Question

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

• A. $\frac{7}{2}a$
• B. $\frac{5}{7}a$
• C. $\frac{7}{5}a$
• D. $\frac{5}{2}a$

Answer 1

Answer: (C)

$\frac{7}{5}a$

Answer 2

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 $a' = \frac{7}{5}a$