Tenth Edition
CHAPTER
11
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Kinematics of Particles
Brian P. Self
California Polytechnic State University
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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Contents
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 RectilinearMotion
Motion of Several Particles:
Relative Motion
Sample Problem 11.4
Motion of Several Particles:
Dependent Motion
Sample Problem 11.5
Graphical Solution of RectilinearMotion 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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11 - 2
Tenth
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Vector Mechanics for Engineers: Dynamics
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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Tenth
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Vector Mechanics for Engineers: Dynamics
Introduction
• Dynamics includes:
Kinematics: study of the geometry of motion.
Relates displacement, velocity, acceleration, and time without reference
to the cause of motion.
Fthrust
Fdrag
Flift
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.
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Tenth
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Vector Mechanics for Engineers: Dynamics
Introduction
• 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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11 - 5
Tenth
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Vector Mechanics for Engineers: Dynamics
Rectilinear Motion: Position, Velocity & Acceleration
• 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.,
2
3
x = 6t − t
or in the form of a graph x vs. t.
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Tenth
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Vector Mechanics for Engineers: Dynamics
Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle which occupies position P
at time t and P’ at t+Dt,
Dx
=
Average velocity
Dt
Dx
=
v
=
lim
Instantaneous velocity
Dt →0 Dt
• Instantaneous velocity may be positive or
negative. Magnitude of velocity is referred
to as particle speed.
• From the definition of a derivative,
Dx dx
v = lim
=
dt
Dt →0 Dt
e.g., x = 6t 2 − t 3
v=
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dx
= 12t − 3t 2
dt
11 - 7
Tenth
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Vector Mechanics for Engineers: Dynamics
Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle with velocity v at time t and
v’ at t+Dt,
Dv
Instantaneous acceleration = a = lim
Dt →0 Dt
• Instantaneous acceleration may be:
- positive: increasing positive velocity
or decreasing negative velocity
- negative: decreasing positive velocity
or increasing negative velocity.
• From the definition of a derivative,
Dv dv d 2 x
a = lim
=
= 2
dt dt
Dt →0 Dt
e.g. v = 12t − 3t 2
dv
a=
= 12 − 6t
dt
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Tenth
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Vector Mechanics for Engineers: Dynamics
Concept Quiz
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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2-9
Tenth
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Vector Mechanics for Engineers: Dynamics
Rectilinear Motion: Position, Velocity & Acceleration
• From our example,
x = 6t 2 − t 3
v=
dx
= 12t − 3t 2
dt
dv d 2 x
a=
=
= 12 − 6t
dt dt 2
• What are x, v, and a at t = 2 s ?
- at t = 2 s,
x = 16 m, v = vmax = 12 m/s, a = 0
• 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 ?
- at t = 4 s,
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x = xmax = 32 m, v = 0, a = -12 m/s2
11 - 10
Tenth
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Vector Mechanics for Engineers: Dynamics
Determination of the Motion of a Particle
• 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
When force is a function of velocity?
function of position?
a spring
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drag
11 - 11
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Vector Mechanics for Engineers: Dynamics
Acceleration as a function of time, position, or velocity
If….
a = a (t )
a = a ( x)
Kinematic relationship
dt =
dx
dv
and a =
v
dt
dv
= a(v)
dt
dv
v = a (v)
dx
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v
t
v0
0
 dv =  a ( t ) dt
dv
= a(t )
dt
v dv = a ( x ) dx
a = a (v)
Integrate
v
x
v0
x0
 v dv =  a ( x ) dx
v
t
dv
v a ( v ) = 0 dt
0
x
v
v dv
a (v)
v0
 dx = 
x0
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.2
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.
Ball tossed with 10 m/s vertical velocity
from window 20 m above ground.
• Solve for t when altitude equals zero
(time for ground impact) and evaluate
corresponding velocity.
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.
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Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.2
SOLUTION:
• Integrate twice to find v(t) and y(t).
dv
= a = −9.81m s 2
dt
v (t )
t
v(t ) − v0 = −9.81t
 dv = −  9.81dt
v0
0
v(t ) = 10
dy
= v = 10 − 9.81t
dt
y (t )
t
 dy =  (10 − 9.81t )dt
y0
0
m 
m
−  9.81 2  t
s 
s 
y (t ) − y0 = 10t − 12 9.81t 2
m
 m 
y(t ) = 20 m + 10 t −  4.905 2 t 2
 s 
s 
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11 - 14
Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.2
• Solve for t when velocity equals zero and evaluate
corresponding altitude.
v(t ) = 10
m 
m
−  9.81 2  t = 0
s 
s 
t = 1.019 s
• Solve for t when altitude equals zero and evaluate
corresponding velocity.
m
 m 
y (t ) = 20 m + 10 t −  4.905 2 t 2
 s 
s 
m

