Mass Moment of Inertia

Dynamics · FE Reference Handbook section

Dynamics

3 formulas

10 exam-style examples

~51 min

All Dynamics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Mass Moment of Inertia within Dynamics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what mass moment of inertia describes physically and when it applies.
State every one of the 3 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight.

Lecture

Why this section exists. Mass Moment of Inertia is the part of Dynamics that lets you connect a particle or rigid body moving under known forces to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as kinematics first, then a work-energy or impulse-momentum shortcut. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Interior of a steel and glass pedestrian bridge showing the structural framing. — Photo 1. Where this shows up in practice: mass moment of inertia.
Wikimedia Commons, CC BY-SA 4.0

Dynamics — Mass Moment of Inertia: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a particle or rigid body moving under known forces. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 3 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The definitions for the mass moments of inertia are
A table listing moment of inertia formulas for some standard shapes is at the end of this section.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Acceleration from Newton's second law — Mass Moment of Inertia

A 5858 lb cart is pulled horizontally by a 858 lb force against rolling resistance μ = 0.25. What is its acceleration?

Given

W = 5858 lb
F = 858 lb
μ = 0.25
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 5858/32.2 = 181.9 s l ug s$
Resistance — F_r = μW = 0.25(5858) = 1,465 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 858 − 1,465 = -606.5 lb
Acceleration — a = ΣF/m = -606.5/181.9 = -3.334 ft/s²

Answer: a ≈ -3.33 ft/s²

Why the other options are there

-0.104 ft/s² (weight used as mass)
4.72 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

Example 2

Angular acceleration of a rotating drum — Mass Moment of Inertia

A drum with mass moment of inertia 10.0 kg·m² is driven by a constant torque of 69 N·m from rest. Find α and the angular speed after 2.0 s.

Given

I = 10.0 kg·m²
M = 69 N·m
t = 2.0 s

Find

α and ω

Start with the thinking

Rotational analogue of F = ma is M = Iα.
Constant torque gives constant angular acceleration.

Step-by-step solution

Rotational equation
$M = I α$
Substituting
$α = 69/10.0 = 6.900 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 6.900 (2.0) = 13.80 r a d / s$
Convert
$ω = 131.8 r p m$

Answer: α ≈ 6.90 rad/s²; ω ≈ 13.8 rad/s (131.8 rpm)

Why the other options are there

690.0 rad/s² (multiplied instead of divided)
2.20 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

Example 3

Parallel-axis transfer of a slender bar's mass moment of inertia — Mass Moment of Inertia

A slender uniform bar of mass 32.5 kg and length 2.9 m rotates about an axis 0.85 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 6.0 rad/s.

Given

m = 32.5 kg, L = 2.9 m
d = 0.85 m
ω = 6.0 rad/s

Find

I_c, I, k and the rotational kinetic energy

Start with the thinking

The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

Formula
$I_{c} = \frac{m L ^{2}}{12}$
Substituting — I_c = 32.5(2.9)²/12 = 22.7771 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 32.5(0.85)² = 23.4813 kg·m²
Substituting — I = 22.7771 + 23.4813 = 46.2583 kg·m²
Formula
$k = I / m$
Substituting
$k = (46.2583/32.5) = 1.193 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (46.2583) (6.0)^{2} = 832.7 J$

Answer: I_c = 22.777 kg·m², I = 46.258 kg·m², k = 1.193 m, T = 832.7 J

Why the other options are there

I = -0.7042 kg·m² (transfer term subtracted)
I = 91.1083 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

Example 4

Acceleration from Newton's second law — Mass Moment of Inertia (2)

A 2450 lb cart is pulled horizontally by a 679 lb force against rolling resistance μ = 0.25. What is its acceleration?

Given

W = 2450 lb
F = 679 lb
μ = 0.25
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 2450/32.2 = 76.09 s l ug s$
Resistance — F_r = μW = 0.25(2450) = 612.5 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 679 − 612.5 = 66.5 lb
Acceleration — a = ΣF/m = 66.5/76.09 = 0.874 ft/s²

Answer: a ≈ 0.87 ft/s²

Why the other options are there

0.027 ft/s² (weight used as mass)
8.92 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

Example 5

Angular acceleration of a rotating drum — Mass Moment of Inertia (2)

A drum with mass moment of inertia 24.5 kg·m² is driven by a constant torque of 188 N·m from rest. Find α and the angular speed after 4.5 s.

Given

I = 24.5 kg·m²
M = 188 N·m
t = 4.5 s

Find

α and ω

Start with the thinking

Rotational analogue of F = ma is M = Iα.
Constant torque gives constant angular acceleration.

Step-by-step solution

Rotational equation
$M = I α$
Substituting
$α = 188/24.5 = 7.673 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 7.673 (4.5) = 34.53 r a d / s$
Convert
$ω = 329.7 r p m$

Answer: α ≈ 7.67 rad/s²; ω ≈ 34.5 rad/s (329.7 rpm)

Why the other options are there

4,606 rad/s² (multiplied instead of divided)
5.50 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

Example 6

Parallel-axis transfer of a slender bar's mass moment of inertia — Mass Moment of Inertia (2)

A slender uniform bar of mass 37.5 kg and length 0.7 m rotates about an axis 0.55 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 14.5 rad/s.

