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Question

(A) For a body rolling down an inclined plane without slipping,the acceleration $(a)$ is given by the formula:$a = \frac{g \sin 	heta}{1 + \frac{K^2}

Vedclass · Accepted Answer

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

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(A) For a body rolling down an inclined plane without slipping,the acceleration $(a)$ 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 sphere.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 value 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 rolls down on an inclined plane of inclination $\theta$. What is the acceleration as the sphere reaches the bottom?

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