 m
y = 20 m + 10 (1.019 s ) −  4.905 2 (1.019 s )2
 s

s 
y = 25.1 m
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11 - 15
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.2
• Solve for t when altitude equals zero and evaluate
corresponding velocity.
m
 m 
y(t ) = 20 m + 10 t −  4.905 2 t 2 = 0
 s 
s 
t = −1.243s (meaningless )
t = 3.28 s
v(t ) = 10
m 
m
−  9.81 2  t
s 
s 
v(3.28 s ) = 10
m 
m
−  9.81 2  (3.28 s )
s 
s 
v = −22.2
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m
s
11 - 16
Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.3
SOLUTION:
a = − kv
• Integrate a = dv/dt = -kv to find v(t).
• Integrate v(t) = dx/dt to find x(t).
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.
• Integrate a = v dv/dx = -kv to find
v(x).
Determine v(t), x(t), and v(x).
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11 - 17
Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.3
SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
v
dv
a=
= −kv
dt
t
dv
v v = −k 0 dt
0
ln
v (t )
= −kt
v0
v(t ) = v0 e −kt
• Integrate v(t) = dx/dt to find x(t).
v (t ) =
dx
= v0e− kt
dt
x
t
0
0
− kt
dx
=
v
e
dt
0

t
 1

x ( t ) = v0  − e − kt 
 k
0
(
v
x(t ) = 0 1 − e −kt
k
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11 - 18
)
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.3
• Integrate a = v dv/dx = -kv to find v(x).
dv
a = v = −kv
dx
dv = −k dx
v
x
v0
0
 dv = −k  dx
v − v0 = −kx
v = v0 − kx
• Alternatively,
with
and
then
(
v
x(t ) = 0 1 − e −kt
k
)
v(t )
v(t ) = v0 e −kt or e −kt =
v0
v  v(t ) 

x(t ) = 0 1 −
k 
v0 
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v = v0 − kx
11 - 19
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
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.
+y
Which integral should you choose?
v
(a)
(b)
t
 dv =  a ( t ) dt
v0
0
x
v
v dv
x dx = v a ( v )
0
0
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(c)
v
x
v0
x0
 v dv =  a ( x ) dx
v
(d)
t
dv
v a ( v ) = 0 dt
0
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Vector Mechanics for Engineers: Dynamics
Concept Question
When will the bowling ball start slowing down?
+y
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
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2 - 21
Tenth
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
SOLUTION:
• Determine the proper kinematic
relationship to apply (is acceleration
a function of time, velocity, or
position?
The car starts from rest and accelerates
according to the relationship
a = 3 − 0.001v
2
• Determine the total distance the car
travels in one-half lap
• Integrate to determine the velocity
after one-half lap
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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11 - 22
Tenth
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Given: a = 3 − 0.001v 2
vo = 0, r = 200 m
Find: v after ½ lap
Maximum speed
Choose the proper kinematic relationship
Acceleration is a function of velocity, and
we also can determine distance. Time is not
involved in the problem, so we choose:
dv
v = a (v)
dx
x
v
v dv
a (v)
v0
 dx = 
x0
Determine total distance travelled
x =  r = 3.14(200) = 628.32 m
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2 - 23
Tenth
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Determine the full integral, including limits
x
v
v dv
dx
=
x
v a ( v )
0
0
628.32

