Chapter 11 : Kinematics of Particles - สำนักวิชาวิศวกรรมศาสตร์

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Nov 14, 2013 (3 years and 6 months ago)

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Chapter 11 : Kinematics of Particles

โดย สาขาวิชาวิศวกรรมเครื่องกล ส านักวิชาวิศวกรรมศาสตร์

Engineering Dynamics

Introduction

Mechanics (
กลศาสตร์
)



เป็นสาขาหนึ่งของวิทยาศาสตร์กายภาพที่
ศึกษาและพิจารณาถึงสภาพของวัตถุเมื่อมีแรง
กระท าเป็นผลให้วัตถุนั


เคลื่อนที่
หรือ
หย

ดนิ่ง


Introduction

Mechanics
แบ

งออกเป็น
2


วน



1
. Statics



2
. Dynamics


Introduction

Statics (
สถิตยศาสตร์
)



พิจารณาถึงการสมดุลของวัตถุที่อยู



บที่
หรือ เคลื่อนที่ด้วยความเร็วคงที่ เมื่อมีแรงมา
กระท า




Introduction

Dynamics (
พลศาสตร์
)



พิจารณาถึงสภาพการเ คลื่อนที่ของวัตถุ
ภายใต้แรงกระท า


Introduction

Dynamics
แบ

งได้เป็น
2


วน



1
. Kinematics



2
. Kinetics




Introduction

Kinematics

(
จลนคณิตศาสตร์
)



เป็นการศึกษาสภาพและเส้นทางการเคลื่อนที่
ของวัตถุ โดยไม

พิจารณาถึงแรงที่กระท าต

อวัตถุ
และสาเหตุที่ท าให้เก

ดการเคลื่อนที่



กล

าวถึงความเชื่อมโยงก

นของ ระยะทาง
ความเร็ว และความเร




Introduction

Kinetics

(
จลนศาสตร์
)



เป็นการศึกษาการเคลื่อนที่ของวัตถุ โดย
พิจารณาถึงแรงที่กระท าต

อวัตถุนั

นด้วย



กล

าวถึงความเชื่อมโยงก

นของ แรง มวล และ
ความเร




แรง

การเคลื่อนที่ของวัตถ


Introduction

ลักษณะของวัตถุที่สนใจ



1
. Particle



2
. Rigid Body


Introduction

Particle (
อนุภาค
)



วัตถุใดๆ ที่มีมวล แต

จะไม

ค านึงถึงขนาด
และรูปร

าง ท าให้เราจะพิจารณาวัตถุนั

นเป็นจุด
หรือพิจารณาที่จุดศูนย์กลางมวลของวัตถุนั




Introduction

Rigid Body (
วัตถุคงรูป
)



วัตถุใดๆ ที่มีมวล และค านึงถึงขนาดและ
รูปร

าง ดังนั

นเมื่อมีการเคลื่อนที่ การวิเคราะห์
ปัญหาจ าเป็นต้องพิจารณาการหมุนรอบจุด
ศูนย์กลางมวลของวัตถุด้วย

Motion of Particles

Motion of Particles


1
. Rectilinear Motion
(
การเคลื่อนที่เชิงเส้น
)


2
. Curvilinear Motion
(
การเคลื่อนที่ในแนวเส้นโค้ง
)

Rectilinear Motion of Particles

Position


Velocity

t
x
v
t
x
t









0
lim
Average velocity

Instantaneous velocity

Rectilinear Motion of Particles

Acceleration


Instantaneous acceleration

t
v
a
t






0
lim
t
v



Average acceleration

Rectilinear Motion of Particles


Consider particle with motion given by

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

= 0,
x

= 0,
v

= 0,
a

= 12 m/s
2


at
t

=
2
s,
x

=
16
m,
v

=
v
max

=
12
m/s,
a

=
0


at
t

=
4
s,
x

=
x
max

=
32
m,
v

=
0
,
a

=
-
12
m/s
2


at
t

= 6 s,
x

= 0,
v

=
-
36 m/s,
a

= 24 m/s
2

Determination of the Motion of a Particles


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
)

