Seventh 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
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Seventh
Edition
Vector Mechanics for Engineers: Dynamics
Introduction
• Rigid Body Dynamics : distance between points is fixed, no size or shape
changes, idealization of flexible body
- particle: mass concentrated at one point, center of mass, rotation of
points w/r to c.m. is neglected
- rigid body: rotation of points w/r to c.m. is taken into account
• Example - Ski jumper :
– model as a particle: if interested in finding
how far she jumps
– model as a rigid body: if interested in finding
the position of the head w/r to torso
• Example - Football player kicking a ball :
– model as particles: if interested in finding how high/far the ball goes
– model as rigid bodies: if interested in finding the forces on the knee
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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, i.e. forces are not considered.
- 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.
•
Newton’s laws:
1.
F=0
2.
F = ma
3.
Action-reaction
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Vector Mechanics for Engineers: Dynamics
Derivatives of Functions
• Let s(u) be a scalar function of scalar variable u,
∆s
ds
s (u + ∆u ) − s (u )
= lim
= lim
du ∆u → 0 ∆u ∆u → 0
∆u
r
• Let P(u ) be a vector function of scalar variable u,
r
r
r
r
dP
P(u + ∆u ) − P(u )
∆P
= lim
= lim
du ∆u →0 ∆u ∆u →0
∆u
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Vector Mechanics for Engineers: Dynamics
Position, Velocity & Acceleration
• Consider a particle moving along a certain path
• 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.
r
r′
• Consider particle which occupies position P defined
r
r
by r at time t and P’ defined by r ′ at t + ∆t,
r
r
∆r dr
r r r
r
v = lim
=
∆r = r ′ − r
dt
∆t →0 ∆t
= instantaneous velocity (vector)
v = lim
∆s
∆t →0 ∆t
=
ds
dt
= instantaneous speed (scalar)
•Velocity vector is always tangent to particle path.
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Vector Mechanics for Engineers: Dynamics
Position, Velocity & Acceleration
r
• Consider velocity v of particle at time t and velocity
r
v′ at t + ∆t,
r
r
∆v dv
r
a = lim
=
dt
∆t →0 ∆t
= instantaneous acceleration (vector)
• In general, acceleration vector is not tangent to
particle path.
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Vector Mechanics for Engineers: Dynamics
Motion of a particle along a straight line
• The motion of a particle is known if position is known for all time t.
• If path is a straight line we have rectilinear motion, we can describe the
motion in terms of a, v and x.
• 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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Vector Mechanics for Engineers: Dynamics
Motion of a particle along a straight line
• 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
0
v
dx
= v(t )
dt
x (t )
dx = v(t ) dt
t
v(t ) − v0 = ∫ f (t ) dt
0
0
t
∫ dx = ∫ v(t ) dt
x0
t
x(t ) − x0 = ∫ v(t ) dt
0
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 )
dt
dx
x
∫ v dv = ∫ f ( x )dx
v0
x0
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1 v ( x )2
2
− 12 v02 =
x
∫ f ( x )dx
x0
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Vector Mechanics for Engineers: Dynamics
Motion of a particle along a straight line
• Acceleration given as a function of velocity, a = f(v):
dv
= a = f (v )
dt
v (t )
dv
= dt
f (v )
dv
t
= ∫ dt
∫
v0 f (v ) 0
dv
=t
∫
v0 f (v )
v
v (t )
dv
= a = f (v )
dx
dx =
v (t )
v dv
x(t ) − x0 = ∫
v f (v )
v dv
f (v )
x (t )
v (t )
v dv
∫ dx = ∫ f (v )
x0
v0
0
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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
x0
0
∫ dx = v ∫ dt
x − x0 = vt
x = x0 + vt
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Vector Mechanics for Engineers: Dynamics
Uniformly Accelerated Rectilinear Motion
For particle in uniformly accelerated rectilinear motion, the acceleration of
the particle is constant.
