Parallel And Perpendicular Axis Theorem

The Parallel and Perpendicular Axis Theorems are rules that help us find how hard it is to spin an object around different lines. The Parallel Axis Theorem helps when we want to spin it around a line that is not through its centre. The Perpendicular Axis Theorem helps for flat objects. An example is figuring out how a wheel turns or how a seesaw moves.

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In this article, we will cover the concept of the moment of inertia of a rectangular plate. This topic falls under the broader category of rotational motion, which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), more than seventeen questions have been asked on this concept. It's also an important topic for NEET point of view.

Let's read this entire article to gain an in-depth understanding of the Parallel and Perpendicular Axis theorem.

What is the Parallel Axis Theorem?

The moment of inertia is equal to the sum of the moments of inertia at two mutually perpendicular axes in its plane, in accordance with the perpendicular axis theorem for an axis perpendicular to the plane. I_a is the moment of inertia about the provided axis that is perpendicular to the object's plane; I_b and I_c are the moments of inertia along two axes that are mutually perpendicular and that run along the object's plane; the three axes also meet on the object's plane.)

$I_{b b^{\prime}}=I_{a a^{\prime}}+M h^2$

$b b^{\prime}$ is an axis parallel to $a a^{\prime} \& a a^{\prime}$ an axis passing through the centre of mass.

What is the Perpendicular Axis theorem?

The MOI of a two-dimensional object around an axis going perpendicular to it is equal to the sum of its MOI about two mutually perpendicular axes located in the object's plane.

According to the previous definition, the perpendicular axis theorem can be expressed as:

$
I_z=I_x+I_y
$

(for a body in XY plane)
Where $I_z=$ moment of inertia about the $z$-axis
$I_x I_y$ : moment of inertia about the $\mathrm{x} \& \mathrm{y}$-axis in the plane of the body respectively.

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Solved Examples Based on the Parallel and Perpendicular Axis Theorem

Example 1: For the given uniform square lamina ABCD, whose centre is O,

1) $I_{A C}=\sqrt{2} I_{E F}$
2) $\sqrt{2} I_{A C}=I_{E F}$
3) $I_{A D}=3 I_{E F}$
(4) $I_{A C}=I_{E F}$

Solution:

Perpendicular Axis theorem

$
I_z=I_x+I_y
$
(for a body in $\mathrm{XY}$ plane )
- wherein
$I_z=$ moment of inertia about the $z$-axis
$I_x , I_y$ : moment of inertia about the $\mathrm{x} \& \mathrm{y}$-axis in the plane of the body respectively.

$ \begin{aligned}
& I_{E F}=M \frac{\left(a^2+b^2\right)}{12} \quad a=b \\
& I_{E F}=M \frac{\left(a^2+a^2\right)}{12}=\frac{M a^2}{6} \\
& I_Z=\frac{M a^2}{6}+\frac{M a^2}{6}=\frac{M a^2}{3}
\end{aligned}$

By the theorem of the perpendicular axis.

$
I_{A C}+I_{B D}=I_Z \Rightarrow I_{A C}=\frac{I_Z}{2}=\frac{M a^2}{6}
$

Similarly
$
\begin{aligned}
& \quad I_{E F}=\frac{I_Z}{2}=\frac{M a^2}{6} \\
& \text { Similarly } \\
& \therefore \quad I_{A C}=I_{E F}
\end{aligned}
$

Example 2: The moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane, is I_o as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the midpoints of the sides. The moment of inertia of the remaining part of the lamina about the same axis is :

1) $\frac{7}{8} I \circ$
2) $\frac{15}{16} \mathrm{I}_{\mathrm{o}}$
3) $\frac{3 \mathrm{I}_0}{4}$
4) $\frac{31 I_0}{32}$

Solution:

According to the theorem of perpendicular axes. Moment of inertia of triangle (ABC)

$
I_0=m R^2
$

Moment of inertia of cavity (DEF)
$
\begin{aligned}
& I_{D E F}=\frac{m}{4}\left(\frac{1}{2}\right)^2=\frac{m R^2}{16} \\
& I_{D E F}=\frac{I_0}{16}
\end{aligned}
$

Therefore, Remaining part
$
=I_{\text {remain }}=I_0-\frac{I_0}{16}=\frac{15 I_0}{16}
$

Example 3: Consider a thin uniform square sheet made of a rigid material. If its side is 'a', mass m and moment of inertia l about one of its diagonals, then :

1) $I>\frac{m a^2}{12}$
2) $\frac{m a^2}{24}<I<\frac{m a^2}{12}$
3) $I=\frac{m a^2}{12}$
4) $I=\frac{m a^2}{24}$

Solution:

Perpendicular Axis theorem
$
I_z=I_x+I_y
$
(for a body in $\mathrm{XY}$ plane )
wherein
$I_z=$ moment of inertia about the z-axis
$I_x , I_y$ : the moment of inertia about the $\mathrm{x} \& \mathrm{y}$-axis in the plane of the body respectively.

For a thin uniform sheet(square)
where $I_1=I_2=I_3=\frac{m a^2}{12}$

Example 4: From a uniform circular disc of radius R and mass 9 M, a small disc of radius is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through the centre of the disc is:

1) $\frac{37}{9} M R^2$
2) $4 \mathrm{MR}^2$
3) $\frac{40}{9} M R^2$
4) $10 M R^2$

Solution:

Moment of inertia for disc
$
I=\frac{1}{2} M R^2
$
wherein
About an axis perpendicular to the plane of the disc \& passing through its centre.

Perpendicular Axis theorem
$
I_z=I_x+I_y
$
(for a body in $X Y$ plane)
wherein
$I_z=$ moment of inertia about the z-axis
$I_x, I_y:$ the moment of inertia about the $\mathrm{x} \& \mathrm{y}$-axis in the plane of the body respectively.

Mass of removed part
let the mass density be $\sigma$ $9 m=\sigma \pi r^2$
mass of removed part $=\sigma \frac{\pi r^2}{3^2}=m$
$
\begin{aligned}
I & =\frac{9}{2} M R^2-\left[\frac{M\left(\frac{R}{3}\right)^2}{2}+M\left(\frac{2 R}{3}\right)^2\right] \\
& =M R^2\left[\frac{9}{2}-\frac{1}{18}-\frac{4}{9}\right] \\
I & =4 M R^2
\end{aligned}
$

Example 5: Consider a uniform square plate of side a and mass m. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

1) $\frac{2}{3} M a^2$
2) $\frac{5}{6} m a^2$
3) $\frac{1}{12} m a^2$
4) $\frac{7}{12} m a^2$

Solution:

According to the parallel axis theorem -

$
I_{b b^{\prime}}=I_{a a^{\prime}}+m R^2
$

Now,

$
i=M\left|\frac{a^2+b^2}{12}\right|
$

for a square plate it a = b

$I=\frac{M a^2}{6}$

From Theorem of Parallel axis.

$\begin{aligned}
& I=I_0+m r^2=\frac{m a^2}{6}+\frac{m a^2}{2} \\
& I=\frac{2}{3} m a^2
\end{aligned}$

Summary

According to the parallel axes theorem the moment of inertia of a body about any axis is equal to its moment of inertia about a parallel axis through its centre of mass plus the product of the mass of the body and the square of the perpendicular distance between the two parallel axis.

And moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of the moments of inertia of the lamina about any two mutually perpendicular axes in its own plane intersecting each other at the point where the perpendicular axis passes through the lamina.