Question

Three identical thin rods, each of mass $m$ and length $l$, are joined to form an equilateral triangular frame. Find the moment of inertia of the frame about an axis parallel to its one side and passing through the opposite vertex. Also, find its radius of gyration about the given axis.

Answer 1

Moment of inertia of each of rod $AC$ and $BC$ about the given axis $O{O}^{\prime }$ is

$I_{AC}=I_{BC}=\frac{m{l}^2}{3}{\sin }^2{60}^o=\frac{m{l}^2}{4}$

and $MI$ of rod $AB$ about the given axis $O{O}^{\prime }$ is

$I_{AB}=m{\left(\frac{l\sqrt{3}}{2}\right)}^2=\frac{3}{4}m{l}^2$

Hence, $I-I_{AC}+I_{BC}+I_{AB}=\frac{m{l}^2}{4}+\frac{m{l}^2}{4}+\frac{3}{4}m{l}^2=\frac{5}{4}m{l}^2$

Now, $I_{AC}=I_{BC}=\frac{m{l}^2}{3}{\sin }^2{60}^o=\frac{m{l}^2}{4}$

and $MI$ of rod $AB$ about the given axis $I_{AB}=m{\left(\frac{l\sqrt{3}}{2}\right)}^2=\frac{3}{4}m{l}^2$$

Hence, $I-I_{AC}+I_{BC}+I_{AB}=\frac{m{l}^2}{4}+\frac{m{l}^2}{4}+\frac{3}{4}m{l}^2=\frac{5}{4}m{l}^2$

Now, $I=M{K}^2\Rightarrow K=\sqrt{\frac{I}{M}}=\sqrt{\frac{\frac{5}{4}m{l}^2}{3m}}=l\sqrt{\frac{5}{12}}$