dv
= a = constant
dt
v
t
v0
0
∫ dv = a ∫ dt
v − v0 = at
v = v0 + at
dx
= v0 + at
dt
x=
x
t
x0
0
∫ dx = ∫ (v0 + at )dt
x − x0 = v0 t + 12 at 2
x0 + v0 t + 12 at 2
v
dv
= a = constant
dx
v
2
= v02
v
x
v0
x0
∫ v dv = a ∫ dx
1
2
(v 2 − v02 ) = a(x − x0 )
+ 2 a ( x − x0 )
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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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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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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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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 ) − v0 = −9.81t
∫ dv = − ∫ 9.81 dt
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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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⎞
⎛ 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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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.2
• Solve for t at which altitude equals zero and
evaluate corresponding velocity.
m⎞
⎛ m⎞ ⎛
y (t ) = 20 m + ⎜10 ⎟t − ⎜ 4.905 2 ⎟t 2 = 0
⎝ s⎠ ⎝
s ⎠
t = −1.243 s (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
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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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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.3
SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
v (t )
t
v(t )
dv
dv
= −kv
= −k ∫ dt
ln
= − kt
a=
∫
v0
v
dt
v
0
0
v(t ) = v0 e − kt
• Integrate v(t) = dx/dt to find x(t).
dx
v(t ) =
= v0 e − kt
dt
t
x (t )
t
⎡ 1 − kt ⎤
− kt
(
)
dx
v
e
dt
x
t
v
e
=
=
−
∫
0∫
0⎢
⎥⎦
⎣ k
0
0
0
(
v
x(t ) = 0 1 − e − kt
k
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.3
• Integrate a = v dv/dx = -kv to find v(x).
a=v
dv
= − 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
v0 ⎛ v(t ) ⎞
⎟
x(t ) = ⎜⎜1 −
k ⎝
v0 ⎟⎠
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v = v0 − kx
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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 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 or 2v A + 2v B + vC = 0
dt
dt
dt
dv A
dv B dvC
2
+2
+
= 0 or 2a A + 2a B + aC = 0
dt
dt
dt
2
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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
relationship and solve for change of block
initial velocity. Knowing that
B position at time t.
velocity of collar A is 12 in./s as it
passes L, determine the change in
• Differentiate motion relation twice to
elevation, velocity, and acceleration
develop equations for velocity and
of block B when block A is at L.
acceleration of block B.
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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.
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
t = 1.333 s
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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 ⎟(1.333 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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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 ⎠
in.
a B = −9 2
s
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Vector Mechanics for Engineers: Dynamics
Rectangular Components of Velocity & Acceleration
• When position vector of particle P is given by its
rectangular components,
r
r
r
r
r = xi + y j + zk
• Velocity vector,
r
r
r
r dx r dy r dz r
v = i + j + k = x&i + y& j + z&k
dt
dt
dt
r
r
r
= vx i + v y j + vz k
• Acceleration vector,
r
r
r
r d 2 xr d 2 y r d 2 z r
a = 2 i + 2 j + 2 k = &x&i + &y& j + &z&k
dt
dt
dt
r
r
r
= ax i + a y j + az k
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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 ) , (v z )0 = 0
x0 = y 0 = z 0 = 0
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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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.7
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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.
r
r
• et and et′ are tangential unit vectors for the
particle path at P and P’. When drawn with
r r r
respect to the same origin, ∆et = et′ − et and
∆θ is the angle between them.
∆et = 2 sin (∆θ 2 )
r
∆et
sin (∆θ 2 ) r
r
lim
en = en
= lim
∆θ →0 ∆θ
∆θ →0 ∆θ 2
r
de
r
en = t
dθ
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Vector Mechanics for Engineers: Dynamics
Tangential and Normal Components
r
r
• With the velocity vector expressed as v = vet
the particle acceleration may be written as
r
r
r
de dv r
de dθ ds
r dv dv r
a=
= et + v
= et + v
dt dt
dt dt
dθ ds dt
but r
det r
ds
= en
ρ dθ = ds
=v
dt
dθ
After substituting,
dv
v2
r dv r v 2 r
a = et + en
at =
an =
ρ
ρ
dt
dt
• 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.
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Vector Mechanics for Engineers: Dynamics
Tangential and Normal Components
• Relations for tangential and normal acceleration
also apply for particle moving along space curve.
r dv r v 2 r
a = et + en
ρ
dt
at =
dv
dt
an =
v2
ρ
• Plane containing tangential and normal unit
vectors is called the osculating plane.