Determination of the Motion of a Particles


Acceleration given as a function of
time
,
a

=
f
(
t
):










































t
t
t
x
x
t
t
t
v
v
dt
t
v
x
t
x
dt
t
v
dx
dt
t
v
dx
t
v
dt
dx
dt
t
f
v
t
v
dt
t
f
dv
dt
t
f
dv
t
f
a
dt
dv
0
0
0
0
0
0
0
0

Acceleration given as a function of
position
,
a

=
f
(
x
):

























x
x
x
x
x
v
v
dx
x
f
v
x
v
dx
x
f
dv
v
dx
x
f
dv
v
x
f
dx
dv
v
a
dt
dv
a
v
dx
dt
dt
dx
v
0
0
0
2
0
2
1
2
2
1
or

or

Determination of the Motion of a Particles


Acceleration given as a function of velocity,
a

=
f
(
v
):














































t
v
v
t
v
v
t
x
x
t
v
v
t
t
v
v
v
f
dv
v
x
t
x
v
f
dv
v
dx
v
f
dv
v
dx
v
f
a
dx
dv
v
t
v
f
dv
dt
v
f
dv
dt
v
f
dv
v
f
a
dt
dv
0
0
0
0
0
0
0
Sample
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.

Sample
11.2

SOLUTION:


Integrate twice to find
v
(
t
) and
y
(
t
).





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















Sample
11.2


Solve for
t

at which velocity equals zero and evaluate
corresponding altitude.



0
s
m
81
.
9
s
m
10
2









t
t
v
s
019
.
1

t






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
Sample
11.2


Solve for
t

at which altitude equals zero and
evaluate corresponding velocity.



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
Sample
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
v
0
,
piston moves and oil is forced through
orifices in piston, causing piston and
cylinder to decelerate at rate proportional
to their velocity; that is a =
-
kv

Determine
v
(
t
),
x
(
t
), and
v
(
x
).

Sample
11.3

SOLUTION:


Integrate
a

=
dv/dt

=
-
kv

to find
v
(
t
).





kt
v
t
v
dt
k
v
dv
kv
dt
dv
a
t
t
v
v









0
0
ln
0


kt
e
v
t
v


0

Integrate
v
(
t
) =
dx/dt

to find
x
(
t
).







t
kt
t
kt
t
x
kt
e
k
v
t
x
dt
e
v
dx
e
v
dt
dx
t
v
0
0
0
0
0
0
1




















kt
e
k
v
t
x



1
0
Sample
11.3


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

Uniform Rectilinear Motion

For particle in uniform rectilinear motion, the acceleration is zero and
the velocity is constant.

vt
x
x
vt
x
x
dt
v
dx
v
dt
dx
t
x
x









0
0
0
0
constant
Uniformly Accelerated Rectilinear Motion

For particle in uniformly accelerated rectilinear motion, the acceleration of
the particle is constant.

at
v
v
at
v
v
dt
a
dv
a
dt
dv
t
v
v









0
0
0
0
constant


2
2
1
0
0
2
2
1
0
0
0
0
0
0
at
t
v
x
x
at
t
v
x
x
dt
at
v
dx
at
v
dt
dx
t
x
x


















0
2
0
2
0
2
0
2
2
1
2
constant
0
0
x
x
a
v
v
x
x
a
v
v
dx
a
dv
v
a
dx
dv
v
x
x
v
v











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.




A
B
A
B
x
x
x
relative position of
B

with respect to
A

A
B
A
B
x
x
x





A
B
A
B
v
v
v
relative velocity of
B

with respect to
A

A
B
A
B
v
v
v





A
B
A
B
a
a
a
relative acceleration of
B

with respect to
A

A
B
A
B
a
a
a


Sample
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.

