A thin circular ring first slips down a smoot | vedclass

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(C) Case $(i)$: Slipping down a smooth incline.In this case,the ring undergoes pure translational motion. The acceleration is $a_1 = g \sin 	heta$.Us

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$\frac{1}{\sqrt{2}}$

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(C) Case $(i)$: Slipping down a smooth incline.In this case,the ring undergoes pure translational motion. The acceleration is $a_1 = g \sin 	heta$.Using the equation of motion $s = ut + \frac{1}{2}at^2$ with initial velocity $u = 0$,we get $s = \frac{1}{2} (g \sin 	heta) t_1^2$,so $t_1 = \sqrt{\frac{2s}{g \sin 	heta}}$.Case $(ii)$: Rolling down a rough incline.In this case,the ring undergoes rolling motion. The acceleration is $a_2 = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$.For a thin circular ring,the moment of inertia $I = MR^2$. Substituting this,$a_2 = \frac{g \sin 	heta}{1 + 1} = \frac{g \sin 	heta}{2}$.Using the same equation of motion,$s = \frac{1}{2} a_2 t_2^2$,so $t_2 = \sqrt{\frac{2s}{a_2}} = \sqrt{\frac{4s}{g \sin 	heta}}$.Ratio of times:$\frac{t_1}{t_2} = \frac{\sqrt{\frac{2s}{g \sin 	heta}}}{\sqrt{\frac{4s}{g \sin 	heta}}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$.

$A$ thin circular ring first slips down a smooth incline then rolls down a rough incline of identical geometry from the same height. The ratio of the time taken in the two motions is ........

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