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1
Kinematics of Particles(Part 1)
Chapter 11
2
• 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.
Introduction
3
11.2 Position, Velocity and Acceleration
RECTILINEAR MOTION OF PARTICLES
• 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.
4
• 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+t,
t
xv
t
x
t
0lim
Average velocity
Instantaneous velocity
• From the definition of a derivative,
dt
dx
t
xv
t
0lim
e.g.,
2
32
312
6
ttdt
dxv
ttx
5
• Consider particle with velocity v at time t and v’ at t+t,
Instantaneous accelerationt
va
t
0lim
tdt
dva
ttv
dt
xd
dt
dv
t
va
t
612
312e.g.
lim
2
2
2
0
• 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.
6
• Consider particle with motion given by326 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
7
11.3 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)
8
• 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
1221
or or
11.3 Determination of the motion of a particle
9
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
• Acceleration given as a function of velocity, a = f(v):
10
Sample problem 11.1
The position of a particle which moves along a straight line is defined by the relation ,
where x is expected in meter and t in seconds. Determine
(a) the time at which the velocity will be zero(b) the position and distance traveled by the particle at that time (c) the acceleration of the particle at that time(d) the distance traveled by the particle from t= 4s to t= 6s
40156 23 tttx
11
Solution
x (m)
v (m/s)
a (m/s )2
t(s)
t(s)
t(s)
12
x (m)
v (m/s)
a (m/s )2
t(s)
t(s)
t(s)
Solution
13
Sample problem 11.2
A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20m above the ground. Knowing that the acceleration of the ball is constant and equal to 9.81 m/s downward,Determine (a) the velocity v and elevation y of the ball above the ground at any time t,(b) the highest elevation reached by the ball and the corresponding value of t,(c) the time when the ball will hit the ground and the corresponding velocity.
(d) Draw the v-t and y-t curves.
2
14
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
• Integrate twice to find v(t) and y(t).
Solutiona.Velocity and Elevation
15
• 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
22
s019.1s
m905.4s019.1
s
m10m20
s
m905.4
s
m10m20
y
ttty
m1.25y
b. Highest Elevation
16
• 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
c. Balls Hits the Ground
17
The brake mechanism used to reduce recoil in certain types of guns consists essentially of piston attached to the barrel and moving in a fixed cylinder filled with oil. As the barrel recoils with an initial velocity v0, the piston moves and oil is forced through orifices in the piston, causing the piston and the barrel to decelerate at a rate proportional to their velocity. that is, a = -kv
Determine v(t), x(t), and v(x).
Draw the corresponding motion curves.
kva
Sample problem 11.3
18
• Integrate a = dv/dt = -kv to find v(t).
ktv
tvdtk
v
dvkv
dt
dva
ttv
v
00
ln0
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
Solution
19
• 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
0
0 or v
tveevtv ktkt
ktek
vtx 10with
and
then
Solution
20
21
22
23
24
25
26
27
28
29
30
11.4 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
31
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
000
attvxx
attvxxdtatvdxatvdt
dx tx
x
020
2
020
221
2
constant00
xxavv
xxavvdxadvvadx
dvv
x
x
v
v
11.5 Uniformly Acceleration Rectilinear Motion
32
11.6 Motion of Several Particles
• 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
33
• 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
Dependent Motions
34
Sample Problem 11.4
A ball is thrown vertically upward from the 12 m level in an elevator shaft with an initial velocity of 18 m/s. At the same instant, an open-platform elevator passes the 5m level, moving upward with a constant velocity at 2 m/s.
Determine
(a) when and where the ball will hits the elevator
(b) the relative velocity of ball with respect to the elevator when the ball hits the elevator.
35
• Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion.
22
21
00
0
905.418m12
81.918
ttattvyy
tatvv
B
B
• Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion.
ttvyy
smv
EE
E
2m5
/2
0
(1)
(2)
(3)(4)
36
Ball hits elevatorWe first note that the same time t and the same origin O were used in writingthe equations of motion of both the ball and the elevator. We see from the figure that when the ball hits the elevator,
BE yy (5)
Substituting for yE and yB from (2) and (4) into (5), we have
2905.4181225 ttt
st 39.0 st 65.3and
Only the root t=3.65s corresponds to a time after the motion has begun,Substituting this value into (4), we have
myE 30.12)65.3(25
Elevation fro ground = 12.30m
37
The relative velocity of the ball with respect to the elevator is
ttvvv EBEB 81.9162)81.918(/
When the ball hits the elevator at time t= 3.65 s , we have
smv
v
EB
EB
/81.19
)65.3(81.916
/
/
The negative sign means that the ball is observed from the elevator to be moving in the negative sense (downward)
38
200 mm
Sample Problem 11.5
Collar A and block B are connected by a cable passing over three pulleys C,D, and E as shown.Pulleys C and E are fixed, while D is attached to a collar which is pulled downward with a constantvelocity of 75mm/s. At t=0, collar A starts moving downward from position K with constant accelerationand no initial velocity.Knowing that the velocity of collar A is 300 mm/s as it passes through point L, determine the change in elevation, the velocity and the acceleration of block B when collar A passes through L.
39
40
XA
41
42
43
44
45
46
47
48
49
50
THE END