Eighth Edition
CHAPTER
11
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Kinematics of Particles
Lecture Notes:
J. Walt Oler
Texas Tech University
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Eighth
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
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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
11 - 2
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Introduction
• Dynamics includes:
- Kinematics: study of the geometry of motion. Kinematics is used to relate
displacement, velocity, acceleration, and time without reference to the cause of
motion.
- 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.
• 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 - 3
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Rectilinear Motion: Position, Velocity & Acceleration
• Particle moving along a straight line is said to
be in rectilinear motion.
• Position coordinate of a particle is defined by
positive or negative distance of particle 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. Motion of the particle may be
expressed in the form of a function, e.g.,
x = 6t 2 − t 3
or in the form of a graph x vs. t.
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11 - 4
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle which occupies position P at
time t and P’ at t+∆t,
∆x
Average velocity =
∆t
∆x
=
v
=
lim
Instantaneous velocity
∆t → 0 ∆t
• Instantaneous velocity may be positive or
negative. Magnitude of velocity is referred to
as particle speed.
• From the definition of a derivative,
∆ x dx
v = lim
=
dt
∆t → 0 ∆t
e.g.,
x = 6t 2 − t 3
v=
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dx
= 12 t − 3t 2
dt
11 - 5
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle with velocity v at time t and v’
at t+∆t,
∆v
Instantaneous acceleration = a = lim
∆t → 0 ∆t
• 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,
∆ v dv d 2 x
=
= 2
a = lim
dt
∆t → 0 ∆t
dt
e.g.
v = 12 t − 3t 2
a=
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dv
= 12 − 6t
dt
11 - 6
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle with motion given by
x = 6t 2 − t 3
dx
v=
= 12 t − 3t 2
dt
dv d 2 x
a=
= 2 = 12 − 6t
dt
dt
• at t = 0,
x = 0, v = 0, a = 12 m/s2
• 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
• at t = 6 s,
x = 0, v = -36 m/s, a = 24 m/s2
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11 - 7
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Determination of the Motion of a Particle
• Recall, motion of a particle is known if position is known for all time t.
• Typically, conditions of motion are specified by the type of acceleration
experienced by the particle. Determination of velocity and position requires two
successive integrations.
• Three classes of motion may be defined for:
- 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)
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11 - 8
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Determination of the Motion of a Particle
• Acceleration given as a function of time, a = f(t):
v (t )
t
dv
= a = f (t )
dv = f (t ) dt
dv = ∫ f (t ) dt
∫
dt
v
0
t
v (t ) − v 0 = ∫ f (t ) dt
0
0
dx
= v (t )
dt
dx = v (t ) dt
x (t )
t
x0
0
t
∫ dx = ∫ v (t ) dt
x (t ) − x 0 = ∫ v (t ) dt
0
• Acceleration given as a function of position, a = f(x):
v=
dx
dx
or dt =
dt
v
v dv = f ( x )dx
a=
v(x )
dv
dv
or a = v
= f (x )
dx
dt
x
∫ v dv = ∫ f ( x )dx
v0
x0
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2
1 v(x )
2
−
1 v2
2 0
x
= ∫ f ( x )dx
x0
11 - 9
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Determination of the Motion of a Particle
• Acceleration given as a function of velocity, a = f(v):
dv
= a = f (v )
dt
dv
= dt
f (v )
v (t )
t
dv
∫ f (v ) = ∫ dt
v
0
0
v (t )
dv
∫ f (v ) = t
v
0
dv
v = a = f (v )
dx
v (t )
x(t ) − x0 = ∫
v0
v dv
dx =
f (v )
x (t )
v (t )
0
0
v dv
∫ dx = ∫ f (v )
x
v
v dv
f (v )
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11 - 10
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.2
SOLUTION:
• Integrate twice to find v(t) and y(t).
