Question

Three identical uniform thin metal rods form the three sides of an equilateral triangle. If the moment of inertia of the system of these three rods about an axis passing through the centroid of the triangle and perpendicular to the plane of the triangle is $n$ times. The moment of inertia of one rod separately about an axis passing through the centre of the rod and perpendicular to its length, the value of $n$ is

• A. 3
• B. 6
• C. 9
• D. 12

Answer 1

Answer: (B)

Let the length of the rod be $l$.

O-centroid of the triangle.
$A{B}^2=B{D}^2+{\left(\frac{l}{2}\right)}^2$
$\Rightarrow l^2=B{D}^2+\frac{l^2}{4}$
or, $(BD{)}^2=l^2-\frac{l^2}{4}=\frac{3}{4}{l}^2$
or, $BD=\frac{\sqrt{3}}{2}\cdot l$
$\Rightarrow OD=\frac{\sqrt{3}}{2}\cdot l\times \frac{1}{3}=\frac{l}{2\sqrt{3}}$
Moment of inertia about an axis passing through centre and perpendicular to the rod, $l=\frac{m{l}^2}{12}$
Now applying parallel axes theorem, we get
Moment of inertia about centroid
$=m{\left(\frac{l}{2\sqrt{3}}\right)}^2+l$
$=\frac{m{l}^2}{4\times 3}+\frac{m{l}^2}{12}$
$=\frac{m{l}^2}{12}+\frac{m{l}^2}{12}$
$=\frac{2m{l}^2}{12}$
Moment of inertia of the system of 3 rods
$=3\times \frac{2m{l}^2}{12}=6\times \frac{m{l}^2}{12}=6l$
According to the question, $\frac{6m{l}^2}{12}=1$
$=n\cdot \left(\frac{m{l}^2}{12}\right)$
$\therefore n=6$