A solid cylinder (i) rolls down and (ii) slid | vedclass

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Question

(B) For a solid cylinder sliding down an inclined plane without friction,the acceleration is $a_{slide} = g \sin 	heta$.For a solid cylinder rolling

Vedclass · Accepted Answer

$2:3$

Vedclass · Answer

(B) For a solid cylinder sliding down an inclined plane without friction,the acceleration is $a_{slide} = g \sin 	heta$.For a solid cylinder rolling down an inclined plane without slipping,the acceleration is $a_{roll} = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$.For a solid cylinder,the moment of inertia about its central axis is $I = \frac{1}{2} MR^2$.Substituting this into the rolling acceleration formula: $a_{roll} = \frac{g \sin 	heta}{1 + \frac{1/2 MR^2}{MR^2}} = \frac{g \sin 	heta}{1 + 0.5} = \frac{g \sin 	heta}{1.5} = \frac{2}{3} g \sin 	heta$.The ratio of the acceleration of rolling to sliding is $\frac{a_{roll}}{a_{slide}} = \frac{\frac{2}{3} g \sin 	heta}{g \sin 	heta} = \frac{2}{3}$.Thus,the ratio of the accelerations is $2:3$.

$A$ solid cylinder $(i)$ rolls down and $(ii)$ slides down an inclined plane. What is the ratio of the accelerations in these conditions?

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