Question

Let I be the moment of inertia, of a uniform square plate about an axis AB that passes through its center and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the center of the plate and makes an angle $\theta$ with AB as shown in the figure. The moment of inertia of the plate about the axis CD is then equal to

• A. I
• B. I $si{n}^2\theta$
• C. I $co{s}^2\theta$
• D. I $co{s}^2\left(\theta /2\right)$

Answer 1

The correct option is A I

Axis $AB$ is perpendicular to $EF$ An axis passing through centre of square and perpendicular to the plane is $GH$

so,$I_{GH}=I_{AB}+I_{EF}$

$\Rightarrow I_{GH}=2I_{AB}=2I$$(as{I}_{AB}=I_{EF})$

By similar reason we can write

$I_{GH}=I_{CD}+I_{IJ}$

$\Rightarrow I_{GH}=2I_{CD}$

$\Rightarrow I_{CD}=\frac{2I}{2}=I$

$IJ$ is axis perpendicular to $CD$

The moment of inertia of a square plate about an axis on its plane is same about the axis which is also in same plane but perpendicular to the former.