kinematicsofparticles.pdf
TRANSCRIPT
-
Introduction to Dynamics (N. Zabaras)
Introduction to Dynamicsa.k.a. Spinning & Shaking
Prof. Nicholas Zabaras
Warwick Centre for Predictive Modelling
University of Warwick
Coventry CV4 7AL
United Kingdom
Email: [email protected]
URL: http://www.zabaras.com/
January 29, 2015
1
-
Introduction to Dynamics (N. Zabaras)
Outline
2
Course Web Site
Recommended textbooks
Homework
Course Coverage
This (review) Lecture: Kinematics of a Particle (Chapter 12) Engr. Mechanics,
Dynamics, R. C. Hibbeler, 13th Edition) Kinematics of Particles (Chapters 11 and 12),
Mechanics of Materials, Dynamics, F. Beer, E.R. Johnston, et al. (6th Edt)
-
Introduction to Dynamics (N. Zabaras)
HomeHoneowork
The course web site is
The web site will include lecture notes, homework, and course announcements. Visit this link often for useful
information.
Homework problems are assigned on the course web site related to each lecture.
Solution to the homework is not mandatory but highly recommended. We consider these homework activities an
essential part of the course.
Solution to a homework set will be provided on the course web site.
Course Web Site, Lectures & Homework
http://www.zabaras.com/Courses/Dynamics/Dynamics.html
3
-
Introduction to Dynamics (N. Zabaras)
Textbook Vector Mechanics for Engineers, F. Beer,
E.R. Johnston and R. Cornwell, 10th
Edition.
Engineering Mechanics: Dynamics, A. Bedford and W. Fowler (course pack)
Engineering Mechanics, Dynamics, R. C. Hibbeler, 13th Edition
Recommended Textbooks
Notes: Our Lectures and Lecture slides will closely follow the textbooks by Beer et al. and
Hibbeler.
-
Introduction to Dynamics (N. Zabaras)
Office Hours
Office Hours :
4:30 6:30 pm on Mondays (during the duration of the lectures)
Location
Building: Eng.
Room: 408B (close to the Chemistry bldg.)
Office Hours
5
-
Introduction to Dynamics (N. Zabaras)
Attendance
We will work with the hope that no lecture is waste of your valuable time.
We recommend that you download and study the lecture notes from the web before attending so you can
maximize your understanding of the class material.
Even with 300+ students, we encourage your questions, in class or during office hrs.
Out of respect for each other, if you come to class you come to learn. Not allowed in class: Phones on, reading
papers, bla bla with each other, playing computer
games, etc.
Attendance
6
-
Introduction to Dynamics (N. Zabaras)
Course coverage
Kinematics of a particle (Hibbeler, Chapter 12) Force and acceleration (Hibbeler, Chapter 13) Work and energy relations (Hibbeler, Chapter 14, B&F Chapter
15)
Impulse and Momentum (Hibbeler, Chapter 15)
Kinematics of a rigid bodies in planar motion (Hibbeler, Chapter 16, B&F Chapter 17)
Force and acceleration of rigid bodies (Hibbeler, Chapter 17) Work and energy relations of rigid bodies (Hibbeler, Chapter 18) Impulse and Momentum of rigid bodies (Hibbeler, Chapter 19)
Vibrations (Hibbeler, Chapter 22, B&F Chapter 21)
Course Coverage
Notes
1. Several lecture notes provide review of things you should know from elementary physics. They are
essential part of the course but will be discussed briefly.
2. The lecture notes follow closely the three recommended textbooks (thus you dont need to purchase them).
7
-
Introduction to Dynamics (N. Zabaras)
Introduction
Dynamics includes:
- Kinematics is used to relate displacement, velocity, acceleration,
and time without reference to the cause of motion.
- 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.
8
-
Introduction to Dynamics (N. Zabaras)
Areas of mechanics (section12.1) Statics: Study of bodies at rest
Dynamics: Study of bodies in motion
Kinematics, is a study the geometry of the motion s, v, a
Kinetics, is a study of forces F that cause the motion
0
0
0
x
y
F
F
M
x x
y y
F ma
F ma
M I
Introduction to Dynamics
9
-
Introduction to Dynamics (N. Zabaras)
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.,
326 ttx
or in the form of a graph x vs. t.
10
-
Introduction to Dynamics (N. Zabaras)
Frame of ReferenceFrame of reference is a place or object that you
assume is fixed
observations of how objects move are defined in relation to that frame of
reference.
Perception of motion depends on the observers frame of reference.
We usually use the earth as the frame of reference.
11
-
Introduction to Dynamics (N. Zabaras)
Rectilinear Motion: Position, Velocity & Acceleration
Instantaneous velocity may be positive or negative. Magnitude of velocity is
referred to as particle speed.