Sample
11.4

SOLUTION:


Substitute initial position and velocity and constant
acceleration of ball into general equations for
uniformly accelerated rectilinear motion.

2
2
2
2
1
0
0
2
0
s
m
905
.
4
s
m
18
m
12
s
m
81
.
9
s
m
18
t
t
at
t
v
y
y
t
at
v
v
B
B





























Substitute initial position and constant velocity of
elevator into equation for uniform rectilinear motion.

t
t
v
y
y
v
E
E
E











s
m
2
m
5
s
m
2
0
Sample
11.4


Write equation for relative position of ball with respect to
elevator and solve for zero relative position, i.e., impact.





0
2
5
905
.
4
18
12
2






t
t
t
y
E
B


s
65
.
3
s
meaningles

s
39
.
0



t
t

Substitute impact time into equations for position of elevator
and relative velocity of ball with respect to elevator.



65
.
3
2
5


E
y
m
3
.
12

E
y




65
.
3
81
.
9
16
2
81
.
9
18





t
v
E
B
s
m
81
.
19


E
B
v
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.



B
A
x
x
2
constant (one degree of freedom)


Positions of three blocks are dependent.




C
B
A
x
x
x
2
2
constant (two degrees of freedom)


For linearly related positions, similar relations hold
between velocities and accelerations.

0
2
2
or
0
2
2
0
2
2
or
0
2
2












C
B
A
C
B
A
C
B
A
C
B
A
a
a
a
dt
dv
dt
dv
dt
dv
v
v
v
dt
dx
dt
dx
dt
dx
Sample
11.5

Pulley
D

is attached to a collar which
is pulled down at
3
cm/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
cm/s as it passes
L
,
determine the change in elevation,
velocity, and acceleration of block
B

when block
A

is at
L
.

Sample 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
0
2
0
2
s
cm
9
cm
8
2
s
cm
12
2











A
A
A
A
A
A
A
a
a
x
x
a
v
v


s

333
.
1
s
cm
9
s
cm
12
2
0




t
t
t
a
v
v
A
A
A
Sample
11.5







cm

4
s
333
.
1
s
cm
3
0
0











D
D
D
D
D
x
x
t
v
x
x

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
.



























0
cm
4
2
cm
8
0
2
2
2
0
0
0
0
0
0
0















B
B
B
B
D
D
A
A
B
D
A
B
D
A
x
x
x
x
x
x
x
x
x
x
x
x
x
x










cm
cm
16
16
0
0
B
B
B
B
x
x
x
x

Pulley
D

has uniform rectilinear motion. Calculate
change of position at time
t
.

Sample
11.5


Differentiate motion relation twice to develop
equations for velocity and acceleration of block
B
.

0
s
cm
3
2
s
cm
12
0
2
constant
2





















B
B
D
A
B
D
A
v
v
v
v
x
x
x




s
cm
s
cm
18
18
B
B
v
v
0
s
cm
9
0
2
2











B
B
D
A
v
a
a
a




2
2
s
cm
s
cm
9
9
B
B
a
a
จบหัวข้อ
11.1
-
11.6

Motion

1
. Rectilinear Motion

2
. Curvilinear Motion


-

Rectilinear Components


-

Tangential & Normal Components


-

Radial & Transverse Components


Curvilinear Motion: Position, Velocity & 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 at time
t

and
P’

defined by at
t +

t
,

r

r
















dt
ds
t
s
v
dt
r
d
t
r
v
t
t
0
0
lim
lim



instantaneous velocity (vector)

instantaneous speed (scalar)

Curvilinear Motion: Position, Velocity & Acceleration








dt
v
d
t
v
a
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.


