Principles of Dynamics Dynamics is the branch of mechanics which
deals with the study of bodies in motion. It is generally
considered to have been begun by Galileo (1564-1642). Its
development was greatly retarded by the lack of precise methods for
measuring time. The experiments which from the foundation of
dynamics required the use of three kinds of units: force, length,
and time. Dynamics was also retarded by the principles of natural
philosophy which were set up by Aristotle and Galileo s time were
regarded as infallible. Galileo s experimental turn of mind led him
to doubt these dogmas of abstract thought. For example, he did not
accept the notion that heavy weights fall more than light ones. His
experiments with dropping weights from the leaning tower of Pisa
exploded this theory but precipitated such bitter arguments that he
was force to leave Pisa. Galileo s experiments with blocks sliding
down inclined planes led to a relation between the force and
acceleration which Sir Isaac Newton generalized and incorporated
into the laws governing the motion of a particle that are named
after him. Newton s law of motion are the basis for extending the
laws of motion from a particle to a body composed of a system of
particles. The term particle usually denotes an object of point
size. The term body denotes a system of particles which from an
object of appreciable size. In other words, a particle is a body so
small that any differences in the motions of its parts can be
neglected. The term particle and body may apply equally to the same
object. Newton s Law of Motion for a Particle Newton s laws of
motion for a particle have been stated in a variety of ways. For
our purposes we shall phrase them as follows. 1) A particle acted
upon by a balanced force system has no acceleration. 2) A particle
acted upon by an unbalanced force system has no acceleration in
line with and directly proportional to the resultant of the force
system. 3) Action and reaction forces between two particles are
always equal and oppositely directed.
Fundamentals Equation of Kinetic for a Particle If the same
particle is now assumed to be in vacuum, the resultant force acting
upon it is its weight W. By experiment, the acceleration produced
by W is found to be the value of the gravitational constant g which
acts in line with W. again applying.
and
Dividing Eq. (a) by eq. (b) we obtain:
Vector and Scalar Quantities A vector quantity has magnitude and
direction, whereas a scalar quantity only has magnitude. Scalar
Energy, time, speed, mass, length, distance Vector Force,
acceleration, displacement, velocity
Scalar quantities can be added using the normal rules of
arithmetic but vectors are added in a different way using a vector
diagram in which each vector is represented by a line, whose length
represents the magnitude of the vector and whose direction is shown
by the direction of the line (with an arrow). The vectors are then
added head to tail and the line from the start of the first vector
to the end of the last vector represents the sum (or resultant) of
the vectors . Rectilinear Translation Definition and
characteristics of Translation. Translation-is defined as the
motion of a rigid body in which a straight line passing through any
two of its particle always remains parallel to its initial
position. Translation may be either rectilinear or curvilinear,
depending upon whether the path describe by any particle is
straight or curved. The motion of a translating body moving in a
straight line is called Rectilinear Translation.Kinematics of
Rectilinear Translation with Constant Velocity
Kinematic characteristics of the translation of a rigid body is
the fact that all the particles travel the same or parallel paths.
It follows that all the particles have the same values of
displacement, velocity, and acceleration, and the motion may be
completely describe by the motion of any particle of the body.
The particle usually selected is the one of the center of the
gravity of the body. Therefore Rectilinear Translationis a
translating body that may be consider as a particle concentrated at
its center of gravity.Rectilinear Motion with Constant
Acceleration.
One of the most common cases of straight-line motion is that in
which the acceleration is constant. The equation may be derive from
the differential equation of Kinematic Equation.
And proceed to the integration process therefore:
Note that a is a consider as a constant. Drawing the figure
involve in the equation to further understand the relationship of
each equation therefore:
A
B
S
The point A is to be measured, there is an initial velocity
reached after a time interval t, the velocity will be v.
