1.Day 1 Rectilinera Motion Part 01.pdf

ENGINEERING
MECHANICS
“Scientists study the world as it is ;
engineers create the world that has
never been.”
Master Trainer, Lecturer & Counsellor
Co-founder & Director U-Turn International(PVT) LTD
B.Sc.Eng(Hons) Electrical & Electronic Engineer, AMIESL. MECSL
Eng.Ruwan Dissanayake

ENG. RUWAN DISSANAYAKE
B.Sc. Eng (Hons) Electrical & Electronic Engineer (Uni.Of Pera), AMIESL
Engineer Corporate Strategy – Sri Lanka Telecom PLC(Former).
Director U-turn international(PVT)LTD ,Lecturer MiET School of Engineering
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VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
CHAPTER
11
Kinematics of Particles

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
Vector Mechanics for Engineers: Dynamics
Tenth
Edition
Contents
11 - 2
Introduction
Rectilinear Motion: Position,
Velocity & Acceleration
Determination of the Motion of a
Particle
Sample Problem 11.2
Sample Problem 11.3
Uniform Rectilinear-Motion
Uniformly Accelerated Rectilinear-
Motion
Motion of Several Particles:
Relative Motion
Sample Problem 11.4
Motion of Several Particles:
Dependent Motion
Sample Problem 11.5
Graphical Solution of Rectilinear-
Motion Problems
Other Graphical Methods
Curvilinear Motion: Position, Velocity
& Acceleration
Derivatives of Vector Functions
Rectangular Components of Velocity
and Acceleration
Motion Relative to a Frame in
Translation
Tangential and Normal Components
Radial and Transverse Components
Sample Problem 11.10
Sample Problem 11.12

Tenth
Edition
2 - 3
Kinematic relationships are used to
help us determine the trajectory of a
golf ball, the orbital speed of a
satellite, and the accelerations
during acrobatic flying.

Tenth
Edition
Introduction
11 - 4
• Dynamics includes:
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.
Kinematics: study of the geometry of motion.
Relates displacement, velocity, acceleration, and time without reference
to the cause of motion.
Fthrust
Flift
Fdrag

Tenth
Edition
Introduction
11 - 5
• Particle kinetics includes:
• 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.

Tenth
Edition
Rectilinear Motion: Position, Velocity & Acceleration
11 - 6
• Rectilinear motion: particle moving
along a straight line
• Position coordinate: defined by
positive or negative distance 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.
• May be expressed in the form of a
function, e.g., 3
2
6 t
t
x −
=
or in the form of a graph x vs. t.

Tenth
Edition
11 - 7
• 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
x
v
t
x
t D
D
=
=
D
D
=
→
D 0
lim
Average velocity
Instantaneous velocity
• From the definition of a derivative,
dt
dx
t
x
v
t
=
D
D
=
→
D 0
lim
e.g.,
2
3
2
3
12
6
t
t
dt
dx
v
t
t
x
−
=
=
−
=

Tenth
Edition
11 - 8
• Consider particle with velocity v at time t and
v’ at t+Dt,
Instantaneous acceleration
t
v
a
t D
D
=
=
→
D 0
lim
t
dt
dv
a
t
t
v
dt
x
d
dt
dv
t
v
a
t
6
12
3
12
e.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.

Tenth
Edition
Concept Quiz
2 - 9
What is true about the kinematics of a particle?
a) The velocity of a particle is always positive
b) The velocity of a particle is equal to the slope of
the position-time graph
c) If the position of a particle is zero, then the
velocity must zero
d) If the velocity of a particle is zero, then its
acceleration must be zero

Tenth
Edition
11 - 10
• From our example,
3
2
6 t
t
x −
=
2
3
12 t
t
dt
dx
v −
=
=
t
dt
x
d
dt
dv
a 6
12
2
2
−
=
=
=
- 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
• What are x, v, and a at t = 2 s ?
• Note that vmax occurs when a=0, and that the
slope of the velocity curve is zero at this point.
• What are x, v, and a at t = 4 s ?

