# KINEMATICS OF RIGID BODIES - Department of Mechanical ...

Mechanics

Nov 14, 2013 (4 years and 7 months ago)

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Chapter 3

KINEMATICS OF RIGID BODIES

Serge Abrate

Department of Mechanical Engineering and Energy Processes

Southern Illinois University

abrate@engr.siu.edu

After studying the dynamics of particles we now
turn to the study of rigid bodies.

A rigid b
ody is
defined as a set of particles in which the distance
between any two points remains constant. This
distinguishes rigid bodies from fluids and
granular materials such as sand for example. The
distinction between rigid bodies and deformable
bodies is

based on the type of analysis that is
performed. All solids are deformable but
,

in the
types of problems covered in dynamics, the
deformations of the body are negligible
compared to the
overall motion.

1
-
TWO
-
DIMENSIONAL MOTION

Considering the motion
of a lamina in the xy
plane,

we examine two simple types of m
otion

(
translation and rotation
)

and
show that the general
motion consists of a combination of the two. Then
we introduce the concept of the instantaneous
center of rotation which is useful in m
any practical
problems.

1.1. Translation

Translation is a

type of motion in which at a

given
instant every point in the body has the same
velocity
V
. That is, the velocity has the same
direction an
d the same magnitude at every

poin
t.
That does not necessarily mean that the body
moves on a straight line since
V
can change with
time

t
V
V

.

1.2. Rotation about a fixed point

1.2.1
-

Velocity of a point in a body rotating about a fixed point

A

B

Figure 1
: In a rigid body the distance
between a
ny two points A and B remains
constant

A

B

C

V

V

V

Figure 2
:
During translation, at
each instant, the velocity vector
is the same for all the points

For a

body rotating about a fixed point O (Fig. 3), the trajectory of an arbitrary point A is a circle
A
r
.

In the radial and tangential coordinate system shown in Fig.

4, the position of A
is given by

r
A
e
r
r

(1)

Taking the derivative with respect to time, the
velocity of A is

dt
e
d
r
e
dt
dr
dt
r
d
v
r
r
A
A

(2)

Since the body is rigid, the radius r does not change
and
0
dt
dr

. Using the chain rule of differentiation,

dt
d
d
e
d
dt
e
d
r
r

(3)

As the angle

increases by an amount

d
,
the unit vector
r
e

becomes
r
e

and
t
r
r
r
e
d
e
e
e
d

. Then
,

t
r
e
d
e
d

(4)

Substituting into E
q. gives

t
A
e
r
v

(5)

Generally

denotes the angular velocity

so the velocity of A can also be written
as

t
A
e
r
v

(6)

As the angle

increases by an amount

d
, point A travels a distance

rd
.

If the magnitude of
the velocity is v, the
distance traveled is a
lso equal to vdt. The equality

dt
v
rd

gives the
following re
lationship between the linear velocity at A an
d

t
he angular velocity of the body

r
dt
d
r
v

(7)

A

Figure 3
:
fixed

point O

x

y

O

A
r
x

O

A
r
A

t
e
r
e
y

Figure 4

so

t
A
e
v
v

(8)

Eqs. 5,6 and 8 indicate that in the case of a rotation about a fixed point, the velocity is always

oriented in the tangential direction and that its magnitude is increases linearly with the radius r
and the angular velocity.

1.2.2
-

Acceleration

of a point in a body rotating about a fixed point

To find the acceleration at point A, take the derivative

with respect to time of the velocity given
by Eq. 6

dt
e
d
r
e
dt
d
r
dt
v
d
a
t
t
A
A

(9)

dt
d

is the angular acceleration which is usually called

and

r
t
t
e
dt
d
d
e
d
dt
e
d

(10)

Then

r
2
t
A
e
r
e
r
a

(11)

1.2.
3
-

Velocity
and acceleration in vector form

Here we introduce the angular velocity vector

which is perpendicular to the plane of motion
and has a magnitude

.

is then in the

z direction (Fig. 3) and is positive
for a rotation in the
counter
-
clockwise direction.

will be in the negative direction if the body rotates clockwise.

