A thin uniform,circular ring is rolling down | vedclass

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(C) The linear acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula:$a = \frac{g \sin 	heta}{1 + \frac{

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$g/4$

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(C) The linear acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula:$a = \frac{g \sin 	heta}{1 + \frac{K^2}{R^2}}$For a thin uniform circular ring,the moment of inertia $I = MR^2$. Since $I = MK^2$,the radius of gyration $K$ is equal to the radius $R$,so $K^2/R^2 = 1$.Given the angle of inclination $	heta = 30^\circ$,we substitute the values into the formula:$a = \frac{g \sin 30^\circ}{1 + 1}$$a = \frac{g 	imes (1/2)}{2}$$a = \frac{g}{4}$

$A$ thin uniform,circular ring is rolling down an inclined plane of inclination $30^\circ$ without slipping. Its linear acceleration along the inclined plane will be

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