Therefore:
, whereas at some other position B
Then when:
And by using variable separable:
When
Then let us consider the remaining differential equation of
kinematics
The limits are written a before, since it is obvious that at
zero displacement the corresponding velocity is , while at a
displacement s it is v integrating and evaluating the limits we
obtain:
Therefore the three Kinematic Equations of motion with constant
acceleration may be summarized as follow:
Signs It is important to observe that these equation involve
only the magnitude of vector quantities. The direction of the
vectors of displacement, velocity, and acceleration is indicated by
the following sign conversion: the initial direction of motion
represent the positive direction for displacement, velocity, and
acceleration.
Free falling Bodies, Air resistance is Neglected.
It has been seen that the acceleration of a body is directly
proportional to the resultant force acting upon it. In case of a
free falling body, this resultant force is its own weight .the
weight . the weight of a force that results from the attraction
between the mass of the body and the mass of the earth ; it varies
inversely proportional as the square of the distance separating the
two center of the mass and is directly proportional to the product
of the masses The initial direction of the motion determines
positive directions of displacement, velocity, and acceleration.
For the most cases over a given earth surface, however, the
gravitational acceleration may be assumed to be constant. For our
latitude, this acceleration is approximately 32.2 ,
9.81
,981
, and is represented by the symbol g. this value of g will be
used throughout
this book except as otherwise indicated.
S A C
D -S
Sample Problems
1. A ball is dropped down a well and 5 seconds later the sound
of the splash is heared. If the velocity of wound is 330 m/sec.,
what is the depth of the well?
S
Solution: T1 = time for the ball to travel a distance S T2 =
time for the sound to travel a distance S 1) T1 + T2 = 5 2) S = gT1
T1 = 2s/g 3) S = 330T2 T2 = S/330 Substitute in equation 1: 2S/9.81
+ S/330 = 5 0.452 S + S/330 5 = 0 S = 149 S 1650 = 0 Let y = S y =
S y + 149y -1650 = 0
y = 10.35 S = 10.35 S = 107.2 m
2. A stone is thrown upward from the ground with the velocity of
15 m/sec. One second later another stone is thrown vertically
upward with a velocity of 30 m/sec. How far above the ground will
the stones be at same level?
15
S
30
Solution: t = time the first stone travelled t 1 = time the
second stone travelled until the stone are at the same level S =
Vot - (g)t S = 15t (9.81)t S = 15t 4.905t equation 1 equation 2
S = 30(t 1) (9.81)(t 1) Equate (1) and (2):
15t 4.905t = 30(t 1) 4.905(t - 2t + 1) 15t 4.905t = 30t 30 -
4.905t + 9.81t 4.905 34.905 = 24.81t
t = 1.4 sec. S = 15(1.4) 4.905(1.4) S = 11.4 m. from
theground
3.A ball is thrown vertically upward with an initial velocity of
3 m/s from the window of a tall building. The ball strikes the
sidewalk at the ground level 4 seconds later. Determine the
velocity with which the ball hits the ground and the height of the
window above the ground level.
h
H S
Solution: Vf = Vo - 2gh = (3) - 2(9.81)h h = 0.459 m. Vf = Vo gt
0 = 3 9.81(t)
t = 0.31 t2 = 4 secs. 0.31 secs. t2 = 3.69 secs. H = g(t2) H =
(9.81)(3.61) H = 66.79 m Height of the window = H h S = 66.79 0.459
= 66.331 V3 = V1 + 2gh V3 = 0 + 2(9.81)(66.79) V3 = 36.20 m/s
Further Application of Kinematics of Translation I. Given:
Displacement as a function of time
II. Given: Velocity as a function of time
III. Given: Acceleration as a function of time
IV. Given: Acceleration as a function of position
V. Given: Acceleration as a function of displacement
Sample Problems 1. The position of a particle is given by .
Where t is in sec.
2. The velocity of a particle moving along x-axis is given by .
Evaluate the position, velocity and acceleration of particle at ,
when t = 0, .
3.the acceleration of a particle is given by v=
when t=0,s=-40m, and
Determine the position and velocity after t=2 sec.
Projectile Motion A projectile is the name given to an
un-powered object which is moving through the air and is subject to
gravity. Usually at higher we assume that all forces other than
gravity can be ignored.
