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All figures taken from Vector Mechanics for Engineers: Dynamics, Beer
and Johnston, 2004
Dynamics
Kinematics of Particles
2
Introduction
• Dynamics includes:
- Kinematics: study of the motion (displacement,
velocity, acceleration, & time) without reference to
the cause of motion (i.e. regardless of forces).
- Kinetics: study of the forces acting on a body,
and the resulting motion caused by the given
forces.
• 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.
3
Rectilinear Motion: Position, Velocity & Acceleration
4
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
(+ or -) 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.
5
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
6
Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle with velocity v at time t and
v’ at t+Dt,
Instantaneous acceleration t
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,
7
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
8
Determining the Motion of a Particle
• Recall, motion is defined if position x is known for all time t.
• If the acceleration is given, we can determine velocity and
position by 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)
dxv
dt
dva
dt
2
2
d xa
dt
dv dv dx dva v
dt dx dt dx
9
Determining the Motion of a Particle
• Acceleration given as a function of time, a = f(t):
0
0
0 0
( ) ( ) ( ) ( )
v t t
v
dva f t dv f t dt dv f t dt v v f t dt
dt
0
0
0 0
x t t
x
dxv dx vdt dx vdt x x vdt
dt
0 0 0
2 2
0
1 1( ) ( ) ( ) ( )
2 2
v x x
v x x
dva f x v vdv f x dx vdv f x dx v v f x dx
dx
•Acceleration given as a function of position, a = f(x):
0 0
x t
x
dx dx dxv dt dt
dt v v
10
Determining the Motion of a Particle
0 00
( )( ) ( ) ( )
v t v
v v
dv dv dv dva f v dt dt t
dt f v f v f v
0 0 0
0( )( ) ( ) ( )
x v v
x v v
dv vdv vdv vdva f v v dx dx x x
dx f v f v f v
• Acceleration given as a function of velocity, a = f(v):
11
Summary
Procedure:
1. Establish a coordinate system & specify an origin
2. Remember: x,v,a,t are related by:
3. When integrating, either use limits (if known) or add
a constant of integration
dxv
dt
dva
dt
2
2
d xa
dt
dv dv dx dva v
dt dx dt dx
12
Sample Problem 11.2
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.
13
Sample Problem 11.2
tvtvdtdv
adt
dv
ttv
v
81.981.9
sm81.9
00
2
0
ttv
2s
m81.9
s
m10
2
21
00
81.91081.910
81.910
0
ttytydttdy
tvdt
dy
tty
y
2
2s
m905.4
s
m10m20 ttty
SOLUTION:
• Integrate twice to find v(t) and y(t).
14
Sample Problem 11.2
• Solve for t at which velocity equals zero and evaluate
corresponding altitude.
0s
m81.9
s
m10
2
ttv
s019.1t
2
2
2
2
s019.1s
m905.4s019.1
s
m10m20
s
m905.4
s
m10m20
y
ttty
m1.25y
15
Sample Problem 11.2
• 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
16
What if the ball is tossed downwards with the same
speed? (The audience is thinking …)
vo= - 10 m/s
17
Uniform Rectilinear Motion
Uniform rectilinear motion acceleration = 0 velocity = constant
vtxx
vtxx
dtvdx
vdt
dx
tx
x
0
0
00
constant
18
Uniformly Accelerated Rectilinear Motion
Uniformly accelerated motion acceleration = 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
Also:
Application: free fall
19
Motion of Several Particles: Relative Motion
• For particles moving along the same
line, displacements should be
measured from the same origin in
the same direction.
ABAB xxx relative position of B
with respect to A ABAB xxx
ABAB vvv relative velocity of B
with respect to A ABAB vvv
ABAB aaa relative acceleration of B
with respect to A ABAB aaa
20
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.
Sample Problem 11.4
21
Sample Problem 11.4 SOLUTION:
• Ball: uniformly accelerated motion
(given initial position and velocity).
• Elevator: constant velocity (given
initial position and velocity)
• 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.