0
v
v
dv
2
3 − 0.001v
0
dx = 
Evaluate the interval and solve for v
1
2 v
628.32 = −
ln 3 − 0.001v 
0
0.002
628.32(−0.002) = ln 3 − 0.001v 2  − ln 3 − 0.001(0) 
ln 3 − 0.001v 2  = −1.2566 + 1.0986= − 0.15802
Take the exponential of each side
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3 − 0.001v 2 = e−0.15802
2 - 24
Tenth
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Solve for v
3 − 0.001v 2 = e−0.15802
3 − e−0.15802
v =
= 2146.2
0.001
2
v = 46.3268 m/s
How do you determine the maximum speed the car can reach?
a = 3 − 0.001v 2
Velocity is a maximum when
acceleration is zero
0.001v 2 = 3
This occurs when
vmax =
3
0.001
vmax = 54.772 m/s
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2 - 25
Tenth
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Vector Mechanics for Engineers: Dynamics
Uniform Rectilinear Motion
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.
For a particle in uniform
rectilinear motion, the
acceleration is zero and
the velocity is constant.
dx
= v = constant
dt
x
t
x0
0
 dx = v  dt
x − x0 = vt
x = x0 + vt
Careful – these only apply to
uniform rectilinear motion!
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11 - 26
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Vector Mechanics for Engineers: Dynamics
Uniformly Accelerated Rectilinear Motion
If forces applied to a body
are constant (and in a
constant direction), then
you have uniformly
accelerated rectilinear
motion.
Another example is freefall when drag is negligible
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11 - 27
Tenth
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Vector Mechanics for Engineers: Dynamics
Uniformly Accelerated Rectilinear Motion
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.
dv
= a = constant
dt
dx
= v0 + at
dt
x
v
t
 dv = a  dt
v0
0
t
 dx =  ( v0 + at ) dt
x0
dv
v = a = constant
dx
v = v0 + at
x = x0 + v0t + 12 at 2
0
v
x
v0
x0
 v dv = a  dx
v 2 = v02 + 2a ( x − x0 )
Careful – these only apply to uniformly
accelerated rectilinear motion!
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11 - 28
Tenth
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Vector Mechanics for Engineers: Dynamics
Motion of Several Particles
We may be interested in the motion of several different particles,
whose motion may be independent or linked together.
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2 - 29
Tenth
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Vector Mechanics for Engineers: Dynamics
Motion of Several Particles: Relative Motion
• 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.
x B A = x B − x A = relative position of B
with respect to A
xB = x A + xB A
v B A = v B − v A = relative velocity of B
with respect to A
vB = v A + vB A
a B A = a B − a A = relative acceleration of B
with respect to A
aB = a A + aB A
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11 - 30
Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.4
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.
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
• 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.
11 - 31
Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.4
SOLUTION:
• Substitute initial position and velocity and constant
acceleration of ball into general equations for
uniformly accelerated rectilinear motion.
v B = v0 + at = 18
m 
m
−  9.81 2 t
s 
s 
m
 m 
y B = y0 + v0t + 12 at 2 = 12 m + 18 t −  4.905 2 t 2
 s 
s 
• Substitute initial position and constant velocity of
elevator into equation for uniform rectilinear motion.
vE = 2
m
s
 m
y E = y0 + v E t = 5 m +  2 t
 s
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11 - 32
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.4
• Write equation for relative position of ball with respect to
elevator and solve for zero relative position, i.e., impact.
yB
E
(
)
= 12 + 18t − 4.905t 2 − (5 + 2t ) = 0
t = −0.39 s (meaningless )
t = 3.65 s
• Substitute impact time into equations for position of elevator
and relative velocity of ball with respect to elevator.
y E = 5 + 2(3.65)
y E = 12.3 m
v B E = (18 − 9.81t ) − 2
= 16 − 9.81(3.65)
vB
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E
= −19.81
m
s
11 - 33
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Vector Mechanics for Engineers: Dynamics
Motion of Several Particles: Dependent Motion
• 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.
x A + 2 x B = constant (one degree of freedom)
• Positions of three blocks are dependent.
2 x A + 2 xB + xC = constant (two degrees of freedom)
• For linearly related positions, similar relations hold
between velocities and accelerations.
dx
dx A
dx
+ 2 B + C = 0 or
dt
dt
dt
dv
dv
dv
2 A + 2 B + C = 0 or
dt
dt
dt
2