Let be a vector function of scalar variable
u
,


Let be a scalar function of scalar variable
u

Derivatives of Vector Functions





u
Q
,
u
P






u
u
P
u
u
P
u
P
du
P
d
u
u
















0
0
lim
lim

Derivative of vector sum,



du
Q
d
du
P
d
du
Q
P
d









du
P
d
f
P
du
df
du
P
f
d






Derivative of product of scalar and vector functions,


Derivative of
scalar

product

and
vector

product,





du
Q
d
P
Q
du
P
d
du
Q
P
d
du
Q
d
P
Q
du
P
d
du
Q
P
d
























u
f
Rectangular Components of Velocity & Acceleration


When position vector of particle
P

is given by its
rectangular components,

k
z
j
y
i
x
r








Velocity vector,

k
v
j
v
i
v
k
z
j
y
i
x
k
dt
dz
j
dt
dy
i
dt
dx
v
z
y
x























Acceleration vector,

k
a
j
a
i
a
k
z
j
y
i
x
k
dt
z
d
j
dt
y
d
i
dt
x
d
a
z
y
x

























2
2
2
2
2
2
Rectangular Components of Velocity & Acceleration


Rectangular components particularly effective
when component accelerations can be integrated
independently, e.g., motion of a projectile,

0
0







z
a
g
y
a
x
a
z
y
x







with initial conditions,







0
,
,
0
0
0
0
0
0
0




z
y
x
v
v
v
z
y
x

Integrating twice yields









0
0
2
2
1
0
0
0
0








z
gt
y
v
y
t
v
x
v
gt
v
v
v
v
y
x
z
y
y
x
x

Motion in horizontal direction is uniform.


Motion in vertical direction is uniformly accelerated.


Motion of projectile could be replaced by two
independent rectilinear motions.

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

B
A
r
r



Vector


joining
A

and
B

defines the position of
B

with respect to the moving frame
Ax’y’z’

and

A
B
r

A
B
A
B
r
r
r






Differentiating twice,


A
B
v

velocity of
B

relative to
A
.

A
B
A
B
v
v
v






A
B
a

acceleration of
B

relative
to
A
.

A
B
A
B
a
a
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
.

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.

t
t
e
e




and

t
t
t
e
e
e




















d
e
d
e
e
e
e
e
t
n
n
n
t
t



















2
2
sin
lim
lim
2
sin
2
0
0
Tangential and Normal Components


With the velocity vector expressed as

the particle acceleration may be written as

dt
ds
ds
d
d
e
d
v
e
dt
dv
dt
e
d
v
e
dt
dv
dt
v
d
a
t
t














After substituting,



2
2
v
a
dt
dv
a
e
v
e
dt
dv
a
n
t
n
t








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.


but

v
dt
ds
ds
d
e
d
e
d
n
t








Tangential and Normal Components



2
2
v
a
dt
dv
a
e
v
e
dt
dv
a
n
t
n
t








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
.

n
t
b
e
e
e






Normal to the osculating plane is found from

binormal
e
normal
principal
e
b
n







Acceleration has no component along binormal.

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
r
e
d
e
d
e
d
e
d











dt
d
e
dt
d
d
e
d
dt
e
d
r
r









dt
d
e
dt
d
d
e
d
dt
e
d
r

















e
r
e
r
e
dt
d
r
e
dt
dr
dt
e
d
r
e
dt
dr
e
r
dt
d
v
r
r
r
r
r


















The particle velocity vector is


Similarly, the particle acceleration vector is

















e
r
r
e
r
r
dt
e
d
dt
d
r
e
dt
d
r
e
dt
d
dt
dr
dt
e
d
dt
dr
e
dt
r
d
e
dt
d
r
e
dt
dr
dt
d
a
r
r
r
r

















2
2
2
2
2
2

















r
e
r
r



Radial and Transverse Components


When particle position is given in cylindrical
coordinates, it is convenient to express the
velocity and acceleration vectors using the unit
vectors

.

and

,
,
k
e
e
R





Position vector,

k
z
e
R
r
R






Velocity vector,

k
z
e
R
e
R
dt
r
d
v
R















Acceleration vector,





k
z
e
R
R
e
R
R
dt
v
d
a
R
























2
2
Sample
11.10

A motorist is traveling on curved
section of highway at
88
m/s. The
motorist applies brakes causing a
constant deceleration rate.