Then we notice that

,
A
r

and
A
v

form a right handed system and that the magnitude of the
velocity is the product of the magnitude of the two other vectors. Therefore,

A
A
r
x
v

(12)

The acceleration is found by taking the derivative with resp
ect to time and

A
A
A
A
A
A
v
x
r
x
dt
r
d
x
r
x
dt
d
dt
v
d
a

(13)

where we have introduced the angular acceleration vector
A
r
x

. Using Eq. 12, we obtain

A
A
A
r
x
x
r
x
a

(14)

Just as
A
A
r
x
v

is acting in the tangential direct
ion, so is
A
r
x

.

Using the following i
dentity
for the triple cross product

We find that

A
A
A
r
.
r
.
r
x
x

. Since the angular velocity is normal to the
plane of motion, the first term on the rhs is zero. Then, the acceleration
in Eq. 14 is seen to
contain one term in the tangential direction and another in the radial direction and in

fact Eq. 14
is identical to Eq. 11.

1.3
-

General two
-
dimensional motion

Consider two points A and B in a body moving in the xy plane
(Figure 5)
. Since

A
/
B
A
B
r
r
r

(15)

the velocity of B is
A
/
B
A
B
v
v
v

(16)

Since the body is rigid, the relative velocity
A
/
B
v

must be a
rotation about A in the xy
-
plane. That is,

A
/
B
A
/
B
r
x
v

(17)

Substituting into Eq. 16 gives

A
/
B
A
B
r
x
v
v

(18)

Therefore, the motion of the body consists of a translation with a velocity
A
v
, the velocity of the
reference point A, plus a rotation about point A.
The
acceleration of point B is

A
/
B
A
/
B
A
B
B
r
dt
d
x
r
x
dt
d
v
dt
d
v
dt
d
a

(19)

or

A
/
B
A
/
B
A
B
v
x
r
x
a
a

(20)

Finally
,

A
/
B
A
/
B
A
B
r
x
x
r
x
a
a

(21)

Eqs.
1
8,

21 describe the velocity and acceleration of a lamina in general 2D motion.
Later we
will see that the same eq
uations apply for the 3D case.

1.4
-

Invariance of the angular velocity
and acceleration
vector
s

x

y

A

Figure 5

: Two
-
dimensional motion

B

A
r

A
/
B
r

T
he
previous section showed that
motion of an arbitrary point B can be written in terms of the
motion of another arbitrary point A, an angular velocity

and an angular acceleration

.
This
brings up the question: Are those vectors

and

unique or do they depend on the choice of
point A?

we w
rite the velocity of B in terms of the velocity of another arbitrary
point C as

C
/
B
C
B
r
x
v
v

(22)

where we allow for the possibility that

is different than

.

We also

express the velocity of C
in terms of the velocity of A

A
/
C
A
C
r
x
v
v

(23)

S
ubstituting
Eq. 23
into Eq.
22 gives

C
/
B
A
/
C
A
B
r
x
r
x
v
v

(24)

Recalling Eq.

18

we find that

A
/
B
A
C
/
B
A
/
C
A
r
x
v
r
x
r
x
v

(25)

or

A
/
C
A
/
B
C
/
B
r
r
x
r
x

(26)

Since

C
/
B
A
/
C
A
/
B
r
r
r

, we conclude that

. In other words
,

the angular velocity is the
same no matter which point in the body is taken as the reference. This is why it is called the
angular velocity of the body. Since

, their
derivatives, the angular accelerations are also
the same.

1.5
-

Instantaneous center of rotation

In two
-
dimensional the instantaneous center of rotation (I.C.) concept is used to solve problems
most efficiently. The main idea is that at each instant the
re is a point with zero velocity and the
body simply rotates about that point. Of course as time goes on, the location of the I.C. changes.