Tenth
Edition
Eng. Ruwan Dissanayake B.Sc. Eng (Hons) Elcectrical & Electronic Engineer
11 - 13

Tenth
Edition
Determination of the Motion of a Particle
11 - 11
• We often determine accelerations from the forces applied
(kinetics will be covered later)
• Generally have three classes of motion
- 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)
• Can you think of a physical example of when force is a
function of position? When force is a function of velocity?
a spring drag

Tenth
Edition
Acceleration as a function of time, position, or velocity
2 - 12
( )
a a t
= ( )
0 0
v t
v
dv a t dt
=
 
( )
dv
a t
dt
=
( )
vdv a x dx
=
( )
0 0
v x
v x
vdv a x dx
=
 
( )
a a x
=
and
dx dv
dt a
v dt
= =
( )
dv
v a v
dx
=
( )
0 0
v t
v
dv
dt
a v
=
 
( )
0 0
x v
x v
vdv
dx
a v
=
 
( )
a a v
=
( )
dv
a v
dt
=
If…. Kinematic relationship Integrate

Tenth
Edition
Sample Problem 11.2
11 - 13
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:
• Integrate twice to find v(t) and y(t).
• Solve for t when velocity equals zero
(time for maximum elevation) and
evaluate corresponding altitude.
• Solve for t when altitude equals zero
(time for ground impact) and evaluate

Tenth
Edition
Sample Problem 11.2
11 - 14
( )
( ) t
v
t
v
dt
dv
a
dt
dv
t
t
v
v
81
.
9
81
.
9
s
m
81
.
9
0
0
2
0
−
=
−
−
=
−
=
=


( ) t
t
v 





−
= 2
s
m
81
.
9
s
m
10
( )
( ) ( ) 2
2
1
0
0
81
.
9
10
81
.
9
10
81
.
9
10
0
t
t
y
t
y
dt
t
dy
t
v
dt
dy
t
t
y
y
−
=
−
−
=
−
=
=


( ) 2
2
s
m
905
.
4
s
m
10
m
20 t
t
t
y 





−






+
=
SOLUTION:
• Integrate twice to find v(t) and y(t).

Tenth
Edition
Sample Problem 11.2
11 - 15
• Solve for t when velocity equals zero and evaluate
corresponding altitude.
( ) 0
s
m
81
.
9
s
m
10 2
=






−
= t
t
v
s
019
.
1
=
t
• Solve for t when altitude equals zero and evaluate
( )
( ) ( )2
2
2
2
s
019
.
1
s
m
905
.
4
s
019
.
1
s
m
10
m
20
s
m
905
.
4
s
m
10
m
20






−






+
=






−






+
=
y
t
t
t
y
m
1
.
25
=
y

Tenth
Edition
Sample Problem 11.2
11 - 16
• Solve for t when altitude equals zero and evaluate
( ) 0
s
m
905
.
4
s
m
10
m
20 2
2
=






−






+
= t
t
t
y
( )
s
28
.
3
s
meaningles
s
243
.
1
=
−
=
t
t
( )
( ) ( )
s
28
.
3
s
m
81
.
9
s
m
10
s
28
.
3
s
m
81
.
9
s
m
10
2
2






−
=






−
=
v
t
t
v
s
m
2
.
22
−
=
v

Tenth
Edition
11 - 20

Tenth
Edition
11 - 21

Tenth
Edition
11 - 22

Tenth
Edition
Sample Problem 11.3
11 - 17
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).
kv
a −
=
SOLUTION:
• 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).

Tenth
Edition
Sample Problem 11.3
11 - 18
SOLUTION:
• Integrate a = dv/dt = -kv to find v(t).
( )
0 0
0
ln
v t
v
v t
dv dv
a kv k dt kt
dt v v
= = − = − = −
 
( ) kt
e
v
t
v −
= 0
• Integrate v(t) = dx/dt to find x(t).
( )
( )
0
0 0
0
0 0
1
kt
t
x t
kt kt
dx
v t v e
dt
dx v e dt x t v e
k
−
− −
= =
 
= = −
 
 
 
( ) ( )
kt
e
k
v
t
x −
−
= 1
0

Tenth
Edition
Sample Problem 11.3
11 - 19
• Integrate a = v dv/dx = -kv to find v(x).
kx
v
v
dx
k
dv
dx
k
dv
kv
dx
dv
v
a
x
v
v
−
=
−
−
=
−
=
−
=
= 

0
0
0
kx
v
v −
= 0
• Alternatively,
( ) ( )








−
=
0
0 1
v
t
v
k
v
t
x
kx
v
v −
= 0
( ) ( )
0
0 or
v
t
v
e
e
v
t
v kt
kt
=
= −
−
( ) ( )
kt
e
k
v
t
x −
−
= 1
0
with
and
then