Types of Projectile Motion Horizontal Motion of a ball rolling
freely along a level surface Horizontal velocity is ALWAYS constant
Vertical Motion of a freely falling object Force due to gravity
Vertical component of velocity changes with time Parabolic Path
traced by an object accelerating only in the vertical direction
while moving at constant horizontal velocity
Derivation of formulas:
Velocity at any point
Therefore:
Displacement
Where x = distance t = time when t approaches 0 and x approaches
0
Then substitute
Therefore Horizontal Component of Velocity
Horizontal Displacement
Highest Point
If Vy = 0 substitute to Equation of the Path of the Particle
SAMPLE PROBLEM:
1. The car shown is just to clear the water filled gap. Find the
take off velocity?
Solution:
(eq.2)
Eq1 to eq. 2
2. refer to figure. Find the to cause the projectile to hit the
point exactly in 4 sec..what is the distance
3. A ball id thrown so that it just clears a 10ft fence 60 ft.
away. If it left the hand 3 ft. above the ground and at an angle of
? Given x=60ft. =60 y=3ft.
10 ft. 60
5 ft.
60 ft.
Solution
MOTION DIAGRAM It is the representation of the change in
velocity, acceleration and displacement. The motion is usually too
complex to analyze mathematically. Instead, a graphical analysis
like that abuot to be describe is used. The method of graphical
analysis os particularly useful when applied to problems involving
variation of diplacement, velocity, or acceleration with time. ci
from a practical viewpoint, this process of graphical
differentiation isquite tedioys and requires drafting skill of a
high order. It is not often used.it has benn described to emphasize
the following relation between these curves:
They state that the slope of the displacement-time curve is
equal to the corresponding coordinate
1.) A train is to commute between station A and station B with a
top speed of 250 kph but cannot accelerate nor decelerate faster
than 4 m/sec. what is the minimum distance between the two stations
in order for the train to be able to reach its top speed?
t 4 t V
-4 a
V t t S S S 2S
2) A train starts from rest at station P and stops at station Q
which is 10 km from P. The maximum possible acceleartion of the
train is 15 km/hr/min and the maximum deceleration when the brakes
are applied is 10 km/hr/min. if the maximum allowable speed is 60
kph., what is the least time the train can go from P to Q?
a
900 600
V
60
60
S
Accelearation = 15(60) = 900km/hr Deceleration = 10(60) = 600
km/hr 900t1 = 60 t1 = = 4 min
600t3 = 60 t3 = S1= = 6 min. (t1) =
S1 = 2 km S2 S1 = 60t2 S2 2 = 60t2
10 S2 = (60/2)(t3) S2 = 10 3 = 7 km S2 S1 = 7 2 = 5 km 5 = 60t2
t2 = 5(60)/60 = 5 min. total time: 6 + 4 + 5 = 15 mins
3. An automobile is to travel a distance from A to B of 540 m.,
in exactly 40 seconds. The auto accelerates and decelerates at 1.8
m/sec., starting from the rest at A and coming to rest at B. find
the maximum speed.
1.8 m/s t1 t2
ACCELERATION DIAGRAM -1.8 m/s
VELOCITY DIAGRAM
t3
V
V
t1
t2 VELOCITY DIAGRAM 1 2 S2 S1
t3 2 S3 = 540
DISTANCE DIAGRAM
1.8t1 = V 1.8t3 = V
eq. 1 eq. 2
From eq. (1) and (2) 1.8t1 = 1.8t3 t1 = t3 eq. 3
t1 + t2 + t3 = 40 2t1 + t = 40 t2 = 40 2t1 eq. 4
S1 = Vt1/2 S2 = S1 + Vt2
eq. 5
S2 = Vt1/2 + Vt2 S3 = S2 + Vt2/2 But t3 = t1 ; S3 = 540 m 540 =
(Vt1/2 + Vt2) + Vt1/2 540 = Vt1 + Vt2 From(4) 540 = Vt1 + V(40 2t1)
540 = Vt1 + 40V 2Vt1 540 = V(40 t1) From (1)
540 = 1.8t1 (40 t1) 300 = 40t t1 t1 - 40t1 + 300 = 0 t1 = 10
secs. t3 = 10 secs. t2 = 40 2(10) = 20 secs. V = 1.8(10) = 18
m/sec. (max. speed)
RECTILINEAR MOTION Definition and characteristics of
Translation. Translation-is defined as the motion of a rigid body
in which a straight line passing through any two of its particle
always remains parallel to its initial position. Translation may be
either rectilinear or curvilinear, depending upon whether the path
describe by any particle is straight or curved. The motion of a
translating body moving in a straight line is called Rectilinear
Translation. Kinematic characteristics of the translation of a
rigid body is the fact that all the particles travel the same or
parallel paths. It follows that all the particles have the same
values of displacement, velocity, and acceleration, and the motion
may be completely describe by the motion of any particle of the
body. The particle usually selected is the one at center of the
center of the gravity of the body. Therefore Rectilinear
Translation is a translating body that may be consider as a
particle concentrated at its center of gravity. Rectilinear Motion
with Constant Acceleration. One of the most common cases of
straight-line motion is that in which the acceleration is constant.