1.5
.1
-

Existence of the instantaneous center of rotation

Given the velocity of an arbitrary point A and the angul
ar velocity of the body we seek to
determine the location of a point O where the velocity is zero at that instant.

Using Eq. 18,

0
r
x
v
v
A
/
O
A
O

(27)

The cross product
O
A
v
x
v

can be written as

0
r
x
x
v
v
x
v
A
/
O
A
A
A

(28)

The first term being zero, the triple cross product is expanded so that Eq. 28 becomes

0
r
.
v
r
.
v
A
/
O
A
A
/
O
A

(29)

Because

is perpendicular to the plane of motion, the se
co
nd term in Eq. 29 is zero and then,
A
/
O
A
r
.
v
.

Th
erefore,

point O is located on a line
that is perpendicular to the velocity vector
A
v
.

T
aking the cross product of

and
O
v

gives

0
r
x
x
v
x
v
x
A
/
O
A
O

(29)

Using Lagrange’s
fo
rmula
for the triple product
we find that

0
r
.
r
.
v
x
A
/
O
A
/
O
A

(30)

Examining each term in this equation we find that, by definition,
0
v
x
A

since we have 2D
motion and

is

perpendicular to the plane of the mo
tion. The second term
2
.

.

Then,
Eq. 30 becomes

A
2
A
/
O
v
x
1
r

(31)

If both

and
A
v

are known, Eq. 31 provides a convenient way for determining the location of
point O.

This equation shows that
such a point O exists provided that
0

(otherwise there is
no rotation) and that

it is located on a line that is perpendicular to the velocity vector
and that
A
/
O
A
r
and
v
,

form a right handed syste
m.

Using Eq. 18, the velocity of an arbitrary point B can
be written as
A
/
B
A
B
r
x
v
v

and since
O
/
A
A
r
x
v

,
O
/
B
A
/
B
O
/
A
B
r
x
r
x
r
x
v

.
Therefore, the line OB is perpendicular to the velocity
vector at B and the body
,
at that instan
t
, is simply
rotating about O. This is why it is called the
i
nstantaneous center of rotation
.

1.5.2
-

Graphical construction

Given the velocity vectors at two points A and B (Fig.
7), draw a line normal to at A and a line normal to at
B. These two li
nes intersect at the I.C. Then, the
angular velocity can be calculated from

A

B

I.C.

A
v

B
v

Figure 7

:
Finding the location of the
I.C. graphically

O

.
OA
v
A

or

.
OB
v
B

(32)

Since there can be only one angular velocity for a rigid body,

OB
v
OA
v
B
A

(33)

sistency requirement between the geometry and the velocities

OB
OA
v
v
B
A

(34
)

For a given geometry, we can find the I.C. using this graphical construction but the specified

velocities must satisfy Eq. (34
).

1.5
.3
-

Special case I: the v
elocity at point A is the same as the velocity at B

In many problems

is unknown but the velocity is know
n at two points A and B
.
Using Eq. 18
we write

0
r
x
v
v
A
/
O
A
O

and

0
r
x
v
v
B
/
O
B
O

(35
)

Subtracting those two

equations gives

0
r
r
x
v
v
B
/
O
A
/
O
B
A

or

0
r
r
x
v
v
O
/
B
O
/
A
B
A

(36
)

or

B
/
A
B
A
r
x
v
v

(37
)

Eq. 37

shows that
,

when two distinct points A and B have the same velocity
,

the angular velocity
is zero and the body is in pure translation
.

