Question

Three identical thin rods each of length $20cm$ and of mass $30g$ are joined in the form of an equilateral triangular frame. The moment of inertia of triangular frame about an axis perpendicular to the plane of frame and passing through its centre is:

• A. $1.5\times {10}^{-4}kg{m}^2$
• B. $3\times {10}^{-4}kg{m}^2$
• C. $6\times {10}^{-4}kg{m}^2$
• D. $1.2\times {10}^{-4}kg{m}^2$

Answer 1

Answer: (C)

$6\times {10}^{-4}kg{m}^2$

For rods of mass $m$ and length $l$,

The distance between center of the rod and axis passing through the centroid is $l/2\sqrt{3}$

Using parallel axis theorem, its MI is $m{l}^2/12+m{l}^2/12=m{l}^2/6$

So total MI of system is $3\times m{l}^2/6=m{l}^2/2$

Here, $m=30\times {10}^{-3}kg;l=0.2m$

so MI is $30\times {10}^{-3}\times {0.2}^2/2=6\times {10}^{-4}kg{m}^2$