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(D) For a solid sphere rolling down an inclined plane,the acceleration of the center of mass is $a_{cm} = \frac{g \sin 	heta}{1 + \frac{I}{mR^2}} = \

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Acceleration of point $C$ is along line $CA$ .

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(D) For a solid sphere rolling down an inclined plane,the acceleration of the center of mass is $a_{cm} = \frac{g \sin 	heta}{1 + \frac{I}{mR^2}} = \frac{g \sin 	heta}{1 + \frac{2}{5}} = \frac{5}{7} g \sin 	heta$. The angular acceleration is $\alpha = \frac{a_{cm}}{R} = \frac{5g \sin 	heta}{7R}$.Any point on the sphere has two components of acceleration: centripetal acceleration $\omega^2 R$ directed towards the center and tangential acceleration $R\alpha$ perpendicular to the radius.At the point of contact $C$,the acceleration is purely centripetal,directed towards the center of the sphere,which is along the line $CA$.

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