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(A) The 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^

Vedclass · Accepted Answer

$ \frac{5}{7}g \sin 	heta $

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(A) The acceleration $a$ of a body rolling down an inclined plane without slipping is given by the formula:$a = \frac{g \sin 	heta}{1 + \frac{K^2}{R^2}}$where $K$ is the radius of gyration and $R$ is the radius of the body.For a solid sphere,the moment of inertia $I = \frac{2}{5}MR^2$.Since $I = MK^2$,we have $K^2 = \frac{2}{5}R^2$,which implies $\frac{K^2}{R^2} = \frac{2}{5}$.Substituting this into the acceleration formula:$a = \frac{g \sin 	heta}{1 + \frac{2}{5}} = \frac{g \sin 	heta}{\frac{7}{5}} = \frac{5}{7}g \sin 	heta$.

$A$ sphere of radius $R$ is rolling down an inclined plane whose angle of inclination is $\theta$. Its acceleration would be

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