Angular Velocity and Angular Acceleration as Vectors

Angular Velocity and Angular Acceleration as Vectors

M05-01 introduced angular position, angular velocity, and angular acceleration for fixed-axis rotation. In that setting, a signed scalar ω was sufficient because the axis was fixed and only one rotational degree of freedom was active.

Three-dimensional rigid body motion requires a more powerful description.

The instantaneous angular velocity is a vector

(1)

whose:

• direction gives the instantaneous axis of rotation,
• sense follows the right-hand rule,
• magnitude gives the instantaneous angular speed.

The angular acceleration vector is

(2)

when the derivative is taken in an inertial frame.

For a material point at position ρ from a point on a fixed rotation axis,

(3)

and

(4)

These vector equations are the foundation for rigid body kinematics in three dimensions.

1 Why angular velocity needs a direction

A scalar angular speed tells how rapidly orientation changes, but not the axis about which it changes.

Consider two disks spinning at the same rate:

• one rotates counterclockwise when viewed from above,
• the other rotates clockwise.

Their angular speed magnitudes are the same, but their rotational senses are opposite.

The angular velocity vector resolves this ambiguity.

For rotation about a fixed axis with unit vector e, write

(5)

The direction of e is chosen by the right-hand rule.

2 Angular velocity is an axial vector

Position, velocity, force, and acceleration are examples of polar vectors. Their directions correspond directly to directed line segments in space.

Angular velocity is an axial vector, also called a pseudovector.

It is associated with an oriented axis rather than a displacement from one point to another.

The distinction matters mainly under spatial reflection. Under ordinary proper rotations of the coordinate system, axial vectors transform with the same component rules as familiar vectors.

For standard mechanics calculations involving rotations of coordinate axes, ω can be manipulated with the usual vector and cross-product operations.

3 Finite rotations are not ordinary vectors

It is tempting to treat a finite rotation angle as an ordinary vector.

That fails in three dimensions because finite rotations generally do not commute.

For example:

• rotate an object 90^∘ about the x axis, then 90^∘ about the y axis,
• reverse the order of those rotations.

The final orientations are different.

Therefore

(6)

4 Infinitesimal rotations behave vectorially

The situation changes for an infinitesimal rotation.

Let

(7)

be a very small rotation vector whose magnitude is the small angle d𝜃 and whose direction follows the right-hand rule.

For a point with position vector ρ from the instantaneous rotation axis, the first-order displacement produced by the rotation is

(8)

Terms of order

(9)

and higher are neglected.

This is why the instantaneous angular velocity can be treated as a vector even though a finite rotation cannot generally be represented by ordinary vector addition.

5 Definition of angular velocity vector

Divide the infinitesimal rotation vector by the elapsed time:

(10)

Its magnitude is

(11)

and has SI units

(12)

For fixed-axis rotation,

(13)

where e is a constant axis direction.

In general three-dimensional motion, both the magnitude and direction of ω can vary with time.

6 Deriving the point-velocity equation

Start with the infinitesimal displacement

(14)

Divide by dt:

(15)

Therefore

(16)

The cross product automatically enforces the correct geometry:

• v is perpendicular to ω,
• v is perpendicular to ρ’s component away from the axis,
• the direction is tangent to the circular path.

7 Recovering the scalar tangential-speed equation

Take the magnitude:

(17)

where ϕ is the angle between ω and ρ.

The perpendicular distance from the point to the rotation axis is

(18)

Therefore

(19)

Thus the familiar scalar equation is the magnitude of the vector cross-product relation.

8 Example 1: velocity from a cross product

Let

(20)

and

(21)

Then

v	= ω ×ρ	(22)
	= 4ez × (3ex + 2ey + 5ez)	(23)
	= 12ey − 8ex.	(24)

Therefore

(25)

The z component of ρ does not contribute because displacement along the rotation axis does not increase the perpendicular radius.

The speed is

(26)

The perpendicular distance is

(27)

so

(28)

in agreement with the vector result.

9 Angular acceleration as a vector derivative

Angular acceleration is defined by

(29)

in an inertial frame.

This derivative can be nonzero for two distinct reasons:

• the magnitude of ω changes,
• the direction of ω changes.

Therefore a body can have constant angular speed while still having nonzero angular acceleration if its instantaneous rotation axis changes direction.

