1 Kinematics of Particles (Part 2) Chapter 11. 2 11.7 Graphical Solution of Rectilinear Motion...
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Transcript of 1 Kinematics of Particles (Part 2) Chapter 11. 2 11.7 Graphical Solution of Rectilinear Motion...
1
Kinematics of Particles(Part 2)
Chapter 11
2
11.7 Graphical Solution of Rectilinear Motion Problem
• 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.
3
• 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.
4
11.8 Other Graphical Methods
• Moment-area method to determine particle position at time t directly from the a-t curve:
1
0
110
01 curve under areav
v
dvtttv
tvxx
using dv = a dt ,
1
0
11001
v
v
dtatttvxx
1
0
1
v
v
dtatt first moment of area under a-t curve with respect to t = t1 line.
Ct
tta-ttvxx
centroid of abscissa
curve under area 11001
5
• Method to determine particle acceleration from v-x curve:
BC
ABdx
dvva
tan
subnormal to v-x curve
6
Sample Problem 11.6
A subway car leaves station A; it gains speed at the rate of 1.2 m/s for 6 s and then at the rate of 1.8 m/s until it has reached the speed of 14.6 m/s.The car maintains the same speed until it approaches station B ; brakes are then applied, giving the car a constant deceleration and bringing it to a stop in 6 s.The total running time for A to B is 40 s.Draw the a-t, v-t , and x-t curves, and determine the distance stations A and B.
2
2
7
Solution
Acceleration Time Curve Since the acceleration is either constant or zero, the a-t curve is made of horizontal straight-linesegments. The values of t2 and a4 are determine as follows :
The acceleration being negative, the corresponding area is below the t-axisthis area represents a decrease in velocity.
8
Velocity Time Curve Since the acceleration is either constant or zero, the v-t curveis made of straight line segments connecting the points determined above.
Solution
Position Time Curve The points determined above should be joined by three arcs ofvparabola and one straight line segment. In constructing the x-t curve is equal to the value of v at that instant.
9
PROBLEMS
10
11
Solution
A1= -12m/s
A2= 8m/s
a=( m/s)
t(s)
2
6
10 14
-2
0
2
12
v=( m/s)
t(s)
4
-4
8
0
A3
A4
A5
A6
A7
12
6 104
14
16
12
4-8
-4
10 12 14
t(s)
x(m)
13
11.9 Position Vector, Velocity, and Acceleration
CURVILINEAR MOTION OF PARTICLES
• 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 + t, r
r
dt
ds
t
sv
dt
rd
t
rv
t
t
0
0
lim
lim
instantaneous velocity (vector)
instantaneous speed (scalar)
14
dt
vd
t
va
t
0
lim
instantaneous acceleration (vector)
• Consider velocity of particle at time t and velocity at t + t,
v
v
• In general, acceleration vector is not tangent to particle path and velocity vector.
15
11.10 Derivatives of Vector Functions
uP
• Let be a vector function of scalar variable u,
u
uPuuP
u
P
du
Pd
uu
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
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• 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
11.11 Rectangular Components of Velocity And Acceleration
17
• 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
02
21
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.
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11.12 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 Ax’y’z’ 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.
19
Sample Problem 11.7
A projectile is fired from the edge of a 150m cliff with an initial velocity of 180 m/s at an angle of 30ºwith the horizontal. Neglecting air resistance, find
(a) the horizontal distance from the gun to the point where the projectile strikes the ground.(b) the greatest elevation above the ground reached by the projectile.
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Solution
The vertical and the horizontal motion will be considered separately.
