A cylinder and a ring of same mass M and radi | vedclass

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(D) For an object rolling down an inclined plane without slipping,the mechanical energy is conserved because the work done by static friction is zero.

Vedclass · Accepted Answer

All of the above

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(D) For an object rolling down an inclined plane without slipping,the mechanical energy is conserved because the work done by static friction is zero.Using the conservation of energy: $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$.Since $v = R\omega$,we have $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I(\frac{v}{R})^2 = \frac{1}{2}Mv^2(1 + \frac{I}{MR^2})$.For a ring,$I = MR^2$,so $Mgh = \frac{1}{2}Mv^2(1 + 1) = Mv^2$,which gives $v_{ring} = \sqrt{gh}$.For a solid cylinder,$I = \frac{1}{2}MR^2$,so $Mgh = \frac{1}{2}Mv^2(1 + \frac{1}{2}) = \frac{3}{4}Mv^2$,which gives $v_{cylinder} = \sqrt{\frac{4gh}{3}} = 2\sqrt{\frac{gh}{3}}$.Thus,all statements are correct.

$A$ cylinder and a ring of same mass $M$ and radius $R$ are placed on the top of a rough inclined plane of inclination $\theta$. Both are released simultaneously from the same height $h$. When these bodies roll down to the foot of the inclined plane,then

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