The equation may be derive from the differential eqution of
Kinematic Equation.
And proceed to the integration process therefore:
Note that a is a consider as a constant. Drawing the figure
involve in the equation to further understand the relationship of
each equation therefore:
A
B
S
The point A is to be measured, there is an initial velocity
reached after a time interval t, the velocity will be v.
Therefore:
, whereas at some other position B
Then when:
And by using variable separable:
When
Then let us consider the remaining differential equation of
kinematics
The limits are written a before, since it is obvious that at
zero displacement the corresponding velocity is , while at a
displacement s it is v integrating and evaluating the limits we
obtain:
Therefore the three Kinematic Equations of motion with constant
acceleration may be summarized as follow:
Signs It is important to observe that these equation involve
only the magnitude of vector quantities. The direction of the
vectors of displacement, velocity, and acceleration is indicated by
the following sign conversion: the initial direction of motion
represent the positive direction for displacement, velocity, and
acceleration.
SAMPLE PROBLEM: 1. On a certain stretch of track, trains run at
60 mph. How for back of a stopped train should a warning torpedo be
placed to signal an on coming train? Assume that the brakes are
applied at once & retard the train at the uniform rate of 2ft
per . Given: a=-2ft/ =-2(2)S S= 1936ft 2. A stone is thrown
vertically upward & returns to earth in 10sec. What its initial
velocity & how high did it go? Given: t=10sec
3. A stone is dropped down a well & 5 sec later the sound of
the splash is heard. If the velocity of sound is 1120ft/sec, what
is the depth of the well? Given: (1)
CURVILINEAR TRANSLATION Translation of a rigid body has been
defined as the motion in which a straight line passing through any
two points of the body always remain parallel to its initial
position. In rectilinear, we chose the origin of motion of the path
so that only the magnitude of displacement vector could change but
no its inclination. In curvilinear motion the displacement vector
will change in both magnitude and inclination. The figure below
shows the curved path traversed by a particle having curvilinear
motion. The displacement of any position is its vector distance
from the origin O. for example, the vector displacements of two
positions A and B are represented by . It s evident that the change
in displacement is due to combination change in the magnitude and
inclination of these displacement vectors.
Y
A
B
y X
For any position, the displacement s of any point maybe
expressed as the vector sum of its X and Y coordinates as follows:
S = x y Recalling that and a = , we obtain by differentiation of
equation:
Velocity in Curvilinear Motion Let the points A and B in
previous figure represents successive portion of a moving particle
after a small elapsed time The change in displacement during this
interval is the chord distance between A and B. if the changes in
displacement is resolved into components parallel to the reference
axes , inspection of the figure shown the geometric relation
between Is given by the following vector equation:
This create a new vector having a different magnitude but the
same direction as the corresponding displacement. Each of these new
vector represent the average velocity in the respective direction
of displacement.
As approaches zero,B approaches A and chord coincides more
completely with the curve of travel so that, in the limit, becomes
ds which is directed along the path at A. hence the term represent
the instantaneous velocity at A directed tangent to the path at
A.