Vertical Motion . Uniformly Acceleration MotionChoosing the positive sence of the y axis upward and placing the origin Oat the gun, we have
Substituting into the equations of uniformly accelerated motion, we have
Horizontal Motion. Uniform MotionChoosing the positive sense of the x axis to the right, we have
Substituting into the equation of uniform motion, we obtain
21
a. Horizontal Distance When the projectile strikes the ground, we have
Carrying this value into Eq. (2) for the vertical motion, we write
t = 19.91 s
Carrying t=19.91 s into Eq. (4) for the horizontal motion, we obtain
x = 3100 m
b. Greatest ElevationWhen the projectile reaches its greatest elevation, we have Vy=0 ;carrying this value into Eq. (3) for the vertical motion, we write
Greatest Elevation above ground = 150 m + 143 m = 563 m
22
Sample Problem 11.8
A projectile is fired with an initial velocity of 224 m/s at a target B located 610 m above the gun A and at a horizontal distance of 3658 m. Neglecting air resistance, determine the value of the firing angle α
23
Solution
The horizontal and the vertical motion will be considered separately.Horizontal motion . Placing the origin of the coordinate axes at the gun,
we have
Substituting into the equation of uniform horizontal motion, we obtain
The time required for the projectile to move through a horizontal distanceof 3658 m is obtained by setting x equal to 3658 m
Vertical Motion
Substituting into the equation of uniformly accelerated vertical motion,we obtain
t2
24
Solution
25
PROBLEMS
26
30º
27
28
29
50 m
3 m
13 m
h
Vo
30
50 m
3 m
13 m
h
Vo
31
50 m
3 m
13 m
h
Vo
32
33
34
35
11.13 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
d
ede
eee
e
tn
nnt
t
2
2sinlimlim
2sin2
00
36
tevv• With the velocity vector expressed as
the particle acceleration may be written as
dt
ds
ds
d
d
edve
dt
dv
dt
edve
dt
dv
dt
vda tt
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.
37
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.
38
• 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
11.14 Radial and Transverse Components
39
• 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
22
40
Sample Problem 11.10
A motorist is traveling on a curved section of highway of radius 762 m at the speed of 96 km/h.The motorist suddenly applies the brake, causing the automobile to slow down at a constant rate.knowing that after 8s the speed has been reduced to 72 km/h, determine the acceleration of the automobile immediately after the brakes have been applied.
VA = 96 km/h
762 m
41
Tangential Component of Acceleration. First the speeds are expressed in m/s
VA = 96 km/h
762 m
m/s7.26s
m
3600
1000x96
h
km96
m/s20s
m
3600
1000x72
h
km72
Since the automobile slows down at a constant rate, we have
2/8375.08
/7.26/20sm
s
smsm
t
vaaveragea tt
Normal component of acceleration. Immediately after the brakes have beenapplied, the speed is still 26.7 m/s, and we have
22
m/s94.0m762
m/s7.26
v
an
2m/s8375.0ta
2m/s94.0na
42
VA = 96 km/h
762 m
2m/s8375.0ta
2m/s94.0na
Magnitude and Direction of Acceleration.The magnitude and direction of the resultant a of the components an and at are
1223.1/8375.0
/94.0tan
2
2
sm
sm
a
a
t
n
3.48
22
/26.13.48sin
/94.0
sinsm
smaa n
43
Sample Problem 11.12
The rotation of the 0.9m arm about OA about O is defined by = 0.15t2 where is expressed in radians and t in seconds. Collar B slides along the arm in such a way that its distance from O is r = 0.9 - 0.12t2 where r is expressed in meters and t in seconds.
After the arm OA has rotated through 30o, determine
(a) the total velocity of the collar,
(b) the total acceleration of the collar,
(c) the relative acceleration of the collar with respect to the arm.
44
Time t at which θ=30º. Substituting θ=30º =0.524 rad into the expression for θ,we obtain
Equation of Motion. Substituting t=1.869 s in the expressions for r,θ,and their first and second derivatives, we have
sttt 869.115.0524.015.0 22
2
2
/240.024.0
/449.024.0
481.012.09.0
smr
smtr
mtr
2
2
/300.030.0
/561.030.0
524.015.0
srad
sradt
radt
a. Velocity of B We obtain the values of vr and vθwhen t = 1.869 s
smrv
smrvr
/270.0)561.0(481.0
/449.0
22 )()( vvv r
0.31/524.0 smv
45
b. Acceleration of B. We obtain
2
rrar22 /391.0)561.0(481.0240.0 sm
rra 22/359.0)561.0)(449.0(2)300.0(481.0 sm
c. Acceleration of B with Respect to Arm OA.. We note that the motionof the collar with respect to the arm is rectilinear and defined by the coordinate r. We write
Otowardssma
smra
OAB
OAB
2/
2/
/240.0
/240.0
46
PROBLEMS
47
Problem 11.136
At the instant shown, race car A is passing race car B with a relative velocity of 1m/s. Knowing that the speedsof both cars are constant and that the relative acceleration of car A with respect to car B is 0.25 m/s directedtoward the center of curvature, determine
(a) the speed of car A(b) the speed of car B
2
48
Bv AvBa
Aa
49
50
51
Problem 11.142
137.2 m
91.4 m
52
53
54
55
56
57
58
59
THE END