Theorem on the Moment of Inertia

Table Of Contents

A] Perpendicular axes theorem on the moment of inertia
B] Parallel axis theorem on the moment of inertia

We need to calculate the moment of inertia of the rigid rotational body to evaluate other important quantities like work, torque, kinetic energy, power, etc. You can check this article on the moment of inertia for a better understanding of it.

The theorem on the moment of inertia helps to calculate the moment of inertia of any rigid body. They are:

Theorem on perpendicular axes
Theorem on a parallel axis

A] Perpendicular axes theorem on the moment of inertia

This theorem states that the moment of inertia of a plane lamina about a perpendicular axis to the plane lamina is equal to the sum of the moment of inertia of plane lamina under two axes on a plane right angle to each other and intersecting at a point through which a perpendicular axis passes.

In mathematical form, If ‘ $I$ ‘ is the moment of inertia of a plane lamina about a perpendicular axis and ‘ $I_x, I_y$ ‘ is the moment of inertia of the same body under two perpendicular axes on a plane then,

$I = I_x + I_y$

proof

perpendicular axes on the moment of inertia — fig: theorem on a perpendicular axis

Consider a perpendicular axis AB on a plane lamina and $I$ be the moment of inertia of a body about AB. Now consider two axes OX and OY on a plane that are at the right angle to each other in such a way that AB passes through the intersection of OX and OY as shown in the figure.

Consider a particle of mass ‘ $m$ ‘ at a distance ‘ $r$ ‘ from an AB such that it is at distance $'y'$ from OX and $'x'$ from OY.

then Moment of Inertia of a particle about an AB axis is given by:

$I = mr^2$

But

$r^2 = x^2 + y^2$

So,

$I = m(x^2 + y^2)$

$I = mx^2 + my^2$

$I = I_y + I_x$

where $I_y \text{and} I_x$ are the moment of inertia of a particle about OX and OY respectively.

Similarly, we can show the same for all particles (whole lamina).

B] Parallel axis theorem on the moment of inertia

This theorem states that the moment of inertia of a lamina about an axis perpendicular to its plane is equal to the moment of inertia about a parallel axis that passes through the center of mass of the lamina plus the product of the mass of lamina and square of perpendicular distance between two axes.

If ‘ $I$ ‘ be the moment of inertia of a plane lamina about an axis AB then:

$I = I_{CM} + MD^2$

where ‘ $I_{CM}$ ‘ is the moment of inertia of lamina about a parallel axis passing through its center of mass and ‘ $D$ ‘ is the distance between two axes.

proof:

As shown in the figure, let’s consider a particle at point ‘P’. The moment of inertia of the whole lamina about the AB axis can be written as:

$I = \sum mr^2$

Where $m$ is the mass of a particle and $r$ is the perpendicular distance from the axis of rotation.

Now consider a parallel axis to the given axis AB but passes through the CM(Center of Mass) such that it is at distance $D$ from AB and $d$ from the considered particle as shown in the figure above.

Then, we can write: $r = D + d$

So, $I = \sum mr^2$ becomes $I = \sum m(D+d)^2$

$I = \sum m\left( D^2 + 2Dd + d^2 \right)$

$I = \sum mD^2 + \sum 2mDd +\sum md^2$

$I = D^2 \sum m + 2D \sum md\ + d^2 \sum m$

Here, $\sum m = M$ and $\sum md = 0$ because the sum of moments of all particles around axis through center of mass is zero.

$I = MD^2 + Md^2$

$I = MD^2 + I_{CM}$ where $I_{CM} = Md^2$ is a moment of inertia of lamina about an axis perpendicular to its plane passing through a center of mass of lamina.

$\boxed{I = I_{CM} + MD^2}$

A] Perpendicular axes theorem on the moment of inertia

B] Parallel axis theorem on the moment of inertia

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