Tenth
Edition
Group Problem Solving
2 - 20
A bowling ball is dropped from a boat so that it
strikes the surface of a lake with a speed of 15 ft/s.
Assuming the ball experiences a downward
acceleration of a =10 - 0.01v2 when in the water,
determine the velocity of the ball when it strikes the
bottom of the lake.
Which integral should you choose?
( )
0 0
v t
v
dv a t dt
=
  ( )
0 0
v x
v x
vdv a x dx
=
 
( )
0 0
v t
v
dv
dt
a v
=
 
( )
0 0
x v
x v
vdv
dx
a v
=
 
(a)
(b)
(c)
(d)
+y

Tenth
Edition
Concept Question
2 - 21
When will the bowling ball start slowing down?
A bowling ball is dropped from a boat so that it
strikes the surface of a lake with a speed of 15 ft/s.
Assuming the ball experiences a downward
acceleration of a =10 - 0.01v2 when in the water,
determine the velocity of the ball when it strikes the
bottom of the lake.
The velocity would have to be high
enough for the 0.01 v2 term to be bigger
than 10
+y

Tenth
Edition
11 - 22
SOLUTION:
• Determine the proper kinematic
relationship to apply (is acceleration
a function of time, velocity, or
position?
• Determine the total distance the car
travels in one-half lap
• Integrate to determine the velocity
after one-half lap
The car starts from rest and accelerates
according to the relationship
2
3 0.001
a v
= −
It travels around a circular track that has
a radius of 200 meters. Calculate the
velocity of the car after it has travelled
halfway around the track. What is the
car’s maximum possible speed?

Tenth
Edition
2 - 23
( )
dv
v a v
dx
=
( )
0 0
x v
x v
vdv
dx
a v
=
 
Choose the proper kinematic relationship
Given: 2
3 0.001
a v
= −
vo = 0, r = 200 m
Find: v after ½ lap
Maximum speed
Acceleration is a function of velocity, and
we also can determine distance. Time is not
involved in the problem, so we choose:
Determine total distance travelled
3.14(200) 628.32 m
x r

= = =

Tenth
Edition
2 - 24
( )
0 0
x v
x v
vdv
dx
a v
=
 
Determine the full integral, including limits
628.32
2
0 0
3 0.001
v
v
dx dv
v
=
−
 
Evaluate the interval and solve for v
2
0
1
628.32 ln 3 0.001
0.002
v
v
 
= − −
 
 
2
628.32( 0.002) ln 3 0.001 ln 3 0.001(0)
v
 
− = − − −
 
2
ln 3 0.001 1.2566 1.0986= 0.15802
v
 
− = − + −
 
2 0.15802
3 0.001v e−
− =
Take the exponential of each side

Tenth
Edition
2 - 25
2 0.15802
3 0.001v e−
− =
Solve for v
0.15802
2 3
2146.2
0.001
e
v
−
−
= = 46.3268 m/s
v =
How do you determine the maximum speed the car can reach?
2
3 0.001
a v
= −
Velocity is a maximum when
acceleration is zero
This occurs when 2
0.001 3
v =
3
0.001
max
v = max 54.772 m/s
v =

10 CHAPTER 12 KINEMATICS OF A PARTICLE
12
(© R.C. Hibbeler)
The car on the left in the photo and in Fig. 12–2 moves in a straight
line such that for a short time its velocity is defined by
v = (3t2
+ 2t) ft>s, where t is in seconds. Determine its position and
acceleration when t = 3 s. When t = 0, s = 0.
s
O
a, v
Fig. 12–2
EXAMPLE 12.1
SOLUTION
Coordinate System. The position coordinate extends from the fixed
origin O to the car, positive to the right.
Position. Since v = f(t), the car’s position can be determined from
v = ds>dt, since this equation relates v, s, and t. Noting that s = 0
when t = 0, we have*
( S
+ ) v =
ds
dt
= (3t2
+ 2t)
L
s
0
ds =
L
t
0
(3t2
+ 2t)dt
s `
s
0
= t3
+ t2
`
t
0
s = t3
+ t2
When t = 3 s,
s = (3)3
+ (3)2
= 36 ft Ans.
Acceleration. Since v = f(t), the acceleration is determined from
a = dv>dt, since this equation relates a, v, and t.
( S
+ ) a =
dv
dt
=
d
dt
(3t2
+ 2t)
= 6t + 2
When t = 3 s,
a = 6(3) + 2 = 20 ft>s2 S Ans.
NOTE: The formulas for constant acceleration cannot be used to solve
this problem, because the acceleration is a function of time.
*The same result can be obtained by evaluating a constant of integration C rather
than using definite limits on the integral. For example, integrating ds = (3t2
+ 2t)dt
yields s = t3
+ t2
+ C. Using the condition that at t = 0, s = 0, then C = 0.