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2v A + 2v B + vC = 0
2a A + 2a B + aC = 0
11 - 34
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.5
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.
Pulley D is attached to a collar which
Calculate change of position at time t.
is pulled down at 3 in./s. At t = 0,
collar A starts moving down from K • Block B motion is dependent on motions
of collar A and pulley D. Write motion
with constant acceleration and zero
initial velocity. Knowing that velocity relationship and solve for change of block
B position at time t.
of collar A is 12 in./s as it passes L,
determine the change in elevation,
• Differentiate motion relation twice to
velocity, and acceleration of block B
develop equations for velocity and
when block A is at L.
acceleration of block B.
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11 - 35
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.5
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.
v 2A = (v A )02 + 2a A x A − ( x A )0 
2
 in. 
12  = 2a A (8 in.)
 s 
aA = 9
in.
s2
v A = (v A )0 + a At
12
in.
in.
=9 2t
s
s
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t = 1.333 s
11 - 36
Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.5
• Pulley D has uniform rectilinear motion. Calculate
change of position at time t.
x D = ( x D )0 + v D t
 in. 
x D − ( x D )0 =  3 (1.333s ) = 4 in.
 s 
• 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,
x A + 2 x D + x B = ( x A )0 + 2( x D )0 + ( x B )0
x A − ( x A )0 + 2xD − ( xD )0 + xB − ( xB )0  = 0
(8 in.) + 2(4 in.) + x B − ( x B )0  = 0
xB − ( xB )0 = −16 in.
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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.5
• Differentiate motion relation twice to develop
equations for velocity and acceleration of block B.
x A + 2 x D + x B = constant
v A + 2v D + v B = 0
 in.   in. 
12  + 2 3  + v B = 0
 s   s 
v B = 18
in.
s
a A + 2a D + a B = 0
 in. 
 9 2  + vB = 0
 s 
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
in.
a B = −9 2
s
11 - 38
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
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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2 - 39
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Given: vA= 6 m/s left Find: vB
xA
This length is constant no
matter how the blocks move
Sketch your system and choose coordinates
yB
Define your constraint equation(s)
xA + 3 yB + constants = L
Differentiate the constraint equation to
get velocity
6 m/s + 3vB = 0
v B = 2 m/s 
Note that as xA gets bigger, yB gets smaller.
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2 - 40
Graphical Solution of Rectilinear-Motion Problems
Engineers often collect position, velocity, and acceleration
data. Graphical solutions are often useful in analyzing
these data.
Data Fideltity / Highest Recorded Punch
180
160
Acceleration data
from a head impact
during a round of
boxing.
140
Acceleration (g)
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
120
100
80
60
40
20
0
47.76
47.77
47.78
47.79
47.8
47.81
Time (s)
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 41
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Graphical Solution of Rectilinear-Motion Problems
• 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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 42
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Graphical Solution of Rectilinear-Motion Problems
• 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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11 - 43
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Other Graphical Methods
• Moment-area method to determine particle position at
time t directly from the a-t curve:
x1 − x0 = area under v − t curve
v1
= v0t1 +  (t1 − t )dv
v0
using dv = a dt ,
x1 − x0 = v0t1 +
v1
 (t1 − t ) a dt
v0
v1
 (t1 − t ) a dt = first moment of area under a-t curve
v0
with respect to t = t1 line.
x1 = x0 + v0t1 + (area under a-t curve )(t1 − t )
t = abscissa of centroid C
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11 - 44
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Other Graphical Methods
• Method to determine particle acceleration from v-x curve:
dv
dx
= AB tan 
= BC = subnormal to v-x curve
a=v
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11 - 45
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Curvilinear Motion: Position, Velocity & Acceleration
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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11 - 46
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Curvilinear Motion: Position, Velocity & Acceleration
• 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 r at time t

