Fixed-Axis Rotation and Particle Kinematics

Fixed-Axis Rotation and Particle Kinematics

A rigid body in fixed-axis rotation provides one of the clearest examples of how a single angular motion generates different linear motions throughout an extended body.

Let the fixed rotation axis be the z axis. A material point P remains at a constant perpendicular distance ρ from the axis and moves on a circle as the body rotates through angle 𝜃(t).

For that point,

(1)

where ρ and z are constant for a material point in fixed-axis rotation.

The entire time dependence is carried by the rotating cylindrical basis vectors e_r and e_𝜃.

The angular variables are

(2)

The particle velocity is

(3)

and the particle acceleration is

(4)

Thus every material point has:

• tangential velocity proportional to distance from the axis,
• inward normal acceleration proportional to ω²,
• tangential acceleration proportional to α.

1 Geometry of fixed-axis rotation

Choose a point O on the fixed axis and let

(5)

be the perpendicular displacement from the axis to point P.

For an axis along e_z,

(6)

The distance

(7)

is constant for a material point of a rigid body.

The height

(8)

along the axis is also constant.

Therefore a material point traces a circle in a plane perpendicular to the axis.

2 Cylindrical basis vectors

In the plane perpendicular to the axis, define

er	= cos 𝜃 ex + sin 𝜃 ey,	(9)
e𝜃	= − sin 𝜃 ex + cos 𝜃 ey.	(10)

The radial vector e_r points outward from the axis.

The transverse vector e_𝜃 points in the direction of increasing 𝜃.

These basis vectors rotate with the body.

3 Time derivatives of the rotating basis

Differentiate

(11)

Because e_x and e_y are fixed in the inertial frame,

er	= − sin 𝜃 𝜃ex + cos 𝜃 𝜃ey	(12)
	= 𝜃e𝜃.	(13)

Therefore

(14)

Similarly,

(15)

These two derivative identities generate the particle velocity and acceleration formulas.

4 Derivation of particle velocity

The position of a material point is

(16)

For fixed-axis rigid body motion,

(17)

Differentiate:

vP	= rP	(18)
	= ρer	(19)
	= ρωe𝜃.	(20)

Thus

(21)

The velocity is tangent to the circular path.

Its magnitude is

(22)

5 Vector form of particle velocity

For a fixed axis with angular-velocity vector

(23)

we have

(24)

The component of r_P∕O parallel to the axis contributes nothing to the cross product.

Therefore only the perpendicular distance matters:

(25)

This is why the correct radius in fixed-axis rotation is the perpendicular distance to the axis, not necessarily the distance to the coordinate origin.

6 Example 1: particle velocity on a rotating shaft assembly

A point lies

(26)

from a fixed shaft axis. At an instant,

(27)

Its speed is

v	= ρω	(28)
	= (0.40)(12)	(29)
	= 4.8 m∕s.	(30)

Thus

(31)

A second point at twice the radius would have twice the speed even though it has exactly the same angular velocity.

7 Derivation of particle acceleration

Start with

(32)

Differentiate:

aP	= ρωe𝜃 + ρωe𝜃	(33)
	= ραe𝜃 − ρω2e r.	(34)

Therefore

(35)

The two terms have distinct physical roles.

8 Normal acceleration

The radial component is

(36)

Its magnitude is

(37)

It points toward the axis.

Since

(38)

we recover the familiar circular-motion form

(39)

9 Tangential acceleration

The transverse component is

(40)

Its magnitude is

(41)

Tangential acceleration changes the magnitude of the tangential velocity.

Normal acceleration changes its direction.

10 Acceleration magnitude and direction

Because e_r and e_𝜃 are perpendicular,

(42)

If α = 0,

(43)

and the acceleration is purely inward.

If ω = 0 but α≠0 at one instant,

(44)

and the acceleration is purely tangential at that instant.

11 Example 2: acceleration components

A point lies at

(45)

on a rotating body. At an instant,

ω	= 8.0 rad∕s,	(46)
α	= −3.0 rad∕s2.	(47)

The normal acceleration magnitude is

an	= ρω2	(48)
	= (0.25)(64)	(49)
	= 16 m∕s2.	(50)

The signed tangential component is

(51)

The total magnitude is

a	=	(52)
	≈ 16.02 m∕s2.	(53)

Thus

(54)

The Normal term dominates strongly because of the square on ω.

