Instantaneous Center of Rotation (Instant Centers)

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 Instantaneous Center of Rotation (Instant Centers) 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 instantaneous center of rotation (instant centers) 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. Instantaneous Center of Rotation (Instant Centers) 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: instantaneous center of rotation (instant centers).
Capstone Studio instructional photograph

Dynamics — Instantaneous Center of Rotation (Instant Centers): 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.

Notation used in this section

^ h	Quantity produced by "^ h" — read its definition and unit from the handbook line directly above the equation.
c	Quantity produced by "c= n n- 1" — 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.

An instantaneous center of rotation (instant center) is a point, common to two bodies, at which each has the same velocity
(magnitude and direction) at a given instant. It is also a point in space about which a body rotates, instantaneously.
2 3 B
1 GROUND
I 14 ∞
I 23
I12 I 34
The figure shows a fourbar slider-crank. Link 2 (the crank) rotates about the fixed center, O2. Link 3 couples the crank to the
slider (link 4), which slides against ground (link 1). Using the definition of an instant center (IC), we see that the pins at O2, A,
and B are ICs that are designated I12, I23, and I34. The easily observable IC is I14, which is located at infinity with its direction
perpendicular to the interface between links 1 and 4 (the direction of sliding). To locate the remaining two ICs (for a fourbar) we
must make use of Kennedy's rule.
Kennedy's Rule: When three bodies move relative to one another they have three instantaneous centers, all of which lie on the
same straight line.
To apply this rule to the slider-crank mechanism, consider links 1, 2, and 3 whose ICs are I12, I23, and I13, all of which lie on a
straight line. Consider also links 1, 3, and 4 whose ICs are I13, I34, and I14, all of which lie on a straight line. Extending the line
through I12 and I23 and the line through I34 and I14 to their intersection locates I13, which is common to the two groups of links
that were considered.
I14 ∞
I14 ∞
2 3 I34
1 GROUND
Similarly, if body groups 1, 2, 4 and 2, 3, 4 are considered, a line drawn through known ICs I12 and I14 to the intersection of a
line drawn through known ICs I23 and I34 locates I24.

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

Angular acceleration of a rotating drum — Instantaneous Center of Rotation (Instant Centers)

A drum with mass moment of inertia 11.0 kg·m² is driven by a constant torque of 94 N·m from rest. Find α and the angular speed after 2.5 s.

Given

I = 11.0 kg·m²
M = 94 N·m
t = 2.5 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
$α = 94/11.0 = 8.545 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 8.545 (2.5) = 21.36 r a d / s$
Convert
$ω = 204.0 r p m$

Answer: α ≈ 8.55 rad/s²; ω ≈ 21.4 rad/s (204.0 rpm)

Why the other options are there

1,034 rad/s² (multiplied instead of divided)
3.40 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Instantaneous Center of Rotation (Instant Centers)

Example 2

Angular acceleration of a rotating drum — Instantaneous Center of Rotation (Instant Centers) (2)

A drum with mass moment of inertia 9.5 kg·m² is driven by a constant torque of 190 N·m from rest. Find α and the angular speed after 2.5 s.

Given

I = 9.5 kg·m²
M = 190 N·m
t = 2.5 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
$α = 190/9.5 = 20.000 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 20.000 (2.5) = 50.00 r a d / s$
Convert
$ω = 477.5 r p m$

Answer: α ≈ 20.00 rad/s²; ω ≈ 50.0 rad/s (477.5 rpm)

Why the other options are there

1,805 rad/s² (multiplied instead of divided)
7.96 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Instantaneous Center of Rotation (Instant Centers)

Example 3

Angular acceleration of a rotating drum — Instantaneous Center of Rotation (Instant Centers) (3)

A drum with mass moment of inertia 8.0 kg·m² is driven by a constant torque of 157 N·m from rest. Find α and the angular speed after 4.5 s.

Given

I = 8.0 kg·m²
M = 157 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
$α = 157/8.0 = 19.625 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 19.625 (4.5) = 88.31 r a d / s$
Convert
$ω = 843.3 r p m$

Answer: α ≈ 19.63 rad/s²; ω ≈ 88.3 rad/s (843.3 rpm)

Why the other options are there

1,256 rad/s² (multiplied instead of divided)
14.06 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Instantaneous Center of Rotation (Instant Centers)

Example 4

Angular acceleration of a rotating drum — Instantaneous Center of Rotation (Instant Centers) (4)

A drum with mass moment of inertia 29.0 kg·m² is driven by a constant torque of 67 N·m from rest. Find α and the angular speed after 3.5 s.

