RIGID BODY DYNAMICS
MENG 233
___________________________________
WEEK 4
LECTURE #1
RIGID BODY DYNAMICS
MENG 233
___________________________________
KINEMATICS OF PARTICLES
CURVILINEAR MOTION
LECTURE #1
Kinematics of Particles
Components
• Particle motion along a curved path “Curvilinear Motion” using
three coordinate systems
– Rectangular Components
• Position vector r = x i + y j + z k
• Velocity
v = vx i + vy j + vz k
• Acceleration a = ax i + ay j +az k
(tangent to path)
(tangent to hodograph)
– Normal and Tangential Components
• Position (particle itself)
• Velocity
v = u ut 2


a


u

un
• Acceleration
t

1  (dy / dx) 

2 3/ 2
d 2 y / dx 2
(tangent to path)
(normal & tangent)
– Polar & Cylindrical Components
3
Kinematics of Particles
Curvilinear Motion : Cylindrical Components
The cylindrical coordinate system is used
in cases where the particle moves along a
3-D curve.
www.cim.mcgill.ca
(spiral motion)
4
Kinematics of Particles
Curvilinear Motion : Cylindrical Components
• Observed and/or guided from origin or from the
center
• Cylindrical component
r ,  , and z
• Polar component “plane motion”
r and 
5
Kinematics of Particles
Application
• Circular motion but observed and/or controlled from
the center
6
Kinematics of Particles
Polar Coordinates
•
•
•
•
•
•
Radial coordinate r
Transverse coordinate
 and r are perpendicular
Theta  in radians
1 rad = 180o/p
Direction ur and u
We can express the location of P in polar coordinates as r = rur. Note that the
radial direction, r, extends outward from the fixed origin, O, and the
transverse coordinate, , is measured counter-clockwise (CCW) from the
horizontal.
7
Kinematics of Particles
VELOCITY: Polar Coordinates
The instantaneous velocity is defined as:
v = dr/dt = d(rur)/dt
dur
.
v = rur + r dt
Using the chain rule:
dur/dt = (dur/d)(d/dt)
.
We can prove that dur/d = uθ so dur/dt = uθ
.
.
Therefore: v = rur + ruθ
8
Kinematics of Particles
VELOCITY: Polar Coordinates
.
Thus, the velocity vector has. two components: r, called
the radial component, and r, called the transverse
component. The speed of the particle at any given
instant is the sum of the squares of both components or
.
.
v = (r  )2  ( r )2
9
Kinematics of Particles
ACCELERATION: Polar Coordinates
The instantaneous acceleration is defined as:
.
a = dv/dt = (d/dt)(ru
.
r
+ ruθ)
After manipulation, the acceleration can be
expressed as
.2
.. . .
..
a = (r. – r )ur + (r + 2r)uθ
..
The term (r – r2) is the radial acceleration or ar.
..
.
.
The term (r + 2r) is the transverse acceleration or a
..
.2 2
..
..
The magnitude of acceleration is a = (r – r ) + (r + 2r)2
10

Kinematics of Particles
CYINDRICAL COORDINATES
If the particle P moves along a space curve,
its position can be written as
rP = rur + zuz
Velocity:
Taking time derivatives and using the chain
rule:
.
.
.
vP = rur + ruθ + zuz
..
..
.. . 2
..
Acceleration: aP = (r – r )ur + (r + 2r)uθ + zuz
11
Kinematics of Particles
Cylindrical Coordinates
• For spiral motion
cylindrical coordinates
is used r, q, and z.
• Position
rp  r u r  z u z
• Velocity
v  r u r  r u   z u z
• Acceleration
a  (r  r 2 ) u r  (r  2r) u   z u z
16
Kinematics of Particles
Time Derivatives
r, r, , and 
• If r = r(t) and   (t)
r  4t 2
  (8t 3  6)
r  8t
  24 t 2
r  8
  48 t
• If r = f() use chain rule
r  5
2
r  10  
r  10 [ ()    () ]
r  10  2  10  
17
Kinematics of Particles
Three-Dimensional Motion
• For• spatial
For spatial
motion
motion
required
required
threethree
dimension.
dimension.
• Binomial
• Binomial
axis baxis
which
b which
is perpendicular
is perpendicular
to ut to ut
and uand
un is used
n is used
• ub=• uut bx=uunt x un
Kinematics of Particles
Cylindrical Coordinates
Problem
• The slotted fork is rotating about O at a constant
rate of 3 rad/s. Determine the radial and
transverse components of velocity and
acceleration of the pin A at the instant  = 360o.
The path is defined by the spiral groove r =
(5+/p) in., where  is in radians.
  360o  2p rad

