Three bodies: a ring (R),a solid cylinder (C) | vedclass

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(C) The velocity $v$ of a body rolling without slipping down an inclined plane of height $h$ is given by the formula:$v = \sqrt{\frac{2gh}{1 + \frac{K

Vedclass · Accepted Answer

$v_{R} < v_{C} < v_{S}$

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(C) The velocity $v$ of a body rolling without slipping down an inclined plane of height $h$ is given by the formula:$v = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$where $K$ is the radius of gyration and $R$ is the radius of the body.For a ring,$K^2 = R^2$,so $v_{R} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$.For a solid cylinder,$K^2 = \frac{R^2}{2}$,so $v_{C} = \sqrt{\frac{2gh}{1+1/2}} = \sqrt{\frac{4gh}{3}} \approx 1.15 \sqrt{gh}$.For a solid sphere,$K^2 = \frac{2}{5}R^2$,so $v_{S} = \sqrt{\frac{2gh}{1+2/5}} = \sqrt{\frac{10gh}{7}} \approx 1.19 \sqrt{gh}$.Comparing the values,we find that $v_{R} < v_{C} < v_{S}$.

Three bodies: a ring $(R)$,a solid cylinder $(C)$,and a solid sphere $(S)$ having the same mass and same radius roll down an inclined plane without slipping. They start from rest. If $v_{R}$,$v_{C}$,and $v_{S}$ are the velocities of the respective bodies on reaching the bottom of the plane,then:

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