Moment of Inertia

1.0What Is the Moment of Inertia?

Moment of inertia (I), also known as rotational inertia, is the rotational equivalent of mass in linear motion. It is a measure of an object's resistance to a change in its rotational motion (i.e., its resistance to angular acceleration). A larger moment of inertia means it is harder to start or stop an object from rotating.

The Moment of Inertia definition is highly dependent on two factors:

• The total mass of the object.
• The mass distribution relative to the axis of rotation.

The SI unit for moment of inertia is ${\text{m}}^2$ and its dimensional formula is $[M{L}^2]$

Moment of Inertia Formulas and Derivation

1. For a System of Point Masses

Where $r_i$ is the perpendicular distance of $M_i$ from axis of rotation 

$y{y}^{\prime }=m_1{r}_1^2+m_2{r}_2^2+m_3{r}_3^2\dots \dots \dots \dots \dots ..$

For a system of discrete particles, the Moment of Inertia formula is the sum of the products of each particle's mass (mi​) and the square of its perpendicular distance (ri​) from the axis of rotation.

$I={\sum }_{i=1}^n{m}_i{r}_i^2$

This formula is fundamental and forms the basis for more complex derivations.

2. For a Rigid Body (Continuous Mass Distribution)

For a rigid body with a continuous mass distribution, the sum is replaced by an integral.

$I=\int r^{2}dm$

The Moment of Inertia derivation for standard shapes is done by evaluating this integral over the entire body. 

For a uniform rod of mass M and length L rotating about its center, the moment of inertia is 

$\frac{1}{12}M{L}^2$

2.0Calculation of Moment of Inertia

The calculation of moment of inertia (MOI) depends on how the mass is distributed relative to the chosen axis of rotation. The general steps are:

1. Divide the body into small mass elements ((dm)).
2. Find the perpendicular distance of each mass element from the axis of rotation.
3. Apply the integration formula:
   $I=\int r^2$, dm
   where (r) = distance of the element from the axis.
4. Evaluate the integral using the body’s geometry and mass distribution.

Examples of Calculations

• Uniform Thin Rod (length (L), mass (M), axis through centre):

 $I=\frac{1}{12}M{L}^2$

• Solid Disc (radius (R), mass (M), axis through center perpendicular to plane):

$I=\frac{1}{2}M{R}^2$

• Solid Sphere (radius (R), mass (M), axis through diameter):

$I=\frac{2}{5}M{R}^2$

Theorems of Moment of Inertia

These two theorems are powerful tools that simplify the calculation of the moment of inertia for a body about a new axis, provided the moment of inertia about a known axis is given.

Parallel Axis Theorem

This theorem is applicable to any rigid body. It states that the moment of inertia (I) about any axis is equal to the moment of inertia about a parallel axis passing through the body's center of mass(I_{cm}, plus the product of the body's total mass (M) and the square of the perpendicular distance (d) between the two axes.

$I=I_{cm}+M{d}^2$

This theorem is extremely useful when calculating the moment of inertia about an axis that does not pass through the center of mass.

Perpendicular Axis Theorem

This theorem is applicable only to planar laminae (flat, 2D objects). It states that the moment of inertia about an axis perpendicular to the plane of the lamina (I_z​) is equal to the sum of the moments of inertia about two mutually perpendicular axes ($I_{x}and{I}_y$​) that lie in the plane of the lamina and intersect at the same point as the perpendicular axis.

$I_z=I_x+I_y$

All three axes must pass through the same point.

Rotational Kinetic Energy

The energy a rigid body possesses due to its rotation is called rotational kinetic energy. This is the rotational equivalent of translational kinetic energy ($KE=\frac{1}{2}m{v}^2$)

The formula for rotational kinetic energy $K{E}_{rot}$ is:

$K{E}_{rot}=\frac{1}{2}I{\omega }^2$

where:

• I is the moment of inertia.
• ω is the angular velocity.

For a body undergoing both rotational and translational motion, such as a rolling object, the total kinetic energy is the sum of the two:

$K{E}_{total}=K{E}_{trans}+K{E}_{rot}=\frac{1}{2}m{v}_{cm}^2+\frac{1}{2}{I}_{cm}{\omega }^2$

Solved Problems

Problem 1: A uniform thin rod of mass M and length L is pivoted at one end. Find its moment of inertia about the pivot.

