Mass Radius of Gyration

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 Mass Radius of Gyration 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 radius of gyration 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. Mass Radius of Gyration 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: mass radius of gyration.
Capstone Studio instructional photograph

Dynamics — Mass Radius of Gyration: 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

rm	Quantity produced by "rm = I m" — read its definition and unit from the handbook line directly above the equation.
RFx	Quantity produced by "RFx = maxc" — read its definition and unit from the handbook line directly above the equation.
RFy	Quantity produced by "RFy = mayc" — read its definition and unit from the handbook line directly above the equation.
RMzc	Quantity produced by "RMzc = Izc a" — 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 radius of gyration is defined as
Without loss of generality, the body may be assumed to be in the x-y plane. The scalar equations of motion may then be written as
where zc indicates the z axis passing through the body's mass center, axc and ayc are the acceleration of the body's mass center in

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 Radius of Gyration

A 2825 lb cart is pulled horizontally by a 785 lb force against rolling resistance μ = 0.30. What is its acceleration?

Given

W = 2825 lb
F = 785 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
$m = W / g = 2825/32.2 = 87.73 s l ug s$
Resistance — F_r = μW = 0.30(2825) = 847.5 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 785 − 847.5 = -62.5 lb
Acceleration — a = ΣF/m = -62.5/87.73 = -0.712 ft/s²

Answer: a ≈ -0.71 ft/s²

Why the other options are there

-0.022 ft/s² (weight used as mass)
8.95 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Radius of Gyration

Example 2

Parallel-axis transfer of a slender bar's mass moment of inertia — Mass Radius of Gyration

A slender uniform bar of mass 38.5 kg and length 2.7 m rotates about an axis 0.50 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 = 38.5 kg, L = 2.7 m
d = 0.50 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 = 38.5(2.7)²/12 = 23.3888 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 38.5(0.50)² = 9.6250 kg·m²
Substituting — I = 23.3888 + 9.6250 = 33.0138 kg·m²
Formula
$k = I / m$
Substituting
$k = (33.0138/38.5) = 0.926 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (33.0138) (5.5)^{2} = 499.3 J$

Answer: I_c = 23.389 kg·m², I = 33.014 kg·m², k = 0.926 m, T = 499.3 J

Why the other options are there

I = 13.7638 kg·m² (transfer term subtracted)
I = 93.5550 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Mass Radius of Gyration

Example 3

Acceleration from Newton's second law — Mass Radius of Gyration (2)

A 3021 lb cart is pulled horizontally by a 880 lb force against rolling resistance μ = 0.10. What is its acceleration?

Given

W = 3021 lb
F = 880 lb
μ = 0.10
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
$m = W / g = 3021/32.2 = 93.82 s l ug s$
Resistance — F_r = μW = 0.10(3021) = 302.1 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 880 − 302.1 = 577.9 lb
Acceleration — a = ΣF/m = 577.9/93.82 = 6.160 ft/s²

Answer: a ≈ 6.16 ft/s²

Why the other options are there

0.191 ft/s² (weight used as mass)
9.38 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Radius of Gyration

Example 4

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

A slender uniform bar of mass 16.0 kg and length 2.9 m rotates about an axis 0.45 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 3.0 rad/s.

Given

m = 16.0 kg, L = 2.9 m
d = 0.45 m
ω = 3.0 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 = 16.0(2.9)²/12 = 11.2133 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 16.0(0.45)² = 3.2400 kg·m²
Substituting — I = 11.2133 + 3.2400 = 14.4533 kg·m²
Formula
$k = I / m$
Substituting
$k = (14.4533/16.0) = 0.950 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (14.4533) (3.0)^{2} = 65.04 J$

Answer: I_c = 11.213 kg·m², I = 14.453 kg·m², k = 0.950 m, T = 65.0 J

Why the other options are there

I = 7.9733 kg·m² (transfer term subtracted)
I = 44.8533 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Mass Radius of Gyration

Example 5

Acceleration from Newton's second law — Mass Radius of Gyration (3)

A 2308 lb cart is pulled horizontally by a 260 lb force against rolling resistance μ = 0.15. What is its acceleration?