10 Fixed-axis special case

If the axis direction e is constant,

(30)

Differentiate:

(31)

Thus

(32)

for fixed-axis rotation, except at instants when one of the vectors is zero.

In this special case, the signed scalar angular acceleration from M05-01 is simply the component of α along the fixed axis.

11 Changing-axis example

Suppose

(33)

where ω₀ and Ω are constants.

The angular speed is constant:

(34)

But

α	=	(35)
	= ω0Ω .	(36)

Therefore

(37)

even though the angular speed never changes.

Also,

(38)

so α is perpendicular to ω in this example.

12 Derivative of a body-fixed vector

Let A be a vector fixed in a rotating rigid body. Its components in the body remain constant, but its inertial direction changes.

During an infinitesimal rotation,

(39)

Divide by dt:

(40)

for a vector whose body-fixed components are constant.

This result is a special case of the rotating-frame transport theorem developed later in the mechanics sequence.

13 Deriving the acceleration equation

For a material point rotating about a fixed point or fixed axis,

(41)

Differentiate in the inertial frame:

a	=	(42)
	= ×ρ + ω ×.	(43)

Since

(44)

we obtain

(45)

The first term is associated with changing angular velocity.

The second term is the inward Normal acceleration.

14 Why the double cross product points inward

Use the vector triple-product identity

(46)

Set

(47)

Then

(48)

Decompose ρ into components parallel and perpendicular to ω:

(49)

Because

(50)

we obtain

(51)

The term points directly toward the rotation axis.

Its magnitude is

(52)

15 Example 2: fixed-axis acceleration

Let

ω	= 4ez rad∕s,	(53)
α	= 2ez rad∕s2,	(54)
ρ	= (3ex + 2ey + 5ez) m.	(55)

Tangential contribution:

α×ρ	= 2ez × (3ex + 2ey + 5ez)	(56)
	= −4ex + 6ey.	(57)

Normal contribution:

ω × (ω ×ρ)	= −ω2(3e x + 2ey)	(58)
	= −48ex − 32ey.	(59)

Therefore

(60)

No acceleration component arises from the point’s z coordinate because displacement along the axis does not contribute to rotational motion around that axis.

16 Relative velocity of two points on a rigid body

Let A and B be two material points on the same rigid body.

Define

(61)

Because r_B∕A is fixed in the body, its inertial derivative is

(62)

But

(63)

Therefore

(64)

This relation is valid even when the body is translating and rotating simultaneously.

17 Relative acceleration of two points on a rigid body

Differentiate the relative-velocity equation:

(65)

Then

aB

= aA + α× rB∕A + ω ×.

(66)

Since

(67)

we obtain

(68)

This is one of the central equations of rigid body kinematics.

18 Example 3: translating and rotating rigid body

At one instant, point A has velocity

(69)

The body has

(70)

and

(71)

Then

ω × rB∕A	= 3ez × (2ex + ey)	(72)
	= −3ex + 6ey.	(73)

Therefore

vB	= (2ex + ey) + (−3ex + 6ey)	(74)
	= −ex + 7ey.	(75)

Thus

(76)

19 The angular velocity vector is common to the whole rigid body

Different material points have different linear velocities, but the same instantaneous angular velocity describes the orientation change of the entire rigid body.

Thus ω is a property of the body’s instantaneous rotational motion, not of one particular particle.

For any pair of body points A and B,

(77)

The same ω must satisfy this relation throughout the rigid body.

20 A useful rigidity check

Because the distance between material points is constant,

(78)

Differentiate:

(79)

Therefore

(80)

The relative velocity of two points on a rigid body is always perpendicular to their separation vector.

This is consistent with

(81)

21 Component form of the velocity equation

Let

(82)

and

(83)

Then

(84)

gives

vx	= ωyz − ωzy,	(85)
vy	= ωzx − ωxz,	(86)
vz	= ωxy − ωyx.	(87)

These equations are often useful for direct component calculations.

22 Skew-symmetric cross-product matrix

The cross product with ω can be represented by a matrix:

(88)

Then

(89)

The matrix is skew-symmetric:

(90)

23 Why the matrix is skew-symmetric

For any vector x,

(91)

The cross product is perpendicular to x, so the quadratic form vanishes.

Skew-symmetric matrices are the natural infinitesimal generators of three-dimensional rotations.

This observation leads directly to rotation matrices and attitude kinematics in more advanced treatments.