Consider particle which occupies position P at time t and P at t+Dt,
t
xv
t
x
t D
D
D
D
D 0lim
Average velocity
Instantaneous velocity
From the definition of a derivative,
dt
dx
t
xv
t
D
D
D 0lim
e.g.
,2
32
312
6
ttdt
dxv
ttx
12
-
Introduction to Dynamics (N. Zabaras)
Rectilinear Motion: Position, Velocity & Acceleration
Consider particle with velocity v at time tand v at t+Dt,
Instantaneous accelerationt
va
t D
D
D 0lim
tdt
dva
ttv
dt
xd
dt
dv
t
va
t
612
312e.g.
lim
2
2
2
0
D
D
D
From the definition of a derivative,
Instantaneous acceleration may be:
- positive: increasing positive velocity
or decreasing negative velocity
- negative: decreasing positive velocity
or increasing negative velocity.
13
-
Introduction to Dynamics (N. Zabaras)
Rectilinear Motion: Position, Velocity & Acceleration
Consider particle with motion given by
326 ttx
2312 ttdt
dxv
tdt
xd
dt
dva 612
2
2
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
14
-
Introduction to Dynamics (N. Zabaras) 15
-
Introduction to Dynamics (N. Zabaras)
Motion 1Motion 1
16
-
Introduction to Dynamics (N. Zabaras)
Motion 2Motion 2
17
-
Introduction to Dynamics (N. Zabaras)
Motion 3
slowing
Motion 3
18
-
Introduction to Dynamics (N. Zabaras)
Motion 4
bottom
Motion 4
19
-
Introduction to Dynamics (N. Zabaras)
Motion 5
speeding up
Motion 5
20
-
Introduction to Dynamics (N. Zabaras)
Motion 6
Still rising
Motion 6
21
-
Introduction to Dynamics (N. Zabaras)
Motion 7
top
Motion 7
22
-
Introduction to Dynamics (N. Zabaras)
1
2
3
4
5
6
7
Velocity Acceleration Velocity Acceleration
bottom 23
-
Introduction to Dynamics (N. Zabaras)
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)
24
-
Introduction to Dynamics (N. Zabaras)
Determination of the Motion of a Particle
Acceleration given as a function of time, a = f(t):
tttx
x
tttv
v
dttvxtxdttvdxdttvdxtvdt
dx
dttfvtvdttfdvdttfdvtfadt
dv
00
0
00
0
0
0
Acceleration given as a function of position, a = f(x):
x
x
x
x
xv
v
dxxfvxvdxxfdvvdxxfdvv
xfdx
dvva
dt
dva
v
dxdt
dt
dxv
000
202
12
21
or or
25
-
Introduction to Dynamics (N. Zabaras)
Determination of the Motion of a Particle
Acceleration given as a function of velocity, a = f(v):
tv
v
tv
v
tx
x
tv
v
ttv
v
vf
dvvxtx
vf
dvvdx
vf
dvvdxvfa
dx
dvv
tvf
dv
dtvf
dvdt
vf
dvvfa
dt
dv
0
00
0
0
0
0
26
-
Introduction to Dynamics (N. Zabaras)
Relation involving x, v, & a (no time t)
dva
dt
dxdt
v
dxv
dt
dvdt
a
dx dv
v a
adx vdv
Acceleration
Velocity
Position x
27
-
Introduction to Dynamics (N. Zabaras)
Example of applying a dx = v dv A car starts from rest and moves along a straight line with an
acceleration of a = ( 3 x -1/3 ) m/s2. where x is in meters. Determine the
cars acceleration when t = 4 s. Rest t = 0 , v = 0
1
33v x
adx vdv
1
3
0 0
3
s v
x dx vdv
2
233 1
(3)2 2
x v
1
33dx
v xdt
1
3 3x dx dt
1
3
0 0
3
x t
x dx dt
2
33
32
x t
3
2(2 )x t
3 9
2 2
19 3
22
(4) (2 4) 2
3(4) 3 2 /
2 2
x
a m s
28
-
Introduction to Dynamics (N. Zabaras)
Example Problem
Determine:
velocity and elevation above ground at time t,
highest elevation reached by ball and corresponding time, and
time when ball will hit the ground and corresponding velocity.
Ball tossed with 10 m/s vertical velocity
from window 20 m above ground.
SOLUTION:
Integrate twice to find v(t) and y(t).
Solve for t at which velocity equals zero (time for maximum elevation) and
evaluate corresponding altitude.
Solve for t at which altitude equals zero (time for ground impact) and
evaluate corresponding velocity.
29
-
Introduction to Dynamics (N. Zabaras)
Example Problem
tvtvdtdv
adt
dv
ttv
v
81.981.9
sm81.9
00
2
0
ttv
2s
m81.9
s
m10
2
21
00
81.91081.910
81.910
0
ttytydttdy
tvdt
dy
tty
y
22s
m905.4
s
m10m20 ttty
SOLUTION:
Integrate twice to find v(t) and y(t).