r  5
p
vr  r 
 7 in
  2p
3
p
  3 rad/s
 3
r   in/s
p p
 0.955 in/s
v  r  7(3)  21 in/s
  0 rad/s 2
r  p  0 in/s 2
ar  r  r 2  0  7(3) 2  63 in/s 2
3
a  r  2r  0  2( )(3)  5.73 in/s 2
p
19
Kinematics of Particles
Cylindrical Coordinates
20
Kinematics of Particles
Cylindrical Coordinates
21
Kinematics of Particles
Cylindrical Coordinates
22
RIGID BODY DYNAMICS
MENG 233
___________________________________
KINEMATICS OF PARTICLES
DEPENDANT MOTION
TRANSLATING FRAMES OF REFERENCE
LECTURE #1
Kinematics of Particles
We should be able to:
1. Relate the positions, velocities, and
accelerations of particles undergoing
dependent motion
2. Understand translating frames of
reference
3. Use translating frames of reference to
analyze relative motion
24
Kinematics of Particles
DEPENDANT MOTION: APPLICATIONS
The cable and pulley system
shown here can be used to
modify the speed of block B
relative to the speed of the
motor. It is important to relate
the various motions in order to
determine the power
requirements for the motor and
the tension in the cable
25
Kinematics of Particles
DEPENDANT MOTION: APPLICATIONS
Rope and pulley
arrangements are often used
to assist in lifting heavy
objects. The total lifting
force required from the
truck depends on the
acceleration of the cabinet
26
Kinematics of Particles
DEPENDANT MOTION
In many kinematics problems, the motion of one object will depend on
the motion of another object
The blocks in this figure are
connected by an inextensible cord
wrapped around a pulley. If block A
moves downward along the inclined
plane, block B will move up the
other incline
The motion of each block can be related mathematically by defining
position coordinates, sA and sB. Each coordinate axis is defined from a
fixed point or datum line, measured positive along each plane in the
direction of motion of each block
27
Kinematics of Particles
DEPENDANT MOTION
In this example, position
coordinates sA and sB can be
defined from fixed datum lines
extending from the center of the
pulley along each incline to blocks
A and B
If the cord has a fixed length, the position coordinates sA and sB are
related mathematically by the equation
sA + lCD + sB = lT
Here lT is the total cord length and lCD is the length of cord passing over
arc CD on the pulley
28
Kinematics of Particles
DEPENDANT MOTION
The velocities of blocks A and B can
be related by differentiating the
position equation. Note that lCD and
lT remain constant,
so dlCD/dt = dlT/dt = 0
dsA/dt + dsB/dt = 0 =>
vB = -vA
The negative sign indicates that as A moves down the incline (positive
sA direction), B moves up the incline (negative sB direction)
Accelerations can be found by differentiating the velocity expression
29
Kinematics of Particles
DEPENDANT MOTION: EXAMPLE
Consider a slightly more complicated
example. Position coordinates (sA and sB) are
defined from fixed datum lines, measured
along the direction of motion of each block
Note that sB is only defined to the center
of the pulley above block B, since this
block moves with the pulley. Also, h is a
constant
The red colored segments of the cord remain constant in length during
motion of the blocks
30
Kinematics of Particles
DEPENDANT MOTION: EXAMPLE (CONT’D)
The position coordinates are related by
2sB + h + sA = l
Where l is the total cord length minus the
lengths of the red segments
Since l and h remain constant during the
motion, the velocities and accelerations can be
related by two successive time derivatives:
2vB = -vA
and
2aB = -aA
When block B moves downward (+sB), block A moves to the left (-sA).
Remember to be consistent with the sign convention!
31
Kinematics of Particles
DEPENDANT MOTION: EXAMPLE (CONT’D)
This example can also be worked by defining
the position coordinate for B (sB) from the
bottom pulley instead of the top pulley
The position, velocity, and
acceleration relations then become
2(h – sB) + h + sA = l
and
2vB = vA
2aB = aA
32
Kinematics of Particles
DEPENDANT MOTION: PROCEDURE OF ANALYSIS
These procedures can be used to relate the dependent motion of particles
moving along rectilinear paths (only the magnitudes of velocity and
acceleration change, not their line of direction)
1)
Define position coordinates from fixed datum lines, along the path
of each particle. Different datum lines can be used for each particle
2) Relate the position coordinates to the cord length. Segments of cord
that do not change in length during the motion may be left out
3) If a system contains more than one cord, relate the position of a point
on one cord to a point on another cord. Separate equations are
written for each cord
4) Differentiate the position coordinate equation(s) to relate velocities
and accelerations. Keep track of signs!
33
Kinematics of Particles
DEPENDANT MOTION
Given: In the figure on the left, the cord at
A is pulled down with a speed of 8 ft/s
Find: The speed of block B
Plan: There are two cords involved in the motion in this
example. The position of a point on one cord must be related to
the position of a point on the other cord. There will be two
position equations (one for each cord)
34
Kinematics of Particles
DEPENDANT MOTION
1)
Define the position coordinates from a fixed datum line. Three
coordinates must be defined: one for point A (sA), one for block B (sB),
and one relating positions on the two cords. Note that pulley C relates
the motion of the two cords
•Define the datum line through the top pulley
(which has a fixed position).
Datum
DATUM
sA
sC
sB
•sA can be defined to the center of the pulley
above point A.
•sB can be defined to the center of the pulley
above B.
•sC is defined to the center of pulley C.
•All coordinates are defined as positive down
and along the direction of motion of each
point/object.
35
Kinematics of Particles
DEPENDANT MOTION
2) Write position/length equations for each cord. Define l1 as the length of the
first cord, minus any segments of constant length. Define l2 in a similar manner
for the second cord:
Cord 1: 2s + 2s = l
A
C
1
Cord 2: sB + (sB – sC) = l2
3) Eliminating sC between the two equations,
2sA + 4sB = l1 + 2l2
Datum
DATUM
sA
sC
sB
4) Relate velocities by differentiating this
expression. Note that l1 and l2 are constant
lengths.
2vA + 4vB = 0
=>
vB = - 0.5vA = - 0.5(8) = - 4 ft/s
The velocity of block B is 4 ft/s
up (negative sB direction).
36
Kinematics of Particles
DEPENDANT MOTION
• Position coordinate
“cord equation”
s A 3 sB  l
• Velocity
 A  3B  0
 A  3 B
 A  3* 6  18 ft/s
• Acceleration
a A  3 aB  0
a A  3 aB
37
Kinematics of Particles
DEPENDANT MOTION
• Position coordinate
“cord equation”
s A  2 sC  l1 and
sB  (sB  sC )  l2
s A  4 sB  2l2  l1  const
• Velocity
 A  4B  0
 A  4B  4 * 6  24 ft/s 
• Acceleration
a A  4 aB  0
a A  4 aB
38
Kinematics of Particles
DEPENDANT MOTION
Example
Position coordinate “cord equation”
sC  sB  l1
( s A  sC )  ( sB  sC )  sB  l2
s A  4 sB  2l2  l1  const
 A  4B  0
1
1
 B    B   * 2  0.5 ft/s 
4
4
39
Kinematics of Particles
DEPENDANT MOTION
40
Kinematics of Particles
DEPENDANT MOTION: EXAMPLE 12.22
• Establish a relation y = f(x)
l  lDA  lCD
lDA  (15) 2  x 2
and lCD  15  y
30  (15) 2  x 2  (15  y )
Cord =30 m
y  225  x 2  15
s  dy / dt and  A  dx / dt
 dx
dy  1
2x
s   

