# Conceptual doubt regarding parallel axis theorem

I recently got to learn about rotational mechanics and learnt a bit about the parallel axis theorem.

It states that the moment of inertia of a rigid body about an axis YY is equal to the moment of inertia about another axis  Y’Y’ passing through the center-of-mass G of the body in a direction parallel to YY, plus the product of total mass $$M$$ of the body and square of the perpendicular distance between the two parallel axes.

However, I am unable to understand the technique used in the solution to this question:

A symmetrical square lamina of mass $$M$$ has uniform semi-circular plates attached to it on its four sides as shown in the figure. Each plate has the same mass $$M$$, and the disc of a square is equal to $$a$$. Calculate the moment of inertia of system about an axis passing through the center $$O$$, perpendicular to the plane of the lamina.

I think that the solution calculates the moment of inertia of the semi circular mass along the center of square. According to the parallel axis theorem, the moment of inertia along the center of square can be calculated if we know the moment of inertia along the center of mass of the semicircular plate, which obviously doesn't lie on the common periphery of the square and the plates. Am I right or am I missing something basic?

• What is "the disc of a square"? Is $a$ the distance of each side on the square ABCD? Oct 6 '21 at 14:49
• There is a bit of trickery here because the MMOI of pie segment (including a semi-circle) is $I = \tfrac{m}{2} R^2$ regardless of how big the pie slice is. But this is not obvious to everyone. Oct 6 '21 at 14:58
• True jAlex, but the (m) depends on the size of the slice. Also, the formula they use for the square does assume that (a) is the length of each side. Oct 6 '21 at 15:18

Your confusion is understandable; the solution incorrectly applies the parallel-axis theorem. Indeed, the moment of inertia of each semi-circular "wing" of the plate about the central point works out to be $$M a^2 \left( \frac{2}{3\pi} + \frac{3}{8} \right)$$ rather than $$\frac{3}{8} M a^2$$ as claimed in the solution. This can be shown either by direct integration over the plate, or by using Wikipedia's pages of centroid locations and second moments of area. (For a plate of uniform density, the moment of inertia about a given axis is equal to the second moment of area times the density of the plate.)

• I was able to confirm the MMOI expression. Oct 6 '21 at 15:08
• @JAlex: Note that it is stated in the problem that the central square and each of the wings have masses of $M$ apiece. Oct 6 '21 at 15:39
• Using direct integration, with a total mass of $m$ I found that $$I_O = m a^2 \left( \frac{3}{8} + \frac{1}{4 (\pi+2)} \right)$$ next I confirmed this value with CAD $\checkmark$ Oct 6 '21 at 15:54

for resources see: https://www.miniphysics.com/uy1-calculation-of-moment-of-inertia-of-uniform-rigid-rod.html and https://en.wikipedia.org/wiki/List_of_moments_of_inertia. just to confirm what you calculate.