30
-
Introduction to Dynamics (N. Zabaras)
Example Problem
Solve for t at which velocity equals zero and evaluate corresponding altitude.
0s
m81.9
s
m10
2
ttv
s019.1t
22
2
2
s019.1s
m905.4s019.1
s
m10m20
s
m905.4
s
m10m20
y
ttty
m1.25y
31
-
Introduction to Dynamics (N. Zabaras)
Example Problem
Solve for t at which altitude equals zero and evaluate corresponding velocity.
0s
m905.4
s
m10m20 2
2
ttty
s28.3
smeaningles s243.1
t
t
s28.3s
m81.9
s
m10s28.3
s
m81.9
s
m10
2
2
v
ttv
s
m2.22v
32
-
Introduction to Dynamics (N. Zabaras)
Example Problem
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 a(x).
kva
SOLUTION:
Integrate a = dv/dt = -kv to find v(t).
Integrate v(t) = dx/dt to find x(t).
Integrate a = v dv/dx = -kv to find v(x).
33
-
Introduction to Dynamics (N. Zabaras)
Example Problem
SOLUTION:
Integrate a = dv/dt = -kv to find v(t).
ktv
tvdtk
v
dvkv
dt
dva
ttv
v
00
ln
0
ktevtv 0
Integrate v(t) = dx/dt to find x(t).
tkt
tkt
tx
kt
ek
vtxdtevdx
evdt
dxtv
00
00
0
0
1
ktek
vtx 10
34
-
Introduction to Dynamics (N. Zabaras)
Example Problem
Integrate a = v dv/dx = -kv to find v(x).
kxvv
dxkdvdxkdvkvdx
dvva
xv
v
0
00
kxvv 0
Alternatively,
0
0 1v
tv
k
vtx
kxvv 0
00 or
v
tveevtv ktkt
ktek
vtx 10with
and
then
35
0a k v kx
-
Introduction to Dynamics (N. Zabaras)
Uniform Rectilinear Motion
For particle in uniform rectilinear motion, the acceleration is
zero and the velocity is constant.
vtxx
vtxx
dtvdx
vdt
dx
tx
x
0
0
00
constant
36
-
Introduction to Dynamics (N. Zabaras)
Uniformly Accelerated Rectilinear Motion
For particle in uniformly accelerated rectilinear motion, the
acceleration of the particle is constant.
atvv
atvvdtadvadt
dv tv
v
0
000
constant
221
00
221
000
00
0
attvxx
attvxxdtatvdxatvdt
dx tx
x
020
2
020
221
2
constant
00
xxavv
xxavvdxadvvadx
dvv
x
x
v
v
37
-
Introduction to Dynamics (N. Zabaras)
Velocity as a Function of Time
c
dva
dt
cdv a dt
0o
v t
c
v
dv a dt
0 cv v a t
-
Introduction to Dynamics (N. Zabaras)
Position as a Function of Time
0 c
dxv v a t
dt
0
0
( )
o
x t
c
s
dx v a t dt
2
0 0
1
2cx x v t a t
-
Introduction to Dynamics (N. Zabaras)
Velocity as a Function of Position
0 0
v x
c
v s
v dv a dx
cv dv a dx
2 2
0 02 ( )cv v a x x
2 2
0 0
1 1( )
2 2cv v a x x
-
Introduction to Dynamics (N. Zabaras)
Example of Constant Acceleration: Free Fall
We throw two balls at the top of a cliff of height H. Ball A is thrown with an initial speed v0 downwards and ball B with the same speed upwards
(as shown).
The speed of the two balls when they hit the ground are vA and vBrespectively. Which of the following is true:
(a) vA < vB (b) vA = vB (c) vA > vB
v0
v0
BA
H
vA vB
41
-
Introduction to Dynamics (N. Zabaras)
Example of Constant Acceleration: Free Fall
Intuition should tell you that v = v0
We can prove this:
v0
B
H
v = v0
2 20 2( ) 0v v g H H
This looks just like ball B was thrown down with speed v0, sothe speed at the bottom shouldbe the same as that of ball A!!
y = 042
-
Introduction to Dynamics (N. Zabaras)
Example of Constant Acceleration: Free Fall
We can also just use the equation directly:
2 20 2( ) 0v v g H A :
v0
v0
A B
y = 0
2 20 2( ) 0v v g H B :same !!
y = H
43
-
Introduction to Dynamics (N. Zabaras)
Summary Time dependent acceleration Constant acceleration
dxv
dt
( )x t
2
2
dv d xa
dt dt
adx vdv
0 cv v a t
2
0 0
1
2cx x v t a t
2 2
0 02 ( )cv v a x x
This applies to a freely falling object:
2 2a g 9.81 / 32.2 /m s ft s
44
-
Introduction to Dynamics (N. Zabaras)
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.