1.5
.
4
-

Special case II: the v
elocity at A is parallel to the velocity at B

Assume that
A
v

is parallel to
B
v

but that the magnitudes are different. What can we conclude?
Since these two vectors are parallel,

0
v
x
v
B
A

(38
)

From Eq. 18 we can write

A
/
B
A
B
r
x
v
v

(39
)

Substituting into Eq. 38

gives

0
r
x
x
v
v
x
v
A
/
B
A
A
A

. The first term is zero and
expanding the triple cross
-
product gives

0
r
.
v
r
.
v
A
/
B
A
A
/
B
A

(40
)

0
.
v
A

for 2D motion and Eq.
40

becomes

0
r
.
v
A
/
B
A

(41
)

Eq. 41

indi
cates that
there are two cases. With case I
, the velocities are perpendicular to the line
AB so
0
r
.
v
A
/
B
A

and the instantaneou
s center is locate
d on AB. When
A
B
v
v

the I.C. will
be located at a finite distance from A and B and the angular velocity will have a finite value

(Fig.
6.a)
. When
A
B
v
v

the I.C. will be located at infinity and the angular velocity will

be zero

(Fig. 6.b)
. This is a case of pure translation.

With case II, the velocities are not perpendicular to
AB:
0
r
.
v
A
/
B
A

.

In that case, Eq. 41

implies that
0

which suggests that the motion should
be a pure transla
tion. However since
A
B
v
v

th
is situation is impossible

(Fig. 6.c)
.

A

B

I.C.

A
v

B
v

A

B

I.C.

A
v

B
v

B

A
v

B
v

(a)

(b)

(c)

Figure 6

: (a) velocities at A and B are parallel and normal to AB

and have different
magnitudes
; (b)

ve
locities at A and B are the same

and
are
normal to AB

(translation)

;

(c)
velocities at A and B are parallel
with different magnitud
es

and
are not
normal to
AB

(impossible case)

2
-

GENERAL MOTION

OF A RIGID BODY

In order to describe the motion of a rigid body in three
dimensions
,

we use a fixed coordinate system XYZ
and another coordinate syste
m O
-
xyz that is fixed to
the body

(Fig. 7)
.
Point O is an arbitrary point in the
body.
The angular velocity and the angular
acceleration are two vectors that can change
orientation and magnitude with time.

The position of P, an arbitrary point in the bo
dy, is
defined by the position vector

O
/
P
O
P
r
r
r

(42)

The velocity of that point P is

dt
r
d
dt
r
d
dt
r
d
v
O
/
P
O
P
P

(43)

The first term is simply the velocity of the origin of the moving frame

O
O
v
dt
r
d

(44)

In th
e moving coordinate system, if
k
z
j
y
i
x
r
O
/
P

then

dt
k
d
z
dt
j
d
y
dt
i
d
x
k
z
j
y
i
x
dt
r
d
O
/
P

(45)

Since the body is rigid,
0
z
y
x

and because the moving frame is rotating with an angular
velocity

,

k
x
dt
k
d
,
j
x
dt
j
d
,
i
x
dt
i
d

(46)

The velocity of P relative to O becomes

O
/
P
O
/
P
O
/
P
r
x
k
z
j
y
i
x
x
dt
r
d
v

(47)

So,

O
/
P
O
P
r
x
v
v

(48)

The acceleration of P is found by taking the derivative with respect to time so

X

Y

Z

x

y

z

O
/
P
r

Figure 7

: Position vector and local
coordinate system

P

O

O
r

dt
r
d
x
r
x
dt
d
dt
v
d
dt
v
d
a
O
/
P
O
/
P
O
P
P

(49)

or

O
/
P
O
/
P
O
P
r
x
x
r
x
a
a

(50)

where
the angular acceleration
dt
d

.

Note that the expressions for the velocity and the acceleration of an arbitrary point in terms of a
reference point and the angular velocity and acceleration of the body are identica
l when written
in vector form.

3
-

MOTION OF A PARTICLE RELATIVE TO A RIGID BODY

As before, we consider the motion of a rigid body
and the xyz coordinate system is fixed to the body
and rotates with an angular velocity

and an
angu
lar acceleration

.
P is an arbitrary point in
the body (Figure 8).

