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Chapter 11
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VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Tenth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Brian P. SelfCalifornia Polytechnic State University
CHAPTER
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11Kinematics of Particles
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Vector Mechanics for Engineers: Dynamics
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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
Sample Problem 11.12
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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.
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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
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Vector Mechanics for Engineers: Dynamics
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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.
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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., 326 ttx
or in the form of a graph x vs. t.
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Rectilinear Motion: Position, Velocity & Acceleration
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
xv
t
x
t D
D
D
D
D 0lim
Average velocity
Instantaneous velocity
• From the definition of a derivative,
dt
dx
t
xv
t
D
D
D 0lim
e.g.,
2
32
312
6
ttdt
dxv
ttx
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Rectilinear Motion: Position, Velocity & Acceleration
11 - 8
• Consider particle with velocity v at time t and
v’ at t+Dt,
Instantaneous accelerationt
va
t D
D
D 0lim
tdt
dva
ttv
dt
xd
dt
dv
t
va
t
612
312e.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.
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Rectilinear Motion: Position, Velocity & Acceleration
11 - 9
• From our example,
326 ttx
2312 ttdt
dxv
tdt
xd
dt
dva 612
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 ?
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Concept Quiz
2 - 10
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
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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
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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 tdt
vdv a x dx
0 0
v x
v x
v dv a x dx a a x
and dx dv
dt av dt
dv
v a vdx
0 0
v t
v
dvdt
a v
0 0
x v
x v
v dvdx
a v
a a v
( )dv
a vdt
If…. Kinematic relationship Integrate
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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
corresponding velocity.
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Sample Problem 11.2
11 - 14
tvtvdtdv
adt
dv
ttv
v
81.981.9
sm81.9
00
2
0
ttv
2s
m81.9
s
m10
2
21
00
81.91081.910
81.910
0
ttytydttdy
tvdt
dy
tty
y
2
2s
m905.4
s
m10m20 ttty
SOLUTION:
• Integrate twice to find v(t) and y(t).
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Sample Problem 11.2
11 - 15
• Solve for t when velocity equals zero and evaluate
corresponding altitude.
0s
m81.9
s
m10
2
ttv
s019.1t
• Solve for t when altitude equals zero and evaluate
corresponding velocity.
22
2
2
s019.1s
m905.4s019.1
s
m10m20
s
m905.4
s
m10m20
y
ttty
m1.25y
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Sample Problem 11.2
11 - 16
• Solve for t when altitude equals zero and evaluate
corresponding velocity.
0s
m905.4
s
m10m20 2
2
ttty
s28.3
smeaningles s243.1
t
t
s28.3s
m81.9
s
m10s28.3
s
m81.9
s
m10
2
2
v
ttv
s
m2.22v
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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).
kva
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).
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Sample Problem 11.3
11 - 18
SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
0 00
lnv t
v
v tdv dva kv k dt kt
dt v v
ktevtv 0
• Integrate v(t) = dx/dt to find x(t).
0
0 0
00 0
1
kt
tx t
kt kt
dxv t v e
dt
dx v e dt x t v ek
ktek
vtx 10
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Sample Problem 11.3
11 - 19
• Integrate a = v dv/dx = -kv to find v(x).
kxvv
dxkdvdxkdvkvdx
dvva
xv
v
0
00
kxvv 0
• Alternatively,
0
0 1v
tv
k
vtx
kxvv 0
00 or
v
tveevtv ktkt
ktek
vtx 10with
and
then
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Group Problem Solving
11 - 20
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
23 0.001a 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?
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Group Problem Solving
2 - 21
dv
v a vdx
0 0
x v
x v
v dvdx
a v
Choose the proper kinematic relationship
Given:23 0.001a 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 mx r
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Group Problem Solving
2 - 22
0 0
x v
x v
v dvdx
a v
Determine the full integral, including limits
628.32
2
0 03 0.001
vv
dx dvv
Evaluate the interval and solve for v
2
0
1628.32 ln 3 0.001
0.002
v
v
2628.32( 0.002) ln 3 0.001 ln 3 0.001(0)v
2ln 3 0.001 1.2566 1.0986= 0.15802v
2 0.158023 0.001v e Take the exponential of each side
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Group Problem Solving
2 - 23
2 0.158023 0.001v e Solve for v
0.158022 3
2146.20.001
ev
46.3268 m/sv
How do you determine the maximum speed the car can reach?
