Question

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

• A. $I$
• B. $Isi{n}^2\theta$
• C. $Ico{s}^2\theta$
• D. $Ico{s}^2\left(\frac{\theta }{2}\right)$

Answer 1

The correct option is A $I$

$I_{AB}=I=\frac{1}{2}\times \frac{M{L}^2}{6}=\frac{M{L}^2}{12}$ by perpendicular axis theorem.

We have used $\frac{M{L}^2}{6}$ as the MI of the square about an axis passing through the center of the square and perpendicular to the plane of the square plate.
Since the plate is symmetric any two axes that are perpendicular to each other and are in the plane of the plate and passes through center, will give the same MI.

$\therefore I_{CD}=I_{AB}=\frac{M{L}^2}{12}=I$