dynamics11lecture.ppt

VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Tenth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Brian P. Self
California Polytechnic State University
CHAPTER
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11
Kinematics of Particles

Vector Mechanics for Engineers: Dynamics
Tenth
Edition
Contents
11 - 2
Introduction
Rectilinear Motion: Position,
Velocity & Acceleration
Determination of the Motion of a
Particle
Sample Problem 11.2
Sample Problem 11.3
Uniform Rectilinear-Motion
Uniformly Accelerated Rectilinear-
Motion
Motion of Several Particles:
Relative Motion
Sample Problem 11.4
Motion of Several Particles:
Dependent Motion
Sample Problem 11.5
Graphical Solution of Rectilinear-
Motion Problems
Other Graphical Methods
Curvilinear Motion: Position, Velocity
& Acceleration
Derivatives of Vector Functions
Rectangular Components of Velocity
and Acceleration
Motion Relative to a Frame in
Translation
Tangential and Normal Components
Radial and Transverse Components
Sample Problem 11.10

Tenth
Edition
2 - 3
Kinematic relationships are used to
help us determine the trajectory of a
golf ball, the orbital speed of a
satellite, and the accelerations
during acrobatic flying.

Tenth
Edition
Introduction
11 - 4
• Dynamics includes:
Kinetics: study of the relations existing between the forces acting on
a body, the mass of the body, and the motion of the body. Kinetics is
used to predict the motion caused by given forces or to determine the
forces required to produce a given motion.
Kinematics: study of the geometry of motion.
Relates displacement, velocity, acceleration, and time without reference
to the cause of motion.
Fthrust
Flift
Fdrag

Tenth
Edition
Introduction
11 - 5
• Particle kinetics includes:
• Rectilinear motion: position, velocity, and acceleration of a
particle as it moves along a straight line.
• Curvilinear motion: position, velocity, and acceleration of a
particle as it moves along a curved line in two or three
dimensions.

Tenth
Edition
Rectilinear Motion: Position, Velocity & Acceleration
11 - 6
• Rectilinear motion: particle moving
along a straight line
• Position coordinate: defined by
positive or negative distance from a
fixed origin on the line.
• The motion of a particle is known if
the position coordinate for particle is
known for every value of time t.
• May be expressed in the form of a
function, e.g., 3
2
6 t
t
x 

or in the form of a graph x vs. t.

Tenth
Edition
11 - 7
• Instantaneous velocity may be positive or
negative. Magnitude of velocity is referred
to as particle speed.
• Consider particle which occupies position P
at time t and P’ at t+Dt,
t
x
v
t
x
t D
D


D
D


D 0
lim
Average velocity
Instantaneous velocity
• From the definition of a derivative,
dt
dx
t
x
v
t

D
D


D 0
lim
e.g.,
2
3
2
3
12
6
t
t
dt
dx
v
t
t
x






Tenth
Edition
11 - 8
• Consider particle with velocity v at time t and
v’ at t+Dt,
Instantaneous acceleration
t
v
a
t D
D



D 0
lim
t
dt
dv
a
t
t
v
dt
x
d
dt
dv
t
v
a
t
6
12
3
12
e.g.
lim
2
2
2
0







D
D


D
• From the definition of a derivative,
• Instantaneous acceleration may be:
- positive: increasing positive velocity
or decreasing negative velocity
- negative: decreasing positive velocity
or increasing negative velocity.

Tenth
Edition
Concept Quiz
2 - 9
What is true about the kinematics of a particle?
a) The velocity of a particle is always positive
b) The velocity of a particle is equal to the slope of
the position-time graph
c) If the position of a particle is zero, then the
velocity must zero
d) If the velocity of a particle is zero, then its
acceleration must be zero

Tenth
Edition
11 - 10
• From our example,
3
2
6 t
t
x 

2
3
12 t
t
dt
dx
v 


t
dt
x
d
dt
dv
a 6
12
2
2




- at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0
- at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2
• What are x, v, and a at t = 2 s ?
• Note that vmax occurs when a=0, and that the
slope of the velocity curve is zero at this point.
• What are x, v, and a at t = 4 s ?

Tenth
Edition
Determination of the Motion of a Particle
11 - 11
• We often determine accelerations from the forces applied
(kinetics will be covered later)
• Generally have three classes of motion
- acceleration given as a function of time, a = f(t)
- acceleration given as a function of position, a = f(x)
- acceleration given as a function of velocity, a = f(v)
• Can you think of a physical example of when force is a
function of position? When force is a function of velocity?
a spring drag

Tenth
Edition
Acceleration as a function of time, position, or velocity
2 - 12
 
a a t
  
0 0
v t
v
dv a t dt

 
( )
dv
a t
dt

 
vdv a x dx

 
0 0
v x
v x
vdv a x dx

 
 
a a x

and
dx dv
dt a
v dt
 
 
dv
v a v
dx

 
0 0
v t
v
dv
dt
a v

 
 
