A non-uniform mass distribution is one where an object's density varies with position, so you can't treat mass as evenly spread. Finding rotational inertia or center of mass requires integration, typically writing dm in terms of a density function like λ(x) and evaluating I = ∫r² dm.
A non-uniform mass distribution means the object's mass is not spread evenly. The density changes from point to point, often given as a function like λ(x) = αx for a rod whose linear density grows along its length. Because the mass isn't uniform, you can't just look up a rotational inertia formula on the equation sheet. Those standard results (like (1/12)ML² for a rod) are derived assuming constant density, and they break the moment density varies.
Instead, you build the answer from scratch with calculus. The recipe is always the same. Slice the object into tiny pieces, write the mass of each piece as dm using the density function (for a thin rod, dm = λ(x) dx), then integrate whatever quantity you need. For rotational inertia that's I = ∫r² dm, where r is each piece's distance from the rotation axis. For center of mass it's x_cm = (1/M)∫x dm, where M itself often comes from integrating dm first. The physics insight behind all of it is that where the mass sits matters. A rod with its mass piled up at the far end resists rotation much more than a uniform rod of the same total mass.
This concept lives in Topic 5.4 (Rotational Inertia) in Unit 5, Torque and Rotational Dynamics. It's also one of the clearest examples of why Physics C is a calculus-based course. The algebra-based AP Physics classes hand you rotational inertia values; Physics C expects you to derive them. Setting up I = ∫r² dm for a rod with λ(x) = αx is a classic Physics C move, and it tests whether you actually understand that rotational inertia measures how mass is distributed relative to an axis, not just how much mass there is.
It also pays off downstream. Any later problem that needs I, like rotational kinetic energy in Unit 6 or the period of a physical pendulum in Unit 7, can start by making you compute I for a non-uniform object first. Mess up the integral and every following part inherits the error.
Keep studying AP® Physics C: Mechanics Unit 5
Mass Distribution (Unit 5)
Non-uniform distribution is the special case where the general idea of mass distribution gets interesting. Rotational inertia always depends on where mass sits relative to the axis; non-uniformity just means you need a density function and an integral instead of a memorized formula.
Center of Mass by Integration (Unit 2)
The same dm = λ(x) dx setup you use for rotational inertia also finds the center of mass of a non-uniform object via x_cm = (1/M)∫x dm. If you can do one of these integrals, you can do the other. Only the power of r changes.
Parallel Axis Theorem (Unit 5)
Once you've integrated to find I about one axis (often the center of mass), the parallel axis theorem I = I_cm + Md² shifts it to any parallel axis without a second integral. For a non-uniform object you may need the integration-based center of mass first, which is why these two skills get tested together.
Rotational Kinetic Energy and the Physical Pendulum (Units 6-7)
Your computed I feeds directly into K = (1/2)Iω² and into the physical pendulum period T = 2π√(I/mgd). A multi-part FRQ can chain these, so the integral you set up in part (a) determines whether parts (b) through (d) come out right.
No released FRQ has used the exact phrase "non-uniform mass distribution," but the underlying skill is a Physics C staple. The classic setup is a rod or rigid object with a given density function, something like λ(x) = αx or λ(x) = λ₀(1 + x/L), where you must derive total mass, center of mass, or rotational inertia symbolically. Expect to (1) express dm in terms of the density function, (2) set up the integral with correct limits, and (3) evaluate it to get an answer in terms of given variables. Multiple-choice questions test the concept more qualitatively, asking which of two objects with equal mass has greater rotational inertia based on where the mass is concentrated. The trap answer is always the standard uniform-body formula. If the problem gives you a density function, that formula does not apply and you must integrate.
Uniform means density is constant everywhere, so dm = (M/L) dx and the standard rotational inertia formulas (like (1/12)ML² for a rod about its center) are valid. Non-uniform means density is a function of position, so those formulas are off the table and you must integrate I = ∫r² dm using the given density function. Quick check on the exam: if the problem hands you λ(x), σ(r), or any density that depends on position, it's non-uniform and the equation sheet won't save you.
A non-uniform mass distribution has density that varies with position, so standard rotational inertia formulas derived for uniform objects do not apply.
To find rotational inertia, write dm in terms of the density function (for a thin rod, dm = λ(x) dx) and evaluate I = ∫r² dm with limits that cover the whole object.
The same dm setup finds total mass with M = ∫dm and center of mass with x_cm = (1/M)∫x dm, so one density function unlocks all three quantities.
The variable r in I = ∫r² dm is the distance from the rotation axis, so the same object has different rotational inertias about different axes.
Concentrating mass farther from the axis increases rotational inertia, which is why a rod weighted at its far end is harder to spin than a uniform rod of equal mass.
After integrating to find I about the center of mass, use the parallel axis theorem I = I_cm + Md² to shift to any parallel axis without a new integral.
It's an object whose density changes with position, usually given as a function like λ(x) = αx for a rod. Because mass isn't spread evenly, you find rotational inertia, total mass, or center of mass by integrating with dm = λ(x) dx instead of using a standard formula.
No. That formula is derived assuming constant density, so it only works for uniform rods. For a non-uniform rod you must set up I = ∫r² dm with the given density function, and the answer will generally differ from the uniform result.
Mass distribution is the general idea of where an object's mass is located relative to an axis, which matters for any rotational inertia problem. Non-uniform is the specific case where density varies with position, which forces you to use integration rather than tabulated formulas.
For continuous non-uniform objects, yes, integration is the only way to get I, M, or x_cm exactly. The one shortcut is the parallel axis theorem, which lets you shift an already-known I_cm to a parallel axis using I = I_cm + Md² without integrating again.
It means the rod's linear density grows in proportion to distance from x = 0, so the rod is lightest at one end and heaviest at the other. A small slice at position x has mass dm = αx dx, and the center of mass sits closer to the heavy end (at 2L/3 for this density, not L/2).
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