23 0.001a v Velocity is a maximum when
acceleration is zero
This occurs when20.001 3v
30.001maxv
max 54.772 m/sv
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Uniform Rectilinear Motion
11 - 24
For a particle in uniform
rectilinear motion, the
acceleration is zero and
the velocity is constant.
vtxx
vtxx
dtvdx
vdt
dx
tx
x
0
0
00
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!
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Uniformly Accelerated Rectilinear Motion
11 - 25
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
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Uniformly Accelerated Rectilinear Motion
11 - 26
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
constantv t
v
dva dv a dt v v at
dt
0
210 0 0 0 2
0
x t
x
dxv at dx v at dt x x v t at
dt
0 0
2 2
0 0constant 2v x
v x
dvv a vdv a dx v v a x x
dx
Careful – these only apply to uniformly
accelerated rectilinear motion!
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Motion of Several Particles
2 - 27
We may be interested in the motion of several different particles,
whose motion may be independent or linked together.
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Motion of Several Particles: Relative Motion
11 - 28
• 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.
ABAB xxx relative position of B
with respect to AABAB xxx
ABAB vvv relative velocity of B
with respect to AABAB vvv
ABAB aaa relative acceleration of B
with respect to AABAB aaa
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Sample Problem 11.4
11 - 29
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.
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Sample Problem 11.4
11 - 30
SOLUTION:
• Substitute initial position and velocity and constant
acceleration of ball into general equations for
uniformly accelerated rectilinear motion.
2
2
221
00
20
s
m905.4
s
m18m12
s
m81.9
s
m18
ttattvyy
tatvv
B
B
• Substitute initial position and constant velocity of
elevator into equation for uniform rectilinear motion.
ttvyy
v
EE
E
s
m2m5
s
m2
0
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Sample Problem 11.4
11 - 31
• Write equation for relative position of ball with respect to
elevator and solve for zero relative position, i.e., impact.
025905.41812 2 ttty EB
s65.3
smeaningles s39.0
t
t
• Substitute impact time into equations for position of elevator
and relative velocity of ball with respect to elevator.
65.325Ey
m3.12Ey
65.381.916
281.918
tv EB
s
m81.19EBv
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Motion of Several Particles: Dependent Motion
11 - 32
• 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.
BA xx 2 constant (one degree of freedom)
• Positions of three blocks are dependent.
CBA xxx 22 constant (two degrees of freedom)
• For linearly related positions, similar relations hold
between velocities and accelerations.
022or022
022or022
CBACBA
CBACBA
aaadt
dv
dt
dv
dt
dv
vvvdt
dx
dt
dx
dt
dx
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Group Problem Solving
2 - 33
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
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Group Problem Solving
2 - 34
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 nts3 aA Bx y L
Define your constraint equation(s)
6 m/s + 3 0Bv
2 m/sB v
Note that as xA gets bigger, yB gets smaller.
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Graphical Solution of Rectilinear-Motion Problems
2 - 35
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)
Accele
rati
on
(g
)
Acceleration data
from a head impact
during a round of
boxing.
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Graphical Solution of Rectilinear-Motion Problems
11 - 36
• 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.
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Graphical Solution of Rectilinear-Motion Problems
11 - 37
• 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.
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Curvilinear Motion: Position, Velocity & Acceleration
11 - 38
The softball and the car both undergo
curvilinear motion.
• A particle moving along a curve other than a
straight line is in curvilinear motion.
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Curvilinear Motion: Position, Velocity & Acceleration
11 - 39
• 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
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Curvilinear Motion: Position, Velocity & Acceleration
11 - 40
0lim
t
s dsv
t dtD
D
D
Instantaneous velocity
(vector)
Instantaneous speed
(scalar)
0lim
t
r drv
t dtD
D
D
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Curvilinear Motion: Position, Velocity & Acceleration
11 - 41
0lim
t
v dva
t dtD
D
Dinstantaneous 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.