22
Sample Problem 11.4
SOLUTION:
• Ball: uniformly accelerated rectilinear motion.
2
2
221
00
20
s
m905.4
s
m18m12
s
m81.9
s
m18
ttattvyy
tatvv
B
B
• Elevator: uniform rectilinear motion.
ttvyy
v
EE
E
s
m2m5
s
m2
0
23
Sample Problem 11.4
• Relative position of ball with respect to elevator:
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
24
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
25
Applications
26
Sample Problem 11.5
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.
27
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.
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
28
Sample Problem 11.5
• 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
29
Sample Problem 11.5
• 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
in.18
sBv
2
2 0
in.9 0
s
A D B
B
a a a
a
2s
in.9Ba
30
Curvilinear Motion
http://news.yahoo.com/photos/ss/441/im:/070123/ids_photos_wl/r2207709100.jpg
A particle moving along a curve other than a straight line
is said to be in curvilinear motion.
31
Curvilinear Motion: Position, Velocity & Acceleration
• 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)
Velocity is tangent to path
32
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.
33
Rectangular Components of Velocity & Acceleration
• Position vector of particle P 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
34
Rectangular Components of Velocity &
Acceleration
0 0 0 0 00 x yx y z v v given
• Rectangular components are useful when
acceleration components can be integrated
independently, ex: motion of a projectile.
00 zagyaxa zyx
with initial conditions,
Therefore:
• Motion in horizontal direction is uniform.
• Motion in vertical direction is uniformly
accelerated.
• Motion of projectile could be replaced by two
independent rectilinear motions.
0 0
2
0 0
1
2
x x y y
x y
v v v v gt
x v t y v t gt
35
Example
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. Find
(a) the range, and (b) maximum height.
x
y
0v v at 21
0 0 2x x v t at
2 2
0 02v v a x x
Remember:
36
Example
Car A is traveling at a constant speed of 36 km/h. As A crosses
intersection, B starts from rest 35 m north of intersection and
moves with a constant acceleration of 1.2 m/s2. Determine the
speed, velocity and acceleration of B relative to A 5 seconds
after A crosses intersection.
37
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,
tt ee and
t t tde e e
t t te e de
tde dFrom geometry:
t nde d et
n
dee
d
d
tde
38
2 2
t n t n
dv v dv va e e a a
dt dt
Tangential and Normal Components
tevv
• With the velocity vector expressed as
the 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,
• 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.
39
rr re
Radial and Transverse Components • If particle position is given in polar coordinates,
we can 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
rr r
d dr dev re e r
dt dt dt • Particle velocity vector:
• Similarly, particle acceleration:
r
rr
r r
da re r e
dt
de dere r r e r e r
dt dt
d dre re r e r e r e
dt dt
rerr
• Particle position vector:
r r
dr dv e r e re r e
dt dt
2 2ra r r e r r e
40
Sample Problem 11.10
A motorist is traveling on curved section of highway at 60 mph. The
motorist applies brakes causing a constant deceleration.
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.
41
Sample Problem 11.10
ft/s66mph45
ft/s88mph60
SOLUTION:
• Calculate tangential and normal components of
acceleration.
2
22
2
s
ft10.3
ft2500
sft88
s
ft75.2
s 8
sft8866
D
D
va
t
va
n
t
• 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
42
Sample Problem 11.11
Determine the minimum radius of curvature of the trajectory
described by the projectile.
2
n
va
Recall:
2
n
v
a
Minimum , occurs for small v and large an
2
155.92480
9.81m
v is min and an is max
a
an
43
Sample Problem 11.12
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.
44
Sample Problem 11.12
SOLUTION:
• Evaluate time t for = 30o.
s 869.1rad524.030
0.15 2
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
45
Sample Problem 11.12
• 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
46
Sample Problem 11.12
• Evaluate acceleration with respect to arm.
Motion of collar with respect to arm is rectilinear
and defined by coordinate r.
2sm240.0 ra OAB
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