0 0
x v
x v
vdv
dx
a v

 
 
a a v

( )
dv
a v
dt

If…. Kinematic relationship Integrate

Tenth
Edition
Sample Problem 11.2
11 - 13
Determine:
• velocity and elevation above ground at
time t,
• highest elevation reached by ball and
corresponding time, and
• time when ball will hit the ground and
corresponding velocity.
Ball tossed with 10 m/s vertical velocity
from window 20 m above ground.
SOLUTION:
• Integrate twice to find v(t) and y(t).
• Solve for t when velocity equals zero
(time for maximum elevation) and
evaluate corresponding altitude.
• Solve for t when altitude equals zero
(time for ground impact) and evaluate

Tenth
Edition
Sample Problem 11.2
11 - 14
 
  t
v
t
v
dt
dv
a
dt
dv
t
t
v
v
81
.
9
81
.
9
s
m
81
.
9
0
0
2
0










  t
t
v 






 2
s
m
81
.
9
s
m
10
 
    2
2
1
0
0
81
.
9
10
81
.
9
10
81
.
9
10
0
t
t
y
t
y
dt
t
dy
t
v
dt
dy
t
t
y
y










  2
2
s
m
905
.
4
s
m
10
m
20 t
t
t
y 














SOLUTION:
• Integrate twice to find v(t) and y(t).

Tenth
Edition
Sample Problem 11.2
11 - 15
• Solve for t when velocity equals zero and evaluate
corresponding altitude.
  0
s
m
81
.
9
s
m
10 2








 t
t
v
s
019
.
1

t
• Solve for t when altitude equals zero and evaluate
 
   2
2
2
2
s
019
.
1
s
m
905
.
4
s
019
.
1
s
m
10
m
20
s
m
905
.
4
s
m
10
m
20






























y
t
t
t
y
m
1
.
25

y

Tenth
Edition
Sample Problem 11.2
11 - 16
• Solve for t when altitude equals zero and evaluate
  0
s
m
905
.
4
s
m
10
m
20 2
2















 t
t
t
y
 
s
28
.
3
s
meaningles
s
243
.
1



t
t
 
   
s
28
.
3
s
m
81
.
9
s
m
10
s
28
.
3
s
m
81
.
9
s
m
10
2
2
















v
t
t
v
s
m
2
.
22


v

Tenth
Edition
Sample Problem 11.3
11 - 17
Brake mechanism used to reduce gun
recoil consists of piston attached to barrel
moving in fixed cylinder filled with oil.
As barrel recoils with initial velocity v0,
piston moves and oil is forced through
orifices in piston, causing piston and
cylinder to decelerate at rate proportional
to their velocity.
Determine v(t), x(t), and v(x).
kv
a 

SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
• Integrate v(t) = dx/dt to find x(t).
• Integrate a = v dv/dx = -kv to find
v(x).

Tenth
Edition
Sample Problem 11.3
11 - 18
SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
 
0 0
0
ln
v t
v
v t
dv dv
a kv k dt kt
dt v v
      
 
  kt
e
v
t
v 
 0
• Integrate v(t) = dx/dt to find x(t).
 
 
0
0 0
0
0 0
1
kt
t
x t
kt kt
dx
v t v e
dt
dx v e dt x t v e
k

 
 
 
  
 
 
 
   
kt
e
k
v
t
x 

 1
0

Tenth
Edition
Sample Problem 11.3
11 - 19
• Integrate a = v dv/dx = -kv to find v(x).
kx
v
v
dx
k
dv
dx
k
dv
kv
dx
dv
v
a
x
v
v









 

0
0
0
kx
v
v 
 0
• Alternatively,
   










0
0 1
v
t
v
k
v
t
x
kx
v
v 
 0
   
0
0 or
v
t
v
e
e
v
t
v kt
kt

 

   
kt
e
k
v
t
x 

 1
0
with
and
then

Tenth
Edition
Group Problem Solving
2 - 20
A bowling ball is dropped from a boat so that it
strikes the surface of a lake with a speed of 15 ft/s.
Assuming the ball experiences a downward
acceleration of a =10 - 0.01v2 when in the water,
determine the velocity of the ball when it strikes the
bottom of the lake.
Which integral should you choose?
 
0 0
v t
v
dv a t dt

   
0 0
v x
v x
vdv a x dx

 
 
0 0
v t
v
dv
dt
a v

 
 
0 0
x v
x v
vdv
dx
a v

 
(a)
(b)
(c)
(d)
+y

Tenth
Edition
Concept Question
2 - 21
When will the bowling ball start slowing down?
A bowling ball is dropped from a boat so that it
strikes the surface of a lake with a speed of 15 ft/s.
Assuming the ball experiences a downward
acceleration of a =10 - 0.01v2 when in the water,
determine the velocity of the ball when it strikes the
bottom of the lake.
The velocity would have to be high
enough for the 0.01 v2 term to be bigger
than 10
+y

Tenth
Edition
11 - 22
SOLUTION:
• Determine the proper kinematic
relationship to apply (is acceleration
a function of time, velocity, or
position?
• Determine the total distance the car
travels in one-half lap
• Integrate to determine the velocity
after one-half lap
The car starts from rest and accelerates
according to the relationship
2
3 0.001
a v
 
It travels around a circular track that has
a radius of 200 meters. Calculate the
velocity of the car after it has travelled
halfway around the track. What is the
car’s maximum possible speed?

