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  4. When a solid sphere rolls without slipping down an inclined plane making an angle θ with the horizontal, the acceleration of its centre of mass is a. If the same sphere slides without friction, its acceleration a' will be

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Q. When a solid sphere rolls without slipping down an inclined plane making an angle $\theta$ with the horizontal, the acceleration of its centre of mass is $a$. If the same sphere slides without friction, its acceleration $a'$ will be

System of Particles and Rotational Motion

Solution:

Acceleration of the solid sphere when it rolls without slipping down an inclined plane is
$ a= \frac{g\, sin \,\theta}{1+\frac{I}{MR^{2}}} $
For a solid sphere, $I= \frac{2}{5}MR^{2}$
$\therefore a = \frac{g \,sin \,\theta}{1+\frac{2}{5}} = \frac{5}{7} g \,sin\, \theta\quad...\left(i\right)$
Acceleration of the same sphere when it slides without friction down an same inclined plane is
$a'= g \,sin \,\theta\quad...\left(ii\right) $
Divide $\left(ii\right)$ by $\left(i\right)$, we get
$ \frac{a'}{a} = \frac{7}{5}$ or $a' = \frac{7}{5}a$