Y
A
X
Show this velocity and its component algebraic expression
Incidentally, if we replace
its direction with the X axis, by by their corresponding
value
. The magnitude of the velocity is given by the . .
It is evident that
is the slope of the velocity is tangent to the path.
CURVILINEAR TRANSLATION 1.) The block shown reaches a velocity
of 12 m/s in 30 m starting from rest. Compute the coefficient of
kinetic friction between the block and the ground.
V1 = 0 REF
W = 70 kg
V2 = 12 m/s
W
P = 24 kg
F N
30 mN=W
F = N
Solution: V2 = V1 + 2aS (12) = 0 + 2a(30) a = 2.4 m/sec P = wa/g
+ F 24 = 70(2.4)/9.81 + F F = 6.87 kg F = (70) 6.87 = (70) =
0.1
2) Determine the magnitude of W so that the 400 N body will have
the acceleration up the plane of 1.2 m/s.
400 = 0.20 W 30
Solution: N = 400 cos 30a = 1.2 ma
40030
T1 F
F = NN
F = 0.20(364.4) = 69.28 T1 = 400 sin 30 + F + ma T1= 200 + 69.28
+ 400(1.2)/9.81 T1 = 318.20 Na/2 T2
T2 = W m(a/2)W
T2 = W W(1.2)/2(9.81) T2 = W 0.06W T2 = 0.94W T2 = 2T1 0.94W =
2(318.20) W = 677 NT2 T1 ma/2 T1
3.) A body takes twice as long to side down a plane 30 to the
horizontal as it would if the plane were smooth. What is the
coefficient of friction? Solution: For rough surface S = V1t +
atW
S = 0 + (a1)tS
Wa1/g
N = Wcos F = N F = (Wcos )W N F
Fx = 0 F + (Wa1)/g = Wsin a1 = (sin 30 - cos 30)g a1 = g(0.5
0.866) For smooth surface S = V1 (t2) + (a2)(t2) T2 = t/2 S = 0 +
(a2)(t/4) S = a2t/8 a1t/2 = a2t/2 a2 = 4a1S
30
Wa2/g
N 30
Fx = 0 Wa2/g = Wsin 30 a2 = 0.5(9.81) a2 = 4.9 m/s a1 = (4.9)/4
= 1.225 m/s 1.225 = 9.81(0.5 0.866)
= 0.433
Dynamics of Rigid Bodies Dynamics of Rigid Bodies
A rigid body is an idealization of a body that does not deform
or change shape. Formally it is defined as a collection of
particles with the property that the distance between particles
remains unchanged during the course of motions of the body. Like
the approximation of a rigid body as a particle, this is never
strictly true. All bodies deform as they move. However, the
approximation remains acceptable as long as the deformations are
negligible relative to the overall motion of the body. Kinematics
of Rigid Bodies As rigid bodies are viewed as collections of
particles, this may appear an insurmountable task, requiring a
description of the motion of each particle. However, the assumption
that the body does not deform is a very strong one, requiring that
the distance between every pair of particles comprising the body
remains unchanged CURVILINEAR TRANSLATION CURVILINEAR TRANSLATION
Translation of a rigid body has been defined as the motion in which
a straight line passing through any two points of the body always
remain parallel to its initial position. In rectilinear, we chose
the origin of motion of the path so that only the magnitude of
displacement vector could change but no its inclination. In
curvilinear motion the displacement vector will change in both
magnitude and inclination. The figure below shows the curved path
traversed by a particle having curvilinear motion. The displacement
of any position is its vector distance from the origin O. for
example, the vector displacements of two positions A and B are
represented by . It s evident that the change in displacement is
due to combination change in the magnitude and inclination of these
displacement vectors.