Tenth
Edition
2 - 23
 
dv
v a v
dx

 
0 0
x v
x v
vdv
dx
a v

 
Choose the proper kinematic relationship
Given: 2
3 0.001
a v
 
vo = 0, r = 200 m
Find: v after ½ lap
Maximum speed
Acceleration is a function of velocity, and
we also can determine distance. Time is not
involved in the problem, so we choose:
Determine total distance travelled
3.14(200) 628.32 m
x r

  

Tenth
Edition
2 - 24
 
0 0
x v
x v
vdv
dx
a v

 
Determine the full integral, including limits
628.32
2
0 0
3 0.001
v
v
dx dv
v


 
Evaluate the interval and solve for v
2
0
1
628.32 ln 3 0.001
0.002
v
v
 
  
 
 
2
628.32( 0.002) ln 3 0.001 ln 3 0.001(0)
v
 
    
 
2
ln 3 0.001 1.2566 1.0986= 0.15802
v
 
    
 
2 0.15802
3 0.001v e
 
Take the exponential of each side

Tenth
Edition
2 - 25
2 0.15802
3 0.001v e
 
Solve for v
0.15802
2 3
2146.2
0.001
e
v


  46.3268 m/s
v 
How do you determine the maximum speed the car can reach?
2
3 0.001
a v
 
Velocity is a maximum when
acceleration is zero
This occurs when 2
0.001 3
v 
3
0.001
max
v  max 54.772 m/s
v 

Tenth
Edition
Uniform Rectilinear Motion
11 - 26
For a particle in uniform
rectilinear motion, the
acceleration is zero and
the velocity is constant.
vt
x
x
vt
x
x
dt
v
dx
v
dt
dx
t
x
x









0
0
0
0
constant
During free-fall, a parachutist
reaches terminal velocity when
her weight equals the drag
force. If motion is in a straight
line, this is uniform rectilinear
motion.
Careful – these only apply to
uniform rectilinear motion!

Tenth
Edition
Uniformly Accelerated Rectilinear Motion
11 - 27
If forces applied to a body
are constant (and in a
constant direction), then
you have uniformly
accelerated rectilinear
motion.
Another example is free-
fall when drag is negligible

Tenth
Edition
Uniformly Accelerated Rectilinear Motion
11 - 28
For a particle in uniformly accelerated rectilinear motion, the
acceleration of the particle is constant. You may recognize these
constant acceleration equations from your physics courses.
0
0
0
constant
v t
v
dv
a dv a dt v v at
dt
    
 
 
0
2
1
0 0 0 0 2
0
x t
x
dx
v at dx v at dt x x v t at
dt
      
 
 
0 0
2 2
0 0
constant 2
v x
v x
dv
v a vdv a dx v v a x x
dx
     
 
Careful – these only apply to uniformly
accelerated rectilinear motion!

Tenth
Edition
Motion of Several Particles
2 - 29
We may be interested in the motion of several different particles,
whose motion may be independent or linked together.

Tenth
Edition
Motion of Several Particles: Relative Motion
11 - 30
• For particles moving along the same line, time
should be recorded from the same starting
instant and displacements should be measured
from the same origin in the same direction.


 A
B
A
B x
x
x relative position of B
with respect to A
A
B
A
B x
x
x 



 A
B
A
B v
v
v relative velocity of B
with respect to A
A
B
A
B v
v
v 



 A
B
A
B a
a
a relative acceleration of B
with respect to A
A
B
A
B a
a
a 


Tenth
Edition
Sample Problem 11.4
11 - 31
Ball thrown vertically from 12 m level
in elevator shaft with initial velocity of
18 m/s. At same instant, open-platform
elevator passes 5 m level moving
upward at 2 m/s.
Determine (a) when and where ball hits
elevator and (b) relative velocity of ball
and elevator at contact.
SOLUTION:
• Substitute initial position and velocity
and constant acceleration of ball into
general equations for uniformly
accelerated rectilinear motion.
• Substitute initial position and constant
velocity of elevator into equation for
uniform rectilinear motion.
• Write equation for relative position of
ball with respect to elevator and solve
for zero relative position, i.e., impact.
• Substitute impact time into equation
for position of elevator and relative
velocity of ball with respect to
elevator.

Tenth
Edition
Sample Problem 11.4
11 - 32
SOLUTION:
• Substitute initial position and velocity and constant
acceleration of ball into general equations for
uniformly accelerated rectilinear motion.
2
2
2
2
1
0
0
2
0
s
m
905
.
4
s
m
18
m
12
s
m
81
.
9
s
m
18
t
t
at
t
v
y
y
t
at
v
v
B
B




























• Substitute initial position and constant velocity of
elevator into equation for uniform rectilinear motion.
t
t
v
y
y
v
E
E
E











s
m
2
m
5
s
m
2
0

Tenth
Edition
Sample Problem 11.4
11 - 33
• Write equation for relative position of ball with respect to
elevator and solve for zero relative position, i.e., impact.
    0
2
5
905
.
4
18
12 2





 t
t
t
y E
B
 
s
65
.
3
s
meaningles
s
39
.
0



t
t
• Substitute impact time into equations for position of elevator
and relative velocity of ball with respect to elevator.
 
