Question

Three identical solid spheres each of mass $M$ and radius $R$ are fixed at three corners of light equilateral triangular frame of side length $3R$ such that centre of spheres coincide with corners of the frame. Find the moment of inertia of these 3 spheres about an axis passing through any side of frame.

• A. $\frac{121M{R}^2}{5}$
• B. $\frac{159M{R}^2}{5}$
• C. $\frac{121M{R}^2}{20}$
• D. $\frac{159M{R}^2}{20}$

Answer 1

Answer: (D)

$\frac{159M{R}^2}{20}$

For the sphere not passing through the axis, the perpendicular distance from the axis is the height of the equilateral triangle of side 3R. This height is $\frac{\sqrt{3}}{2}\times 3R$. So by Parallel Axis Theorem, its MI about the axis will be given by :

$\frac{2}{5}M{R}^2+M{\left(\frac{3\sqrt{3}}{2}R\right)}^2$

Total M.I for all 3 spheres is given by :

$I_{A{A}^1}=2\left(\frac{2}{5}M{R}^2\right)+(\frac{2}{5}M{R}^2+M{\left(\frac{3\sqrt{3}}{2}R\right)}^2)$

$=\frac{6}{5}M{R}^2+\frac{27}{4}M{R}^2$ = $\frac{24+135}{20}M{R}^2$

$=\frac{159}{20}M{R}^2$