Question

$AB$ and $CD$ are two identical rods each of length $l$ and mass $m$ joined to form a cross. The moment of inertia of these two rods about a bisector $(XY)$ of angle between the rods is

• A. $\frac{m{l}^2}{12}$
• B. $\frac{m{l}^2}{3}$
• C. $\frac{2m{l}^2}{3}$
• D. $\frac{m{l}^2}{6}$

Answer 1

Answer: (A)

Moment of inertia of rods $AB$ and $CD$ about an axis
passing through their point of intersection $O$ and perpendícular
to the plane of the rods is
$I_O=\frac{m{l}^2}{12}+\frac{m{l}^2}{12}=\frac{m{l}^2}{6}$

In figure, $XY$ and $X^{\prime }Y$ are 'two
mutually perpendicular axes passing
through $O$ lying in the plane of the two
rods. By symmetry
$I_{XY}=I_{X^{\prime }{Y}^{\prime }}=I,$ say
From the theorem of perpendicular axes
$I_{XY}+I_{X^{\prime }{Y}^{\prime }}=I_O$
$\text{ or }2I=\frac{m{l}^2}{6}\text{ or }I=\frac{m{l}^2}{12}$