A ring,a solid sphere,a disc,and a solid cyli | vedclass

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

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(A) The acceleration $a$ of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{k^2}{R^2}}$,where $k$ is the radius of gyration and $R$ is the radius of the body.For a given inclined plane,$a \propto \frac{1}{1 + \frac{k^2}{R^2}}$.The body with the largest value of $\frac{k^2}{R^2}$ will have the smallest acceleration and thus take the longest time to reach the bottom.Values of $\frac{k^2}{R^2}$ for different bodies:$1$. Ring: $k^2 = R^2 \implies \frac{k^2}{R^2} = 1$$2$. Disc: $k^2 = \frac{R^2}{2} \implies \frac{k^2}{R^2} = 0.5$$3$. Solid sphere: $k^2 = \frac{2}{5}R^2 \implies \frac{k^2}{R^2} = 0.4$$4$. Solid cylinder: $k^2 = \frac{R^2}{2} \implies \frac{k^2}{R^2} = 0.5$Since the ring has the largest value of $\frac{k^2}{R^2}$ $(1)$,it has the minimum acceleration and will reach the bottom last.

$A$ ring,a solid sphere,a disc,and a solid cylinder of the same radii roll down an inclined plane. Which one would reach the bottom last?

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