dt  2 225  x 2  dt
x
S 
A
225  x 2
41
RIGID BODY DYNAMICS
MENG 233
___________________________________
KINEMATICS OF PARTICLES
DEPENDANT MOTION
TRANSLATING FRAMES OF REFERENCE
LECTURE #1
Kinematics of Particles
Relative motion analysis of two particles using translating axis
44
Kinematics of Particles
Relative motion analysis of two particles using translating axis
When you try to hit a moving object, the position, velocity, and
acceleration of the object must be known. Here, the boy
on the ground is at d = 10 ft when the girl in the window throws the ball
to him
If the boy on the ground is running at a constant speed of 4 ft/s,
how fast should the ball be thrown?
45
Kinematics of Particles
Relative motion analysis of two particles using translating axis
When fighter jets take off or land on an
aircraft carrier, the velocity of the carrier
becomes an issue
If the aircraft carrier travels at a forward velocity of 50 km/hr and
plane A takes off at a horizontal air speed of 200 km/hr (measured by
someone on the water), how do we find the velocity of the plane
relative to the carrier?
How would you find the same thing for airplane B?
How does the wind impact this sort of situation?
46
Kinematics of Particles
Relative motion analysis of two particles using translating axis
Relative Motion (Translating Axes)
- Consider two particles A and B. Both are
moving with respect to fixed axes X-Y
- The translating axes x-y is moving with
particle B
- The position vector of the particle A
measured relative to the x-y axes is
rA  rB  r A / B
(A/B means A with respect to B)
rA B
 x iˆ  y ˆj
47
Kinematics of Particles
Relative motion analysis of two particles using translating axis
If rB is the absolute position of B, then
rA  rB  rA B
Take derivative with respect to time :
rA  rB  rA B
or
VA  VB  VA B
and
rA
 rB  rA B
or
aA  aB  aA B
48
Kinematics of Particles
Relative motion analysis of two particles using translating axis
Relative position
The absolute position of two
particles A and B with respect to
the fixed x, y, z reference frame are
given by rA and rB. The position of
B relative to A is represented by
rB/A = rB – rA
Therefore, if rB = (10 i + 2 j ) m
and
rA = (4 i + 5 j ) m
then
rB/A = (6 i – 3 j ) m
49
Kinematics of Particles
Relative motion analysis of two particles using translating axis
Relative velocity
To determine the relative velocity
of B with respect to A, the time
derivative of the relative position
equation is taken.
vB/A = vB – vA
or
vB = vA + vB/A
In these equations, vB and vA are called absolute velocities and vB/A is the
relative velocity of B with respect to A.
Note that vB/A = - vA/B .
50
Kinematics of Particles
Relative motion analysis of two particles using translating axis
Relative Acceleration
The time derivative of the relative velocity
equation yields a similar vector relationship
between the absolute and relative
accelerations of particles A and B
aB/A = aB – aA
or
aB = aA + aB/A
51
Kinematics of Particles
Relative motion analysis of two particles using translating axis
Solving Problems
Since the relative motion equations are vector equations, problems
involving them may be solved in one of two ways.
For instance, the velocity vectors in vB = vA + vB/A could be written as
Cartesian vectors and the resulting scalar equations solved for up to two
unknowns.
Alternatively, vector problems can be solved “graphically” by use of
trigonometry. This approach usually makes use of the law of sines or the
law of cosines.
52
Laws of sines and cosines
C
b
a
B
A
c
Law of Sines:
Since vector addition or subtraction
forms a triangle, sine and cosine laws can
be applied to solve for relative or
absolute velocities and accelerations.
For review, their formulations are
provided below.
a
sin A
Law of Cosines:
b