12 Cartesian description of a rotating particle

For a point at radius ρ,

x	= ρ cos 𝜃,	(55)
y	= ρ sin 𝜃.	(56)

Differentiate:

ẋ	= −ρω sin 𝜃,	(57)
ẏ	= ρω cos 𝜃.	(58)

Therefore

(59)

Differentiate again:

ẍ	= −ρα sin 𝜃 − ρω2 cos 𝜃,	(60)
ÿ	= ρα cos 𝜃 − ρω2 sin 𝜃.	(61)

Thus

(62)

13 Example 3: Cartesian particle kinematics

Let

ρ	= 0.50 m,	(63)
𝜃	= 30∘,	(64)
ω	= 4.0 rad∕s,	(65)
α	= 2.0 rad∕s2.	(66)

Using

(67)

we obtain

vx	= −(0.50)(4)(0.5) = −1.0 m∕s,	(68)
vy	= (0.50)(4) ≈ 1.73 m∕s.	(69)

Therefore

(70)

The acceleration components are

ax	= −(0.50)(2)(0.5) − (0.50)(16)	(71)
	≈−7.43 m∕s2,	(72)
ay	= (0.50)(2) − (0.50)(16)(0.5)	(73)
	≈−3.13 m∕s2.	(74)

Thus

(75)

14 Fixed-axis kinematics from cross products

M05-02 introduced

(76)

Differentiate while O remains fixed in the inertial frame:

aP	= × rP∕O + ω ×	(77)
	= α× rP∕O + ω × vP .	(78)

Substituting

(79)

gives

(80)

The first term is tangential.

The second term points toward the axis.

15 Using the vector triple product

Use

(81)

Then

(82)

Decompose

(83)

where r_∥ is parallel to the axis and r_⊥ is perpendicular to it.

Because

(84)

we obtain

(85)

This makes the inward direction explicit.

16 Two particles on the same rigid body

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

For fixed-axis rotation,

(86)

Similarly,

(87)

These equations relate any two material points without first referring both to the axis.

17 Rigidity check from relative velocity

The distance between two material points remains fixed:

(88)

Differentiate:

(89)

Therefore

(90)

The relative velocity of two points on a rigid body is perpendicular to the line joining them.

This is exactly consistent with

(91)

18 Velocity field of a rotating rigid body

At any instant, the entire fixed-axis velocity field is determined by

(92)

For each material point,

(93)

This field has several properties:

• velocity vanishes on the rotation axis,
• speed grows linearly with perpendicular radius,
• velocity is tangent to circles around the axis,
• neighboring points have velocities that differ linearly with separation.

19 Acceleration field of a rotating rigid body

At the same instant,

(94)

The tangential part grows linearly with radius:

(95)

The normal part also grows linearly with radius at fixed ω:

(96)

Thus points farther from the axis experience proportionally larger linear accelerations.

20 Three-dimensional point positions

A point need not lie in the plane z = 0.

Let

(97)

The axial coordinate z remains constant.

Because

(98)

the axial component does not contribute to

(99)

Therefore

(100)

regardless of the value of z.

Likewise,

(101)

The kinematics depend only on perpendicular distance from the axis.

21 Example 4: a point offset along the axis

A material point has Cartesian coordinates

(102)

The body rotates about the z axis with

ω	= 5ez rad∕s,	(103)
α	= 2ez rad∕s2.	(104)

The perpendicular radius is

(105)

Therefore

(106)

The tangential acceleration magnitude is

(107)

and the normal acceleration magnitude is

(108)

The 1.20 m axial offset does not change these magnitudes.

22 Example 5: relative velocity between two points

Let

(109)

and

(110)

Then

vB − vA	= 6ez × (0.20ex + 0.10ey)	(111)
	= 1.20ey − 0.60ex.	(112)

Thus

(113)

Check rigidity:

rB∕A ⋅ (vB − vA)	= (0.20)(−0.60) + (0.10)(1.20)	(114)
	= 0.	(115)

The relative velocity is perpendicular to the separation vector, as required.

