Instantaneous Center of Rotation (Instant Centers)

Dynamics · FE Reference Handbook section

Dynamics

2 formulas

10 exam-style examples

~49 min

All Dynamics lectures

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
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
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
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
$M = I\alpha$
Substituting
$\alpha = 94/11.0 = 8.545 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 8.545(2.5) = 21.36 rad/s$
Convert
$\omega = 204.0 rpm$

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
$M = I\alpha$
Substituting
$\alpha = 190/9.5 = 20.000 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 20.000(2.5) = 50.00 rad/s$
Convert
$\omega = 477.5 rpm$

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
$M = I\alpha$
Substituting
$\alpha = 157/8.0 = 19.625 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 19.625(4.5) = 88.31 rad/s$
Convert
$\omega = 843.3 rpm$

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
$M = I\alpha$
Substituting
$\alpha = 67/29.0 = 2.310 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 2.310(3.5) = 8.09 rad/s$
Convert
$\omega = 77.2 rpm$

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
$M = I\alpha$
Substituting
$\alpha = 44/20.5 = 2.146 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 2.146(5.5) = 11.80 rad/s$
Convert
$\omega = 112.7 rpm$

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
$M = I\alpha$
Substituting
$\alpha = 112/27.0 = 4.148 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 4.148(2.5) = 10.37 rad/s$
Convert
$\omega = 99.0 rpm$

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
$M = I\alpha$
Substituting
$\alpha = 61/11.5 = 5.304 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 5.304(10.0) = 53.04 rad/s$
Convert
$\omega = 506.5 rpm$

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
$M = I\alpha$
Substituting
$\alpha = 135/24.5 = 5.510 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 5.510(6.5) = 35.82 rad/s$
Convert
$\omega = 342.0 rpm$

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
$M = I\alpha$
Substituting
$\alpha = 22/11.5 = 1.913 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 1.913(8.0) = 15.30 rad/s$
Convert
$\omega = 146.1 rpm$

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
$M = I\alpha$
Substituting
$\alpha = 159/15.0 = 10.600 rad/s^{2}$
Angular speed — ω = ω₀ + αt
Substituting
$\omega = 0 + 10.600(6.0) = 63.60 rad/s$
Convert
$\omega = 607.3 rpm$

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)