ABAB xxx relative position of B
with respect to AABAB xxx
ABAB vvv relative velocity of B
with respect to AABAB vvv
ABAB aaa relative acceleration of
B with respect to AABAB aaa
45
-
Introduction to Dynamics (N. Zabaras)
A B
160 km/h 180 km/h
A B A/B
A/B A B
V V V
V V - V 160 180 20 /km h
B A B/A
B/A B A
V V V
V V - V 180 160 20 /km h
Relative Motion
46
-
Introduction to Dynamics (N. Zabaras)
The passenger aircraft B is flying east with velocity vB = 800 km/h. A
jet is traveling south with velocity vA = 1200 km/h. What velocity does
A appear to a passenger in B ?
BABA VVV
Solution i800vB
j1200vA
x
y
BAv
BAvi800j1200
jiV BA1200800
Example
47
-
Introduction to Dynamics (N. Zabaras)
22/ 1200800 BAAbsolute value :
h/km1442
The direction of BAV
1200
800tan
iVB800
jVA1200
x
y
BAv
7.33 West of
South
Example
48
-
Introduction to Dynamics (N. Zabaras)
T
A
60 mi/h
45 mi/h
T A T/AV V V
o o
T/A60 i (45 cos 45 i 45 sin 45 j) V
T/AV { 28.2 i - 31.8 j} mi/h
2 2
/ (28.2) ( 31.8) 42.5 mi/hT A
/
/
( ) 31.8tan
( ) 28.2
T A y
T A x
o5.48
A train travels at a constant speed of 60 mi/
h and crosses over a road. If the automobile
A is traveling at 45 mi /h along the road
determine the magnitude and direction of
the velocity of the train relative to the
automobile.
Example
-
Introduction to Dynamics (N. Zabaras)
At the instant shown, the car at A is traveling at 10 m/s around
the curve while increasing its speed at 2 m/s2. The car at B is
traveling at 18.5 m/s along the straightway and increasing its
speed at 5 m/s2. Determine the relative velocity and relative
acceleration of A with respect to B at this instant
A B A/BV V V
A/B10 cos 45i -10 sin 45 j 18.5 i V
2 2o o
A
(10) (10)a {2 cos 45 i 2 sin 45 j- cos 45 i - sin45 j }
100 100
Ba 5 i
A B A/Ba a a
A/BV { -11.42 i 7.07 j} m/s
Example
50
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
Ball thrown vertically from 12 m level
in elevator shaft with initial velocity
of 18 m/s. At same instant, open-
platform elevator passes 5 m level
moving upward at 2 m/s.
Determine (a) when and where ball
hits elevator and (b) relative velocity
of ball and elevator at contact.
SOLUTION:
Substitute initial position and velocity and constant acceleration of ball into
general equations for uniformly
accelerated rectilinear motion.
Substitute initial position and constant velocity of elevator into
equation for uniform rectilinear
motion.
Write equation for relative position of ball with respect to elevator and
solve for zero relative position, i.e.,
impact.
Substitute impact time into equation for position of elevator and relative
velocity of ball with respect to
elevator.
51
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
SOLUTION:
Substitute initial position and velocity and constant acceleration of ball into general equations for
uniformly accelerated rectilinear motion.
2
2
221
00
20
s
m905.4
s
m18m12
s
m81.9
s
m18
ttattvyy
tatvv
B
B
Substitute initial position and constant velocity of elevator into equation for uniform rectilinear
motion.
ttvyy
v
EE
E
s
m2m5
s
m2
0
52
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact.
025905.41812 2 ttty EB
s65.3
smeaningles s39.0
t
t
Substitute impact time into equations for position of elevator and relative velocity of ball with respect to
elevator.
65.325 Ey m3.12Ey
65.381.916
281.918
tv EB
s
m81.19EBv
53
-
Introduction to Dynamics (N. Zabaras)
Absolute Dependent Motion Analysis of Two Particles
The motion of one particle will depend on the
corresponding motion of
another particle
Inextensible cords connection
-
Introduction to Dynamics (N. Zabaras)
Position
How to establish the position coordinate Reference from a fixed point
(O) or fixed datum.
Measures along each inclined plane in the
direction of motion
Has positive sense from C to A and from D to B.