23 Kinematics of a rotating line segment

Consider two material points A and B connected by a rigid line segment.

If the segment rotates about a fixed axis, then the direction of

(116)

changes while its magnitude remains constant.

The relative velocity

(117)

is perpendicular to the segment.

The relative acceleration has two parts:

(118)

The first changes the relative tangential speed.

The second bends the relative velocity direction inward.

24 Angular displacement from linear displacement

If a material point moves through tangential arc length s at fixed radius ρ,

(119)

Likewise, if its tangential speed is known,

(120)

for ρ≠0.

If tangential acceleration is known,

(121)

These inverse relations let a measured particle motion reveal the angular motion of the entire rigid body.

25 Example 6: recovering body angular motion from a particle

A marked point on a rotating wheel lies

(122)

from the axis. At an instant its tangential speed is

(123)

and its tangential acceleration magnitude is

(124)

Then

(125)

and

(126)

The directions or signs require the observed directions of motion and acceleration.

26 Connection to particle kinetic energy

For one material particle of mass m_i at perpendicular radius ρ_i,

(127)

Its kinetic energy is therefore

Ki	= mivi2	(128)
	= miρi2ω2.	(129)

Thus

(130)

Summing this expression over all particles naturally introduces the moment of inertia.

That is the subject of M05-04.

27 Common mistakes

1. Using distance from the origin instead of perpendicular distance to the rotation axis.
2. Treating e_r and e_𝜃 as fixed basis vectors when differentiating.
3. Forgetting e_r = ωe_𝜃.
4. Forgetting e_𝜃 = −ωe_r.
5. Saying velocity points radially outward rather than tangent to the path.
6. Omitting the normal acceleration term when angular speed is constant.
7. Omitting the tangential acceleration term when angular speed changes.
8. Using a_n = ρω instead of a_n = ρω².
9. Assuming points at different radii have the same linear speed because they have the same angular speed.
10. Forgetting that axial position z does not affect fixed-axis speed or acceleration magnitude.
11. Mixing signed tangential acceleration ρα with the nonnegative magnitude ρ|α|.
12. Forgetting that ω × (ω × r) points toward the rotation axis.
13. Applying fixed-axis formulas when the axis itself translates or changes direction.
14. Confusing particle kinematics with force or torque dynamics.

28 Practice exercises

1. Starting from e_r = cos 𝜃e_x + sin 𝜃e_y, derive e_r = ωe_𝜃.
2. Derive e_𝜃 = −ωe_r.
3. Starting from r = ρe_r + ze_z with constant ρ and z, derive the fixed-axis velocity.
4. Differentiate the velocity to derive the radial and tangential acceleration terms.
5. A point at ρ = 0.35 m rotates with ω = 9 rad∕s. Find its speed and normal acceleration.
6. If the same point has α = −2 rad∕s², find its tangential acceleration and total acceleration magnitude.
7. Derive the Cartesian velocity components for fixed-axis rotation.
8. Derive the Cartesian acceleration components for fixed-axis rotation.
9. Show that the vector formula v = ω × r reduces to v = ρ|ω|.
10. Use the vector triple-product identity to prove that the double-cross-product acceleration points toward the axis.
11. Derive the relative velocity relation between two points on the same rigid body.
12. Show that r_B∕A ⋅ (v_B − v_A) = 0.
13. A point is located at (0.4, 0.3, 2.0) m relative to a z-axis rotation. Find its perpendicular radius.
14. If ω = 5 rad∕s and α = 1.5 rad∕s² for the preceding point, find its speed and both acceleration magnitudes.
15. Explain how the particle kinetic-energy expression leads naturally to the definition of moment of inertia.

29 Summary

For fixed-axis rotation,

(131)

with constant ρ and z for each material point.

The rotating basis satisfies

(132)

Therefore

(133)

and

(134)

Equivalently,

(135)

and

(136)

For two points on the same rigid body,

(137)

The particle kinetic-energy relation

(138)

provides the direct bridge to M05-04 and the moment of inertia.

References

References