A ring takes times t_1 and t_2 to reach the b | vedclass

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(B) For sliding down an inclined plane of length $L$ and angle of inclination $	heta$,the acceleration is $a_1 = g \sin 	heta$.The time taken to sli

Vedclass · Accepted Answer

$1 : \sqrt{2}$

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(B) For sliding down an inclined plane of length $L$ and angle of inclination $	heta$,the acceleration is $a_1 = g \sin 	heta$.The time taken to slide is $t_1 = \sqrt{\frac{2L}{a_1}} = \sqrt{\frac{2L}{g \sin 	heta}}$.For rolling down the same inclined plane,the acceleration is $a_2 = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$.For a ring,the moment of inertia $I = MR^2$,so $a_2 = \frac{g \sin 	heta}{1 + 1} = \frac{g \sin 	heta}{2}$.The time taken to roll is $t_2 = \sqrt{\frac{2L}{a_2}} = \sqrt{\frac{2L 	imes 2}{g \sin 	heta}} = \sqrt{2} 	imes \sqrt{\frac{2L}{g \sin 	heta}}$.Therefore,the ratio $\frac{t_1}{t_2} = \frac{\sqrt{\frac{2L}{g \sin 	heta}}}{\sqrt{2} 	imes \sqrt{\frac{2L}{g \sin 	heta}}} = \frac{1}{\sqrt{2}}$.Thus,the ratio $t_1 : t_2 = 1 : \sqrt{2}$.

$A$ ring takes times $t_1$ and $t_2$ to reach the bottom of an inclined plane of length $L$ by sliding and rolling,respectively. What is the ratio of $t_1$ to $t_2$?

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