Question

Let $I$ be the moment of inertia of 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 and it passes through the centre of the plate, making an angle $\theta$ with $AB$. The moment of inertia of the plate about the axis $CD$ is equal to

• A. $I$
• B. $I{\sin }^2\theta$
• C. $I{\cos }^2\theta$
• D. $I{\cos }^2\left(\frac{\theta }{2}\right)$

Answer 1

Answer: (A)

$I$

Consider the figure as shown.
We see that $I_{AB}=I_{A^{\prime }{B}^{\prime }}$
Similarly using symmetry we see that $I_{CD}=I_{C^{\prime }{D}^{\prime }}$
Using perpendicular axis theorem we get $I_Z=I_{AB}+I_{A^{\prime }{B}^{\prime }}=I_{CD}+I_{C^{\prime }{D}^{\prime }}$
Thus we get $I_{AB}=I_{CD}=I$