Establish a total cord length equation
A CD B Ts l s l
55
-
Introduction to Dynamics (N. Zabaras)
Velocity
Velocity is the time derivative of the position
From the total cord length equation
Velocity
A CD B Ts l s l
0A B B Ads ds
ordt dt
A B T CDs s l l const
56
-
Introduction to Dynamics (N. Zabaras)
Acceleration
Acceleration is the time derivative of the velocity
From the velocity equation
Acceleration
B A
B Aa a
B Ad d
dt dt
57
-
Introduction to Dynamics (N. Zabaras)
Example Problem
Position coordinate
Velocity
Acceleration
2 B As h s l
2 B A
2 B Aa a
2 .B As s l h const
1
2B A
58
-
Introduction to Dynamics (N. Zabaras)
Example Problem
Position coordinate
Velocity
Acceleration
Positive signs, why?
lshsh AB )(2
2 B A
2 B Aa a
2 3B As s l h Const
59
-
Introduction to Dynamics (N. Zabaras)
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.
BA xx 2 constant (one degree of freedom)
Positions of three blocks are dependent.
CBA xxx 22 constant (two degrees of freedom)
For linearly related positions, similar relations hold between velocities and accelerations.
022or022
022or022
CBACBA
CBACBA
aaadt
dv
dt
dv
dt
dv
vvvdt
dx
dt
dx
dt
dx
60
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
Pulley D is attached to a collar
which is pulled down at 3 in./s. At t
= 0, collar A starts moving down
from K with constant acceleration
and zero initial velocity. Knowing
that velocity of collar A is 12 in./s as
it passes L, determine the change in
elevation, velocity, and acceleration
of block B when block A is at L.
SOLUTION:
Define origin at upper horizontal surface with positive displacement downward.
Collar A has uniformly accelerated rectilinear motion. Solve for acceleration
and time t to reach L.
Pulley D has uniform rectilinear motion. Calculate change of position at time t.
Block B motion is dependent on motions of collar A and pulley D. Write motion
relationship and solve for change of
block B position at time t.
Differentiate motion relation twice to develop equations for velocity and
acceleration of block B.61
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
SOLUTION:
Define origin at upper horizontal surface with positive displacement downward.
Collar A has uniformly accelerated rectilinear motion. Solve for acceleration and time t to reach
L.
2
2
020
2
s
in.9in.82
s
in.12
2
AA
AAAAA
aa
xxavv
s 333.1s
in.9
s
in.12
2
0
tt
tavv AAA
62
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
Pulley D has uniform rectilinear motion. Calculate change of position at time t.
in. 4s333.1s
in.30
0
DD
DDD
xx
tvxx
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,
0in.42in.8
02
22
0
000
000
BB
BBDDAA
BDABDA
xx
xxxxxx
xxxxxx
in.160 BB xx
63
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
Differentiate motion relation twice to develop equations for velocity and acceleration of block B.
0s
in.32
s
in.12
02
constant2
B
BDA
BDA
v
vvv
xxx
s
in.18Bv
0s
in.9
02
2
B
BDA
v
aaa
2s
in.9Ba
64
-
Introduction to Dynamics (N. Zabaras)
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 + Dt,
r
r
D
D
D
D
D
D
dt
ds
t
sv
dt
rd
t
rv
t
t
0
0
lim
lim
instantaneous velocity (vector)
instantaneous speed (scalar)
65
-
Introduction to Dynamics (N. Zabaras)
Curvilinear Motion: Position, Velocity & Acceleration
D
D
D dt
vd
t
va
t
0lim
instantaneous acceleration
(vector)
Consider velocity of particle at time t and velocity at t + Dt,
v
v
In general, acceleration vector is not tangent to particle path and velocity vector.
66
-
Introduction to Dynamics (N. Zabaras)
Derivatives of Vector Functions
uP
Let be a vector function of scalar variable u,
u
uPuuP
u
P
du
Pd
uu D
D
D
D
DD
00limlim
Derivative of vector sum,
du
Qd
du
Pd
du
QPd
du
PdfP
du
df
du
Pfd
Derivative of product of scalar and vector functions,
Derivative of scalar product and vector product,
du
QdPQ
du
Pd
du
QPd
du
QdPQ
du
Pd
du
QPd
67
-
Introduction to Dynamics (N. Zabaras)
Curvilinear MotionWe will investigate particle motion along a curved path
using three coordinate systems
Rectangular Components
Normal and Tangential Components
Polar & Cylindrical Components
68
-
Introduction to Dynamics (N. Zabaras)
Rectangular Components of Velocity & Acceleration
When position vector of particle P is given by its rectangular components,
kzjyixr
Velocity vector,
kvjviv
kzjyixkdt
dzj
dt
dyi
dt
dxv
zyx
Acceleration vector,
kajaia
kzjyixkdt
zdj
dt
ydi
dt
xda
zyx
2
2
2
2
2
2
69
-
Introduction to Dynamics (N. Zabaras)
Rectangular Components of Velocity & Acceleration
Rectangular components particularly effective when component accelerations can be integrated
independently, e.g., motion of a projectile,
00 zagyaxa zyx
with initial conditions,
0,,0 000000 zyx vvvzyx
Integrating twice yields
0
0
221
00
00
zgtyvytvx
vgtvvvv
yx
zyyxx
Motion in horizontal direction is uniform.