• Solve for t at which 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 at which 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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11 - 11
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.2
SOLUTION:
• Integrate twice to find v(t) and y(t).
dv
= a = −9 .81 m s 2
dt
v (t )
t
v (t ) − v 0 = −9 .81t
∫ dv = − ∫ 9.81 dt
0
v0
m 
m
v (t ) = 10 −  9.81 2  t
s 
s 
dy
= v = 10 − 9.81 t
dt
y (t )
t
∫ dy = ∫ (10 − 9.81 t )dt
y0
0
y (t ) − y 0 = 10 t − 12 9 .81 t 2
m
 m 
y (t ) = 20 m +  10 t −  4.905 2 t 2
 s 
s 
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11 - 12
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.2
• Solve for t at which 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 at which altitude equals zero and evaluate
corresponding velocity.
m 2
 m 
y (t ) = 20 m +  10 t −  4 .905 2 t
 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 - 13
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.2
• Solve for t at which altitude equals zero and evaluate
corresponding velocity.
m 2
 m 
y (t ) = 20 m +  10 t −  4 .905 2 t = 0
 s 
s 
t = − 1.243 s (meaningles s )
t = 3 .28 s
v (t ) = 10
m 
m
−  9.81 2  t
s 
s 
m 
m
v (3 .28 s ) = 10 −  9 .81 2  (3 .28 s )
s 
s 
m
v = −22 .2
s
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11 - 14
Eighth
Edition
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).
• Integrate a = v dv/dx = -kv to find v(x).
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).
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11 - 15
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.3
SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
v (t )
t
dv
dv
a=
= − kv
= − k ∫ dt
∫
dt
v v
0
0
v (t )
ln
= − kt
v0
v (t ) = v 0 e − kt
• Integrate v(t) = dx/dt to find x(t).
dx
= v 0 e − kt
v (t ) =
dt
t
x (t )
t
1


− kt
x (t ) = v 0  − e − kt 
∫ dx = v 0 ∫ e dt
 k
0
0
0
x (t ) =
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(
v0
1 − e − kt
k
11 - 16
)
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.3
• Integrate a = v dv/dx = -kv to find v(x).
v
dv
a=v
= − kv
dx
x
∫ dv = − k ∫ dx
dv = − k dx
v0
0
v − v 0 = − kx
v = v 0 − kx
• Alternatively,
(
)
with
v0
1 − e − kt
x (t ) =
k
and
v (t ) = v 0 e − kt or e − kt =
then
v  v (t ) 

x (t ) = 0  1 −
k 
v0 
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v (t )
v0
v = v 0 − kx
11 - 17
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Uniform Rectilinear Motion
For particle in uniform rectilinear motion, the acceleration is zero and the
velocity is constant.
dx
= v = constant
dt
x
t
∫ dx = v ∫ dt
x0
0
x − x0 = vt
x = x0 + vt
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11 - 18
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Uniformly Accelerated Rectilinear Motion
For particle in uniformly accelerated rectilinear motion, the acceleration of the
particle is constant.
v
dv
= a = constant
dt
t
∫ dv = a ∫ dt
v − v0 = at
0
v0
v = v0 + at
dx
= v0 + at
dt
x
t
x0
0
∫ dx = ∫ (v0 + at )dt
x − x0 = v0 t + 12 at 2
x = x0 + v0t + 12 at 2
dv
v = a = constant
dx
v
x
∫ v dv = a ∫ dx
v0
x0
1
2
(v 2 − v02 ) = a(x − x0 )
v 2 = v02 + 2a( x − x0 )
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11 - 19
Eighth
Edition
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 - 20
Eighth
Edition
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.
© 2007 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 - 21
Eighth
Edition
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.
m 
m
v B = v 0 + at = 18 −  9.81 2 t
s 
s 
m
 m 
y B = y 0 + v 0 t + 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.
m
vE = 2
s
 m
y E = y 0 + v E t = 5 m +  2 t
 s
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11 - 22
Eighth
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.
(
)
y B E = 12 + 18t − 4.905 t 2 − (5 + 2t ) = 0
t = −0 .39 s (meaningles s )
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
vB
E
= (18 − 9.81t ) − 2
= 16 − 9.81(3.65 )
vB
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E
= −19 .81
m
s
11 - 23
Eighth
Edition
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 x B + 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
dt
dt
dt
dv
dv
dv
2 A +2 B + C =0
dt
dt
dt
2
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
or
2 v A + 2v B + vC = 0
or
2 a A + 2 a B + aC = 0
11 - 24
Eighth
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.
• Pulley D has uniform rectilinear motion.
Pulley D is attached to a collar which is
Calculate change of position at time t.
pulled down at 3 in./s. At t = 0, collar A
• Block B motion is dependent on motions of
starts moving down from K with
collar A and pulley D. Write motion
constant acceleration and zero initial
relationship and solve for change of block B
velocity. Knowing that velocity of
position at time t.
collar A is 12 in./s as it passes L,
determine the change in elevation,
• Differentiate motion relation twice to develop
velocity, and acceleration of block B
equations for velocity and acceleration of
when block A is at L.
block B.
