Question

Three identical rods, each of mass $m$ and length $l$, form an equilateral triangle. Moment of inertia about one of the sides is

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

Answer 1

The correct option is D $\frac{m{l}^2}{2}$

Given, $m_1=m_2=m_3=m\text{kg}$
$l_1=l_2=l_3=l\text{m}$
Moment of inertia of rod $\text{AB}$ about axis along $\text{BC}$
$I_{AB}=\frac{m{l}^2}{3}{\sin }^2{60}^{\circ }$
$=\frac{m{l}^2}{4}$
(because, effective length of rod $=l\sin {60}^{\circ }$)
Similarly,
Moment of inertia of rod $\text{AC}$ about axis along $\text{BC}$
$I_{AC}=\frac{m{l}^2}{3}{\sin }^2{60}^{\circ }$
$=\frac{m{l}^2}{4}$

Moment of inertia of rod $\text{BC}$ about axis along $\text{BC}$
$I_{BC}=0$

Total moment of inertia $\left(I\right)=I_{AB}+I_{AC}+I_{BC}$
$=\frac{m{l}^2}{4}+\frac{m{l}^2}{4}+0=\frac{m{l}^2}{2}$