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

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(D) For an object of mass $M$ and radius $R$ rolling down an inclined plane of angle $	heta$,the acceleration is given by $a = \frac{g \sin 	heta}{1

Vedclass · Accepted Answer

Both $(B)$ and $(C)$.

Vedclass · Answer

(D) For an object of mass $M$ and radius $R$ rolling down an inclined plane of angle $	heta$,the acceleration is given by $a = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$.For a ring,$I = MR^2$,so $a_{ring} = \frac{g \sin 	heta}{2}$. The friction force $f_{ring} = Mg \sin 	heta - Ma_{ring} = \frac{1}{2} Mg \sin 	heta$.For a solid cylinder,$I = \frac{1}{2} MR^2$,so $a_{cyl} = \frac{g \sin 	heta}{1 + 0.5} = \frac{2}{3} g \sin 	heta$. The friction force $f_{cyl} = Mg \sin 	heta - Ma_{cyl} = \frac{1}{3} Mg \sin 	heta$.Comparing the friction forces,$f_{ring} = 0.5 Mg \sin 	heta$ and $f_{cyl} = 0.33 Mg \sin 	heta$. Thus,$f_{ring} > f_{cyl}$.Also,the condition for pure rolling is $f \le \mu N = \mu Mg \cos 	heta$. Since $f_{ring} > f_{cyl}$,if the friction is sufficient for the ring to roll,it is definitely sufficient for the cylinder to roll. Therefore,both $(B)$ and $(C)$ are correct.

Column-$I$	Column-$II$
$(A)$ Ring	$(I)$ $\frac{Mg \sin \theta}{3.5}$
$(B)$ Solid sphere	$(II)$ $\frac{Mg \sin \theta}{2}$
$(C)$ Solid cylinder	$(III)$ $\frac{Mg \sin \theta}{3}$
$(D)$ Hollow cylinder	$(IV)$ $\frac{Mg \sin \theta}{2.5}$

$A$ 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 $h$. Identify the correct statement$(s)$.

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