Question

Find the moment of inertia of 3 rods each of mass $M$ and length $L$, connected to form an equilateral triangle about $MN$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

We cannot use parallel axis or perpendicular axis theorem instead we can use over simple formula,

$I$ = $\int r^{2}dm$

Let us consider rod AC.

Let $\lambda$ = $\frac{M}{L}$

If we take a differential element of mass $dm$ and length $dx$ at a distance $x$ from $A$,

$\therefore dm$ = $\lambda dx$

$r$ = $xsin{60}^{\circ }$

$I$ = $\int r^{2}dm$ = $\underset{0}{\overset{L}{\int }}{x}^{2}si{n}^2{60}^{\circ }\lambda dx$

= $\lambda \frac{3}{4}\underset{0}{\overset{L}{\int }}{x}^{2}dx$

= $\frac{3}{4}\frac{M}{L}\frac{L^3}{3}=\frac{M{L}^2}{4}$

= $\frac{M{L}^2}{4}+\frac{M{L}^2}{4}+\frac{3}{4}M{L}^2$

= $\frac{5}{4}M{L}^2$