Parallel-Axis Theorem

Dynamics · FE Reference Handbook section

Dynamics

5 formulas

10 exam-style examples

~55 min

All Dynamics lectures

Learning objectives

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

This chapter section covers Parallel-Axis Theorem 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 parallel-axis theorem describes physically and when it applies.
State every one of the 5 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. Parallel-Axis Theorem 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.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: parallel-axis theorem.
Capstone Studio instructional photograph

Dynamics — Parallel-Axis Theorem: 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 5 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.

Notation used in this section

Inew	Quantity produced by "Inew = Ic + md 2" — read its definition and unit from the handbook line directly above the equation.
Ic	Quantity produced by "Ic = mass moment of inertia about an axis that is parallel to the above specified axis but passes through the" — read its definition and unit from the handbook line directly above the equation.
m	Quantity produced by "m = mass of the body" — read its definition and unit from the handbook line directly above the equation.
d	Quantity produced by "d = normal distance from the body's mass center to the above-specified axis" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The mass moments of inertia may be calculated about any axis through the application of the above definitions. However, once
the moments of inertia have been determined about an axis passing through a body's mass center, it may be transformed to
another parallel axis. The transformation equation is
where
body's mass center

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

Parallel-axis transfer of a slender bar's mass moment of inertia — Parallel-Axis Theorem

A slender uniform bar of mass 21.5 kg and length 2.3 m rotates about an axis 1.05 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 4.5 rad/s.

Given

m = 21.5 kg, L = 2.3 m
d = 1.05 m
ω = 4.5 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 = 21.5(2.3)²/12 = 9.4779 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 21.5(1.05)² = 23.7038 kg·m²
Substituting — I = 9.4779 + 23.7038 = 33.1817 kg·m²
Formula
$k = I / m$
Substituting
$k = (33.1817/21.5) = 1.242 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (33.1817) (4.5)^{2} = 336.0 J$

Answer: I_c = 9.478 kg·m², I = 33.182 kg·m², k = 1.242 m, T = 336.0 J

Why the other options are there

I = -14.2258 kg·m² (transfer term subtracted)
I = 37.9117 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

Example 2

Parallel-axis transfer of a slender bar's mass moment of inertia — Parallel-Axis Theorem (2)

A slender uniform bar of mass 27.0 kg and length 1.5 m rotates about an axis 0.70 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 4.5 rad/s.

Given

m = 27.0 kg, L = 1.5 m
d = 0.70 m
ω = 4.5 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 = 27.0(1.5)²/12 = 5.0625 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 27.0(0.70)² = 13.2300 kg·m²
Substituting — I = 5.0625 + 13.2300 = 18.2925 kg·m²
Formula
$k = I / m$
Substituting
$k = (18.2925/27.0) = 0.823 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (18.2925) (4.5)^{2} = 185.2 J$

Answer: I_c = 5.063 kg·m², I = 18.292 kg·m², k = 0.823 m, T = 185.2 J

Why the other options are there

I = -8.1675 kg·m² (transfer term subtracted)
I = 20.2500 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

Example 3

Parallel-axis transfer of a slender bar's mass moment of inertia — Parallel-Axis Theorem (3)

A slender uniform bar of mass 14.0 kg and length 2.7 m rotates about an axis 0.20 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 10.5 rad/s.

Given

m = 14.0 kg, L = 2.7 m
d = 0.20 m
ω = 10.5 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 = 14.0(2.7)²/12 = 8.5050 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 14.0(0.20)² = 0.5600 kg·m²
Substituting — I = 8.5050 + 0.5600 = 9.0650 kg·m²
Formula
$k = I / m$
Substituting
$k = (9.0650/14.0) = 0.805 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (9.0650) (10.5)^{2} = 499.7 J$

Answer: I_c = 8.505 kg·m², I = 9.065 kg·m², k = 0.805 m, T = 499.7 J

Why the other options are there

I = 7.9450 kg·m² (transfer term subtracted)
I = 34.0200 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

Example 4

Parallel-axis transfer of a slender bar's mass moment of inertia — Parallel-Axis Theorem (4)

A slender uniform bar of mass 40.0 kg and length 3.0 m rotates about an axis 0.40 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 = 40.0 kg, L = 3.0 m
d = 0.40 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 = 40.0(3.0)²/12 = 30.0000 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 40.0(0.40)² = 6.4000 kg·m²
Substituting — I = 30.0000 + 6.4000 = 36.4000 kg·m²
Formula
$k = I / m$
Substituting
$k = (36.4000/40.0) = 0.954 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (36.4000) (9.0)^{2} = 1, 474 J$

Answer: I_c = 30.000 kg·m², I = 36.400 kg·m², k = 0.954 m, T = 1,474 J

Why the other options are there

I = 23.6000 kg·m² (transfer term subtracted)
I = 120.0 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

Example 5

Parallel-axis transfer of a slender bar's mass moment of inertia — Parallel-Axis Theorem (5)

A slender uniform bar of mass 4.5 kg and length 1.1 m rotates about an axis 1.10 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 14.0 rad/s.

