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(B) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$,where $I$ is the moment of in

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Solid cylinder and disc

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(B) The acceleration of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$,where $I$ is the moment of inertia about the center of mass,$M$ is the mass,and $R$ is the radius.For a solid cylinder,$I = \frac{1}{2}MR^2$,so $a = \frac{g \sin 	heta}{1 + 0.5} = \frac{2}{3} g \sin 	heta$.For a disc,$I = \frac{1}{2}MR^2$,so $a = \frac{g \sin 	heta}{1 + 0.5} = \frac{2}{3} g \sin 	heta$.Since the acceleration depends only on the factor $\frac{I}{MR^2}$ (the shape factor),bodies with the same shape factor will have the same acceleration and reach the bottom at the same time.Both the solid cylinder and the disc have $I = \frac{1}{2}MR^2$,meaning their shape factors are identical.Therefore,the solid cylinder and the disc will reach the bottom simultaneously.

$A$ solid sphere,a solid cylinder,a disc,and a ring are rolling down an inclined plane. Which of these bodies will reach the bottom simultaneously?

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