and P’ defined by r  at t + Dt,
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 47
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Curvilinear Motion: Position, Velocity & Acceleration
Instantaneous velocity
(vector)
Dr dr
v = lim
=
Dt →0 Dt
dt
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Instantaneous speed
(scalar)
Ds ds
=
Dt →0 Dt
dt
v = lim
11 - 48
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Curvilinear Motion: Position, Velocity & Acceleration


• Consider velocity v of a particle at time t and velocity v at t + Dt,
Dv dv
=
= instantaneous acceleration (vector)
Dt →0 Dt
dt
a = lim
• In general, the acceleration vector is not tangent
to the particle path and velocity vector.
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11 - 49
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Derivatives of Vector Functions

• Let P(u ) be a vector function of scalar variable u,




dP
DP
P(u + Du ) − P(u )
= lim
= lim
du Du →0 Du Du →0
Du
• Derivative of vector sum,

 

d (P + Q ) dP dQ
=
+
du
du du
Delete or put in
“bonus” slides
• Derivative of product of scalar and vector functions,


d ( f P ) df 
dP
=
P+ f
du
du
du
• Derivative of scalar product and vector product,

 

d (P • Q ) dP   dQ
=
•Q + P•
du
du
du

 



d (P  Q ) dP
dQ
=
Q + P
du
du
du
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 50
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Rectangular Components of Velocity & Acceleration
• When position vector of particle P is given by its
rectangular components,

 

r = xi + y j + zk
• Velocity vector,



 dx  dy  dz 
v = i + j + k = xi + y j + zk
dt
dt
dt



= vx i + v y j + vz k
• Acceleration vector,

 
 d 2 x d 2 y  d 2 z 
a = 2 i + 2 j + 2 k = xi + y j + zk
dt
dt
dt



= ax i + a y j + az k
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11 - 51
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Rectangular Components of Velocity & Acceleration
• Rectangular components particularly effective
when component accelerations can be integrated
independently, e.g., motion of a projectile,
a x = x = 0
a y = y = − g
a z = z = 0
with initial conditions,
(v x )0 , v y 0 , (v z )0 = 0
x0 = y0 = z0 = 0
( )
Integrating twice yields
v x = (v x )0
x = (v x )0 t
( )
( )0
v y = v y − gt
0
y = v y y − 12 gt 2
vz = 0
z=0
• 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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11 - 52
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.7
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
•
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.
Determine time t for projectile to hit the
ground, use this to find the horizontal
distance
Maximum elevation occurs when vy=0
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 53
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.7
SOLUTION:
Given:
(v)o =180 m/s
(a)y = - 9.81 m/s2
(y)o =150 m
(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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11 - 54
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.7
SOLUTION:
Horizontal distance
Projectile strikes the ground at:
Substitute into equation (1) above
Solving for t, we take the positive root
Substitute t into equation (4)
Maximum elevation occurs when vy=0
Maximum elevation above the ground =
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 55
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Quiz
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)
b)
c)
d)
A launch angle of 45
A launch angle less than 45
A launch angle greater than 45
It depends on the launch velocity
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2 - 56
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
SOLUTION:
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.
• 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
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 57
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Given: x= 40 ft, yo = 5 ft,
yf= 42 in.
Find: vo
Analyze the motion in
the y-direction
y f = y0 + (0)t −
1 2
gt
2
1 2
3.5 = 5 − gt
2
1
−1.5 ft = − (32.2 ft/s2 )t 2
2
Analyze the motion in
the x-direction
x = 0 + (vx )0 t = v0t
40 ft = (v0 )(0.305234 s)
v0 = 131.047 ft/s = 89.4 mi/h
t = 0.305234 s
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2 - 58
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Motion Relative to a Frame in Translation
A soccer player must consider
the relative motion of the ball
and her teammates when
making a pass.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
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.
2 - 59
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Motion Relative to a Frame in Translation
• 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 rA and rB .

r
• Vector B A joining A and B defines the position of
B with respect to the moving frame Ax’y’z’ and

 
rB = rA + rB A
• Differentiating twice,




vB = v A + vB A vB



a B = a A + aB A
A
= velocity of B relative to A.

