A solid cylinder of radius R and mass M rolls | vedclass

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(D) For a body rolling down an inclined plane of height $h$ without slipping,the conservation of energy gives: $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\o

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$A$ hollow cylinder of same mass and same radius

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(D) For a body rolling down an inclined plane of height $h$ without slipping,the conservation of energy gives: $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$. Since $v = R\omega$,we have $\omega = v/R$. Substituting this,$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I(v/R)^2 = \frac{1}{2}Mv^2(1 + \frac{I}{MR^2})$. For a solid cylinder,the moment of inertia $I = \frac{1}{2}MR^2$,so $Mgh = \frac{1}{2}Mv^2(1 + 1/2) = \frac{3}{4}Mv^2$,which gives $v = \sqrt{4gh/3}$. The speed $v$ depends on the factor $k^2/R^2$ where $I = Mk^2$. Specifically,$v = \sqrt{\frac{2gh}{1 + k^2/R^2}}$. For a hollow cylinder,$I = MR^2$,so $k^2/R^2 = 1$. Thus,$v_{hollow} = \sqrt{\frac{2gh}{1+1}} = \sqrt{gh}$. Since $\sqrt{gh} < \sqrt{4gh/3}$,the speed is less for a hollow cylinder. The speed is independent of mass $M$ and radius $R$ for the same geometry.

$A$ solid cylinder of radius $R$ and mass $M$ rolls down an inclined plane without slipping and reaches the bottom with a speed $v$. The speed would be less than $v$ if we use:

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