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(A) Using the principle of conservation of mechanical energy,the potential energy at the top is converted into translational and rotational kinetic en

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$4$

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(A) Using the principle of conservation of mechanical energy,the potential energy at the top is converted into translational and rotational kinetic energy at the bottom: $Mgh = \frac{1}{2} Mv^2 + \frac{1}{2} I\omega^2$. Since the object rolls without slipping,$\omega = \frac{v}{R}$,so $Mgh = \frac{1}{2} Mv^2 (1 + \frac{k^2}{R^2})$,where $k$ is the radius of gyration. Thus,$v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$. For a ring,$I = MR^2$,so $k^2 = R^2$ and $v_{	ext{ring}} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$. For a solid sphere,$I = \frac{2}{5}MR^2$,so $k^2 = \frac{2}{5}R^2$ and $v_{	ext{sphere}} = \sqrt{\frac{2gh}{1 + 2/5}} = \sqrt{\frac{10gh}{7}}$. The ratio of velocities is $\frac{v_{	ext{ring}}}{v_{	ext{sphere}}} = \frac{\sqrt{gh}}{\sqrt{10gh/7}} = \sqrt{\frac{7}{10}}$. To match the form $\sqrt{\frac{x}{5}}$,we write $\sqrt{\frac{7}{10}} = \sqrt{\frac{3.5}{5}}$. Rounding to the nearest integer,$x = 4$.

$A$ circular ring and a solid sphere having the same radius roll down an inclined plane from rest without slipping. The ratio of their velocities when they reach the bottom of the plane is $\sqrt{\frac{x}{5}}$,where $x=$ . . . . . . .

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