Motion in vertical direction is uniformly accelerated.
Motion of projectile could be replaced by two independent rectilinear motions.
70
-
Introduction to Dynamics (N. Zabaras)
Example Problem The position of a particle is described by rA= {2t i +(t
2-1) j} ft. where t is
in seconds.
Determine the position of the point and the speed at 2 seconds.
22(2) (2 1) 4 3Ar i j i j
2 2 2 4AAdr
v i t j i jdt
2 22 4 4.47 ft/sAspeed v
71
-
Introduction to Dynamics (N. Zabaras)
Acceleration
Acceleration is the first time derivative of v
Where
Magnitude of acceleration
Direction is not tangent to the path
x y z
dva a i a j a k
dt
x xa x
2 2 2
x y za a a a
y ya y
z za z
72
-
Introduction to Dynamics (N. Zabaras)
Position
8(2) 16ftx
2(16) /10 25.6 fty
2 2(16) (25.6) 30.2 ftr
Velocity
(8 ) 8ft/sxd
x tdt
2( /10) 2 /10 2(16)(8) /10 25.6ft/syd
y x xxdt
2 2(8) (25.6) 26.8 ft
1 1 25.6tan tan 72.68
y o
x
Acceleration
(8) 0ft/sxd
a xdt
2(2 /10) 2( ) /10 2 ( ) /10 12.8ft/syd
a y xx x x x xdt
2 2 2(0) (12.8) 12.8 ft/sa
1 1 12.8tan tan 900
y o
a
x
a
a
A x=8t
At any instant the horizontal postition of the weather
balloon is defined by x = (8t) ft, where t is in second . If the equation of the path is y=x2/10 determine the distance
of the balloon from A, the magnitude and direction of the
velocity and the acceleration when t=2sec.
Example Problem
73
-
Introduction to Dynamics (N. Zabaras)
Example
74
r(0.75) 0.5sin(1.5)i 0.5cos(1.5) j 0.2(0.75)k
)180( orad
0.5sin(2 )i 0.5cos(2 ) j 0.2 kr t t t
t in seconds, arguments in radians
At t = 0.75 s find location, velocity, and
acceleration
r 0.499i 0.0354j 0.15k
2 2 2(0.499) (0.0354) ( 0.15) 0.522 mr
0.499 0.0354 0.15i j k 0.955i 0.0678j-0.287k
0.522 0.522 0.522ru
1cos (0.955) 17.2o
1cos (0.0678) 86.1o
1cos ( 0.287) 107o
Take derivatives of r(t) w.r.t. time to
define similar eqs for v and a
-
Introduction to Dynamics (N. Zabaras)
Motion of a Projectile
Projectile: a body that is given an initial velocity and then follows a path determined by gravitational acceleration
and air resistance.
Trajectory path followed by a projectile
75
-
Introduction to Dynamics (N. Zabaras)
Horizontal motion is uniform motion
Notice that the horizontal motion is in no way affected by the vertical motion.
76
-
Introduction to Dynamics (N. Zabaras)
Projectile
77
-
Introduction to Dynamics (N. Zabaras)
Vertical velocity decreases at a constant rate
due to the influence of gravity.
Horizontal and vertical components of velocity are independent.
Vertical velocity decreases at a constant rate due to the influence
of gravity.
Horizontal and vertical components of velocity
78
-
Introduction to Dynamics (N. Zabaras)
Horizontal Motion
Acceleration : ax= 0
Horizontal velocity remains constant
Equal distance covered in equal time intervals
0 0( ) ( )c x xv v a t v v
2
0 0 0 0
1( ) ( )
2c xx x v t a t x x v t
22
0 00( ) 2 ( ) ( )c x xv a s s v vv
0( )x
x xv
t
79
-
Introduction to Dynamics (N. Zabaras)
Vertical Motion ac= -g = 9.81 m/s
2 = 32.2 ft/s2
Equal increments of speed gained in equal increments of time
Distance increases in each time interval
0 0( ) ( )c y yv v a t v v gt
2 2
0 0 0 0
1 1( ) ( )
2 2c yy y v t a t y y v t gt
22 22
0 00( ) 2 ( ) 2 ( )
0c y yv a s s g y yvv v
80
-
Introduction to Dynamics (N. Zabaras)
Assumptions:
(1) free-fall acceleration
(2) neglect air resistance
Choosing the y direction as positive upward:
ax = 0; ay = - g (a constant)
Take x0= y0 = 0 at t = 0
Initial velocity v0 makes an
angle 0 with the horizontalv0
x
y
0 0 0 0 0 0cos sin
x yv v v v
Projectile Motion
81
-
Introduction to Dynamics (N. Zabaras)
At the peak of its trajectory, vy = 0.
