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. $\frac{m{l}^2}{6}$
• B. $m{l}^2$
• C. $\frac{3m{l}^2}{4}$
• D. $\frac{2m{l}^2}{3}$

Answer 1

The correct option is A $\frac{m{l}^2}{6}$

$M.I$ of $AB=0$

$M.I$ of $BC=\frac{m{l}^2}{3}{\sin }^{2}30$

$=\frac{m{l}^2}{4}\times \frac{1}{3}$

$M.I$ of $AC=\frac{m{l}^2}{12}$

$M.I=\frac{2m{l}^2}{12}$

$=\frac{m{l}^2}{6}$

Putting in first equation,

$\Rightarrow mg-T=m\frac{2T}{M}$

$\Rightarrow T+\frac{2Tmn}{M}=mg$

$\Rightarrow T\left(\frac{M+2m}{M}\right)=mg$

$\Rightarrow T=\frac{mgM}{M+2m}$

Hence, the answer is not given in options.