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

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Solid Sphere

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(C) The acceleration $a$ of a body rolling down an inclined plane is given by the formula: $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$,where $I$ is the moment of inertia,$M$ is the mass,and $R$ is the radius.For a given inclined plane,$g$ and $	heta$ are constant. Therefore,the acceleration $a$ is inversely proportional to the factor $(1 + \frac{I}{MR^2})$.The body with the smallest value of $\frac{I}{MR^2}$ will have the largest acceleration and will reach the base first.- For a ring: $I = MR^2$,so $\frac{I}{MR^2} = 1$.- For a disc: $I = \frac{1}{2}MR^2$,so $\frac{I}{MR^2} = 0.5$.- For a solid sphere: $I = \frac{2}{5}MR^2$,so $\frac{I}{MR^2} = 0.4$.- For a cylinder: $I = \frac{1}{2}MR^2$,so $\frac{I}{MR^2} = 0.5$.Comparing the values,the solid sphere has the smallest value $(0.4)$.Thus,the solid sphere has the highest acceleration and reaches the base first.

From an inclined plane,a ring,a disc,a sphere,and a cylinder of equal radius are rolled without slipping. Which one reaches the base first?

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