a B A = acceleration of B relative
to 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.
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11 - 60
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.9
SOLUTION:
• Define inertial axes for the system
• Determine the position, speed, and
acceleration of car A at t = 5 s
• Determine the position, speed, and
acceleration of car B at t = 5 s
Automobile A is traveling east at the • Using vectors (Eqs 11.31, 11.33, and
11.34) or a graphical approach, determine
constant speed of 36 km/h. As
the relative position, velocity, and
automobile A crosses the intersection
acceleration
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 61
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.9
SOLUTION:
Given:
• Define axes along the road
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
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 62
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.9
SOLUTION:
Determine motion of Automobile B:
We have uniform acceleration for B so:
At t = 5 s
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11 - 63
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.9
SOLUTION:
We can solve the problems geometrically, and apply the arctangent relationship:
Or we can solve the problems using vectors to obtain equivalent results:
rB = rA + rB/ A
20 j = 50i + rB/ A
rB/ A = 20 j − 50i (m)
v B = v A + v B/ A
−6 j = 10i + v B/ A
v B/ A = −6 j − 10i (m/s)
aB = a A + aB/ A
−1.2 j = 0i + a B/ A
a B/ A = −1.2 j (m/s 2 )
Physically, a rider in car A would “see” car B travelling south and west.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 64
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Quiz
If you are sitting in train
B looking out the window,
it which direction does it
appear that train A is
moving?
a)
25o
c)
b)
25o
d)
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 65
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Tangential and Normal Components
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.
2 - 66
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Tangential and Normal Components
y
r= the instantaneous
radius of curvature
v = v et
en
v= vt et
dv
v2
a = e t + en
dt
r
et
x
• The tangential direction (et) is tangent to the path of the
particle. This velocity vector of a particle is in this direction
• The normal direction (en) is perpendicular to et and points
towards the inside of the curve.
• The acceleration can have components in both the en and et directions
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 67
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Tangential and Normal Components
• To derive the acceleration vector in tangential
and normal components, define the motion of a
particle as shown in the figure.


• et and et are tangential unit vectors for the
particle path at P and P’. When drawn with
  
respect to the same origin, Det = et − et and
D is the angle between them.
Det = 2 sin (D 2)

Det
sin (D 2) 

lim
= lim
en = e n
D →0 D
D →0 D 2

det

en =
d
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 68
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Tangential and Normal Components


• With the velocity vector expressed as v = vet
the particle acceleration may be written as



de dv 
de d ds
 dv dv 
a=
= et + v
= et + v
dt dt
dt dt
d ds dt
but 
det 
ds
= en
r d = ds
=v
d
dt
After substituting,
dv
v2
 dv  v 2 
a = et + en
at =
an =
dt
r
dt
r
• 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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 69
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Tangential and Normal Components
• Relations for tangential and normal acceleration
also apply for particle moving along a space curve.
 dv  v 2 
a = et + en
dt
r
dv
at =
dt
an =
v2
r
• The plane containing tangential and normal unit
vectors is called the osculating plane.
• The normal to the osculating plane is found from

 
eb = et  en

en = principal normal

eb = binormal
• Acceleration has no component along the binormal.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 70
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.8
SOLUTION:
• Define your coordinate system
• Calculate the tangential velocity and
tangential acceleration
• Calculate the normal acceleration
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.
• Determine overall acceleration magnitude
after the brakes have been applied
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 71
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.8
SOLUTION:
• Define your coordinate system
• Determine velocity and acceleration in
the tangential direction
et
en
• The deceleration constant, therefore
• Immediately after the brakes are applied,
the speed is still 88 ft/s
a = an2 + at2 = 2.752 + 3.102
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 72
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Tangential and Normal Components
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
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 73
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
SOLUTION:
• Define your coordinate system
• Calculate the tangential velocity and
tangential acceleration
• Calculate the normal acceleration
The tangential acceleration of the
centrifuge cab is given by
• Determine overall acceleration
magnitude
at = 0.5 t (m/s2 )
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 74
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Define your coordinate system
In the side view, the tangential
direction points into the “page”
en
Determine the tangential velocity
at = 0.5 t
t
2 t
0
0
vt =  0.5 t dt = 0.25t
= 0.25t 2
Top View
vt = 0.25 (10 ) = 25 m/s
2
et
en
Determine the normal acceleration
2
v
( t ) 252
2
an =
r
=
8
= 78.125 m/s
Determine the total acceleration magnitude
amag = an2 + at2 = 78.1252 + (0.5)(10)
2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
amag = 78.285 m/s 2
11 - 75
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
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?
a)
b)
c)
d)
The accelerations would remain the same
The an would increase and the at would decrease
The an and at would both increase
The an would decrease and the at would increase
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 76
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Radial and Transverse Components
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.
2 - 77
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Radial and Transverse Components
• 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 = rer
• The particle velocity vector is
v = r er + r e
• The particle acceleration vector is
(
)
(
)
a = r − r 2 er + r + 2r e
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 78
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Radial and Transverse Components
• We can derive the velocity and acceleration
relationships by recognizing that the unit vectors
change direction.