From
Time t1 to reach the peak
Substituting into:
0
1
yvt
g
2
0
max2
yvh y
g
0 0y y oyv v gt v gt
2
0
1
2yy v t gt
Maximum Height
82
-
Introduction to Dynamics (N. Zabaras)
The optimal angle of projection is dependent on the goal of the activity.
For maximal height the optimal angle is 90o.
For maximal distance the optimal angle is 45o.
Projection Angle
83
-
Introduction to Dynamics (N. Zabaras)
Projection angle = 10 degrees
84
-
Introduction to Dynamics (N. Zabaras)
10 degrees
30 degrees
40 degrees
45 degrees
Projection angle = 45 degrees
85
-
Introduction to Dynamics (N. Zabaras)
10 degrees
30 degrees
40 degrees
45 degrees
60 degrees
Projection angle = 60 degrees
86
-
Introduction to Dynamics (N. Zabaras)
10 degrees
30 degrees
40 degrees
45 degrees
60 degrees
75 degrees
Angle that maximizes range
optimal = 45 degrees
Projection angle = 75 degrees
87
-
Introduction to Dynamics (N. Zabaras)
Example Problem 1/4 A ball is given an initial velocity of V0 = 37 m/s at an angle of = 53.1.
Find the position of the ball, and the magnitude and direction of its velocity, when t = 2.00 s.
Find the time when the ball reaches the highest point of its flight, and find its height h at this point
The initial velocity of the ball has components:
v0x = v0 cos 0 = (37.0 m/s) cos 53.1 = 22.2 m/s
v0y = v0 sin 0 = (37.0 m/s) sin 53.1= 29.6 m/s
a) position
x = v0xt = (22.2 m/s)(2.00 s) = 44.4 m
y = v0yt - gt2
= (29.6 m/s)(2.00 s) (9.80 m/s2)(2.00 s)2
= 39.6 m
88
-
Introduction to Dynamics (N. Zabaras)
Example Problem 2/4
Velocity vx = v0x = 22.2 m/s vy = v0y gt = 29.6 m/s (9.80 m/s
2)(2.00 s) = 10.0 m/s
22 2 222.2 / (10.0 / )
24.3 /
10.0 /arctan arctan 0.450 24.2
22.2 /
x yv v v m s m s
m s
m s
m s
89
-
Introduction to Dynamics (N. Zabaras)
Example Problem 3/4
b) Find the time when the ball reaches the highest point of its flight, and find its height H at this point.
0 1
0
1 2
0
29.6 /3.02
9.80 /
y y
y
v v gt
v m st s
g m s
1
2
0 1
2 2
1
2
1(29.6 / )(3.02 ) (9.80 / )(3.02 )
2
44.7
yH v t gt
m s s m s s
m
90
-
Introduction to Dynamics (N. Zabaras)
c) Find the horizontal range R, (e.g. the horizontal distance from the starting point to the point at which
the ball hits the ground.)
0 2 (22.2 / )(6.04 ) 134xR v t m s s m
2
0 2 2 2 0 2
1 10 ( )
2 2y yy v t gt t v gt
0
2 2 2
2 2(29.6 / )0 6.04
9.80 /
yv m st and t s
g m s
Example Problem 4/4
91
-
Introduction to Dynamics (N. Zabaras)
A ball is traveling at 25 m/s drive off of the edge of a cliff 50 m high. Where does it land?
25 m/s
Vertically
v = v0-gt
y = y0 + v0t + 1/2gt2 .
v2 = v02 - 2g(y-y0).
Horizontally
x = x0 + (v0)x t x = 25 *3.19 = 79.8 m
79.8 m
Initial
Conditions
vx = 25 m/s
vy0 = 0 m/s
a =- 9.8 m/s2
t = 0
y0 = 0 m
y =- 50 m
x0 =0 m
-50 = 0+0+1/2(-9.8)t2 t = 3.19 s
Example Problem
92
-
Introduction to Dynamics (N. Zabaras)
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 . and BA rr
Vector joining A and B defines the position of B with respect to the moving frame Axyz and
ABr
ABAB rrr
Differentiating twice,
ABv
velocity of B relative to A.ABAB vvv
ABa
acceleration of B relative
to A.ABAB aaa
Absolute motion of B can be obtained by combining motion of A with relative motion of B with respect to
moving reference frame attached to A.
93
-
Introduction to Dynamics (N. Zabaras)
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.
are tangential unit vectors for the particle path at P and P. When drawn with
respect to the same origin, and
is the angle between them.
tt ee and
ttt eee
DD
d
ede
eee
e
tn
nnt
t
D
D
D
D
DD
DD 2
2sinlimlim
2sin2
00
94
-
Introduction to Dynamics (N. Zabaras)
Tangential and Normal Components
tevv
With the velocity vector expressed asthe particle acceleration may be written as
t tt t
de dedv dv dv d dsa e v e v
dt dt dt dt d ds dt
but
vdt
dsdsde
d
edn
t
After substituting,
22 va
dt
dvae
ve
dt
dva ntnt
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.
