Question

Three identical uniform rods each of length $1m$ and $2kg$ are arranged to from an equilateral triangle. What is the moment of inertia of the system about an axis passing through one corner and perpendicular to the plane of the triangle?

Answer 1

Mol of rod passing through the center = $\frac{1}{12}M{L}^2$

mol of rod passing through one side = $\frac{1}{3}M{L}^2$

AP = $\frac{\sqrt{3}}{2}$ L

MoI of BC through the perpendicular A

= $\frac{1}{12}M{L}^2+M(\frac{\sqrt{3}}{2}L{)}^2$

= $\frac{1}{12}M{L}^2+\frac{\sqrt{3}}{4}M{L}^2$

= $\frac{5}{6}M{L}^2$

By the principal of superposition

= $\frac{1}{3}M{L}^2+\frac{1}{3}M{L}^2+\frac{5}{6}M{L}^2$

= $\frac{9}{6}M{L}^2$

= $\frac{3}{2}M{L}^2$

radius of gyration

K = $\frac{\sqrt{3}}{2}\frac{M{L}^2}{3M}$

= $\frac{L}{\sqrt{2}}$