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11 - 25
Eighth
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 + 2 a A [x A − ( x A )0 ]
2
 in. 
 12  = 2 a A (8 in. )
s 

in.
aA = 9 2
s
v A = (v A )0 + a At
12
in.
in.
=9 2t
s
s
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t = 1.333 s
11 - 26
Eighth
Edition
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 .333 s ) = 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[x D − ( x D )0 ] + [x B − ( x B )0 ] = 0
(8 in. ) + 2(4 in. ) + [x B − ( x B )0 ] = 0
x B − ( x B )0 = −16 in.
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11 - 27
Eighth
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 
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
in.
a B = −9 2
s
11 - 28
Eighth
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.
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11 - 29
Eighth
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 - 30
Eighth
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 − x 0 = area under v − t curve
v1
= v 0 t1 + ∫ (t1 − t )dv
v0
using dv = a dt ,
v1
x1 − x 0 = v 0 t1 + ∫ (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 = x 0 + v 0 t1 + (area under a-t curve )(t1 − t )
t = abscissa of centroid C
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11 - 31
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Other Graphical Methods
• Method to determine particle acceleration from
v-x curve:
dv
a=v
dx
= AB tan θ
= BC = subnormal to v-x curve
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11 - 32
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Curvilinear Motion: Position, Velocity & Acceleration
• Particle moving along a curve other than a straight line is in
curvilinear motion.
• 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 particle which occupies position P defined by
at time
t
and
P’
defined
by
at
t
+
r
∆′ t,
r
∆r dr
v = lim
=
dt
∆t → 0 ∆ t
= instantaneous velocity (vector)
∆ s ds
v = lim
=
dt
∆t → 0 ∆t
= instantaneous speed (scalar)
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11 - 33
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Curvilinear Motion: Position, Velocity & Acceleration
• Consider velocity
v ′at t + ∆t,
∆v
=
a = lim
∆t → 0 ∆t
vof particle at time t and velocity
dv
dt
= instantaneous acceleration (vector)
• In general, acceleration vector is not tangent to
particle path and velocity vector.
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11 - 34
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Derivatives of Vector Functions
• Let P (u )be a vector function of scalar variable u,
dP
∆P
P (u + ∆ u ) − P (u )
= lim
= lim
du ∆u → 0 ∆ u ∆u → 0
∆u
• Derivative of vector sum,
d (P + Q ) dP dQ
=
+
du
du du
• 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 ) d P dQ
=
•Q + P•
du
du
du
d (P × Q ) d P
dQ
=
×Q + P×
du
du
du
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11 - 35
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Rectangular Components of Velocity & Acceleration
• When position vector of particle P is given by its
rectangular components,
r = x i + y j + zk
• Velocity vector,
dx dy dz v= i +
j + k = x i + y j + zk
dt
dt
dt
= vx i + v y j + vz k
• Acceleration vector,
d 2x d 2 y d 2z a = 2 i + 2 j + 2 k = xi + y j + zk
dt
dt
dt
= ax i + a y j + azk
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11 - 36
Eighth
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,
x0 = y 0 = z 0 = 0
(v x )0 , (v y )0 , (v z )0 = 0
Integrating twice yields
v x = (v x )0
x = (v x )0 t
( )0 − gt
y = (v y ) y − 12 gt 2
0
vy = vy
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.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 37
Eighth
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
rA and rB .
fixed frame of reference Oxyz are
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,
v B = v A + v B A v B A = velocity of B relative to A.
a B = a A + a B A aB 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.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 38
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Tangential and Normal Components
• Velocity vector of particle is tangent to path of
particle. In general, acceleration vector is not.
Wish to express acceleration vector in terms of
tangential and normal components.
• et and et′ are tangential unit vectors for the particle
path at P and P’. When drawn with respect to the
same origin,
∆ et = et′ − eand
t
∆ θ is the angle between them.
∆ et = 2 sin (∆ θ 2 )
∆ et
sin (∆ θ 2 ) lim
= lim
en = en
∆θ → 0 ∆θ
∆θ → 0 ∆θ 2
d
e
en = t
dθ
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 39
Eighth
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
d
v
dv
d
e
dv
d
e
dθ ds
a=
=
et + v
=
et + v
dt dt
dt dt
dθ ds dt
but d et ds
= en
ρ d θ = ds
=v
dθ
dt
After substituting,
dv
v2
dv v 2 a=
et +
en
at =
an =
dt
ρ
dt
ρ
• Tangential component of acceleration reflects
change of speed and normal component reflects
change of direction.
