1. No category

Chapter 11
Kinematics of Particles
(Part 2)
1
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.
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• 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.
3
11.8 Other Graphical Methods
• Moment-area method to determine particle position at
time t directly from the a-t curve:
x1  x0  area under v  t curve
v1
 v0t1   t1  t dv
v0
using dv = a dt ,
v1
x1  x0  v0t1   t1  t  a dt
v0
v1
 t1  t  a dt  first moment of area under a-t curve
v0
with respect to t = t1 line.
x1  x0  v0t1  area under a-t curve t1  t 
t  abscissa of centroid C
4
• Method to determine particle acceleration
from v-x curve:
dv
dx
 AB tan
 BC  subnormal to v-x curve
av
5
Sample Problem 11.6
2
A subway car leaves station A; it gains speed at the rate of 1.2 m/s for 6 s and then
2
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.
6
Solution
Acceleration Time Curve Since the acceleration is either constant or zero,
the a-t curve is made of horizontal straight-line
segments. The values of t2 and a4 are determine as follows :
The acceleration being negative, the corresponding area is below the t-axis
this area represents a decrease in velocity.
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Solution
Velocity Time Curve Since the acceleration is either constant or zero, the v-t curve
is made of straight line segments connecting the points determined above.
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.
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PROBLEMS
9
10
Solution
a=( m/s)
2
2
A2= 8m/s
6
0
-2
A1= -12m/s
10
14
t(s)
11
v=( m/s)
8
4
A7
A3
4
6
10
t(s)
0
12
A5
-4
A4
14
A6
x(m)
16
12
10
-4
-8
4
12
14
t(s)
12
CURVILINEAR MOTION OF PARTICLES
11.9 Position Vector, Velocity, and 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 r at time t and P’ defined by r  at t + Dt,


Dr dr

v  lim

dt
Dt 0 Dt
 instantaneous velocity (vector)
Ds ds

dt
Dt 0 Dt
v  lim
 instantaneous speed (scalar)
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
• Consider velocity v of particle at time t and velocity

v  at t + Dt,


Dv d v

a  lim

dt
D t  0 Dt
 instantaneous acceleration (vector)
• In general, acceleration vector is not tangent to
particle path and velocity vector.
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11.10 Derivatives of Vector Functions

• Let Pu  be a vector function of scalar variable u,




dP
DP
Pu  Du   Pu 
 lim
 lim
du Du 0 Du Du 0
Du
• Derivative of vector sum,

 

d P  Q  dP dQ


du
du du
• Derivative of product of scalar and vector functions,



d  f P  df
dP

P f
du
du
du
• Derivative of scalar product and vector product,

 


 dQ
d P  Q  dP

Q  P
du
du
du

 



d P  Q  dP
dQ

Q  P
du
du
du
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11.11 Rectangular Components of Velocity And Acceleration
• When position vector of particle P is given by its
rectangular components,




r  xi  y j  zk
• Velocity vector,



 dx  dy  dz 
v  i  j  k  xi  y j  zk
dt
dt
dt



 vx i  v y j  vz k
• Acceleration vector,



 d 2 x d 2 y  d 2 z 
a  2 i  2 j  2 k  xi  y j  zk
dt
dt
dt



 ax i  a y j  az k
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• Rectangular components particularly effective
when component accelerations can be integrated
independently, e.g., motion of a projectile,
a x  x  0
a y  y   g
a z  z  0
with initial conditions,
v x 0 , v y  , v z 0  0
x0  y 0  z 0  0
0
Integrating twice yields
v x  v x 0
x  v x 0 t
v y  v y   gt
0
y  v y  y  12 gt 2
0
vz  0
z0
• 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 rA and rB .

r
• Vector B A joining A and B defines the position of
B with respect to the moving frame Ax’y’z’ and

 
rB  rA  rB A
• Differentiating twice,




vB  v A  vB A vB A  velocity of B relative to A.




a B  a A  a B A a B A  acceleration of B relative
to A.
• 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.
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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 30º
with 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 Motion
Choosing the positive sence of the y axis upward and placing the origin O
at the gun, we have
Substituting into the equations of uniformly accelerated motion, we have
Horizontal Motion. Uniform Motion
Choosing the positive sense of the x axis to the right, we have
Substituting into the equation of uniform motion, we obtain
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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 Elevation
When 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
21
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
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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 distance
of 3658 m is obtained by setting x equal to 3658 m
Vertical Motion
Substituting into the equation of uniformly accelerated vertical motion,
we obtain
t
2
23
Solution
24
PROBLEMS
25
30º
26
27
28
Vo
h
13 m
50 m
3m
29
Vo
h
13 m
50 m
3m
30
Vo
h
13 m
50 m
3m
31
32
33
34
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.


