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(B) For an object rolling down an inclined plane without slipping,the acceleration $a$ is given by the formula:$a = \frac{g \sin 	heta}{1 + \frac{I}{

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$\frac{g \sin 	heta}{1 + \frac{I}{M R^2}}$

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(B) For an object rolling down an inclined plane without slipping,the acceleration $a$ is given by the formula:$a = \frac{g \sin 	heta}{1 + \frac{I}{M R^2}}$Here,$I$ is the moment of inertia about the central axis,$M$ is the mass,and $R$ is the radius.Since the object rolls without slipping,the torque equation $	au = I \alpha$ and the force equation $F = Ma$ lead to the derivation where the term $\frac{I}{M R^2}$ is added to the denominator of the translational acceleration $g \sin 	heta$.

$A$ spherical object of mass $M$ and radius $R$ has a moment of inertia $I$ about its axis. If this object rolls down an inclined plane of angle $\theta$ without slipping,its acceleration is given by:

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