General Plane Motion of a Rigid Body

General Plane Motion of a Rigid Body

A rigid body undergoes general plane motion when:

• every material point moves in a plane parallel to one fixed plane,
• the body can translate,
• the body can rotate about an axis perpendicular to that plane.

The motion can always be decomposed into:

(1)

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

(2)

where

(3)

is fixed in the body.

Differentiating gives the fundamental relative-velocity relation

(4)

Differentiating again gives the relative-acceleration relation

(5)

These two equations are the central kinematic tools for planar rigid body mechanisms.

1 What makes the motion planar?

Choose the xy plane as the plane of motion.

Then the angular velocity and angular acceleration are perpendicular to that plane:

ω	= ωez,	(6)
α	= αez.	(7)

Position, velocity, and acceleration of all material points lie in the xy plane.

Thus the three-dimensional vector equations reduce to a very useful two-dimensional geometry.

2 The rigid body constraint

For any two material points A and B,

(8)

Although

(9)

can rotate in an inertial frame, its magnitude remains fixed.

That is the kinematic meaning of rigidity.

3 Translation plus rotation about an arbitrary point

The reference point A can be any material point on the rigid body.

Then

(10)

The body motion is described by:

• translation of A through r_A(t),
• orientation angle 𝜃(t) of the body.

For plane motion, a rigid body has three configuration variables:

(11)

These are the three degrees of freedom of an unconstrained rigid body in a plane.

4 Derivation of the relative-velocity relation

Differentiate

(12)

Then

(13)

Because

(14)

is fixed in the rotating body,

(15)

Therefore

(16)

The second term is the velocity of B relative to A due to rigid body rotation.

5 Geometric properties of relative velocity

Because

(17)

the relative velocity is perpendicular to

(18)

Thus

(19)

This is an immediate kinematic consequence of rigidity.

If the two points had relative velocity along the line joining them, their separation would be changing.

6 Planar component form

Let

(20)

and

(21)

Then

ω × rB∕A	= ωez × (xex + yey)	(22)
	= −ωy ex + ωxey.	(23)

Therefore

(24)

(25)

These equations are often convenient for mechanisms.

7 Example 1: velocity of a second point

A rigid bar contains points A and B with

(26)

At an instant,

(27)

and

(28)

Then

ω × rB∕A	= 2ez × (0.60ex + 0.80ey)	(29)
	= −1.60ex + 1.20ey.	(30)

Thus

vB	= 3.0ex − 1.60ex + 1.20ey	(31)
	= (1.40ex + 1.20ey) m∕s.	(32)

8 Derivation of the relative-acceleration relation

Differentiate the velocity relation:

(33)

Then

aB

= aA + × rB∕A + ω ×.

(34)

Use

(35)

and

(36)

Therefore

(37)

9 Tangential and normal relative acceleration

The relative acceleration can be separated into:

(38)

The tangential term is

(39)

Its magnitude is

(40)

The Normal term is

(41)

For planar motion,

(42)

Thus the normal component points from B toward A.

10 Planar acceleration component form

Let

(43)

(44)

Then

(45)

and

(46)

Therefore

(47)

(48)

11 Example 2: acceleration of a second point

Use

(49)

Suppose

(50)

(51)

Tangential relative acceleration:

α× rB∕A	= 3ez × (0.60ex + 0.80ey)	(52)
	= −2.40ex + 1.80ey.	(53)

Normal relative acceleration:

− ω2r B∕A	= −4(0.60ex + 0.80ey)	(54)
	= −2.40ex − 3.20ey.	(55)

Hence

aB	= (1.0, 0.5) + (−2.4, 1.8) + (−2.4,−3.2)	(56)
	= (−3.8ex − 0.9ey) m∕s2.	(57)

12 Choosing the reference point

Any material point can be used as the reference point.

Common choices include:

• a point with known velocity,
• a point constrained to move along a known path,
• the center of mass,
• a fixed pin,
• a contact point in a rolling problem.

For dynamics and energy, the center of mass is usually especially convenient.

For mechanism kinematics, a constrained endpoint is often better.

13 Special case: pure translation

If

(58)

then

(59)

If also

(60)

then

(61)

All points share the same velocity and acceleration.

The body orientation remains constant.

14 Special case: fixed-axis rotation

If point A lies on a fixed axis and is stationary,

(62)

Then

(63)

and

(64)

Thus fixed-axis rotation is a special case of general plane motion.