Solution:

The moment of inertia of a rod about its center of mass is $I_{cm}=\frac{1}{12}M{L}^2$ The distance from the center of mass to the pivot (the end) is d=L/2. Using the Parallel Axis Theorem:

$I=I_{cm}+M{d}^2=\frac{1}{12}M{L}^2+M{\left(\frac{L}{2}\right)}^2=\frac{1}{12}M{L}^2+\frac{1}{4}M{L}^2=\left(\frac{1+3}{12}\right)M{L}^2=\frac{1}{3}M{L}^2$

Problem 2: The moment of inertia of a uniform circular disc about an axis passing through its center and perpendicular to its plane is $\frac{1}{2}M{R}^2$. Find the moment of inertia about a diameter.

Solution:

This is a classic application of the Perpendicular Axis Theorem. Let the diameter axes be $I_{x}and{I}_y$​. Due to symmetry, I_x​=I_y​. Let the axis perpendicular to the plane be I_z​.

According to the theorem, $I_z=I_x+I_y$

Given $I_z=\frac{1}{2}M{R}^2,and{I}_x=I_y$, we have:

$\frac{1}{2}M{R}^2=I_x+I_x=2I_x$

$I_x=\frac{1}{4}M{R}^2$

Thus, the moment of inertia about a diameter is \frac{1}{4}MR^2.

3.0Moment of Inertia of Standard Rigid Bodies

Thin Rod (about perpendicular bisector):

$I=\frac{1}{12}M{L}^2$

Thin Rod (about one end):

$I=\frac{1}{3}M{L}^2$

Ring (about central axis perpendicular to plane):

$I=M{R}^2$

Disc (about central axis perpendicular to plane):

$I=\frac{1}{2}M{R}^2$

Solid Cylinder (about its own axis): 

$I=\frac{1}{2}M{R}^2$

Hollow Cylinder (about central axis):

$I=M{R}^2$

Solid Sphere (about diameter):

$I=\frac{2}{5}M{R}^2$

Hollow Sphere (about diameter):

$I=\frac{2}{3}M{R}^2$

Rectangular Plate (about axis through center perpendicular to plane):

$I=\frac{1}{12}M(a^2+b^2)$

Where a,b = sides of rectangle

MOI of Rectangular Plate about an axis passing through centre and perpendicular to the plane of rectangular plate

Let us consider x, y and z Axisto be three mutually perpendicular axes.

The moment of inertia of the rectangular lamina about the

centroidal axis is $I_z$

The moment of inertia about x axis is $I_x=\frac{M{b}^2}{12}$

The moment of inertia about x axis is $I_y=\frac{M{L}^2}{12}$

From perpendicular axes theorem $I=I_x+I_y=\frac{M{L}^2+M{b}^2}{12}=\frac{M(L^2+b^2)}{12}$

Factors on which Moment of Inertia Depends

The Moment of Inertia depends on the following:

• Mass of the body: A body with more mass has a higher moment of inertia.
• Mass Distribution: This is the most critical factor. A body with mass concentrated farther from the axis of rotation has a greater moment of inertia than one with the same mass concentrated closer to the axis.
• Axis of Rotation: The moment of inertia changes with the choice of the axis. This is why you must always specify the axis when giving a value for moment of inertia.

Solved Examples

Example 1: Using the Parallel Axis Theorem Find the moment of inertia of a thin circular disc of mass M and radius R about an axis tangent to its edge and perpendicular to its plane.

Solution:
The moment of inertia of a disc about an axis through its center and perpendicular to its plane is $I_{cm}=\frac{1}{2}M{R}^2$. The parallel axis is tangent to the edge, so the distance d from the center of mass is the radius, R. Using the parallel axis theorem, $I=I_{cm}+M{d}^2$. 

$I=\frac{1}{2}M{R}^2+M(R{)}^2=\frac{3}{2}M{R}^2$

Example 2: Moment of Inertia of a System of Particles Four particles, each of mass m, are placed at the vertices of a square with side length a. Find the moment of inertia about an axis passing through the center and perpendicular to the plane of the square.

Solution: The distance of each particle from the center is half the diagonal, which is $r=\frac{\sqrt{2}a}{2}=\frac{a}{\sqrt{2}}$

$$\begin{gathered} I=\sum m_i{r}_i^2=m{\left(\frac{a}{\sqrt{2}}\right)}^2+m{\left(\frac{a}{\sqrt{2}}\right)}^2+m{\left(\frac{a}{\sqrt{2}}\right)}^2+m{\left(\frac{a}{\sqrt{2}}\right)}^2 \\ I=4m\left(\frac{a^2}{2}\right)=2m{a}^2 \end{gathered}$$.