Upload
hoangkhanh
View
509
Download
11
Embed Size (px)
Citation preview
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
1 - 32
Introduction to the course
• Scope
• Key concepts
Dynamics
Particles
Rigid body
Deal
with Motions under
applied force
Kinematics
Kinetics
Particle A moving body, on which two arbitrary points keeps fixed relative
position. (A geometric point with mass. The position of an arbitrary
point represents the whole body’s position)
Rigid body A moving body, on which the relative position of two arbitrary points
changes.
Kinematics Geometry of motion (The relation between displacement, velocity,
acceleration). Don’t refer to the cause of motion.
Kinetics (1) Relation between motion and applied forces.
(2) To predict the motion by given forces or to determine the forces to
produce a given motion
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
Dr. Gangyi Zhou
UC Irvine
Jul 3rd, 2006
CHAPTER
11 Kinematics of Particles
Part 1: Rectilinear motion
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
3 - 32
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 (Displacement) 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.
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
4 - 32
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
06
ttdt
dxv
ttx
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
5 - 32
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,
• Instantaneous acceleration may be:
- positive: increasing positive velocity
or decreasing negative velocity
- negative: decreasing positive velocity
or increasing negative velocity.
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
6 - 32
Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle with motion given by
326 ttx
2312 ttdt
dxv
tdt
xd
dt
dva 612
2
2
Dis
pla
cem
ent
Vel
oci
ty
Acc
eler
atio
n
Po
siti
on
0 7 7 -40
20
0 40
20
40
-30
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
7 - 32
Concept comparison
Displacement
Velocity
Acceleration
Average
velocity
Average
acceleration
Instant Course
X
X
X
X
X
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
8 - 32
Determination of the Motion of a Particle
Motion of
a particle
Acceleration
experienced
by the particle
a function
of time: a = f(t)
a function
of displacement:
a = f(x)
a function
of velocity:
a = f(v)
Displacement
function
Expressed by
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
9 - 32
Determination of the Motion of a Particle
• Acceleration given as a function of time, a = f(t):
tfadt
dvBy the definition of acceleration: dtdt
yields: dttfdv
Integrating both sides yields:
ttv
v
dfdv00
0vtv
So, velocity can be expressed by:
t
dfvtv0
0
Because of , by the same procedure, we can get: dt
dxv
t
dvxtx0
0
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
10 - 32
Determination of the Motion of a Particle
• Acceleration given as a function of position, a = f(x):
By the definition of acceleration: xfadt
dv
dt
dx
dx
dv
=v
yields: xfdx
dvv
Multiplying dx to both sides, yields: dxxfvdv
Integrating both sides, we get:
x
x
xv
xv
xdxfvdv
00
~~ 2
0
2
2
1
2
1xvxv
So, velocity can be expressed by:
x
xdxfxvxv0
2
0
2 ~~2
Not uniquely determined
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
11 - 32
Determination of the Motion of a Particle
• Acceleration given as a function of velocity, a = f(v):
By the definition of acceleration: vfadt
dv
vfdt
1
dt
vf
1
Integrating both sides, yields
yields:
dtvf
dv
tv
v
t
dvf
dv
0 0
t
For the displacement, because of , we have : vfadx
dvv
dx
vf
vdv
Integrating both sides, yields:
000
xtxdxvf
vdvtx
x
tv
v
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
12 - 32
t
dfvtv0
0
Determination of the Motion of a Particle: summary
a = f(t):
a = f(x):
a = f(v)
Velocity Displacement
NA
t
dvxtx0
0
tv
vvf
vdvxtx
0
0
x
xdxfxvxv0
2
0
2 ~~2
Not uniquely determined
tv
v
t
dvf
dv
0 0
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
13 - 32
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.
SOLUTION: (a problem of a=f(t))
• 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.
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
14 - 32
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).
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
15 - 32
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
• Solve for t at which altitude equals zero and evaluate
corresponding velocity.
22
2
2
s019.1s
m905.4s019.1
s
m10m20
s
m905.4
s
m10m20
y
ttty
m1.25y
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
16 - 32
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
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
17 - 32
Sample Problem 11.3
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).
kva
SOLUTION: (a problem of a=f(v))
• 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).
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
18 - 32
Sample Problem 11.3
SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
kt
v
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
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
19 - 32
Sample Problem 11.3
• 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
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
20 - 32
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
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
21 - 32
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
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
22 - 32
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 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
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
23 - 32
Sample Problem 11.4
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.
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
24 - 32
Sample Problem 11.4 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
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
25 - 32
Sample Problem 11.4
• 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.
65.325Ey
m3.12Ey
65.381.916
281.918
tv EB
s
m81.19EBv
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
26 - 32
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
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
27 - 32
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.
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.
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
28 - 32
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
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
29 - 32
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
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
30 - 32
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
s
in.18Bv
0s
in.9
02
2
B
BDA
v
aaa
2s
in.9Ba
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
31 - 32
Motion
determination
Uniform
rectilinear
motion and
Uniform
accelerated
motion
Particle and
rigid body
Relative
motion and
Dependent
motions
Course summary
Position
(displacement)
, velocity, and
acceleration
Vector Mechanics for Engineers: Dynamics
Se
ve
nth
E
ditio
n
32 - 32
Homework
• 11.5(7th) or 11.3(8th);
• 11.20(7th) or 11.21(8th);
• 11.21(7th) or 11.25(8th);
• 11.39(7th) or 11.43(8th),
• 11.45(7th) or 11.47(8th),
• 11.48(7th) or 11.50(8th),
Due, next Wednesday, 07/12/2006!