• Tangential component may be positive or
negative. Normal component always points
toward center of path curvature.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 40
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Tangential and Normal Components
• Relations for tangential and normal acceleration also
apply for particle moving along space curve.
dv v 2 a=
et +
en
dt
ρ
dv
at =
dt
an =
v2
ρ
• Plane containing tangential and normal unit vectors
is called the osculating plane.
• Normal to the osculating plane is found from
eb = et × e n
e n = principal normal
eb = binormal
• Acceleration has no component along binormal.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 41
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Radial and Transverse Components
• When particle position is given in polar coordinates, it
is convenient to express velocity and acceleration with
components parallel and perpendicular to OP.
r = rer
der = eθ
dθ
• The particle velocity vector is
de r dr dr dθ d =
eθ
v = (re r ) =
er + r
er + r
dt
dt
dt
dt
dt
= r e r + r θ eθ
deθ
= − er
dθ
der der dθ dθ
=
= eθ
dt
dθ dt
dt
deθ deθ dθ
dθ
=
= − er
dt
dθ dt
dt
• Similarly, the particle acceleration vector is
dθ 
d  dr a =  er + r
eθ 
dt  dt
dt

2
2
d
e
d r
dr de r dr dθ d θ dθ θ
= 2 er +
+
eθ + r 2 eθ + r
dt dt
dt dt
dt dt
dt
dt
= r − rθ 2 e r + (rθ + 2 rθ )eθ
(
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
)
11 - 42
Eighth
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
e R , eθ , and k .
• Position vector,
r = R e R+z k
• Velocity vector,
dr
v=
= R e R + R θ eθ + z k
dt
• Acceleration vector,
dv
2 a=
= R − R θ e R + (R θ + 2 R θ )eθ + z k
dt
(
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
)
11 - 43
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.10
SOLUTION:
• Calculate tangential and normal
components of acceleration.
• Determine acceleration magnitude and
direction with respect to tangent to curve.
A motorist is traveling on curved
section of highway at 60 mph. The
motorist applies brakes causing a
constant deceleration rate.
Knowing that after 8 s the speed has
been reduced to 45 mph, determine the
acceleration of the automobile
immediately after the brakes are
applied.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 44
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.10
SOLUTION:
• Calculate tangential and normal components of
acceleration.
ft
∆ v (66 − 88 ) ft s
at =
= −2 .75 2
=
∆t
8s
s
an =
60 mph = 88 ft/s
45 mph = 66 ft/s
v2
ρ
(88 ft s )2
ft
=
= 3.10 2
2500 ft
s
• Determine acceleration magnitude and direction with
respect to tangent to curve.
ft
2
2
2
2
a
=
4
.
14
a = a t + a n = (− 2.75 ) + 3.10
s2
α = tan
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
−1
an
3.10
= tan −1
at
2 .75
α = 48 .4°
11 - 45
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.12
SOLUTION:
• Evaluate time t for θ = 30o.
• Evaluate radial and angular positions,
and first and second derivatives at time t.
Rotation of the arm about O is defined by
θ = 0.15t2 where θ 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.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 46
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.12
SOLUTION:
• Evaluate time t for θ = 30o.
θ = 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
θ = 0 .15 t 2 = 0.524 rad
θ = 0 .30 t = 0 .561 rad s
θ = 0.30 rad s 2
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
11 - 47
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 11.12
• Calculate velocity and acceleration.
vr = r = − 0 .449 m s
vθ = rθ = (0.481 m )(0 .561 rad s ) = 0 .270 m s
v=
vr2
2
β = tan
+ vθ
−1 vθ
vr
v = 0.524 m s
β = 31 .0 °
ar = r − rθ 2
= − 0 .240 m s 2 − (0 .481 m )(0 .561 rad s )2
= − 0 .391 m s 2
aθ = rθ + 2 rθ
(
)
= (0 .481 m ) 0 .3 rad s 2 + 2(− 0.449 m s )(0 .561 rad s )
= − 0 .359 m s 2
a=
a r2 + aθ2
a
γ = tan −1 θ
ar
a = 0 .531 m s
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
γ = 42 .6°
11 - 48
Eighth
Edition
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
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11 - 49