Q.
Two thin discs each of mass $M$ and radius $R$ are placed at either end of a rod of mass $m$, length $l$ and radius $r$. Moment of inertia of the system about an axis passing through the centre of rod and perpendicular to its length is
A) $\frac{mL^{2}}{12} + \frac{1}{4}MR^{2} + \frac{1}{4}ML^{2}$
B) $\frac{ML^{2}}{12} + \frac{1}{2}mR^{2} + \frac{1}{2}mL^{2}$
C) $\frac{1}{2}mL^{2} + \frac{MR^{2}}{2} + \frac{ML^{2}}{12}$
D) $\frac{mL^{2}}{12} + MR^{2} + \frac{1}{2}ML^{2}$
Explanation
(a) Moment of inertia of rod about the given axis $= \frac{ML^{2}}{12}$
Moment of inertia of each disc about its diameter
$= \frac{MR^{2}}{4}$
Using theorem of parallel axes, moment of inertia of each disc about the given axis
$= \frac{MR^{2}}{4} + M\left( \frac{L}{2} \right)^{2} = \frac{MR^{2}}{4} + \frac{ML^{2}}{4}$
$\therefore$ For theorem of parallel axes, moment of inertia about the given axis is
$I = \frac{mL^{2}}{12} + \left( \frac{MR^{2}}{4} + \frac{ML^{2}}{4} \right)$
$I = \frac{mL^{2}}{12} + \frac{MR^{2}}{4} + \frac{ML^{2}}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.