Q. Four particles each of mass $m$ are lying symmetrically on the rim of a disc of mass $M$ and radius $R$. Moment of inertia of this system about an axis passing through one of the particles and perpendicular to plane of disc is
A) $16\ mR^{2}$
B) $(3M + 16m)\frac{R^{2}}{2}$
C) $(3M + 12m)\frac{R^{2}}{2}$
D) Zero
Explanation
(b) According to the theorem of || axes, Moment of inertia of disc about an axis passing through $K$ and $\bot$ to plane of disc,
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$= \frac{1}{2}MR^{2} + MR^{2} = \frac{3}{2}MR^{2}$
Total moment of inertia of the system
$= \frac{3}{2}MR^{2} + m(2R)^{2} + m\left( \sqrt{2}R \right)^{2} + m\left( \sqrt{2}R \right)^{2}$
$= 3(M + 16m)\frac{R^{2}}{12}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.