95
-
Introduction to Dynamics (N. Zabaras)
Tangential and Normal Components
22 va
dt
dvae
ve
dt
dva ntnt
Relations for tangential and normal acceleration also apply for particle moving along space curve.
Plane containing tangential and normal unit vectors is called the osculating plane.
ntb eee
Normal to the osculating plane is found from
binormale
normalprincipal e
b
n
Acceleration has no component along binormal.
96
-
Introduction to Dynamics (N. Zabaras)
Radius of curvature ()
For the Circular motion : () = radius of the circle
For y = f(x):
2 3/2
2 2
1 ( / )
/
dy dx
d y dx
97
-
Introduction to Dynamics (N. Zabaras)
Example Problem
Find the radius of curvature of the parabolic path in the
figure at x = 150 ft.
21 200200
y x
1/
100dy dx x
2 2 1/100
d y dx 3/2
2
3/22
2 2
150
11
1001 ( / )585.9 ft
1/
100
x
xdy dx
d y dx
98
-
Introduction to Dynamics (N. Zabaras)
Example Problem A skier travels with a constant
speed of 20 ft/s along the parabolic
path shown. Determine the velocity
at x = 150 ft.
21 200200
y x
1/
100dy dx x
150
1/ 1.5
100 xdy dx x
1tan 1.5 56.3o
15020 ft/s
99
-
Introduction to Dynamics (N. Zabaras)
Special case
1- Straight line motion
2- Constant speed curve motion (centripetal acceleration)
0na
ta a
0ta
2
na a
100
-
Introduction to Dynamics (N. Zabaras)
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.
rr e
d
ede
d
ed
dt
de
dt
d
d
ed
dt
ed rr
dt
de
dt
d
d
ed
dt
edr
erer
edt
dre
dt
dr
dt
edre
dt
drer
dt
dv
r
rr
rr
The particle velocity vector is
Similarly, the particle acceleration vector is
errerr
dt
ed
dt
dre
dt
dre
dt
d
dt
dr
dt
ed
dt
dre
dt
rd
edt
dre
dt
dr
dt
da
r
rr
r
22
2
2
2
2
rerr
101
-
Introduction to Dynamics (N. Zabaras)
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 . and ,, keeR
Position vector,
kzeRr R
Velocity vector,
kzeReRdt
rdv R
Acceleration vector,
kzeRReRRdt
vda R
2
2
102
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
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.
SOLUTION:
Calculate tangential and normal components of acceleration.
Determine acceleration magnitude and direction with respect to tangent
to curve.
103
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
ft/s66mph45
ft/s88mph60
SOLUTION:
Calculate tangential and normal components of acceleration.
2
22
2
66 88 ft s ft2.75
8 s s
88ft s ft3.10
2500ft s
t
n
va
t
va
D
D
Determine acceleration magnitude and direction with respect to tangent to curve.
2222 10.375.2 nt aaa 2s
ft14.4a
75.2
10.3tantan 11
t
n
a
a 4.48
104
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
Rotation of the arm about O is
defined by = 0.15t2 where is in radians and t in seconds. Collar B
slides along the arm such that r =
0.9 - 0.12t2 where r is in meters.
After the arm has rotated through
30o, determine (a) the total velocity
of the collar, (b) the total
acceleration of the collar, and (c)
the relative acceleration of the collar
with respect to the arm.
SOLUTION:
Evaluate time t for = 30o.
Evaluate radial and angular positions, and first and second
derivatives at time t.
Calculate velocity and acceleration in cylindrical
coordinates.
Evaluate acceleration with respect to arm.
105
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
SOLUTION:
Evaluate time t for = 30o.
20.15
30 0.524rad 1.869 s
t
t
Evaluate radial and angular positions, and first and second derivatives at time t.
2
2
sm24.0
sm449.024.0
m 481.012.09.0
r
tr
tr
2
2
srad30.0
srad561.030.0
rad524.015.0
t
t
106
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
Calculate velocity and acceleration.
rr
r
v
vvvv
rv
srv
122 tan
sm270.0srad561.0m481.0
m449.0
0.31sm524.0 v
rr
r
a
aaaa
rra
rra
122
2
2
2
22
2
tan
sm359.0
srad561.0sm449.02srad3.0m481.0
2
sm391.0
srad561.0m481.0sm240.0
6.42sm531.0 a
107
-
Introduction to Dynamics (N. Zabaras)
Sample Problem
Evaluate acceleration with respect to arm.
Motion of collar with respect to arm is
rectilinear and defined by coordinate r.
2sm240.0 ra OAB
108