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(A) For a body rolling down an inclined plane without slipping,the linear speed $v$ at the bottom is given by the formula: $v = \sqrt{\frac{2gh}{1 + \

Vedclass · Accepted Answer

$5.29$

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(A) For a body rolling down an inclined plane without slipping,the linear speed $v$ at the bottom is given by the formula: $v = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$.For a solid cylinder,the radius of gyration $K$ is related to the radius $R$ by $K^2 = \frac{1}{2}R^2$,so $\frac{K^2}{R^2} = \frac{1}{2}$.Given: height $h = 2.0 \ m$ and acceleration due to gravity $g = 10 \ m/s^2$.Substituting the values: $v = \sqrt{\frac{2 	imes 10 	imes 2.0}{1 + 0.5}} = \sqrt{\frac{40}{1.5}} = \sqrt{26.66} \approx 5.29 \ m/s$.

$A$ solid cylinder $30 \ cm$ in diameter at the top of an inclined plane $2.0 \ m$ high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom is .......... $m/s$.

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