Given

m = 37.5 kg, L = 0.7 m
d = 0.55 m
ω = 14.5 rad/s

Find

I_c, I, k and the rotational kinetic energy

Start with the thinking

The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

Formula
$I_{c} = \frac{m L ^{2}}{12}$
Substituting — I_c = 37.5(0.7)²/12 = 1.5313 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 37.5(0.55)² = 11.3438 kg·m²
Substituting — I = 1.5313 + 11.3438 = 12.8750 kg·m²
Formula
$k = I / m$
Substituting
$k = (12.8750/37.5) = 0.586 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (12.8750) (14.5)^{2} = 1, 353 J$

Answer: I_c = 1.531 kg·m², I = 12.875 kg·m², k = 0.586 m, T = 1,353 J

Why the other options are there

I = -9.8125 kg·m² (transfer term subtracted)
I = 6.1250 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

Example 7

Acceleration from Newton's second law — Mass Moment of Inertia (3)

A 4984 lb cart is pulled horizontally by a 332 lb force against rolling resistance μ = 0.20. What is its acceleration?

Given

W = 4984 lb
F = 332 lb
μ = 0.20
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 4984/32.2 = 154.8 s l ug s$
Resistance — F_r = μW = 0.20(4984) = 996.8 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 332 − 996.8 = -664.8 lb
Acceleration — a = ΣF/m = -664.8/154.8 = -4.295 ft/s²

Answer: a ≈ -4.30 ft/s²

Why the other options are there

-0.133 ft/s² (weight used as mass)
2.14 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

Example 8

Angular acceleration of a rotating drum — Mass Moment of Inertia (3)

A drum with mass moment of inertia 25.0 kg·m² is driven by a constant torque of 147 N·m from rest. Find α and the angular speed after 5.5 s.

Given

I = 25.0 kg·m²
M = 147 N·m
t = 5.5 s

Find

α and ω

Start with the thinking

Rotational analogue of F = ma is M = Iα.
Constant torque gives constant angular acceleration.

Step-by-step solution

Rotational equation
$M = I α$
Substituting
$α = 147/25.0 = 5.880 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 5.880 (5.5) = 32.34 r a d / s$
Convert
$ω = 308.8 r p m$

Answer: α ≈ 5.88 rad/s²; ω ≈ 32.3 rad/s (308.8 rpm)

Why the other options are there

3,675 rad/s² (multiplied instead of divided)
5.15 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

Example 9

Parallel-axis transfer of a slender bar's mass moment of inertia — Mass Moment of Inertia (3)

A slender uniform bar of mass 23.0 kg and length 1.7 m rotates about an axis 1.15 m from its mass centre. Compute the centroidal mass moment of inertia, the inertia about the offset axis, the radius of gyration, and the kinetic energy at 9.0 rad/s.

Given

m = 23.0 kg, L = 1.7 m
d = 1.15 m
ω = 9.0 rad/s

Find

I_c, I, k and the rotational kinetic energy

Start with the thinking

The transfer term m d² is always positive: the centroidal axis gives the smallest possible inertia.
The parallel-axis theorem only transfers between parallel axes, and one of them must be centroidal.

Step-by-step solution

Formula
$I_{c} = \frac{m L ^{2}}{12}$
Substituting — I_c = 23.0(1.7)²/12 = 5.5392 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 23.0(1.15)² = 30.4175 kg·m²
Substituting — I = 5.5392 + 30.4175 = 35.9567 kg·m²
Formula
$k = I / m$
Substituting
$k = (35.9567/23.0) = 1.250 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (35.9567) (9.0)^{2} = 1, 456 J$

Answer: I_c = 5.539 kg·m², I = 35.957 kg·m², k = 1.250 m, T = 1,456 J

Why the other options are there

I = -24.8783 kg·m² (transfer term subtracted)
I = 22.1567 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

Example 10

Acceleration from Newton's second law — Mass Moment of Inertia (4)

A 2732 lb cart is pulled horizontally by a 502 lb force against rolling resistance μ = 0.30. What is its acceleration?

Given

W = 2732 lb
F = 502 lb
μ = 0.30
g = 32.2 ft/s²

Find

Acceleration a

Start with the thinking

Convert weight to mass before applying F = ma.
Net force is applied force minus resistance.

Step-by-step solution

Mass
$m = W / g = 2732/32.2 = 84.84 s l ug s$
Resistance — F_r = μW = 0.30(2732) = 819.6 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 502 − 819.6 = -317.6 lb
Acceleration — a = ΣF/m = -317.6/84.84 = -3.743 ft/s²

Answer: a ≈ -3.74 ft/s²

Why the other options are there

-0.116 ft/s² (weight used as mass)
5.92 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Moment of Inertia

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a particle or rigid body moving under known forces, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Mass Moment of Inertia contains 3 relations; you must be able to find this page in under 15 seconds.
Exam style: kinematics first, then a work-energy or impulse-momentum shortcut.
Unit rule: g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.