Q. A round uniform body of radius $R$, mass $M$ and moment of inertia $I$ , rolls down (without slipping ) an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is
A) $\frac{g\ sin\ \theta}{1 + I/{MR}^{2}}$
B) $\frac{g\ sin\ \theta}{1 + {MR}^{2}/I}$
C) $\frac{g\ sin\ \theta}{1 - I/{MR}^{2}}$
D) $\frac{g\ sin\ \theta}{1 - {MR}^{2}/I}$
Explanation
(a) Assuming that no energy is used up against friction, the loss in potential energy is equal to the total gain in the kinetic energy.
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Thus, $Mgh = \frac{1}{2}I\left( \frac{v^{2}}{R^{2}} \right) + \frac{1}{2}Mv^{2}$
or $\frac{1}{2}v^{2}\left( M + \frac{I}{R^{2}} \right) = Mgh$
or $v^{2} = \frac{2Mgh}{M + \frac{I}{R^{2}}} = \frac{2gh}{1 + \frac{I}{MR^{2}}}$
If $s$ be the distance covered along the plane,
$h = s\sin\theta$
$v^{2} = \frac{2\ gs\sin\theta}{1 + \frac{I}{MR^{2}}}$
Now, $v^{2} = 2as$
$\therefore$ $2as = \frac{2\ gs\sin\theta}{1 + \frac{I}{MR^{2}}}$ or $a = \frac{g\sin\theta}{1 + \frac{I}{MR^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.