Y
A
B y X
For any position, the displacement s of any point maybe
expressed as the vector sum of its X and Y coordinates as follows:
S = x y Recalling that and a = , we obtain by differentiation of
equation:
Velocity in Curvilinear Motion Let the points A and B in
previous figure represents successive portion of a moving particle
after a small elapsed time The change in displacement during this
interval is the chord distance between A and B. if the changes in
displacement is resolved into components parallel to the reference
axes , inspection of the figure shown the geometric relation
between Is given by the following vector equation: This create a
new vector having a different magnitude but the same direction as
the corresponding displacement. Each of these new vector represent
the average velocity in the respective direction of
displacement.
For any position, the displacement s of any point maybe
expressed as the vector sum of its X and Y coordinates as follows:
S = x y Recalling that and a = , we obtain by differentiation of
equation:
As approaches zero,B approaches A and chord coincides more
completely with the curve of travel so that, in the limit, becomes
ds which is directed along the path at A. hence the term represent
the instantaneous velocity at A directed tangent to the path at
A
A
Show this velocity and its component
its direction with the X axis, by Incidentally, if we replace
corresponding value by their . is the slop
It is evident that
m
s
g
e of the velocity is tangent to the path
g
!
"
.
The block s hown reaches a velocity of 12 m/s in 30 m s tarting
from res t. C ompute the coefficient of kinetic friction between
the block and the g round.
X . The magnitude of the velocity is given by the algebraic
expression .
Solution: V2 = V1 + 2aS (12) = 0 + 2a(30) a = 2.4 m/sec P = wa/g
+ F 24 = 70(2.4)/9.81 + F F = 6.87 kg F = (70) 6.87 = (70) =
0.1
#
30
&4
%
& 3&
2' 1 0
)
& &
)( '
2) ete mine t e magnitu e o lane o 1.2 m/s.
so t at t e 400
o
will a e t e accele ation u t e
&
( 4 % $
0. 0 3
.
9.
0 0 s in 3 0 9.
ma
00
00
.
9.
a
W W W
m a W
.
9.
0 .0
W
ma
0 .9 W
W
A body takes twice as long to side down a plane 30 to the
horizontal as it would if the plane were smooth. What is the
coefficient of friction?
o lu tio n
F
t
at
W
0
a
t
W cos
W cos
0 g W s in
W
Wa
a
g 0.
0.
w
a
s in 3 0
cos 30 g
30
76
0 .9 W
3
. 0
B6
98
B6
A@
3
. 0
W
5
76
W
5 VVP E Q A RS IA E F S 6A E A 6 F E A6 P U E A6 QSR IA @QA SI U
E A 6 QA @ I U E A 6 5 A RS E S 6 SR @QA SI F T RA P T A ES 6 TGT F
E S6 RA P E QF FP I A E G 5H E G FE5 C00 cos 30
W
w v p r
i hgfa b e d cb `a a ` Y X
wW
u wr v t s p X u t s rq p X
p wr
W
v p y x p y
vpr v sy
D
o lu tio n
30
p y
px
For smooth surface S = V1 (t2) + (a2)(t2) Fx = 0 T2 = t/2 Wa2/g
= Wsin 30 S = 0 + (a2)(t/4) a2 = 0.5(9.81) S = a2t/8 a2 = 4.9 m/s
a1t/2 = a2t/2 a1 = (4.9)/4 = 1.225 m/s a2 = 4a1 1.225 = 9.81(0.5
0.866) = 0.433
CURVILINEAR TRANSLATION Translation of a rigid body has been
defined as the motion in which a straight line passing through any
two points of the body always remain parallel to its initial
position. In rectilinear, we chose the origin of motion of the path
so that only the magnitude of displacement vector could change but
no its inclination. In curvilinear motion the displacement vector
will change in both magnitude and inclination. The figure below
shows the curved path traversed by a particle having curvilinear
motion. The displacement of any position is its vector distance
from the origin O. for example, the vector displacements of two
positions A and B are represented by . It s evident that the change
in displacement is due to combination change in the magnitude and
inclination of these displacement vectors.