Given

I = 29.0 kg·m²
M = 67 N·m
t = 3.5 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
$α = 67/29.0 = 2.310 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 2.310 (3.5) = 8.09 r a d / s$
Convert
$ω = 77.2 r p m$

Answer: α ≈ 2.31 rad/s²; ω ≈ 8.1 rad/s (77 rpm)

Why the other options are there

1,943 rad/s² (multiplied instead of divided)
1.29 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Instantaneous Center of Rotation (Instant Centers)

Example 5

Angular acceleration of a rotating drum — Instantaneous Center of Rotation (Instant Centers) (5)

A drum with mass moment of inertia 20.5 kg·m² is driven by a constant torque of 44 N·m from rest. Find α and the angular speed after 5.5 s.

Given

I = 20.5 kg·m²
M = 44 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
$α = 44/20.5 = 2.146 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 2.146 (5.5) = 11.80 r a d / s$
Convert
$ω = 112.7 r p m$

Answer: α ≈ 2.15 rad/s²; ω ≈ 11.8 rad/s (112.7 rpm)

Why the other options are there

902.0 rad/s² (multiplied instead of divided)
1.88 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Instantaneous Center of Rotation (Instant Centers)

Example 6

Angular acceleration of a rotating drum — Instantaneous Center of Rotation (Instant Centers) (6)

A drum with mass moment of inertia 27.0 kg·m² is driven by a constant torque of 112 N·m from rest. Find α and the angular speed after 2.5 s.

Given

I = 27.0 kg·m²
M = 112 N·m
t = 2.5 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
$α = 112/27.0 = 4.148 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 4.148 (2.5) = 10.37 r a d / s$
Convert
$ω = 99.0 r p m$

Answer: α ≈ 4.15 rad/s²; ω ≈ 10.4 rad/s (99 rpm)

Why the other options are there

3,024 rad/s² (multiplied instead of divided)
1.65 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Instantaneous Center of Rotation (Instant Centers)

Example 7

Angular acceleration of a rotating drum — Instantaneous Center of Rotation (Instant Centers) (7)

A drum with mass moment of inertia 11.5 kg·m² is driven by a constant torque of 61 N·m from rest. Find α and the angular speed after 10.0 s.

Given

I = 11.5 kg·m²
M = 61 N·m
t = 10.0 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
$α = 61/11.5 = 5.304 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 5.304 (10.0) = 53.04 r a d / s$
Convert
$ω = 506.5 r p m$

Answer: α ≈ 5.30 rad/s²; ω ≈ 53.0 rad/s (506.5 rpm)

Why the other options are there

701.5 rad/s² (multiplied instead of divided)
8.44 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Instantaneous Center of Rotation (Instant Centers)

Example 8

Angular acceleration of a rotating drum — Instantaneous Center of Rotation (Instant Centers) (8)

A drum with mass moment of inertia 24.5 kg·m² is driven by a constant torque of 135 N·m from rest. Find α and the angular speed after 6.5 s.

Given

I = 24.5 kg·m²
M = 135 N·m
t = 6.5 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
$α = 135/24.5 = 5.510 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 5.510 (6.5) = 35.82 r a d / s$
Convert
$ω = 342.0 r p m$

Answer: α ≈ 5.51 rad/s²; ω ≈ 35.8 rad/s (342.0 rpm)

Why the other options are there

3,308 rad/s² (multiplied instead of divided)
5.70 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Instantaneous Center of Rotation (Instant Centers)

Example 9

Angular acceleration of a rotating drum — Instantaneous Center of Rotation (Instant Centers) (9)

A drum with mass moment of inertia 11.5 kg·m² is driven by a constant torque of 22 N·m from rest. Find α and the angular speed after 8.0 s.

Given

I = 11.5 kg·m²
M = 22 N·m
t = 8.0 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
$α = 22/11.5 = 1.913 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 1.913 (8.0) = 15.30 r a d / s$
Convert
$ω = 146.1 r p m$

Answer: α ≈ 1.91 rad/s²; ω ≈ 15.3 rad/s (146.1 rpm)

Why the other options are there

253.0 rad/s² (multiplied instead of divided)
2.44 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Instantaneous Center of Rotation (Instant Centers)

Example 10

Angular acceleration of a rotating drum — Instantaneous Center of Rotation (Instant Centers) (10)

A drum with mass moment of inertia 15.0 kg·m² is driven by a constant torque of 159 N·m from rest. Find α and the angular speed after 6.0 s.

Given

I = 15.0 kg·m²
M = 159 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
$α = 159/15.0 = 10.600 r a d / s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$ω = 0 + 10.600 (6.0) = 63.60 r a d / s$
Convert
$ω = 607.3 r p m$

Answer: α ≈ 10.60 rad/s²; ω ≈ 63.6 rad/s (607.3 rpm)

Why the other options are there

2,385 rad/s² (multiplied instead of divided)
10.12 rad/s (revolutions confused with radians)

Reference: FE Reference Handbook — Dynamics → Instantaneous Center of Rotation (Instant Centers)

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

Instantaneous Center of Rotation (Instant Centers) 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.