• Normal to the osculating plane is found from
r
r r
eb = et × en
r
en = principal normal
r
eb = binormal
• Acceleration has no component along binormal.
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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.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.10
SOLUTION:
• Calculate tangential and normal components of
acceleration.
∆v (66 − 88) ft s
ft
= −2.75 2
at =
=
∆t
8s
s
an =
60 mph = 88 ft/s
45 mph = 66 ft/s
v2
ρ
=
(88 ft s )2
2500 ft
= 3.10
ft
s2
• Determine acceleration magnitude and direction
with respect to tangent to curve.
ft
a = at2 + an2 = (− 2.75)2 + 3.10 2 a = 4.14 2
s
α = tan −1
© 2004 The McGraw-Hill Companies, Inc. All rights reserved.
an
3.10
= tan −1
at
2.75
α = 48.4°
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.11
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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
r
r = re r
r
de r r
= eθ
dθ
• The particle velocity vector is
r
de
dr r
dr r
dθ r
r d r
v = ( re r ) = e r + r r = e r + r
eθ
dt
dt
dt
dt
dt
r
r
= r& er + rθ& eθ
r
deθ
r
= − er
dθ
r
r
der der dθ r dθ
=
= eθ
dt
dt
dθ dt
r
r
deθ deθ dθ
r dθ
=
= − er
dt
dt
dθ dt
• Similarly, the particle acceleration vector is
dθ r ⎞
r d ⎛ dr r
a = ⎜ er + r
eθ ⎟
dt ⎝ dt
dt ⎠
r
r
d 2 r r dr der dr dθ r
d 2θ r
dθ deθ
= 2 er +
+
eθ + r 2 eθ + r
dt dt
dt dt dt dt
dt
dt
r
r
= &r& − rθ& 2 er + (rθ&& + 2r&θ& )eθ
(
)
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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
r
r r
vectors eR , eθ , and k .
• Position vector,
r
r
r
r = R e R +z k
• Velocity vector,
r
r
r
r dr & r
v=
= R eR + Rθ& eθ + z& k
dt
• Acceleration vector,
r
r
r dv
&& − Rθ& 2 erR + (Rθ&& + 2 R& θ& )erθ + &z& k
= R
a=
dt
(
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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.
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Seventh
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
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Seventh
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Vector Mechanics for Engineers: Dynamics
Sample Problem 11.12
• Calculate velocity and acceleration.
vr = r& = −0.449 m s
vθ = rθ& = (0.481m )(0.561rad s ) = 0.270 m s
v
β = tan −1 θ
v = vr2 + vθ2
vr
v = 0.524 m s
β = 31.0°
ar = &r& − rθ
&2
= −0.240 m s 2 − (0.481m )(0.561rad s )2
= −0.391m s 2
aθ = rθ&& + 2r&θ&
(
)
= (0.481m ) 0.3 rad s 2 + 2(− 0.449 m s )(0.561rad s )
= −0.359 m s
a = ar2 + aθ2
2
a
γ = tan −1 θ
ar
a = 0.531 m s
γ = 42.6°
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39 - 49
Seventh
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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Seventh
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
r
r
the fixed frame of reference Oxyz are rA and rB .
r
• Vector rB A joining A and B defines the position of
B with respect to the moving frame Ax’y’z’ and
r
r r
rB = rA + rB A
• Differentiating twice,
r
r
r
r
vB = v A + vB A vB
r
r
r
a B = a A + aB
A
A = velocity of B relative to A.
r
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.
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Seventh
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 − x A = relative position of B
with respect to A
xB = x A + xB A
xB
A
= v B − v A = relative velocity of B
with respect to A
vB = v A + vB A
vB
A
= a B − a A = relative acceleration of B
with respect to A
aB = a A + aB A
aB
A
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Seventh
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.
• 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.
© 2004 The McGraw-Hill Companies, Inc. All rights reserved.
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Seventh
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.
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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Seventh
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)
vB
E
y E = 12.3 m
= (18 − 9.81t ) − 2
= 16 − 9.81(3.65)
vB
© 2004 The McGraw-Hill Companies, Inc. All rights reserved.
E
= −19.81
m
s
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