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 $I = Mk^2$,we have $k^2 = R^2$,which implies $\frac{k^2}{R^2} = 1$.Given the inclination angle $	heta = 30^o$,we substitute the values into the formula:$a = \frac{g \sin 30^o}{1 + 1}$Since $\sin 30^o = 1/2$,we get:$a = \frac{g(1/2)}{2} = \frac{g}{4}$

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

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