Kinetics of a Rigid Body

Dynamics · FE Reference Handbook section

Dynamics

4 formulas

10 exam-style examples

~53 min

All Dynamics lectures

Learning objectives

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

This chapter section covers Kinetics of a Rigid Body 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 kinetics of a rigid body describes physically and when it applies.
State every one of the 4 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. Kinetics of a Rigid Body 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: kinetics of a rigid body.
Capstone Studio instructional photograph

Dynamics — Kinetics of a Rigid Body: 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 4 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

ΣF	Quantity produced by "ΣF = mac" — read its definition and unit from the handbook line directly above the equation.
ΣMc	Quantity produced by "ΣMc = Ic 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.

In general, Newton's second law for a rigid body, with constant mass and mass moment of inertia, in plane motion may be
written in vector form as
angular acceleration both about an axis normal to the plane of motion, Ic is the mass moment of inertia about the normal axis
through the mass center, and ρpc is a vector from point p to point c.

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

Work–energy: force needed to stop a mass — Kinetics of a Rigid Body

A 7380 lb vehicle travelling 59 ft/s must stop in 234 ft. What constant resisting force is required?

Given

W = 7380 lb
v₁ = 59 ft/s
d = 234 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 7380/32.2 = 229.2 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(229.2)(59²) = 398,910 ft·lb
Work–energy — F·d = KE
Substituting
$F = 398, 910/234 = 1, 705 l b$

Answer: F ≈ 1,705 lb

Why the other options are there

54,893 lb (weight used as mass)
398,910 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

Example 2

Angular acceleration of a rotating drum — Kinetics of a Rigid Body

A drum with mass moment of inertia 14.5 kg·m² is driven by a constant torque of 173 N·m from rest. Find α and the angular speed after 6.0 s.

Given

I = 14.5 kg·m²
M = 173 N·m
t = 6.0 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
$α = 173/14.5 = 11.931 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 11.931 (6.0) = 71.59 r a d / s$
Convert
$ω = 683.6 r p m$

Answer: α ≈ 11.93 rad/s²; ω ≈ 71.6 rad/s (683.6 rpm)

Why the other options are there

2,509 rad/s² (multiplied instead of divided)
11.39 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

Example 3

Work–energy: force needed to stop a mass — Kinetics of a Rigid Body (2)

A 3355 lb vehicle travelling 40 ft/s must stop in 193 ft. What constant resisting force is required?

Given

W = 3355 lb
v₁ = 40 ft/s
d = 193 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 3355/32.2 = 104.2 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(104.2)(40²) = 83,354 ft·lb
Work–energy — F·d = KE
Substituting
$F = 83, 354/193 = 431.9 l b$

Answer: F ≈ 431.9 lb

Why the other options are there

13,907 lb (weight used as mass)
83,354 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

Example 4

Angular acceleration of a rotating drum — Kinetics of a Rigid Body (2)

A drum with mass moment of inertia 24.5 kg·m² is driven by a constant torque of 14 N·m from rest. Find α and the angular speed after 3.0 s.

Given

I = 24.5 kg·m²
M = 14 N·m
t = 3.0 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
$α = 14/24.5 = 0.571 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 0.571 (3.0) = 1.71 r a d / s$
Convert
$ω = 16.4 r p m$

Answer: α ≈ 0.57 rad/s²; ω ≈ 1.7 rad/s (16 rpm)

Why the other options are there

343.0 rad/s² (multiplied instead of divided)
0.27 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

Example 5

Work–energy: force needed to stop a mass — Kinetics of a Rigid Body (3)

A 2522 lb vehicle travelling 33 ft/s must stop in 156 ft. What constant resisting force is required?

Given

W = 2522 lb
v₁ = 33 ft/s
d = 156 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 2522/32.2 = 78.32 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(78.32)(33²) = 42,647 ft·lb
Work–energy — F·d = KE
Substituting
$F = 42, 647/156 = 273.4 l b$

Answer: F ≈ 273.4 lb

Why the other options are there

8,803 lb (weight used as mass)
42,647 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

Example 6

Angular acceleration of a rotating drum — Kinetics of a Rigid Body (3)

A drum with mass moment of inertia 19.5 kg·m² is driven by a constant torque of 191 N·m from rest. Find α and the angular speed after 9.5 s.

Given

I = 19.5 kg·m²
M = 191 N·m
t = 9.5 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
$α = 191/19.5 = 9.795 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 9.795 (9.5) = 93.05 r a d / s$
Convert
$ω = 888.6 r p m$

Answer: α ≈ 9.79 rad/s²; ω ≈ 93.1 rad/s (888.6 rpm)

Why the other options are there

3,725 rad/s² (multiplied instead of divided)
14.81 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

Example 7

Work–energy: force needed to stop a mass — Kinetics of a Rigid Body (4)

A 2991 lb vehicle travelling 50 ft/s must stop in 157 ft. What constant resisting force is required?

Given

W = 2991 lb
v₁ = 50 ft/s
d = 157 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 2991/32.2 = 92.89 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(92.89)(50²) = 116,110 ft·lb
Work–energy — F·d = KE
Substituting
$F = 116, 110/157 = 739.6 l b$

Answer: F ≈ 739.6 lb

Why the other options are there

23,814 lb (weight used as mass)
116,110 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

Example 8

Angular acceleration of a rotating drum — Kinetics of a Rigid Body (4)

A drum with mass moment of inertia 8.5 kg·m² is driven by a constant torque of 43 N·m from rest. Find α and the angular speed after 9.0 s.

Given

I = 8.5 kg·m²
M = 43 N·m
t = 9.0 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
$α = 43/8.5 = 5.059 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 5.059 (9.0) = 45.53 r a d / s$
Convert
$ω = 434.8 r p m$

Answer: α ≈ 5.06 rad/s²; ω ≈ 45.5 rad/s (434.8 rpm)

Why the other options are there

365.5 rad/s² (multiplied instead of divided)
7.25 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

Example 9

Work–energy: force needed to stop a mass — Kinetics of a Rigid Body (5)

A 3218 lb vehicle travelling 54 ft/s must stop in 167 ft. What constant resisting force is required?

Given

W = 3218 lb
v₁ = 54 ft/s
d = 167 ft

Find

Resisting force F

Start with the thinking

Work done by the resisting force equals the change in kinetic energy.
Mass, not weight, appears in kinetic energy.

Step-by-step solution

Mass
$m = 3218/32.2 = 99.94 s l ug s$
Kinetic energy
$K E = ½ m v^{2}$
Substituting — KE = 0.5(99.94)(54²) = 145,709 ft·lb
Work–energy — F·d = KE
Substituting
$F = 145, 709/167 = 872.5 l b$

Answer: F ≈ 872.5 lb

Why the other options are there

28,095 lb (weight used as mass)
145,709 lb (energy reported as force)

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

Example 10

Angular acceleration of a rotating drum — Kinetics of a Rigid Body (5)

A drum with mass moment of inertia 25.5 kg·m² is driven by a constant torque of 15 N·m from rest. Find α and the angular speed after 4.5 s.

Given

I = 25.5 kg·m²
M = 15 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
$α = 15/25.5 = 0.588 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 0.588 (4.5) = 2.65 r a d / s$
Convert
$ω = 25.3 r p m$

Answer: α ≈ 0.59 rad/s²; ω ≈ 2.6 rad/s (25 rpm)

Why the other options are there

382.5 rad/s² (multiplied instead of divided)
0.42 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Kinetics of a Rigid Body

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

Kinetics of a Rigid Body contains 4 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.