65
.
3
2
5

E
y
m
3
.
12

E
y
 
 
65
.
3
81
.
9
16
2
81
.
9
18




 t
v E
B
s
m
81
.
19


E
B
v

Tenth
Edition
Motion of Several Particles: Dependent Motion
11 - 34
• Position of a particle may depend on position of one
or more other particles.
• Position of block B depends on position of block A.
Since rope is of constant length, it follows that sum of
lengths of segments must be constant.

 B
A x
x 2 constant (one degree of freedom)
• Positions of three blocks are dependent.


 C
B
A x
x
x 2
2 constant (two degrees of freedom)
• For linearly related positions, similar relations hold
between velocities and accelerations.
0
2
2
or
0
2
2
0
2
2
or
0
2
2












C
B
A
C
B
A
C
B
A
C
B
A
a
a
a
dt
dv
dt
dv
dt
dv
v
v
v
dt
dx
dt
dx
dt
dx

Tenth
Edition
Sample Problem 11.5
11 - 35
Pulley D is attached to a collar which
is pulled down at 3 in./s. At t = 0,
collar A starts moving down from K
with constant acceleration and zero
initial velocity. Knowing that velocity
of collar A is 12 in./s as it passes L,
determine the change in elevation,
velocity, and acceleration of block B
when block A is at L.
SOLUTION:
• Define origin at upper horizontal surface
with positive displacement downward.
• Collar A has uniformly accelerated
rectilinear motion. Solve for acceleration
and time t to reach L.
• Pulley D has uniform rectilinear motion.
Calculate change of position at time t.
• Block B motion is dependent on motions
of collar A and pulley D. Write motion
relationship and solve for change of block
B position at time t.
• Differentiate motion relation twice to
develop equations for velocity and
acceleration of block B.

Tenth
Edition
Sample Problem 11.5
11 - 36
SOLUTION:
• Define origin at upper horizontal surface with
positive displacement downward.
• Collar A has uniformly accelerated rectilinear
motion. Solve for acceleration and time t to reach L.
   
 
  2
2
0
2
0
2
s
in.
9
in.
8
2
s
in.
12
2











A
A
A
A
A
A
A
a
a
x
x
a
v
v
 
s
333
.
1
s
in.
9
s
in.
12 2
0




t
t
t
a
v
v A
A
A

Tenth
Edition
Sample Problem 11.5
11 - 37
• Pulley D has uniform rectilinear motion. Calculate
change of position at time t.
 
    in.
4
s
333
.
1
s
in.
3
0
0











D
D
D
D
D
x
x
t
v
x
x
• Block B motion is dependent on motions of collar
A and pulley D. Write motion relationship and
solve for change of block B position at time t.
Total length of cable remains constant,
     
 
   
   
 
     
  0
in.
4
2
in.
8
0
2
2
2
0
0
0
0
0
0
0















B
B
B
B
D
D
A
A
B
D
A
B
D
A
x
x
x
x
x
x
x
x
x
x
x
x
x
x
  in.
16
0 

 B
B x
x

Tenth
Edition
Sample Problem 11.5
11 - 38
• Differentiate motion relation twice to develop
equations for velocity and acceleration of block B.
0
s
in.
3
2
s
in.
12
0
2
constant
2





















B
B
D
A
B
D
A
v
v
v
v
x
x
x
s
in.
18

B
v
0
s
in.
9
0
2
2











B
B
D
A
v
a
a
a
2
s
in.
9


B
a

Tenth
Edition
2 - 39
Slider block A moves to the left with a
constant velocity of 6 m/s. Determine the
velocity of block B.
Solution steps
• Sketch your system and choose
coordinate system
• Write out constraint equation
• Differentiate the constraint equation to
get velocity

Tenth
Edition
2 - 40
Given: vA= 6 m/s left Find: vB
xA
yB
This length is constant no
matter how the blocks move
Sketch your system and choose coordinates
Differentiate the constraint equation to
get velocity
const nts
3 a
A B
x y L
  
Define your constraint equation(s)
6 m/s + 3 0
B
v 
2 m/s
B  
v
Note that as xA gets bigger, yB gets smaller.

Tenth
Edition
Graphical Solution of Rectilinear-Motion Problems
2 - 41
Engineers often collect position, velocity, and acceleration
data. Graphical solutions are often useful in analyzing
these data.
Data Fideltity / Highest Recorded Punch
0
20
40
60
80
100
120
140
160
180
47.76 47.77 47.78 47.79 47.8 47.81
Time (s)
Acceleration
(g)
Acceleration data
from a head impact
during a round of
boxing.

Tenth
Edition
11 - 42
• Given the x-t curve, the v-t curve is
equal to the x-t curve slope.
• Given the v-t curve, the a-t curve is
equal to the v-t curve slope.