15 Special case: rolling

For pure rolling on a stationary surface, the contact point has

(65)

Therefore at that instant,

(66)

This makes the contact point an instantaneous center of zero velocity.

Again, this statement concerns velocity, not necessarily acceleration.

16 Instantaneous center of zero velocity

For any planar rigid body motion with

(67)

there is generally an instantaneous point I in the plane whose velocity is zero:

(68)

Then for every material point P,

(69)

Therefore

(70)

At that instant, the velocity field looks exactly like pure rotation about I.

17 Constructing the instantaneous center

If velocity directions are known at two different points A and B:

1. draw a line through A perpendicular to v_A,
2. draw a line through B perpendicular to v_B,
3. their intersection is the instantaneous center I.

Why?

For instantaneous rotation about I,

(71)

and

(72)

Thus I must lie on both perpendicular lines.

18 Instantaneous center for pure translation

If the body is in pure translation,

(73)

All point velocities are parallel and equal.

The perpendicular lines used to locate the instantaneous center are parallel and do not meet at a finite point.

One may interpret the instantaneous center as lying at infinity.

This is a useful geometric way to see pure translation as a limiting form of planar motion.

19 Example 3: sliding bar

A rigid bar of length L has endpoint A constrained to move horizontally and endpoint B constrained to move vertically.

Suppose at one instant:

(74)

with

(75)

The velocity of A is horizontal.

Therefore the line from A to the instantaneous center is vertical.

The velocity of B is vertical.

Therefore the line from B to the instantaneous center is horizontal.

The instantaneous center is at

(76)

Thus

(77)

The speed relations are

(78)

(79)

Hence

(80)

20 Example 4: numerical sliding bar

Let

(81)

and at one instant

(82)

Suppose endpoint A moves to the right at

(83)

Then

(84)

Therefore

(85)

The direction of v_B follows from the rigid body rotation sense.

21 Instantaneous center is a velocity concept

The instantaneous center satisfies

(86)

at one instant.

It does not generally satisfy

(87)

Therefore one generally cannot write

(88)

unless the acceleration of I is also properly included.

The correct relation is

(89)

The instantaneous center is therefore extremely useful for velocities but must be used cautiously for accelerations.

22 Velocity field of a planar rigid body

Choose reference point A.

Then

(90)

This is an affine velocity field.

Every point velocity is the sum of:

• one common translational velocity,
• a rotational velocity that grows linearly with distance from the reference point.

Differences in point velocities are determined entirely by ω:

(91)

23 Acceleration field of a planar rigid body

Similarly,

(92)

Unlike velocity, acceleration contains:

• a translational contribution,
• a tangential rotational contribution,
• a normal rotational contribution.

The normal term is quadratic in angular speed.

24 Determining angular velocity from two point velocities

From

(93)

the relative velocity magnitude is

(94)

because the relative velocity is perpendicular to r_B∕A.

Thus, if the full velocity vectors are known,

(95)

The sign follows from the rotational sense.

25 Determining angular acceleration from two point accelerations

The acceleration difference is

(96)

for planar motion.

Project along r_B∕A:

(97)

Project perpendicular to r_B∕A:

(98)

Therefore, once ω is known, acceleration data can determine α.

26 Center of mass as reference point

Choosing

(99)

gives

(100)

and

(101)

This form naturally separates translation of the center of mass from rotation about the center of mass.

27 Kinetic energy of general plane motion

For a rigid body undergoing general plane motion,

(102)

The first term is translational kinetic energy of the center of mass.

The second is rotational kinetic energy about the center of mass.

This decomposition is valid for arbitrary plane motion, not only rolling.

28 Example 5: kinetic energy of a moving bar

A slender bar has

(103)

Its center of mass speed is

(104)

and its angular speed is

(105)

For a slender bar about its center,

(106)

Thus

K	= (6.0)(3.0)2 + (2.0)(4.0)2	(107)
	= 27 + 16	(108)
	= 43 J.	(109)

29 Constraints in mechanisms

General plane motion often appears in mechanisms.

Examples include:

• connecting rods,
• sliders,
• crank mechanisms,
• robotic links,
• doors and linkages,
• rolling wheels.

Constraints provide known velocity or acceleration directions.

The rigid body relations then connect the constrained points.

A typical workflow is:

1. identify the rigid link,
2. choose two useful points,
3. write the known direction of each constrained velocity,
4. use the relative-velocity relation to solve for unknown speeds and ω,
5. use the relative-acceleration relation to solve for accelerations and α.