,

there is a particle located at point P’
.
The position of P’ is given by

P
/
'
P
O
/
P
o
'
P
r
r
r
r

(51)

Taking the derivative with respect to time
, the
velocity of P’ is

xyz
/
'
P
P
/
'
P
O
/
P
o
'
P
v
r
x
r
x
v
v

(52)

Taking the derivative with respect to time again
gives the acce
leration of P’

dt
v
d
v
x
dt
r
d
x
r
x
dt
d
dt
r
d
x
r
x
dt
d
a
a
xyz
/
'
P
xyz
/
'
P
P
/
'
P
P
/
'
P
O
/
P
O
/
P
o
'
P

(53)

xyz
/
'
P
xyz
/
'
P
xyz
/
'
P
P
/
'
P
O
/
P
O
/
P
o
'
P
a
v
x
v
x
r
x
r
x
x
r
x
a
a

(54)

Equations 52 and 54 simplify a little bit when P’ is at P

0
r
P
/
'
P

.

In that case, the velocity and
acceleration of the moving partic
le are

xyz
/
'
P
O
/
P
o
'
P
v
r
x
v
v

(55)

xyz
/
'
P
xyz
/
'
P
O
/
P
O
/
P
o
'
P
a
v
x
2
r
x
x
r
x
a
a

(56)

X

Y

Z

x

y

z

O
/
P
r

Figure 8

: Position vector and local
coordinate system

P

O

O
r

P’

Equation 55 says that the velocity of the moving particle is equal the velocity of the point on the

rigid body that occupies the same location at that instant

O
/
P
o
r
x
v

plus the velocity of the
particle relative to the body

xyz
/
'
P
v
. In Equation 56, we have the acceleration of point P

O
/
P
O
/
P
o
r
x
x
r
x
a

, the accelerati
on of P’ relative to the body

xyz
/
'
P
a
, and then an

xyz
/
'
P
v
x
2

which is called Coriolis’ acceleration.
It is named
after
Gaspard
-
Gustave Coriolis

(1792
-
1843)
, a French scientist who in 1835, observed an apparent

deflection
of moving objects from a straight path when they are viewed from a
rotating frame
.

Appendix: Vector algebra

Cross product

The dot product of two vectors
k
U
j
U
i
U
U
z
y
x

and
k
V
j
V
i
V
V
z
y
x

can be written in
terms of the comp
onents of the two vectors as

z
z
y
y
x
x
V
U
V
U
V
U
V
.
U

or in terms of the magnitude of the two vectors and the angle between them as

cos
V
U
V
.
U

Then it becomes obvious that when the two vectors
V
and
U

are orthogonal,
0
V
.
U

.

Cross product

The cross product of two vectors
V
and
U

is a vector
W
of magnitude

sin
V
U
W
.
W
and
V
,
U
form a right angled system. The components of
W

can be determined from the
determinant

x
y
y
x
x
z
z
x
y
z
z
y
z
y
x
z
y
x
V
U
V
U
k
V
U
V
U
j
V
U
V
U
i
V
V
V
U
U
U
k
j
i
V
x
U
W

A straightforward expansion will show that
U
x
V
V
x
U

. Similarly,

0
U
c
x
U

so, the
dot product

of
two vectors that are parallel is zero.

Triple

product

The cross product of a vector with a cross product is called a triple cross product. The expansion
formula of the triple cross product (Lagrange’s formula) is

c
b
.
a
b
c
.
a
c
x
b
x
a

Translation

a fixed point

Two
-
dimensional motion

3D motion

I.C. of rotation

A
A
r
x
v

A
A
A
r
x
x
r
x
a

A
/
B
A
B
r
x
v
v

A
/
B
A
/
B
A
B
r
x
x
r
x
a
a

O
/
P
O
P
r
x
v
v

O
/
P
O
/
P
O
P
r
x
x
r
x
a
a

Figure 8