Moment of Inertia Formulas

Why This Matters

Moment of inertia is the rotational equivalent of mass. It determines how an object responds to torques, whether that's a spinning flywheel, a swinging crane arm, or a rolling wheel. On exams, you need to select the correct formula for a given geometry, apply axis theorems to shift reference points, and understand why mass distribution matters more than total mass for rotation.

The core principle: mass farther from the rotation axis contributes more to rotational inertia. That's why a hollow cylinder resists angular acceleration more than a solid one of the same mass, and why the parallel axis theorem always adds inertia when you move away from the centroid. Don't just memorize the fractions in front of each formula. Understand what each one represents geometrically and when to apply the shortcut theorems.

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Fundamental Axis Theorems

These two theorems let you calculate moment of inertia about any axis once you know the centroidal value, saving you from integrating every time.

Parallel Axis Theorem

• $I = I_{cm} + md^2$ shifts the rotation axis parallel to the centroidal axis by distance $d$
• Always adds inertia because $md^2$ is always positive; moving away from the centroid increases rotational resistance
• Essential for composite bodies where you must reference all parts to a common axis before summing

Perpendicular Axis Theorem

• $I_z = I_x + I_y$ relates the out-of-plane moment of inertia to two in-plane axes
• Only valid for planar (2D) bodies like thin plates, discs, and laminae
• Useful as a shortcut when one in-plane inertia is known or easily found by symmetry

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Cylindrical and Circular Geometries

Rotating machinery, shafts, wheels, and pipes all fall here. The key variable is how mass is distributed radially from the central axis.

Solid Cylinder About Central Axis

• $I = \frac{1}{2}mr^2$ is the baseline formula for shafts, rollers, and solid wheels
• The coefficient of 1/2 reflects that mass is distributed from center to edge, so the average $r^2$ contribution is half the outer value
• The same formula applies to a solid disc rotating about the axis perpendicular to its face

Hollow Cylinder About Central Axis

• $I = \frac{1}{2}m(R_{outer}^2 + R_{inner}^2)$ accounts for the missing inner mass
• This gives higher inertia than a solid cylinder of the same total mass because the remaining mass sits farther from the axis
• Critical for pipes, tubes, and flywheels where you want high inertia without excessive weight
• For a thin-walled ring or hoop where $R_{outer} \approx R_{inner} = R$, this simplifies to $I = mR^2$

Circular Disc About Diameter

• $I = \frac{1}{4}mr^2$ describes rotation about an axis in the plane of the disc
• This is half the central-axis value because mass is distributed closer to an in-plane axis on average
• You can verify with the perpendicular axis theorem: $I_z = I_x + I_y$ gives $\frac{1}{2}mr^2 = 2 \times \frac{1}{4}mr^2$ ✓

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Slender Rods and Beams

Thin rods are idealized as having all mass along a line. The axis location relative to the rod's length dramatically changes the result.

Thin Rod About Center

• $I = \frac{1}{12}mL^2$ describes rotation about the centroidal axis perpendicular to the rod's length
• This is the smallest possible value for a rod because mass is symmetrically distributed about the axis
• Use this as the starting point for parallel axis calculations when the pivot is elsewhere

Thin Rod About End

• $I = \frac{1}{3}mL^2$ describes rotation about an axis through one end of the rod
• This is four times the centroidal value because you've shifted the axis by $d = L/2$
• Derived via the parallel axis theorem: $\frac{1}{12}mL^2 + m\left(\frac{L}{2}\right)^2 = \frac{1}{12}mL^2 + \frac{1}{4}mL^2 = \frac{1}{3}mL^2$

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Rectangular Plates

Flat rectangular shapes appear in structural analysis, doors, and panels. Pay attention to which dimension defines the rotation axis.

Rectangular Plate About Centroidal Axis

• $I = \frac{1}{12}m(b^2 + h^2)$ applies to rotation about the axis perpendicular to the plate's face (the $z$-axis through the center)
• Both dimensions contribute because mass extends in both the $b$ and $h$ directions from the central axis
• For rotation about an in-plane centroidal axis parallel to width $b$, use $I = \frac{1}{12}mh^2$. About the axis parallel to height $h$, use $I = \frac{1}{12}mb^2$.

Rectangular Plate About Edge

• $I = \frac{1}{3}mh^2$ describes rotation about an edge parallel to the width
• The same 1/3 factor as the rod about its end appears here because the geometry is analogous in that direction
• Derived from the parallel axis theorem: $\frac{1}{12}mh^2 + m\left(\frac{h}{2}\right)^2 = \frac{1}{3}mh^2$

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Spheres and Triangular Shapes

These geometries appear less frequently but are still fair game, especially for conceptual questions about mass distribution.

Solid Sphere About Diameter

• $I = \frac{2}{5}mr^2$ is valid for rotation about any diameter due to symmetry
• The coefficient of 2/5 is smaller than a solid cylinder's 1/2 because a sphere's three-dimensional shape concentrates more mass near the axis
• Important for rolling motion problems where both translation and rotation occur simultaneously

Hollow (Thin-Walled) Sphere About Diameter

• $I = \frac{2}{3}mr^2$ applies to a thin spherical shell
• Higher than the solid sphere because all mass sits at the outer radius

Triangular Plate About Base

• $I = \frac{1}{6}mh^2$ describes rotation about the base of a uniform triangular lamina, where $h$ is the height measured perpendicular to the base
• The centroidal axis (parallel to the base, through the centroid at $h/3$ from the base) gives $I = \frac{1}{18}mh^2$
• Less common on exams but tests your understanding of how non-uniform shape affects inertia

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Quick Reference Table

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Self-Check Questions

1. A solid cylinder and a hollow cylinder have the same mass and outer radius. Which has the greater moment of inertia about the central axis, and why?

2. You know the moment of inertia of a thin rod about its center is $\frac{1}{12}mL^2$. Using the parallel axis theorem, derive the moment of inertia about one end.

3. The perpendicular axis theorem applies to which type of bodies: 3D solids, planar laminae, or both? What's the key limitation?

4. Compare: a circular disc rotating about its central axis versus rotating about a diameter. Which has greater inertia, and by what factor?

5. A composite body is made of a rectangular plate with a rod attached at its edge. To find the total moment of inertia about the plate's opposite edge, what strategy would you use? (Hint: you'll need the parallel axis theorem for each component separately, then sum.)