Question

The moments of inertia of a thin square plate $ABCD$ of uniform thickness an axis passing through the centre $O$ and perpendicular to the plate are
(where $I_1,I_2,I_3,I_4$ are respectively, the moments of inertia about axes 1,2,3,4 where axes are in the plane of the plate)

• A. $I_1+I_2$
• B. $I_3+I_4$
• C. $I_1+I_3$
• D. $I_1+I_2+I_3+I_4$

Answer 1

Answer: (A), (B)

The correct options are
A $I_1+I_2$
B $I_3+I_4$

From the given figure it can be seen that

axis of $I_1$ and $I_2$ and that of $I_3$ and $I_4$ are perpendicular to each other.

Applying the theorem of perpendicular axis

$I=I_1+I_2..........\left(i\right)$

As it is a square of uniform thickness, the x and y axis will be symmetrical

$I_1=I_2$ and $I_3=I_4$

from (i),

$I_1=\frac{I}{2}=I_2.........\left(ii\right)$

Similarly, $I_3+I_4=I$

$⟹$$I_3=I_4=\frac{I}{2}..........\left(iii\right)$

So,

from (ii) and (iii),

$I_3=I_1$

Hence the moment ofinertia of the thin square plate ABCD of uniform thickness an axis passing through the centre O and perpendicular to the plate are $I=I_1+I_2=I_3+I_4=I_1+I_3$