A solid cylinder of mass M and radius R rolls | vedclass

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(B) The acceleration $a$ of a body rolling without slipping down an inclined plane is given by the formula:$a = \frac{g \sin 	heta}{1 + \frac{k^2}{R^

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$\frac{2}{3} g \sin 	heta$

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(B) The acceleration $a$ of a body rolling without slipping down an inclined plane is given by the formula:$a = \frac{g \sin 	heta}{1 + \frac{k^2}{R^2}}$where $k$ is the radius of gyration.For a solid cylinder,the moment of inertia $I = \frac{1}{2} M R^2$.Since $I = M k^2$,we have $M k^2 = \frac{1}{2} M R^2$,which implies $k^2 = \frac{R^2}{2}$ or $\frac{k^2}{R^2} = \frac{1}{2}$.Substituting this value into the acceleration formula:$a = \frac{g \sin 	heta}{1 + \frac{1}{2}} = \frac{g \sin 	heta}{\frac{3}{2}} = \frac{2}{3} g \sin 	heta$.

$A$ solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is:

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