12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 11
12
A small projectile is fired vertically downward into a fluid medium with
an initial velocity of 60 m>s. Due to the drag resistance of the fluid the
projectile experiences a deceleration of a = (-0.4v3
) m>s2
,where v is in
m>s. Determine the projectile’s velocity and position 4 s after it is fired.
SOLUTION
Coordinate System. Since the motion is downward, the position
coordinate is positive downward, with origin located at O, Fig. 12–3.
Velocity. Here a = f(v) and so we must determine the velocity as a
function of time using a = dv>dt, since this equation relates v, a, and t.
(Why not use v = v0 + act?) Separating the variables and integrating,
with v0 = 60 m>s when t = 0, yields
(+ T) a =
dv
dt
= -0.4v3
L
v
60 m>s
dv
-0.4v3
=
L
t
0
dt
1
-0.4
a
1
-2
b
1
v2
`
60
v
= t - 0
1
0.8
c
1
v2
-
1
(60)2
d = t
v = e c
1
(60)2
+ 0.8td
-1>2
f m>s
Here the positive root is taken, since the projectile will continue to
move downward.When t = 4 s,
v = 0.559 m>sT Ans.
Position. Knowing v = f(t), we can obtain the projectile’s position
from v = ds>dt, since this equation relates s, v, and t. Using the initial
condition s = 0, when t = 0, we have
(+ T) v =
ds
dt
= c
1
(60)2
+ 0.8td
-1>2
L
s
0
ds =
L
t
0
c
1
(60)2
+ 0.8td
-1>2
dt
s =
2
0.8
c
1
(60)2
+ 0.8td
1>2
`
0
t
s =
1
0.4
e c
1
(60)2
+ 0.8td
1>2
-
1
60
f m
When t = 4 s,
s = 4.43 m Ans.
EXAMPLE 12.2
s
O
Fig. 12–3

12.2 RECTILINEAR KINEMATICS: CONTINUOUS MOTION 13
12
A metallic particle is subjected to the influence of a magnetic field as
it travels downward through a fluid that extends from plate A to
plate B,Fig.12–5.If the particle is released from rest at the midpoint C,
s = 100 mm, and the acceleration is a = (4s) m>s2
, where s is in
meters, determine the velocity of the particle when it reaches plate B,
s = 200 mm, and the time it takes to travel from C to B.
SOLUTION
Coordinate System. As shown in Fig. 12–5, s is positive downward,
measured from plate A.
Velocity. Since a = f(s), the velocity as a function of position can
be obtained by using v dv = a ds. Realizing that v = 0 at s = 0.1 m,
we have
(+ T) v dv = a ds
L
v
0
v dv =
L
s
0.1 m
4s ds
1
2
v2
`
0
v
=
4
2
s2
`
0.1 m
s
v = 2(s2
- 0.01)1>2
m>s (1)
At s = 200 mm = 0.2 m,
vB = 0.346 m>s = 346 mm>s T Ans.
The positive root is chosen since the particle is traveling downward,
i.e., in the +s direction.
Time. The time for the particle to travel from C to B can be obtained
using v = ds>dt and Eq. 1, where s = 0.1 m when t = 0. From
Appendix A,
(+ T) ds = v dt
= 2(s2
- 0.01)1>2
dt
L
s
0.1
ds
(s2
- 0.01)1>2
=
L
t
0
2 dt
ln12s2
- 0.01 + s2 `
0.1
s
= 2t `
0
t
ln12s2
- 0.01 + s2 + 2.303 = 2t
At s = 0.2 m,
t =
ln12(0.2)2
- 0.01 + 0.22 + 2.303
2
= 0.658 s Ans.
NOTE: The formulas for constant acceleration cannot be used here
because the acceleration changes with position, i.e., a = 4s.
EXAMPLE 12.4
A
200 mm
100 mm
B
s
C
Fig. 12–5