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Rectangular Components of Velocity & Acceleration
11 - 42
• When position vector of particle P is given by its
rectangular components,
kzjyixr
• Velocity vector,
kvjviv
kzjyixkdt
dzj
dt
dyi
dt
dxv
zyx
• Acceleration vector,
kajaia
kzjyixkdt
zdj
dt
ydi
dt
xda
zyx
2
2
2
2
2
2
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Rectangular Components of Velocity & Acceleration
11 - 43
• Rectangular components particularly effective
when component accelerations can be integrated
independently, e.g., motion of a projectile,
00 zagyaxa zyx
with initial conditions,
0,,0 000000 zyx vvvzyx
Integrating twice yields
0
0
221
00
00
zgtyvytvx
vgtvvvv
yx
zyyxx
• Motion in horizontal direction is uniform.
• Motion in vertical direction is uniformly accelerated.
• Motion of projectile could be replaced by two
independent rectilinear motions.
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Sample Problem 11.7
11 - 44
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
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Sample Problem 11.7
11 - 45
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
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Sample Problem 11.7
11 - 46
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 =
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Group Problem Solving
11 - 47
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
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Group Problem Solving
2 - 48
Analyze the motion in
the y-direction
Given: x= 40 ft, yo = 5 ft,
yf= 42 in.
Find: vo
20
1(0)
2fy y t gt
2 211.5 ft (32.2 ft/s )
2t
213.5 5
2gt
0.305234 st
Analyze the motion in
the x-direction
0 00 ( )xx v t v t
040 ft ( )(0.305234 s)v
0 131.047 ft/s 89.4 mi/hv
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Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Motion Relative to a Frame in Translation
2 - 49
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.
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Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Motion Relative to a Frame in Translation
11 - 50
• 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 BA rr
• Vector joining A and B defines the position of
B with respect to the moving frame Ax’y’z’ andABr
ABAB rrr
• Differentiating twice,
ABv
velocity of B relative to A.ABAB vvv
ABa
acceleration of B relative
to A.ABAB aaa
• 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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Tangential and Normal Components
2 - 51
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).
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Tangential and Normal Components
11 - 52
• 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
2dv v
dt r t na e e
v tv e
• The acceleration can have components in both the en and et directions
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Sample Problem 11.8
11 - 53
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
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Sample Problem 11.8
11 - 54
SOLUTION: • Define your coordinate system
eten
• 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 22.75 3.10n ta a a
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Group Problem Solving
11 - 55
SOLUTION:
• Define your coordinate system
• 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.
20.5 (m/s )ta t
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Group Problem Solving
11 - 56
In the side view, the tangential
direction points into the “page”
Define your coordinate system
et
en
en
Top View
Determine the tangential velocity
0.5ta t
2 2
000.5 0.25 0.25
t t
tv t dt t t
2
0.25 10 25 m/stv
Determine the normal acceleration
2 2
22578.125 m/s
8
t
n
va
r
Determine the total acceleration magnitude
22 2 278.125 + (0.5)(10) mag n ta a a 278.285 m/smaga
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Radial and Transverse Components
2 - 57
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Radial and Transverse Components
11 - 58
• 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 eq is perpendicular to er
rv r e r eqq
• The particle velocity vector is
• The particle acceleration vector is
2 2ra r r e r r eqq q q
rerr
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Sample Problem 11.12
11 - 59
Rotation of the arm about O is defined
by q = 0.15t2 where q 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 q = 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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Sample Problem 11.12
11 - 60
SOLUTION:
• Evaluate time t for q = 30o.
s 869.1rad524.030
0.15 2
t
tq
• Evaluate radial and angular positions, and first
and second derivatives at time t.
2
2
sm24.0
sm449.024.0
m 481.012.09.0
r
tr
tr
2
2
srad30.0
srad561.030.0
rad524.015.0
q
q
q
t
t
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Sample Problem 11.12
11 - 61
• Calculate velocity and acceleration.
rr
r
v
vvvv
rv
srv
q
q
122 tan
sm270.0srad561.0m481.0
m449.0
0.31sm524.0 v
rr
r
a
aaaa
rra
rra
q
q
122
2
2
2
22
2
tan
sm359.0
srad561.0sm449.02srad3.0m481.0
2
sm391.0
srad561.0m481.0sm240.0
6.42sm531.0 a
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Te
nth
Ed
ition
Sample Problem 11.12
11 - 62
• Evaluate acceleration with respect to arm.
Motion of collar with respect to arm is rectilinear
and defined by coordinate r.
2sm240.0 ra OAB