Tenth
Edition
11 - 43
• Given the a-t curve, the change in velocity between t1 and t2 is
equal to the area under the a-t curve between t1 and t2.
• Given the v-t curve, the change in position between t1 and t2 is
equal to the area under the v-t curve between t1 and t2.

Tenth
Edition
11 - 44
• Moment-area method to determine particle position at
time t directly from the a-t curve:
 
 





1
0
1
1
0
0
1 curve
under
area
v
v
dv
t
t
t
v
t
v
x
x
using dv = a dt ,
 
 



1
0
1
1
0
0
1
v
v
dt
a
t
t
t
v
x
x
  


1
0
1
v
v
dt
a
t
t first moment of area under a-t curve
with respect to t = t1 line.
  
C
t
t
t
a-t
t
v
x
x
centroid
of
abscissa
curve
under
area 1
1
0
0
1






Tenth
Edition
11 - 45
• Method to determine particle acceleration from v-x curve:




BC
AB
dx
dv
v
a

tan
subnormal to v-x curve

Tenth
Edition
Curvilinear Motion: Position, Velocity & Acceleration
11 - 46
The softball and the car both undergo
curvilinear motion.
• A particle moving along a curve other than a
straight line is in curvilinear motion.

Tenth
Edition
11 - 47
• The position vector of a particle at time t is defined by a vector between
origin O of a fixed reference frame and the position occupied by particle.
• Consider a particle which occupies position P defined by at time t
and P’ defined by at t + Dt,
r

r


Tenth
Edition
11 - 48
0
lim
t
s ds
v
t dt
D 
D
 
D
Instantaneous velocity
(vector)
Instantaneous speed
(scalar)
0
lim
t
r dr
v
t dt
D 
D
 
D

Tenth
Edition
11 - 49
0
lim
t
v dv
a
t dt
D 
D
  
D
instantaneous acceleration (vector)
• Consider velocity of a particle at time t and velocity at t + Dt,
v

v


• In general, the acceleration vector is not tangent
to the particle path and velocity vector.

Tenth
Edition
Derivatives of Vector Functions
11 - 50
 
u
P

• Let be a vector function of scalar variable u,
   
u
u
P
u
u
P
u
P
du
P
d
u
u D

D


D
D


D

D




0
0
lim
lim
• Derivative of vector sum,
 
du
Q
d
du
P
d
du
Q
P
d







 
du
P
d
f
P
du
df
du
P
f
d





• Derivative of product of scalar and vector functions,
• Derivative of scalar product and vector product,
 
 
du
Q
d
P
Q
du
P
d
du
Q
P
d
du
Q
d
P
Q
du
P
d
du
Q
P
d






















Delete or put in
“bonus” slides

Tenth
Edition
Rectangular Components of Velocity & Acceleration
11 - 51
• When position vector of particle P is given by its
rectangular components,
k
z
j
y
i
x
r







• Velocity vector,
k
v
j
v
i
v
k
z
j
y
i
x
k
dt
dz
j
dt
dy
i
dt
dx
v
z
y
x






















• Acceleration vector,
k
a
j
a
i
a
k
z
j
y
i
x
k
dt
z
d
j
dt
y
d
i
dt
x
d
a
z
y
x
























 2
2
2
2
2
2

Tenth
Edition
Rectangular Components of Velocity & Acceleration
11 - 52
• Rectangular components particularly effective
when component accelerations can be integrated
independently, e.g., motion of a projectile,
0
0 





 z
a
g
y
a
x
a z
y
x 





with initial conditions,
      0
,
,
0 0
0
0
0
0
0 


 z
y
x v
v
v
z
y
x
Integrating twice yields
   
    0
0
2
2
1
0
0
0
0








z
gt
y
v
y
t
v
x
v
gt
v
v
v
v
y
x
z
y
y
x
x
• Motion in horizontal direction is uniform.
• Motion in vertical direction is uniformly accelerated.
• Motion of projectile could be replaced by two
independent rectilinear motions.

Tenth
Edition
Sample Problem 11.7
11 - 53
A projectile is fired from the edge
of a 150-m cliff with an initial
velocity of 180 m/s at an angle of
30°with the horizontal. Neglecting
air resistance, find (a) the horizontal
distance from the gun to the point
where the projectile strikes the
ground, (b) the greatest elevation
above the ground reached by the
projectile.
SOLUTION:
• Consider the vertical and horizontal motion
separately (they are independent)
• Apply equations of motion in y-direction
• Apply equations of motion in x-direction
• Determine time t for projectile to hit the
ground, use this to find the horizontal
distance
• Maximum elevation occurs when vy=0

Tenth
Edition
Sample Problem 11.7
11 - 54
SOLUTION:
Given: (v)o =180 m/s (y)o =150 m
(a)y = - 9.81 m/s2
(a)x = 0 m/s2
Vertical motion – uniformly accelerated:
Horizontal motion – uniformly accelerated:
Choose positive x to the right as shown

Tenth
Edition
Sample Problem 11.7
11 - 55
SOLUTION:
Horizontal distance
Projectile strikes the ground at:
Solving for t, we take the positive root
Maximum elevation occurs when vy=0
Substitute into equation (1) above
Substitute t into equation (4)
Maximum elevation above the ground =