24 Angular acceleration in matrix form

Define

(92)

in an inertial basis.

For a point fixed relative to a rotation center,

(93)

Thus

(94)

This compact matrix form is especially useful in computation.

25 Instantaneous axis interpretation

At any instant for which

(95)

there is an instantaneous rotation axis parallel to ω.

Points lying on that axis have no velocity due to the instantaneous rotation about that axis because

(96)

implies

(97)

For a translating and rotating body, however, points on the instantaneous rotational axis can still have the translational velocity of the chosen reference point.

26 Angular speed versus angular velocity

Angular speed is the magnitude

(98)

Differentiate its square:

(99)

Then

(100)

For ω≠0,

(101)

Only the component of α parallel to ω changes the angular speed.

The perpendicular component changes the direction of the angular velocity vector.

27 Parallel and perpendicular components of angular acceleration

Write

(102)

The parallel component is

(103)

The perpendicular component is

(104)

Then:

• α_∥ changes |ω|,
• α_⊥ changes the direction of ω.

This decomposition is useful in gyroscope and precession problems later in the M05 sequence.

28 Example 4: separating magnitude and direction change

Suppose

(105)

and

(106)

The parallel component is

(107)

The perpendicular component is

(108)

The instantaneous rate of change of angular speed is

(109)

The 3e_x component changes the axis direction but does not instantaneously change |ω|.

29 Common mistakes

1. Treating a finite three-dimensional rotation as an ordinary vector that can be freely added in any order.
2. Forgetting that angular velocity is associated with an axis and a rotational sense.
3. Confusing angular speed ω with angular velocity ω.
4. Assuming α must always be parallel to ω.
5. Assuming constant angular speed implies α = 0 when the axis direction changes.
6. Using v = ω ×ρ with the cross-product order reversed.
7. Using the full distance |ρ| instead of the perpendicular radius when applying v = ωr_⊥.
8. Forgetting the double-cross-product normal-acceleration term.
9. Assigning an outward sign to ω × (ω ×ρ).
10. Using a different angular velocity vector for different points of the same rigid body.
11. Applying the fixed-center equation v = ω × ρ to a translating body without adding the reference-point velocity.
12. Forgetting that v_B − v_A is perpendicular to r_B∕A for a rigid body.
13. Confusing the angular velocity cross-product matrix with an ordinary symmetric transformation matrix.
14. Using a body-frame derivative of ω when the definition of α requires an inertial derivative.

30 Practice exercises

1. Explain why finite rotations about different axes cannot generally be added as ordinary vectors.
2. Starting from an infinitesimal rotation, derive dr = d𝜃 ×ρ to first order.
3. Derive v = ω ×ρ.
4. Show that the magnitude of ω ×ρ is ωr_⊥.
5. Let ω = 2e_z mrad∕s and ρ = 4e_x + 3e_y mm. Find v.
6. A body has constant |ω| but its axis direction changes. Explain why α can be nonzero.
7. For ω = ω₀[cos(Ωt)e_x + sin(Ωt)e_y], derive α and show that ω ⋅α = 0.
8. Derive a = α×ρ + ω × (ω ×ρ).
9. Use the vector triple-product identity to show that the normal term is −ω²ρ _⊥.
10. Derive the rigid body relative-velocity equation v_B = v_A + ω × r_B∕A.
11. Derive the corresponding relative-acceleration equation.
12. Show directly from the rigid body distance constraint that r_B∕A ⋅ (v_B − v_A) = 0.
13. Construct [ω×] for ω = ω_xe_x + ω_ye_y + ω_ze_z and verify it is skew-symmetric.
14. Prove that ω = (ω ⋅α)∕ω for nonzero ω.
15. Decompose an arbitrary α into components parallel and perpendicular to ω and explain the physical meaning of each.

31 Summary

Angular velocity is the instantaneous axial vector describing the rotational motion of a rigid body:

(110)

in the infinitesimal-rotation sense.

Finite three-dimensional rotations do not generally commute, but infinitesimal rotations do combine vectorially to first order.

For a point fixed in the body relative to a rotation center,

(111)

Angular acceleration is

(112)

Point acceleration is

(113)

For any two points A and B on the same rigid body,

(114)

and

(115)

The cross-product matrix

(116)

provides a compact matrix representation of the same geometry and prepares the way for full three-dimensional rigid body kinematics.

References

References