Q. Four spheres of diameter $2\ a$ and mass $M$ are placed with their centres on the four corners of a square of side $b$. Then the moment of inertia of the system about an axis along one of the sides of the square is
A) $\frac{4}{5}M\ a^{2} + 2M\ b^{2}$
B) $\frac{8}{5}M\ a^{2} + 2M\ b^{2}$
C) $\frac{8}{5}M\ a^{2}$
D) $\frac{4}{5}M\ a^{2} + M\ b^{2}$
Explanation
(b) We calculate moment of inertia of the system about $AD$
[image:7_image122.emf]
Moment of inertia of each of the sphere $A$ and $D$ about
$AD = \frac{2}{5}\ Ma^{2}$
Moment of inertia of each of the sphere $B$ and $C$ about $AD$
$= \left( \frac{2}{5}Ma^{2} + Mb^{2} \right)$
Using theorem of parallel axes
$\therefore$ Total moment of inertia
$I = \left( \frac{2}{5}Ma^{2} \right) \times 2 + \left( \frac{2}{5}Ma^{2} + Mb^{2} \right) \times 2$
$= \frac{8}{5}Ma^{2} + 2Mb^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.