Tenth
Edition
Concept Quiz
2 - 56
If you fire a projectile from 150
meters above the ground (see
Ex Problem 11.7), what launch
angle will give you the greatest
horizontal distance x?
a) A launch angle of 45
b) A launch angle less than 45
c) A launch angle greater than 45
d) It depends on the launch velocity

Tenth
Edition
11 - 57
A baseball pitching machine
“throws” baseballs with a
horizontal velocity v0. If you
want the height h to be 42 in.,
determine the value of v0.
SOLUTION:
• Consider the vertical and horizontal motion
separately (they are independent)
• Apply equations of motion in y-direction
• Apply equations of motion in x-direction
• Determine time t for projectile to fall to 42
inches
• Calculate v0=0

Tenth
Edition
2 - 58
Analyze the motion in
the y-direction
Given: x= 40 ft, yo = 5 ft,
yf= 42 in.
Find: vo
2
0
1
(0)
2
f
y y t gt
  
2 2
1
1.5 ft (32.2 ft/s )
2
t
  
2
1
3.5 5
2
gt
 
0.305234 s
t 
Analyze the motion in
the x-direction
0 0
0 ( )
x
x v t v t
  
0
40 ft ( )(0.305234 s)
v

0 131.047 ft/s 89.4 mi/h
v  

Tenth
Edition
Motion Relative to a Frame in Translation
2 - 59
A soccer player must consider
the relative motion of the ball
and her teammates when
making a pass.
It is critical for a pilot to
know the relative motion
of his aircraft with respect
to the aircraft carrier to
make a safe landing.

Tenth
Edition
Motion Relative to a Frame in Translation
11 - 60
• Designate one frame as the fixed frame of reference.
All other frames not rigidly attached to the fixed
reference frame are moving frames of reference.
• Position vectors for particles A and B with respect to
the fixed frame of reference Oxyz are .
and B
A r
r


• Vector joining A and B defines the position of
B with respect to the moving frame Ax’y’z’ and
A
B
r

A
B
A
B r
r
r





• Differentiating twice,

A
B
v

velocity of B relative to A.
A
B
A
B v
v
v






A
B
a

acceleration of B relative
to A.
A
B
A
B a
a
a





• Absolute motion of B can be obtained by combining
motion of A with relative motion of B with respect to
moving reference frame attached to A.

Tenth
Edition
Sample Problem 11.9
11 - 61
Automobile A is traveling east at the
constant speed of 36 km/h. As
automobile A crosses the intersection
shown, automobile B starts from rest
35 m north of the intersection and
moves south with a constant
acceleration of 1.2 m/s2. Determine
the position, velocity, and
acceleration of B relative to A 5 s
after A crosses the intersection.
SOLUTION:
• Define inertial axes for the system
• Determine the position, speed, and
acceleration of car A at t = 5 s
• Using vectors (Eqs 11.31, 11.33, and
11.34) or a graphical approach, determine
the relative position, velocity, and
acceleration
• Determine the position, speed, and
acceleration of car B at t = 5 s

Tenth
Edition
Sample Problem 11.9
11 - 62
SOLUTION: • Define axes along the road
Given: vA=36 km/h, aA= 0, (xA)0 = 0
(vB)0= 0, aB= - 1.2 m/s2, (yA)0 = 35 m
Determine motion of Automobile A:
We have uniform motion for A so:
At t = 5 s

Tenth
Edition
Sample Problem 11.9
11 - 63
SOLUTION:
Determine motion of Automobile B:
We have uniform acceleration for B so:
At t = 5 s

Tenth
Edition
Sample Problem 11.9
11 - 64
SOLUTION:
We can solve the problems geometrically, and apply the arctangent relationship:
Or we can solve the problems using vectors to obtain equivalent results:
 
B A B/A
r r r  
B A B/A
v v v  
B A B/A
a a a
20 50
20 50 (m)
 
 
B/A
B/A
j i r
r j i
6 10
6 10 (m/s)
  
  
B/A
B/A
j i v
v j i
2
1.2 0
1.2 (m/s )
  
 
B/A
B/A
j i a
a j
Physically, a rider in car A would “see” car B travelling south and west.

Tenth
Edition
Concept Quiz
2 - 65
If you are sitting in train
B looking out the window,
it which direction does it
appear that train A is
moving?
a)
b)
c)
d)
25o
25o

Tenth
Edition
2 - 66
If we have an idea of the path of a vehicle, it is often convenient
to analyze the motion using tangential and normal components
(sometimes called path coordinates).