Given

W = 2308 lb
F = 260 lb
μ = 0.15
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
$m = W / g = 2308/32.2 = 71.68 s l ug s$
Resistance — F_r = μW = 0.15(2308) = 346.2 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 260 − 346.2 = -86.2 lb
Acceleration — a = ΣF/m = -86.2/71.68 = -1.203 ft/s²

Answer: a ≈ -1.20 ft/s²

Why the other options are there

-0.037 ft/s² (weight used as mass)
3.63 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Radius of Gyration

Example 6

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

A slender uniform bar of mass 36.0 kg and length 2.9 m rotates about an axis 0.50 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 = 36.0 kg, L = 2.9 m
d = 0.50 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 = 36.0(2.9)²/12 = 25.2300 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 36.0(0.50)² = 9.0000 kg·m²
Substituting — I = 25.2300 + 9.0000 = 34.2300 kg·m²
Formula
$k = I / m$
Substituting
$k = (34.2300/36.0) = 0.975 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (34.2300) (2.5)^{2} = 107.0 J$

Answer: I_c = 25.230 kg·m², I = 34.230 kg·m², k = 0.975 m, T = 107.0 J

Why the other options are there

I = 16.2300 kg·m² (transfer term subtracted)
I = 100.9 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Mass Radius of Gyration

Example 7

Acceleration from Newton's second law — Mass Radius of Gyration (4)

A 2795 lb cart is pulled horizontally by a 1012 lb force against rolling resistance μ = 0.20. What is its acceleration?

Given

W = 2795 lb
F = 1012 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
$m = W / g = 2795/32.2 = 86.80 s l ug s$
Resistance — F_r = μW = 0.20(2795) = 559.0 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 1012 − 559.0 = 453.0 lb
Acceleration — a = ΣF/m = 453.0/86.80 = 5.219 ft/s²

Answer: a ≈ 5.22 ft/s²

Why the other options are there

0.162 ft/s² (weight used as mass)
11.66 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Radius of Gyration

Example 8

Parallel-axis transfer of a slender bar's mass moment of inertia — Mass Radius of Gyration (4)

A slender uniform bar of mass 31.0 kg and length 1.8 m rotates about an axis 0.25 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 = 31.0 kg, L = 1.8 m
d = 0.25 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 = 31.0(1.8)²/12 = 8.3700 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 31.0(0.25)² = 1.9375 kg·m²
Substituting — I = 8.3700 + 1.9375 = 10.3075 kg·m²
Formula
$k = I / m$
Substituting
$k = (10.3075/31.0) = 0.577 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (10.3075) (14.0)^{2} = 1, 010 J$

Answer: I_c = 8.370 kg·m², I = 10.308 kg·m², k = 0.577 m, T = 1,010 J

Why the other options are there

I = 6.4325 kg·m² (transfer term subtracted)
I = 33.4800 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Mass Radius of Gyration

Example 9

Acceleration from Newton's second law — Mass Radius of Gyration (5)

A 2342 lb cart is pulled horizontally by a 612 lb force against rolling resistance μ = 0.15. What is its acceleration?

Given

W = 2342 lb
F = 612 lb
μ = 0.15
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
$m = W / g = 2342/32.2 = 72.73 s l ug s$
Resistance — F_r = μW = 0.15(2342) = 351.3 lb
Net force — ΣF = F − F_r
Substituting — ΣF = 612 − 351.3 = 260.7 lb
Acceleration — a = ΣF/m = 260.7/72.73 = 3.584 ft/s²

Answer: a ≈ 3.58 ft/s²

Why the other options are there

0.111 ft/s² (weight used as mass)
8.41 ft/s² (resistance ignored)

Reference: FE Reference Handbook — Dynamics → Mass Radius of Gyration

Example 10

Parallel-axis transfer of a slender bar's mass moment of inertia — Mass Radius of Gyration (5)

A slender uniform bar of mass 19.5 kg and length 1.8 m rotates about an axis 0.95 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 11.0 rad/s.

Given

m = 19.5 kg, L = 1.8 m
d = 0.95 m
ω = 11.0 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 = 19.5(1.8)²/12 = 5.2650 kg·m²
Formula
$I = I_{c} + m d^{2}$
Transfer term — m d² = 19.5(0.95)² = 17.5988 kg·m²
Substituting — I = 5.2650 + 17.5988 = 22.8638 kg·m²
Formula
$k = I / m$
Substituting
$k = (22.8638/19.5) = 1.083 m$
Kinetic energy
$T = ½ I ω^{2} = 0.5 (22.8638) (11.0)^{2} = 1, 383 J$

Answer: I_c = 5.265 kg·m², I = 22.864 kg·m², k = 1.083 m, T = 1,383 J

Why the other options are there

I = -12.3337 kg·m² (transfer term subtracted)
I = 21.0600 kg·m² (end-axis formula misapplied)

Reference: FE Reference Handbook — Dynamics → Mass Radius of Gyration

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 Radius of Gyration 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.