratio $(0.5 < 1)$,therefore,the solid cylinder will reach the bottom first.

$A$ solid cylinder and a hollow cylinder,both of the same mass and same external diameter,are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

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