r = rer

der 
= e
d
• The particle velocity vector is

der dr 
dr 
d 
 d 
v = (rer ) = er + r
= er + r
e
dt
dt
dt
dt
dt


= r er + r e

de

= −er
d


der der d  d
=
= e
dt
d dt
dt


de de d
 d
=
= −er
dt
d dt
dt
• Similarly, the particle acceleration vector is
d  
 d  dr 
a =  er + r
e 
dt  dt
dt 


d 2 r  dr der dr d 
d 2 
d de
= 2 er +
+
e + r 2 e + r
dt dt dt dt
dt dt
dt
dt


= r − r 2 er + (r + 2r )e
(
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
)
11 - 79
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Quiz
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
2 - 80
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Radial and Transverse Components
• When particle position is given in cylindrical
coordinates, it is convenient to express the
velocity and acceleration
vectors using the unit

 
vectors eR , e , and k .
• Position vector,



r = R e R +z k
• Velocity vector,


 dr  


v=
= R eR + R e + z k
dt
• Acceleration vector,


 dv


2




 − R eR + (R + 2 R  )e + z k
a=
= R
dt
(
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
)
11 - 81
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.12
SOLUTION:
• Evaluate time t for  = 30o.
• Evaluate radial and angular positions,
and first and second derivatives at
time t.
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.
• Calculate velocity and acceleration in
cylindrical coordinates.
• Evaluate acceleration with respect to
arm.
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.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.12
SOLUTION:
• Evaluate time t for  = 30o.
 = 0.15 t 2
= 30 = 0.524 rad
t = 1.869 s
• Evaluate radial and angular positions, and first
and second derivatives at time t.
r = 0.9 − 0.12 t 2 = 0.481 m
r = −0.24 t = −0.449 m s
r = −0.24 m s 2
 = 0.15 t 2 = 0.524 rad
 = 0.30 t = 0.561rad s
 = 0.30 rad s 2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.12
• Calculate velocity and acceleration.
vr = r = −0.449 m s
v = r = (0.481m )(0.561rad s ) = 0.270 m s
v
 = tan −1 
v = vr2 + v2
vr
v = 0.524 m s
 = 31.0
ar = r − r 2
= −0.240 m s 2 − (0.481m )(0.561rad s )2
= −0.391m s 2
a = r + 2r
(
)
= (0.481m ) 0.3 rad s 2 + 2(− 0.449 m s )(0.561rad s )
= −0.359 m s 2
a = ar2 + a2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
a
 = tan −1 
ar
a = 0.531m s
 = 42.6
11 - 84
Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.12
• Evaluate acceleration with respect to arm.
Motion of collar with respect to arm is rectilinear
and defined by coordinate r.
a B OA = r = −0.240 m s 2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 85
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
SOLUTION:
• Define your coordinate system
• Calculate the angular velocity after
three revolutions
• Calculate the radial and transverse
accelerations
The angular acceleration of the
centrifuge arm varies according to
• Determine overall acceleration
magnitude
 = 0.05 (rad/s 2 )
where  is measured in radians. If the
centrifuge starts from rest, determine the
acceleration magnitude after the gondola
has travelled two full rotations.
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 86
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Define your coordinate system
In the side view, the transverse
direction points into the “page”
er
Determine the angular velocity
 = 0.05 (rad/s 2 )
Acceleration is a function
of position, so use:
 d = d
Evaluate the integral
(2)(2 )

0
er
Top View
e

0.05 d =   d
0.05
2
0
2 2(2 )
0
=

2
2

 = 0.05  2(2 ) 
2
2
0
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 87
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Determine the angular velocity
2
 2 = 0.05  2(2 ) 
er
 = 2.8099 rad/s
Determine the angular acceleration
 = 0.05 = 0.05(2)(2 ) = 0.6283 rad/s2
Find the radial and transverse accelerations
(
) (
)
= ( 0 − (8)(2.8099) ) e + ( (8)(0.6283) + 0 ) e
a = r − r 2 er + r + 2r e
2

r
= −63.166 er + 5.0265 e (m/s 2 )
Magnitude:
amag = ar2 + a2 = (−63.166)2 + 5.0265
2
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
amag = 63.365 m/s 2
11 - 88
Tenth
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
r
What would happen if you
designed the centrifuge so
that the arm could extend
from 6 to 10 meters?
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).
(
a = r − r
2
) e + ( r + 2r ) e
r
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
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