30 Velocity polygons

The equation

(110)

can be represented graphically as a vector triangle.

Since

(111)

the direction of the relative-velocity vector is known once the body orientation is known.

Velocity polygons were historically important for mechanism design and remain useful for geometric intuition.

31 Acceleration analysis is harder than velocity analysis

Velocity relative to a reference point has only one rotational term:

(112)

Acceleration has two:

(113)

and

(114)

Thus acceleration analysis requires both:

• angular acceleration,
• angular speed.

A mechanism can have zero angular acceleration at an instant while still having substantial normal relative acceleration because ω≠0.

32 Example 6: zero angular acceleration does not mean equal accelerations

Suppose a rigid link rotates instantaneously with

(115)

and

(116)

Then

(117)

Its magnitude is

|aB − aA|	= ω2r B∕A	(118)
	= 25(0.40)	(119)
	= 10 m∕s2.	(120)

Even though

(121)

the point accelerations are not equal.

33 Motion taxonomy

The major planar rigid body motions fit into one hierarchy.

• Pure translation:

  (122)

• Fixed-axis rotation: one point or axis is fixed.
• Rolling Without Slipping: the contact point is an instantaneous center of zero velocity.
• General plane motion: translation and rotation occur simultaneously without requiring a permanently fixed point.

34 Common mistakes

1. Treating general plane motion as pure translation.
2. Assuming every point on the body has the same velocity.
3. Forgetting that v_B − v_A is perpendicular to r_B∕A.
4. Using scalar speed addition instead of vector addition.
5. Forgetting the normal acceleration term −ω²r _B∕A.
6. Assuming α = 0 means all material points have equal acceleration.
7. Treating the instantaneous center as a permanently fixed material point.
8. Assuming the instantaneous center has zero acceleration because it has zero velocity.
9. Using instantaneous-center rotation formulas blindly for acceleration.
10. Choosing an awkward reference point when another point has a known velocity or acceleration.
11. Confusing the center of mass with the instantaneous center.
12. Forgetting that pure translation corresponds to ω = 0.
13. Forgetting that fixed-axis rotation is a special case with a stationary reference point.
14. Ignoring constraint directions in mechanism problems.
15. Using a velocity relation to infer acceleration without differentiating the rotating relative-position vector.

35 Practice exercises

1. Starting from r_B = r_A + r_B∕A, derive the relative-velocity equation.
2. Derive the relative-acceleration equation by differentiating the velocity relation.
3. Prove that r_B∕A ⋅ (v_B − v_A) = 0 for a rigid body.
4. Derive the planar component equations for v_B,x and v_B,y.
5. Derive the planar component equations for a_B,x and a_B,y.
6. A rigid link has r_B∕A = (0.3e_x + 0.4e_y) m, v_A = 2e_x m∕s, and ω = 3 rad∕s. Find v_B.
7. For the previous geometry, let a_A = 0, ω = 3 rad∕s, and α = 2 rad∕s². Find a _B.
8. Explain how to construct the instantaneous center from two known velocity directions.
9. Explain why the instantaneous center is at infinity for pure translation.
10. A 5 m sliding bar has endpoints at (4, 0) and (0, 3) m. If the horizontal endpoint speed is 6 m/s, find ω and the speed of the vertical endpoint.
11. Show why the instantaneous center is generally unsuitable as a fixed pivot for acceleration calculations.
12. Derive the kinetic-energy decomposition K = Mv_CM² + I_CMω².
13. A bar has M = 8 kg, L = 3 m, v_CM = 2 m/s, and ω = 3 rad/s. Find its total kinetic energy.
14. Show that if α = 0 but ω≠0, two points can still have different accelerations.
15. Classify pure translation, fixed-axis rotation, rolling, and general plane motion within one common rigid body framework.

36 Summary

For any two material points A and B on a planar rigid body,

(123)

The relative velocity is perpendicular to the line joining the points:

(124)

Acceleration is

(125)

For planar motion,

(126)

When ω≠0, an instantaneous center of zero velocity can often simplify velocity analysis:

(127)

But the instantaneous center is not generally a fixed pivot for acceleration analysis.

The kinetic energy of general plane motion is

(128)

This completes the planar kinematics foundation before M05-09 introduces the inertia tensor and full three-dimensional Rotational Inertia.

References

References