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(A) The time taken by a body to roll down an inclined plane is given by the formula $t = \sqrt{\frac{2L(1 + \frac{k^2}{R^2})}{g \sin 	heta}}$,where $

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The hollow cylinder

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(A) The time taken by a body to roll down an inclined plane is given by the formula $t = \sqrt{\frac{2L(1 + \frac{k^2}{R^2})}{g \sin 	heta}}$,where $k$ is the radius of gyration and $R$ is the radius of the object.For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $k^2 = \frac{1}{2}R^2$,which gives $\frac{k^2}{R^2} = 0.5$.For a hollow cylinder,the moment of inertia $I = MR^2$,so $k^2 = R^2$,which gives $\frac{k^2}{R^2} = 1$.Since the hollow cylinder has a larger value of $\frac{k^2}{R^2}$,it will have a larger value of $t$.Therefore,the hollow cylinder takes more time to reach the bottom.

$A$ hollow cylinder and a solid cylinder start rolling down an inclined plane. Which one will take more time to reach the bottom of the plane?

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