Knowing that after
8
s the speed has
been reduced to
66
m/s, determine
the acceleration of the automobile
immediately after the brakes are
applied.

SOLUTION:


Calculate tangential and normal
components of acceleration.


Determine acceleration magnitude and
direction with respect to tangent to
curve.

Sample
11.10

SOLUTION:


Calculate tangential and normal components of
acceleration.





2
2
2
2
s
m
10
.
3
m
2500
s
m
88
s
m
75
.
2
s

8
s
m
88
66











v
a
t
v
a
n
t

Determine acceleration magnitude and direction
with respect to tangent to curve.



2
2
2
2
10
.
3
75
.
2





n
t
a
a
a
2
s
m
14
.
4

a
75
.
2
10
.
3
tan
tan
1
1




t
n
a
a



4
.
48

Sample 11.12

The rotation of the 0.9 m arm OA about O is
defined by the relation
 
0.15t
2
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
-
0.12t
2
, where r is
expressed in meters and t in seconds. After the
arm OA has rotated through 30
o
, 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

Sample 11.12

SOLUTION:


Evaluate time
t

for


=
30
o
.

s

869
.
1
rad
524
.
0
30
0.15
2





t
t


Evaluate radial and angular positions, and first
and second derivatives at time
t
.

2
2
s
m
24
.
0
s
m
449
.
0
24
.
0
m

481
.
0
12
.
0
9
.
0









r
t
r
t
r



2
2
s
rad
30
.
0
s
rad
561
.
0
30
.
0
rad
524
.
0
15
.
0











t
t
Sample 11.12


Calculate velocity and acceleration.





r
r
r
v
v
v
v
v
r
v
s
r
v





1
2
2
tan
s
m
270
.
0
s
rad
561
.
0
m
481
.
0
m
449
.
0















0
.
31
s
m
524
.
0

v












r
r
r
a
a
a
a
a
r
r
a
r
r
a







1
2
2
2
2
2
2
2
2
tan
s
m
359
.
0
s
rad
561
.
0
s
m
449
.
0
2
s
rad
3
.
0
m
481
.
0
2
s
m
391
.
0
s
rad
561
.
0
m
481
.
0
s
m
240
.
0




























6
.
42
s
m
531
.
0

a
Sample
11.12


Evaluate acceleration with respect to arm.


Motion of collar with respect to arm is rectilinear
and defined by coordinate
r
.

2
s
m
240
.
0



r
a
OA
B


จบหัวข้อ
11.9
-
11.14

จบเนื้อหาของบทที่
11

Quiz
1

มอเตอร์ไซด์เริ่มจากหย

ดนิ่งที่
s =
0
โดยมีการเปลี่ยนแปลงความเร็วเมื่อเวลาเปลี่ยนไป
แสดงดังกราฟ จงหาความเร่งและต าแหน่งของรถคันนี้เมื่อเวลา
t =
8
และ
t =
12
s


Quiz
2

ความสัมพันธ์ระหว

างความเร

งและระยะทาง
(a
-
s)
ของรถจี๊ปที่เคลื่อนที่เป็นแนวเส้นตรง
ในช

วง
300
m
แรกแสดงในรูปด้านล

าง จงหาสมการของความเร็วที่เป็นฟังก

ชันของ
ระยะทาง และวาดกราฟแสดงความสัมพันธ์ระหว

างความเร็วและระยะทาง
(v
-
s)
ของ
รถจี๊ปคันนี


Quiz
3

Block C
เคลื่อนที่ลงด้วยความเร็วคงที่
0
.
6
m/s
จงหา

(a)

ความเร็วของ
Block A

(b)
ความเร็วของ
Block D