Tenth
Edition
11 - 67
• The tangential direction (et) is tangent to the path of the
particle. This velocity vector of a particle is in this direction
x
y
et
en
• The normal direction (en) is perpendicular to et and points
towards the inside of the curve.
v= vt et
r= the instantaneous
radius of curvature
2
dv v
dt r
 
t n
a e e
v
 t
v e
• The acceleration can have components in both the en and et directions

Tenth
Edition
11 - 68
• To derive the acceleration vector in tangential
and normal components, define the motion of a
particle as shown in the figure.
• are tangential unit vectors for the
particle path at P and P’. When drawn with
respect to the same origin, and
is the angle between them.
t
t e
e



and
t
t
t e
e
e






D

D
 
 







d
e
d
e
e
e
e
e
t
n
n
n
t
t







D
D

D
D
D

D

D

D 2
2
sin
lim
lim
2
sin
2
0
0

Tenth
Edition
11 - 69
t
e
v
v



• With the velocity vector expressed as
the particle acceleration may be written as
dt
ds
ds
d
d
e
d
v
e
dt
dv
dt
e
d
v
e
dt
dv
dt
v
d
a t
t













but
v
dt
ds
ds
d
e
d
e
d
n
t 

 
r



After substituting,
r
r
2
2
v
a
dt
dv
a
e
v
e
dt
dv
a n
t
n
t 






• The tangential component of acceleration
reflects change of speed and the normal
component reflects change of direction.
• The tangential component may be positive or
negative. Normal component always points
toward center of path curvature.

Tenth
Edition
11 - 70
r
r
2
2
v
a
dt
dv
a
e
v
e
dt
dv
a n
t
n
t 






• Relations for tangential and normal acceleration
also apply for particle moving along a space curve.
• The plane containing tangential and normal unit
vectors is called the osculating plane.
n
t
b e
e
e





• The normal to the osculating plane is found from
binormal
e
normal
principal
e
b
n




• Acceleration has no component along the binormal.

Tenth
Edition
Sample Problem 11.8
11 - 71
A motorist is traveling on a curved
section of highway of radius 2500 ft
at the speed of 60 mi/h. The motorist
suddenly applies the brakes, causing
the automobile to slow down at a
constant rate. Knowing that after 8 s
the speed has been reduced to 45
mi/h, determine the acceleration of
the automobile immediately after the
brakes have been applied.
SOLUTION:
• Define your coordinate system
• Calculate the tangential velocity and
tangential acceleration
• Determine overall acceleration magnitude
after the brakes have been applied
• Calculate the normal acceleration

Tenth
Edition
Sample Problem 11.8
11 - 72
SOLUTION: • Define your coordinate system
et
en
• Determine velocity and acceleration in
the tangential direction
• The deceleration constant, therefore
• Immediately after the brakes are applied,
the speed is still 88 ft/s
2 2 2 2
2.75 3.10
n t
a a a
   

Tenth
Edition
2 - 73
In 2001, a race scheduled at the Texas Motor Speedway was
cancelled because the normal accelerations were too high and
caused some drivers to experience excessive g-loads (similar to
fighter pilots) and possibly pass out. What are some things that
could be done to solve this problem?
Some possibilities:
Reduce the allowed speed
Increase the turn radius
(difficult and costly)
Have the racers wear g-suits

Tenth
Edition
11 - 74
SOLUTION:
• Calculate the tangential velocity and
tangential acceleration
• Determine overall acceleration
magnitude
• Calculate the normal acceleration
The tangential acceleration of the
centrifuge cab is given by
where t is in seconds and at is in
m/s2. If the centrifuge starts from
fest, determine the total acceleration
magnitude of the cab after 10
seconds.
2
0.5 (m/s )
t
a t


Tenth
Edition
11 - 75
In the side view, the tangential
direction points into the “page”
Define your coordinate system
et
en
en
Top View
Determine the tangential velocity
0.5
t
a t

2 2
0
0
0.5 0.25 0.25
t t
t
v t dt t t
  

 
2
0.25 10 25 m/s
t
v  
Determine the normal acceleration
 
2 2
2
25
78.125 m/s
8
t
n
v
a
r
  
Determine the total acceleration magnitude
 
2
2 2 2
78.125 + (0.5)(10)
mag n t
a a a
   2
78.285 m/s
mag
a 

Tenth
Edition
11 - 76
a) The accelerations would remain the same
b) The an would increase and the at would decrease
c) The an and at would both increase
d) The an would decrease and the at would increase
Notice that the normal
acceleration is much higher than
the tangential acceleration.
What would happen if, for a
given tangential velocity and
acceleration, the arm radius was
doubled?

Tenth
Edition
2 - 77
By knowing the distance to the aircraft and the
angle of the radar, air traffic controllers can
track aircraft.
Fire truck ladders can rotate as well as extend;
the motion of the end of the ladder can be
analyzed using radial and transverse
components.

Tenth
Edition
11 - 78
• The position of a particle P is
expressed as a distance r from the
origin O to P – this defines the
radial direction er. The transverse
direction e is perpendicular to er
r
v r e r e

 
• The particle velocity vector is
• The particle acceleration vector is
   
2
2
r
a r r e r r e
  
   
r
e
r
r




Tenth
Edition
11 - 79
• We can derive the velocity and acceleration
relationships by recognizing that the unit vectors
change direction.
r
r e
d
e
d
e
d
e
d 










dt
d
e
dt
d
d
e
d
dt
e
d r
r 

 





dt
d
e
dt
d
d
e
d
dt
e
d
r




 





 




e
r
e
r
e
dt
d
r
e
dt
dr
dt
e
d
r
e
dt
dr
e
r
dt
d
v
r
r
r
r
r

















• The particle velocity vector is
• Similarly, the particle acceleration vector is
    











e
r
r
e
r
r
dt
e
d
dt
d
r
e
dt
d
r
e
dt
d
dt
dr
dt
e
d
dt
dr
e
dt
r
d
e
dt
d
r
e
dt
dr
dt
d
a
r
r
r
r

















2
2
2
2
2
2

















r
e
r
r




Tenth
Edition
Concept Quiz
2 - 80
If you are travelling in a perfect
circle, what is always true about
radial/transverse coordinates and
normal/tangential coordinates?
a) The er direction is identical to the en direction.
b) The e direction is perpendicular to the en direction.
c) The e direction is parallel to the er direction.

