Question

Three identical thin rods each of mass $m$ and length $L$ are joined together to form an equilateral triangular frame. The moment of inertia of frame about an axis perpendicular to the plane of frame and passing through a corner is:

• A. $\frac{2m{L}^2}{3}$
• B. $\frac{3m{L}^2}{2}$
• C. $\frac{4m{L}^2}{3}$
• D. $\frac{3m{L}^2}{4}$

Answer 1

Answer: (B)

$\frac{3m{L}^2}{2}$

Sum of MI of the two rods, at vertex of which the axis passes, is $2\times m{L}^2/3=2m{L}^2/3$

For the third rod, the distance between center of the rod and the axis is $d=\sqrt{3}L/2$

Using parallel axis theorem, its MI is $m{L}^2/12+3m{L}^2/4=5m{L}^2/6$

So total MI of the system is $2m{L}^2/3+5m{L}^2/6=3m{L}^2/2$