Given

m = 4.5 kg, L = 1.1 m
d = 1.10 m
ω = 14.0 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 = 4.5(1.1)²/12 = 0.4538 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 4.5(1.10)² = 5.4450 kg·m²
Substituting — I = 0.4538 + 5.4450 = 5.8988 kg·m²
Formula
$k = I / m$
Substituting
$k = (5.8988/4.5) = 1.145 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (5.8988) (14.0)^{2} = 578.1 J$

Answer: I_c = 0.454 kg·m², I = 5.899 kg·m², k = 1.145 m, T = 578.1 J

Why the other options are there

I = -4.9913 kg·m² (transfer term subtracted)
I = 1.8150 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

Example 6

Parallel-axis transfer of a slender bar's mass moment of inertia — Parallel-Axis Theorem (6)

A slender uniform bar of mass 8.5 kg and length 3.0 m rotates about an axis 1.05 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 6.5 rad/s.

Given

m = 8.5 kg, L = 3.0 m
d = 1.05 m
ω = 6.5 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 = 8.5(3.0)²/12 = 6.3750 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 8.5(1.05)² = 9.3713 kg·m²
Substituting — I = 6.3750 + 9.3713 = 15.7463 kg·m²
Formula
$k = I / m$
Substituting
$k = (15.7463/8.5) = 1.361 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (15.7463) (6.5)^{2} = 332.6 J$

Answer: I_c = 6.375 kg·m², I = 15.746 kg·m², k = 1.361 m, T = 332.6 J

Why the other options are there

I = -2.9963 kg·m² (transfer term subtracted)
I = 25.5000 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

Example 7

Parallel-axis transfer of a slender bar's mass moment of inertia — Parallel-Axis Theorem (7)

A slender uniform bar of mass 32.5 kg and length 2.3 m rotates about an axis 0.30 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 4.0 rad/s.

Given

m = 32.5 kg, L = 2.3 m
d = 0.30 m
ω = 4.0 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(2.3)²/12 = 14.3271 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 32.5(0.30)² = 2.9250 kg·m²
Substituting — I = 14.3271 + 2.9250 = 17.2521 kg·m²
Formula
$k = I / m$
Substituting
$k = (17.2521/32.5) = 0.729 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (17.2521) (4.0)^{2} = 138.0 J$

Answer: I_c = 14.327 kg·m², I = 17.252 kg·m², k = 0.729 m, T = 138.0 J

Why the other options are there

I = 11.4021 kg·m² (transfer term subtracted)
I = 57.3083 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

Example 8

Parallel-axis transfer of a slender bar's mass moment of inertia — Parallel-Axis Theorem (8)

A slender uniform bar of mass 17.0 kg and length 2.4 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 2.5 rad/s.

Given

m = 17.0 kg, L = 2.4 m
d = 0.55 m
ω = 2.5 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 = 17.0(2.4)²/12 = 8.1600 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 17.0(0.55)² = 5.1425 kg·m²
Substituting — I = 8.1600 + 5.1425 = 13.3025 kg·m²
Formula
$k = I / m$
Substituting
$k = (13.3025/17.0) = 0.885 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (13.3025) (2.5)^{2} = 41.57 J$

Answer: I_c = 8.160 kg·m², I = 13.302 kg·m², k = 0.885 m, T = 41.6 J

Why the other options are there

I = 3.0175 kg·m² (transfer term subtracted)
I = 32.6400 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

Example 9

Parallel-axis transfer of a slender bar's mass moment of inertia — Parallel-Axis Theorem (9)

A slender uniform bar of mass 32.0 kg and length 0.9 m rotates about an axis 1.05 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 5.5 rad/s.

Given

m = 32.0 kg, L = 0.9 m
d = 1.05 m
ω = 5.5 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 = 32.0(0.9)²/12 = 2.1600 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 32.0(1.05)² = 35.2800 kg·m²
Substituting — I = 2.1600 + 35.2800 = 37.4400 kg·m²
Formula
$k = I / m$
Substituting
$k = (37.4400/32.0) = 1.082 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (37.4400) (5.5)^{2} = 566.3 J$

Answer: I_c = 2.160 kg·m², I = 37.440 kg·m², k = 1.082 m, T = 566.3 J

Why the other options are there

I = -33.1200 kg·m² (transfer term subtracted)
I = 8.6400 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

Example 10

Parallel-axis transfer of a slender bar's mass moment of inertia — Parallel-Axis Theorem (10)

A slender uniform bar of mass 10.0 kg and length 2.6 m rotates about an axis 0.65 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 15.0 rad/s.

Given

m = 10.0 kg, L = 2.6 m
d = 0.65 m
ω = 15.0 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 = 10.0(2.6)²/12 = 5.6333 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 10.0(0.65)² = 4.2250 kg·m²
Substituting — I = 5.6333 + 4.2250 = 9.8583 kg·m²
Formula
$k = I / m$
Substituting
$k = (9.8583/10.0) = 0.993 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (9.8583) (15.0)^{2} = 1, 109 J$

Answer: I_c = 5.633 kg·m², I = 9.858 kg·m², k = 0.993 m, T = 1,109 J

Why the other options are there

I = 1.4083 kg·m² (transfer term subtracted)
I = 22.5333 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Parallel-Axis Theorem

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

Parallel-Axis Theorem contains 5 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.