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(A) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + k^2/R^2}$, where $k$ is the radius of gyration

Vedclass · Accepted Answer

$14/15$

Vedclass · Answer

(A) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + k^2/R^2}$, 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 $k^2 = R^2/2$, which gives $k^2/R^2 = 1/2$.Thus, $a_c = \frac{g \sin 	heta}{1 + 1/2} = \frac{g \sin 	heta}{3/2} = \frac{2}{3} g \sin 	heta$.For a solid sphere, the moment of inertia $I = \frac{2}{5} MR^2$, so $k^2 = 2/5 R^2$, which gives $k^2/R^2 = 2/5$.Thus, $a_s = \frac{g \sin 	heta}{1 + 2/5} = \frac{g \sin 	heta}{7/5} = \frac{5}{7} g \sin 	heta$.The ratio of the acceleration of the cylinder to that of the sphere is $a_c/a_s = (2/3) / (5/7) = (2/3) 	imes (7/5) = 14/15$.

$A$ solid cylinder and a solid sphere having the same mass and radius roll down on the same smooth inclined plane. The ratio of the acceleration of the cylinder $(a_c)$ to that of the sphere $(a_s)$ is

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