sin B

c
sin C
a 2  b 2  c 2  2 bc cos A
b  a  c  2 ac cos B
2
2
2
c  a  b  2 ab cos C
2
2
2
Kinematics of Particles
Relative motion analysis of two particles using translating axis
A
160 km/h
B
180 km/h
VA  VB  VA/B
VA/B  VA - VB  160  180  20 km/ h
VB  VA  VB/A
VB/A  VB - VA  180  160  20 km/ h
54
Kinematics of Particles
Relative motion analysis of two particles using translating axis
The passenger aircraft B is flying
east with velocity vB = 800 km/h. A
jet is traveling south with velocity vA
= 1200 km/h. What velocity does A
appear to a passenger in B ?
y
Solution

vB  800î

x

vA B

vA  1200 ĵ

VA  VB  VA B

 1200 ĵ  800 î  v A B
VA B   800iˆ  1200 ˆj
55
Kinematics of Particles
Relative motion analysis of two particles using translating axis
The passenger aircraft B is flying
east with velocity vB = 800 km/h. A
jet is traveling south with velocity vA
= 1200 km/h. What velocity does A
appear to a passenger in B ?
( 800)2  ( 1200)2
Absolute value : A / B 
 1442 km / h
The direction of V A B
y
VB  800iˆ

x

vA B
VA  1200 ˆj
tan  
800
1200
  33.7 West of South
56
Kinematics of Particles
Relative motion analysis of two particles using translating axis
T  60 mi/h
 A  45 mi/h
Determine the magnitude and direction of relative velocity
of the Train with respect to the Automobile
VT  VA  VT/A
60 i  (45 cos 45o i  45 sin 45o j)  VT/A
VT/A  { 28.2 i - 31.8 j} mi/h
T / A  (28.2) 2  (31.8) 2  42.5 mi/h
(T / A ) y
31.8
tan  

(T / A ) x 28.2

  48.5o
57
Kinematics of Particles
Relative motion analysis of two particles using translating axis
 A  18 m/s , a A  2 m/s 2
 B  12 m/s , a B  3 m/s 2
Determine the velocity and accelerati on of B with respect to A
VB  VA  VB/A
- 12 j  (-18 cos 60o i  18 sin 60o j)  VB/A
VB/A  { 9 i  3.588 j} m/s
B / A 
(9)  (3.558)  9.69 m/s   tan 1
2
2
( B / A ) y
( B / A ) x
 tan 1 (
3.588
)
9
  21.7 o
a A  {2 cos 60o i  2 sin 60o j}
2
(12) 2
aB  an i  at j  - i - 3 j  i  3 j  {1.44 i  3 j} m / s 2

100
a B  a A  a B/A
a B/A  {-2.44 i - 4.732 j }
aB / A  (2.44) 2  (4.732) 2  5.32 m/s 2
tan  
( B / A ) y
( B / A ) x

4.732
2.44
  62.7 o
58
RIGID BODY DYNAMICS
MENG 233
___________________________________
KINEMATICS OF PARTICLES
REVISION
LECTURE #1
Kinematics of Particles
REVISION
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Kinematics of Particles
REVISION
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QUESTIONS
THANK YOU
FOR YOUR
INTEREST