Mass Moments of Inertia

NCEES FE Civil spec: Dynamics, B. Mass moments of inertia. FE Reference Handbook: Dynamics chapter, mass moment of inertia table.

The mass moment of inertia measures how hard it is to change a body's rotation, just as mass measures resistance to changing linear motion. It depends on both the mass and how far that mass sits from the axis. On the FE Civil exam you will mostly use standard shapes, the parallel axis theorem and the radius of gyration.

What the NCEES spec covers

Section B, "Mass moments of inertia", covers I for rods, disks, rings, spheres and plates, composite bodies, transfer to a parallel axis, and the radius of gyration. These values then feed rigid-body kinetics (section C) and rotational kinetic energy (section D).

Key concepts

Definition. $I = \int r^{2} d m$ , with units of kg·m² (or slug·ft²). Do not confuse it with the area moment of inertia (m⁴) from statics.
Parallel axis theorem. $I = I_{G} + m d^{2}$ , where $I_{G}$ is about a parallel axis through the center of mass. The smallest I for any direction is the one through the center of mass.
Composite bodies. Add the I of each part about the same axis. Subtract holes.
Point masses contribute $m r^{2}$ about an axis at distance r.
Radius of gyration k is the distance at which the whole mass could be concentrated to give the same I.

Core formulas

Slender rod: I_{G} = \frac{m L ^{2}}{12}, I_{end} = \frac{m L ^{2}}{3}

Solid disk or cylinder: I = \frac{1}{2} m r^{2} Thin ring: I = m r^{2} Solid sphere: I = \frac{2}{5} m r^{2}

I = I_{G} + m d^{2} k = \frac{I}{m}

Where it is in the FE Reference Handbook

The Dynamics chapter has a table of mass moments of inertia for common homogeneous bodies, along with the parallel axis theorem and radius of gyration. Check which axis each table entry is about: the slender rod is listed about both its center and its end.

Worked example

A uniform 8 kg slender rod 2 m long is pinned at one end. A 3 kg mass (treat it as a point) is fixed at the free end. Find the mass moment of inertia about the pin and the radius of gyration. Then find the angular acceleration just after the rod is released from rest in a horizontal position.

Step 1: rod about its end.

I_{rod} = \frac{8 ( 2 ) ^{2}}{3} = 10.67 kg \cdotp m^{2}

Step 2: point mass.

I_{mass} = 3 (2)^{2} = 12.00 kg \cdotp m^{2} I_{O} = 10.67 + 12.00 = 22.67 kg \cdotp m^{2}

Step 3: radius of gyration (total mass 11 kg):

k = \frac{22.67}{11} = 1.44 m

Step 4: angular acceleration. Gravity moments about the pin when horizontal:

\sum M_{O} = [8 (1) + 3 (2)] (9.81) = 137.3 N \cdotp m α = \frac{137.3}{22.67} = 6.06 rad/s^{2}

The point mass is less than 30% of the total mass but supplies more than half of the inertia, because it sits at the full 2 m radius.

Common mistakes

Using $m L^{2} /12$ when the axis is at the end of the rod.
Applying the parallel axis theorem from an axis that is not through the center of mass.
Confusing a solid disk (½mr²) with a thin ring (mr²).
Using weight instead of mass, especially in US units (slugs = lb / 32.2 ft/s²).
Mixing area moment of inertia formulas (bh³/12) into a dynamics problem.

See the slender rod example and compare with the area moment of inertia lesson.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A uniform slender rod has mass 10 kg and length 2 m. Its mass moment of inertia about its center, perpendicular to the rod, is: See the worked solution or practice the Dynamics & Seismic deck
A uniform slender rod has mass 10 kg and length 2 m. Its mass moment of inertia about an axis through one end, perpendicular to the rod, is: See the worked solution or practice the Dynamics & Seismic deck
A solid disk has mass 20 kg and radius 0.30 m. Its mass moment of inertia about its central axis is: See the worked solution or practice the Dynamics & Seismic deck
A thin hoop has mass 5 kg and radius 0.40 m. Its mass moment of inertia about its central axis is: See the worked solution or practice the Dynamics & Seismic deck
A solid disk has mass 10 kg and radius 0.20 m. Its mass moment of inertia about an axis tangent to its edge and parallel to its central axis is: See the worked solution or practice the Dynamics & Seismic deck

Related Calculators and Examples

Frequently Asked Questions

What is the mass moment of inertia of a rod about its end?

For a uniform slender rod of mass m and length L, it is m L^2 / 3. About its center it is m L^2 / 12, and the parallel axis theorem links the two with d = L / 2.

What is the radius of gyration?

It is k = sqrt(I / m), the distance from the axis at which the entire mass could be placed to give the same moment of inertia.

How is mass moment of inertia different from area moment of inertia?

Mass moment of inertia has units of mass times length squared and resists angular acceleration. Area moment of inertia has units of length to the fourth and describes a cross section resisting bending.

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