12
A particle moves along a horizontal path with a velocity of
v = (3t2
- 6t) m>s, where t is the time in seconds. If it is initially
located at the origin O, determine the distance traveled in 3.5 s, and the
particle’s average velocity and average speed during the time interval.
SOLUTION
Coordinate System. Here positive motion is to the right, measured
from the origin O, Fig. 12–6a.
Distance Traveled. Since v = f(t), the position as a function of time
may be found by integrating v = ds>dt with t = 0, s = 0.
( S
+ ) ds = v dt
= (3t2
- 6t) dt
L
s
0
ds =
L
t
0
(3t2
- 6t) dt
s = (t3
- 3t2
) m (1)
In order to determine the distance traveled in 3.5 s, it is necessary
to investigate the path of motion. If we consider a graph of the
velocity function, Fig. 12–6b, then it reveals that for 0 6 t 6 2 s the
velocity is negative, which means the particle is traveling to the left,
and for t 7 2 s the velocity is positive, and hence the particle is
traveling to the right. Also, note that v = 0 at t = 2 s. The particle’s
position when t = 0, t = 2 s, and t = 3.5 s can be determined from
Eq. 1.This yields
s兩 t=0 = 0 s兩 t=2 s = -4.0 m s兩 t=3.5 s = 6.125 m
The path is shown in Fig. 12–6a. Hence, the distance traveled in 3.5 s is
sT = 4.0 + 4.0 + 6.125 = 14.125 m = 14.1 m Ans.
Velocity. The displacement from t = 0 to t = 3.5 s is
s = s兩 t=3.5 s - s兩 t=0 = 6.125 m - 0 = 6.125 m
and so the average velocity is
vavg =
s
t
=
6.125 m
3.5 s - 0
= 1.75 ms S Ans.
The average speed is defined in terms of the distance traveled sT . This
positive scalar is
(vsp)avg =
sT
t
=
14.125 m
3.5 s - 0
= 4.04 ms Ans.
NOTE: In this problem, the acceleration is a = dvdt = (6t - 6) ms2
,
which is not constant.
EXAMPLE 12.5
O
s 4.0 m s 6.125 m
t 2 s t 0 s t 3.5 s
(a)
(0, 0)
v (m/s)
v 3t2
6t
(2 s, 0)
t (s)
(1 s, 3 m/s)
(b)
Fig. 12–6

12 12.3 Rectilinear Kinematics: Erratic
Motion
When a particle has erratic or changing motion then its position, velocity,
and acceleration cannot be described by a single continuous mathematical
function along the entire path. Instead, a series of functions will be
required to specify the motion at different intervals. For this reason, it is
convenient to represent the motion as a graph. If a graph of the motion
that relates any two of the variables s,v, a, t can be drawn, then this graph
can be used to construct subsequent graphs relating two other variables
since the variables are related by the differential relationships v = dsdt,
a = dvdt, or a ds = v dv. Several situations occur frequently.
The s–t, v–t, and a–t Graphs. To construct the v9t graph given
the s–t graph, Fig. 12–7a, the equation v = dsdt should be used, since it
relates the variables s and t to v.This equation states that
ds
dt
= v
slope of
s9t graph
= velocity
For example, by measuring the slope on the s–t graph when t = t1, the
velocity is v1, which is plotted in Fig. 12–7b. The v9t graph can be
constructed by plotting this and other values at each instant.
The a–t graph can be constructed from the v9t graph in a similar
manner, Fig. 12–8, since
dv
dt
= a
slope of
v9t graph
= acceleration
Examples of various measurements are shown in Fig. 12–8a and plotted
in Fig. 12–8b.
If the s–t curve for each interval of motion can be expressed by a
mathematical function s = s(t), then the equation of the v9t graph for
the same interval can be obtained by differentiating this function with
respect to time since v = ds/dt. Likewise, the equation of the a–t graph
for the same interval can be determined by differentiating v = v(t) since
a = dvdt. Since differentiation reduces a polynomial of degree n to that
of degree n – 1, then if the s–t graph is parabolic (a second-degree curve),
the v9t graph will be a sloping line (a first-degree curve), and the
a–t graph will be a constant or a horizontal line (a zero-degree curve).
t
O
v0 t 0
(a)
s
ds
dt
v1 t1
s1
t1 t2 t3
s2
s3
ds
dt
v2 t2
ds
dt
v3 t3
ds
dt
t
O
(b)
v0
v
v1
v3
v2
t1 t2
t3
Fig. 12–7
a0
v
t
t1 t2 t3
v1
v2
v3
v0
a1
a2
O
(a)
a3 t3
dv
dt
t2
dv
dt
t 0
dv
dt
t1
dv
dt
t
a
a0 0
a1 a2
a3
t1 t2 t3
O
(b)
Fig. 12–8

1.Day 1 Rectilinera Motion Part 01.pdf

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1.Day 1 Rectilinera Motion Part 01.pdf