Tenth
Edition
11 - 81
• When particle position is given in cylindrical
coordinates, it is convenient to express the
velocity and acceleration vectors using the unit
vectors .
and
,
, k
e
eR




• Position vector,
k
z
e
R
r R





• Velocity vector,
k
z
e
R
e
R
dt
r
d
v R











 

• Acceleration vector,
    k
z
e
R
R
e
R
R
dt
v
d
a R



















 


 2
2

Tenth
Edition
11 - 82
Rotation of the arm about O is defined
by  = 0.15t2 where  is in radians and t
in seconds. Collar B slides along the
arm such that r = 0.9 - 0.12t2 where r is
in meters.
After the arm has rotated through 30o,
determine (a) the total velocity of the
collar, (b) the total acceleration of the
collar, and (c) the relative acceleration
of the collar with respect to the arm.
SOLUTION:
• Evaluate time t for  = 30o.
• Evaluate radial and angular positions,
and first and second derivatives at
time t.
• Calculate velocity and acceleration in
cylindrical coordinates.
• Evaluate acceleration with respect to
arm.

Tenth
Edition
11 - 83
SOLUTION:
• Evaluate time t for  = 30o.
s
869
.
1
rad
524
.
0
30
0.15 2





t
t

• Evaluate radial and angular positions, and first
and second derivatives at time t.
2
2
s
m
24
.
0
s
m
449
.
0
24
.
0
m
481
.
0
12
.
0
9
.
0









r
t
r
t
r



2
2
s
rad
30
.
0
s
rad
561
.
0
30
.
0
rad
524
.
0
15
.
0










 t
t

Tenth
Edition
11 - 84
• Calculate velocity and acceleration.
  
r
r
r
v
v
v
v
v
r
v
s
r
v





1
2
2
tan
s
m
270
.
0
s
rad
561
.
0
m
481
.
0
m
449
.
0














 0
.
31
s
m
524
.
0 
v
  
     
r
r
r
a
a
a
a
a
r
r
a
r
r
a







1
2
2
2
2
2
2
2
2
tan
s
m
359
.
0
s
rad
561
.
0
s
m
449
.
0
2
s
rad
3
.
0
m
481
.
0
2
s
m
391
.
0
s
rad
561
.
0
m
481
.
0
s
m
240
.
0



























 6
.
42
s
m
531
.
0 
a

Tenth
Edition
11 - 85
• Evaluate acceleration with respect to arm.
Motion of collar with respect to arm is rectilinear
and defined by coordinate r.
2
s
m
240
.
0


 r
a OA
B 


Tenth
Edition
11 - 86
SOLUTION:
• Calculate the angular velocity after
three revolutions
• Determine overall acceleration
magnitude
• Calculate the radial and transverse
accelerations
The angular acceleration of the
centrifuge arm varies according to
where  is measured in radians. If the
centrifuge starts from rest, determine the
acceleration magnitude after the gondola
has travelled two full rotations.
2
0.05 (rad/s )
 


Tenth
Edition
11 - 87
In the side view, the transverse
direction points into the “page”
Define your coordinate system
e
er
er
Top View
Determine the angular velocity
Evaluate the integral
2
0.05 (rad/s )
 

d d
   

Acceleration is a function
of position, so use:
2(2 )
2 2
0 0
0.05
2 2


 

(2)(2 )
0 0
0.05 d d
 
   

 
 
2
2
0.05 2(2 )
 


Tenth
Edition
11 - 88
er
Determine the angular velocity
Determine the angular acceleration
2.8099 rad/s
 
 
2
2
0.05 2(2 )
 

2
0.05 = 0.05(2)(2 ) 0.6283 rad/s
  
 
Find the radial and transverse accelerations
   
   
2
2
2
2
0 (8)(2.8099) (8)(0.6283) 0
63.166 5.0265 (m/s )
r
r
r
a r r e r r e
e e
e e



  
   
   
  
 
2
2 2 2
( 63.166) + 5.0265
mag r
a a a
    2
63.365 m/s
mag
a 
Magnitude:

Tenth
Edition
11 - 89
You could now have additional acceleration terms. This might
give you more control over how quickly the acceleration of the
gondola changes (this is known as the G-onset rate).
What would happen if you
designed the centrifuge so
that the arm could extend
from 6 to 10 meters?
r
   
2
2
r
a r r e r r e
  
   

dynamics11lecture.ppt

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