• et and et are tangential unit vectors for the
particle path at P and P’. When drawn with
  
respect to the same origin, Det  et  et and
D is the angle between them.
Det  2 sin D 2 

Det
sin D 2  

lim
 lim
en  en
D 0 D
D 0 D 2

det

en 
d
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

• With the velocity vector expressed as v  vet
the particle acceleration may be written as



de dv 
de d ds
 dv dv 
a
 et  v
 et  v
dt dt
dt dt
d ds dt
but 
det 
ds
 en
 d  ds
v
d
dt
After substituting,
dv
v2
 dv  v 2 
a  et  en
at 
an 
dt

dt

• 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.
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• Relations for tangential and normal acceleration
also apply for particle moving along space curve.
 dv  v 2 
a  et  en
dt

dv
at 
dt
an 
v2

• Plane containing tangential and normal unit
vectors is called the osculating plane.
• Normal to the osculating plane is found from

 
eb  et  en

en  principalnormal

eb  binormal
• Acceleration has no component along binormal.
37
11.14 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.


r  re r

der 
 e
d
• The particle velocity vector is

der dr 
dr 
d 
 d 
v  rer   er  r
 er  r
e
dt
dt
dt
dt
dt


 r er  r e

de

  er
d


der der d  d

 e
dt
d dt
dt


de de d
 d

  er
dt
d dt
dt
• Similarly, the particle acceleration vector is
d  
 d  dr 
a   er  r
e 
dt  dt
dt 


d 2 r  dr der dr d 
d 2 
d de
 2 er 

e  r 2 e  r
dt dt dt dt
dt dt
dt
dt


 r  r 2 er  r  2r e


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• When particle position is given in cylindrical
coordinates, it is convenient to express the
velocity and acceleration
vectors using the unit

 
vectors eR , e , and k .
• Position vector,



r  R e R z k
• Velocity vector,


 dr  


v
 R eR  R e  z k
dt
• Acceleration vector,


 dv


2




  R eR  R  2 R  e  z k
a
 R
dt


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Sample Problem 11.10
VA = 96 km/h
762 m
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.
40
Tangential Component of Acceleration. First the speeds are expressed in m/s
96
km 96 x1000 m

 26.7 m/s
h
3600 s
72
km 72 x1000 m

 20m/s
h
3600 s
VA = 96 km/h
762 m
Since the automobile slows down at a constant rate, we have
at  average at 
at  0.8375m/s 2
Dv 20m / s  26.7m / s

 0.8375m / s 2
Dt
8s
Normal component of acceleration. Immediately after the brakes have been
applied, the speed is still 26.7 m/s, and we have
an 
v2


26.7m/s
 0.94m/s 2
762m
an  0.94m/s 2
41
Magnitude and Direction of Acceleration.
The magnitude and direction of the resultant a of the components
an and at are
VA = 96 km/h
an
0.94m / s 2
tan  

 1.1223
at 0.8375m / s 2
  48.3
762 m
at  0.8375m/s 2
an
0.94m / s 2
a

 1.26m / s 2
sin 
sin 48.3
an  0.94m/s 2
42
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.
43
Time t at which θ=30º. Substituting θ=30º =0.524 rad into the expression for θ,
we obtain
  0.15t 2
0.524  0.15t 2
t  1.869s
Equation of Motion. Substituting t=1.869 s in the expressions for r,θ,and their
first and second derivatives, we have
r  0.9  0.12t 2  0.481m

  0.15t 2  0.524rad

r  0.24t  0.449m / s
  0.30t  0.561rad / s


r  0.24  0.240m / s
a. Velocity of B
2
  0.30  0.300rad / s 2
We obtain the values of vr and vθwhen t = 1.869 s

vr  r  0.449m / s

v  r   0.481(0.561)  0.270m / s
v  (vr ) 2  (v ) 2
v  0.524m / s
  31.0
44
b. Acceleration of B.
We obtain
2
ar  r  r 
 0.240  0.481(0.561) 2  0.391m / s 2

 
a  r   2 r 
 0.481(0.300)  2(0.449)(0.561)  0.359m / s 2
c. Acceleration of B with Respect to Arm OA.. We note that the motion
of the collar with respect to the arm is rectilinear and defined by the coordinate r.
We write

aB / OA  r  0.240m / s 2
aB / OA  0.240m / s 2
towards O
45
PROBLEMS
46
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 speeds
of both cars are constant and that the relative acceleration of car A with respect to car B is 0.25 m/s 2 directed
toward the center of curvature,
determine
(a) the speed of car A
(b) the speed of car B
47
aB
vB
vA
aA
48
49
50
Problem 11.142
91.4 m
137.2 m
51
52
53
54
55
56
57
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THE END
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