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2.5.3.2. Common Moments of Inertia

Table of Contents

Standard Shapes and Their Rotational Inertia

The moment of inertia tells us how mass is distributed relative to an axis of rotation. In this chapter, the focus is not on how to derive the general idea, but on the most frequently used results for common objects. These standard formulas are extremely useful in mechanics because many real systems can be approximated by simple shapes.

For a given object, the moment of inertia depends on two things, the shape of the object and the location of the rotation axis. The same object can have different moments of inertia about different axes.

The moment of inertia is not determined by mass alone.
It depends on how far the mass lies from the axis. In general,
$$
I = \sum m_i r_i^2
$$
for discrete particles, or
$$
I = \int r^2 \, dm
$$
for a continuous body.

Why Standard Results Matter

Instead of calculating an integral every time, physicists often use known formulas for idealized bodies such as rods, rings, disks, and spheres. These results appear repeatedly in problems involving rolling, spinning, torque, and rotational energy.

A useful pattern is easy to notice. If more of the mass is farther from the axis, the moment of inertia is larger. A ring has a larger moment of inertia than a disk of the same mass and radius, because the ring places essentially all its mass at the outer edge.

Table of Common Moments of Inertia

The following table lists important standard results. Here, $M$ is the total mass and $R$ or $L$ represent the relevant size of the object.

ObjectAxis of rotationMoment of inertia
Point particle at distance $r$Through axis, perpendicular distance $r$$I = mr^2$
Thin rod of length $L$Through center, perpendicular to rod$I = \frac{1}{12}ML^2$
Thin rod of length $L$Through one end, perpendicular to rod$I = \frac{1}{3}ML^2$
Thin circular hoop or ring, radius $R$Through center, perpendicular to plane$I = MR^2$
Solid disk, radius $R$Through center, perpendicular to plane$I = \frac{1}{2}MR^2$
Solid cylinder, radius $R$Along central axis$I = \frac{1}{2}MR^2$
Hollow thin cylinder, radius $R$Along central axis$I = MR^2$
Solid sphere, radius $R$Through center$I = \frac{2}{5}MR^2$
Thin spherical shell, radius $R$Through center$I = \frac{2}{3}MR^2$
Rectangular plate, sides $a$ and $b$Through center, perpendicular to plate$I = \frac{1}{12}M(a^2+b^2)$

Thin Rod

A thin rod is one of the most common examples. Its moment of inertia depends strongly on where the axis is placed.

If the axis passes through the center and is perpendicular to the rod, then
$$
I = \frac{1}{12}ML^2.
$$

If the axis passes through one end and is perpendicular to the rod, then
$$
I = \frac{1}{3}ML^2.
$$

The second value is larger because, on average, more of the mass is farther from the axis.

Thin rod with two possible axes

Hoop and Ring

For a thin hoop or ring of radius $R$, all the mass is at the same distance $R$ from the center. That makes the formula especially simple:
$$
I = MR^2.
$$

This is larger than the moment of inertia of a solid disk with the same mass and radius, because none of the mass lies closer to the center.

Thin ring about its central axis

Solid Disk and Solid Cylinder

A solid disk of radius $R$ rotating about its central axis has moment of inertia
$$
I = \frac{1}{2}MR^2.
$$

A solid cylinder about its long central axis has the same expression:
$$
I = \frac{1}{2}MR^2.
$$

This happens because the mass is spread throughout the radius, not concentrated entirely at the edge.

For the same mass $M$ and radius $R$,
$$
I_{\text{ring}} = MR^2 > I_{\text{disk}} = \frac{1}{2}MR^2.
$$
Mass farther from the axis gives a larger moment of inertia.

Solid disk about central axis

Hollow Cylinder

A thin hollow cylinder, or cylindrical shell, rotating about its central axis has
$$
I = MR^2.
$$

This matches the ring result, because the mass is effectively located at the same radius from the axis.

Solid Sphere and Spherical Shell

For a solid sphere of radius $R$ about an axis through its center,
$$
I = \frac{2}{5}MR^2.
$$

For a thin spherical shell of the same mass and radius,
$$
I = \frac{2}{3}MR^2.
$$

Again, the shell has the larger value because more mass is farther from the axis.

Sphere and spherical shell

Rectangular Plate

A flat rectangular plate with sides $a$ and $b$, rotating about an axis through its center and perpendicular to its surface, has
$$
I = \frac{1}{12}M(a^2+b^2).
$$

This formula is useful for doors, panels, and rigid plates in mechanics problems.

Rectangular plate about center axis

Comparing Different Shapes

It is often helpful to compare objects with the same mass and radius.

ShapeMoment of inertia about center
Ring$MR^2$
Disk$\frac{1}{2}MR^2$
Solid sphere$\frac{2}{5}MR^2$
Thin spherical shell$\frac{2}{3}MR^2$

These formulas show how geometry matters. A ring resists rotational acceleration more than a disk of the same $M$ and $R$, and a spherical shell resists more than a solid sphere.

Choosing the Correct Formula

When using a standard moment of inertia, always check the object type and the axis.

A thin ring is not the same as a solid disk. A rod about its center is not the same as a rod about its end. A sphere is not the same as a spherical shell.

A moment of inertia formula is valid only for the specified axis.
Changing the axis changes the value of $I$.

Physical Interpretation

Common moments of inertia make the idea of rotational inertia more concrete. If an object has a larger $I$, it is harder to start or stop its rotation about that axis. This is the rotational analogue of mass in linear motion.

These standard results will later be used in rotational dynamics, rolling motion, and energy calculations, where expressions such as
$$
K_{\text{rot}} = \frac{1}{2}I\omega^2
$$
and
$$
\tau = I\alpha
$$
depend directly on the correct value of $I$.

Final Memory Guide

A few formulas are especially worth remembering because they appear very often:

Common formulas to memorize:
$$
I_{\text{rod, center}} = \frac{1}{12}ML^2
$$
$$
I_{\text{rod, end}} = \frac{1}{3}ML^2
$$
$$
I_{\text{ring}} = MR^2
$$
$$
I_{\text{disk}} = \frac{1}{2}MR^2
$$
$$
I_{\text{solid sphere}} = \frac{2}{5}MR^2
$$

If you remember the general trend, mass farther from the axis means larger moment of inertia, the formulas become much easier to understand and recall.

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