Rotational inertia (also called moment of inertia) measures how much an object resists changes in its spinning motion, just as regular mass measures resistance to straight-line acceleration. The core formula is simple: for a point mass, rotational inertia equals the mass multiplied by the square of its distance from the axis of rotation (I = mr²). The SI unit is kg·m². For real objects with distributed mass, finding rotational inertia requires building on that foundation with summation, integration, or a few powerful shortcuts.
The Basic Formula for Point Masses
Every rotational inertia calculation starts from one idea: each tiny bit of mass contributes to the total based on how far it sits from the rotation axis. For a single point mass, the formula is:
I = m · r²
where m is the mass and r is the perpendicular distance from the axis of rotation. If you have a system of several point masses (like balls on the ends of a rod, or planets orbiting a star), you sum each contribution:
I = m₁r₁² + m₂r₂² + m₃r₃² + …
Notice that distance is squared. That means a mass twice as far from the axis contributes four times as much to the rotational inertia. This is why figure skaters spin faster when they pull their arms in: they’re reducing r for much of their body mass, which drops I dramatically.
Continuous Objects: Using Integration
Real objects aren’t collections of point masses. A solid disk, a rod, or a sphere has mass spread continuously throughout its volume. To handle this, you replace the summation with an integral:
I = ∫ r² dm
Here, dm is an infinitesimally small piece of mass, and r is its distance from the rotation axis. The practical challenge is expressing dm in terms of a geometric variable you can actually integrate over. For a uniform object, you relate dm to the object’s density and a small element of length, area, or volume.
For example, to find the rotational inertia of a thin uniform rod of mass M and length L about an axis through its center, you’d set up coordinates along the rod’s length, express dm as (M/L)dx, and integrate x² · (M/L)dx from −L/2 to L/2. The result is (1/12)ML². If the axis is at one end instead, the same process yields (1/3)ML², a larger value because more mass is farther from the axis.
These integrals have been solved for every common shape. Rather than re-derive them each time, most textbooks and reference sheets provide a table of standard results: solid cylinder about its central axis is (1/2)MR², solid sphere is (2/5)MR², thin-walled hollow sphere is (2/3)MR², and so on. Memorizing or having quick access to these saves enormous time on homework and exams.
The Parallel Axis Theorem
Standard tables give rotational inertia about an axis through the center of mass. But what if your object rotates about a different, parallel axis? That’s where the parallel axis theorem comes in:
I = I_cm + md²
I_cm is the known rotational inertia about the center of mass, m is the total mass of the object, and d is the perpendicular distance between the center-of-mass axis and the new parallel axis. This only works when the two axes are parallel to each other, and one of them passes through the center of mass.
A quick example: a solid sphere of mass M and radius R has I_cm = (2/5)MR² about an axis through its center. If it instead rotates about an axis tangent to its surface (so d = R), the rotational inertia becomes (2/5)MR² + MR² = (7/5)MR². The md² term always adds to the center-of-mass value, so rotational inertia is always smallest about the center of mass.
The Perpendicular Axis Theorem
This theorem applies only to flat (planar) objects. If you know the rotational inertia about two perpendicular axes that lie in the plane of the object and intersect at the same point, the rotational inertia about a third axis perpendicular to the plane through that same point is simply their sum:
I_z = I_x + I_y
This is especially handy for objects like flat plates, rings, and disks. If you already know I_x and I_y from a table or a previous calculation, you get I_z for free without doing another integral.
Composite Bodies: Adding and Subtracting Parts
Many real objects aren’t simple geometric shapes. A wheel with a hub, a bracket with bolt holes, or an L-shaped beam all require the composite body method. The process has four steps:
- Break the object into simple shapes whose individual rotational inertias you can look up (cylinders, rectangles, spheres, etc.).
- Find each piece’s rotational inertia about its own center of mass using a standard table.
- Use the parallel axis theorem to shift each piece’s value to the common axis of rotation.
- Add the adjusted values together. If a piece represents a hole or cutout, treat its mass as negative and subtract its contribution instead.
The negative-mass trick for holes is the key insight that makes this method powerful. A disk with a circular hole through it, for instance, is just a full disk minus a smaller disk. Calculate each one’s rotational inertia about the desired axis, then subtract the smaller from the larger.
Radius of Gyration
Sometimes you’ll see rotational inertia expressed through the radius of gyration, k. This is defined by:
I = mk²
Solving for k gives you the distance from the axis at which you could concentrate all the object’s mass and get the same rotational inertia. It’s a useful shorthand in engineering contexts because it lets you compare how “spread out” different shapes are without worrying about their total mass. A long thin rod spinning about its center has a smaller radius of gyration than the same rod spinning about one end, reflecting the fact that mass is, on average, closer to the center axis.
Choosing the Right Approach
Which method you use depends on what you’re given and what you’re solving for. If the object is a collection of discrete masses at known distances, sum I = Σmr² directly. If it’s a single continuous shape, look up its standard formula or set up the integral I = ∫r² dm. If the rotation axis isn’t through the center of mass, apply the parallel axis theorem. And if the object is built from multiple simple shapes (or has holes), use the composite body method.
A common mistake is forgetting that rotational inertia always depends on the choice of axis. The same object has a different value of I for every possible axis of rotation. Always identify the axis first before plugging into any formula. Another frequent error is applying the parallel axis theorem between two arbitrary axes, when it actually requires that one of them pass through the center of mass. If neither axis goes through the center of mass, you need to step through the center of mass as an intermediate: shift from the first axis back to the center of mass (by subtracting md₁²), then out to the second axis (by adding md₂²).