Y
A
B
yX
For any position, the displacement s of any point maybe
expressed as the vector sum of its X and Y coordinates as follows:
S = x y Recalling that and a = , we obtain by differentiation of
equation:
Velocity in Curvilinear Motion Let the points A and B in
previous figure represents successive portion of a moving particle
after a small elapsed time The change in displacement during this
interval is the chord distance between A and B. if the changes in
displacement is resolved into components parallel to the reference
axes , inspection of the figure shown the geometric relation
between Is given by the following vector equation: This create a
new vector having a different magnitude but the same direction as
the corresponding displacement. Each of these new vector represent
the average velocity in the respective direction of displacement.
As approaches zero,B approaches A and chord coincides more
completely with the curve of travel so that, in the limit, becomes
ds which is directed along the path at A. hence the term represent
the instantaneous velocity at A directed tangent to the path at
A
A
Show this velocity and its component
its direction with the X axis, by Incidentally, if we replace
corresponding value by their .
It is evident that
X . The magnitude of the velocity is given by the algebraic
expression . is the slope of the velocity is tangent to the
path
1 e lo s o n e c e a e cit o 1 m in3 mstatin o e o tet e .) ck w
a s lo 2 /s 0 g m st. mu co icie to k e ictio e e nt e lo a t egon.
e n in tic n twe ck n u
V=0 1 F
=7 k 0g
V=1 m 2 2 /s P=2 k 4g
Solution: V2 = V1 + 2aS (12) = 0 + 2a(30) a = 2.4 m/sec P = wa/g
+ F 24 = 70(2.4)/9.81 + F F = 6.87 kg F = (70) 6.87 = (70) =
0.1
2) etem t e m ine agnitu e o lane o 1.2 m /s.
sot at t e 400
o w a e t e acceleationu t e ill
30
S o lu tio n
= 400 cos 30
F =
F = 0 .2 0 ( 3 6 4 .4 ) = 6 9 .2 8 T 1 = 4 0 0 s in 3 0 + F + m
a T 1 = 2 0 0 + 6 9 .2 8 + 4 0 0 ( 1 .2 ) / 9 .8 1 T 1 = 3 1 8 .2 0
T2 = W T2 = W T2 = W T2
a/2
m (a /2 ) W W ( 1 .2 ) / 2 ( 9 .8 1 ) 0 .0 6 W T1 m a/2 T1
T 2 = 0 .9 4 W T2 = 2T1 0 .9 4 W = 2 ( 3 1 8 .2 0 )
T2 W = 6
A body takes twice as long to side down a plane 30 to the
horizontal as it would if the plane were smooth. What is the
coefficient of friction?S o lu tio n
F o r r o u h su r fa c e S = V 1 t + a t W S = 0 + (a 1 )t = W
cos
F =
F = (W c os Fx = 0
)
W
F + (W a 1 )/g = W s in a 1 = (s in 3 0 a 1 = g (0 .5
c o s 3 0 ) g
0 .8 6 6 )
30
For smooth surface S = V1 (t2) + (a2)(t2) Fx = 0 T2 = t/2 Wa2/g
= Wsin 30 S = 0 + (a2)(t/4) a2 = 0.5(9.81) S = a2t/8 a2 = 4.9 m/s
a1t/2 = a2t/2 a1 = (4.9)/4 = 1.225 m/s a2 = 4a1 1.225 = 9.81(0.5
0.866) = 0.433
30
KINEMATICS OF ROTATION Rotation is defined as that motion of a
body in which the particles move in circular paths with their
center of fixed straight line that is called the axis of rotation.
The plane of the circles in which the particles move are
perpendicular to the axis of rotation.
If any radius to any point A is permitted to rotate throught
radians, point A moves through the arc distance . Since the body is
rigid, angle AOB cannot change; hence the radius to any other point
B will also rotate through radians and point B will move through
the arc distance . It was shown that in the motion of translation
all the particle have identical values of linear displacement,
linear velocity and the linear acceleration. The motion of rotation
has a similar characteristics; all the particles have the same
values of angular displacement, angular velocity, and angular
acceleration. Kinematic Differential Equation of Rotation Consider
a pulley free to rotate an axle O under the action of a weight W
suspended in the cord wound around the pulley. Assume that the
weight descends s ft., as shown in the figure. This will unwind
from the pulley a length of cord equal to s ft. so that the point B
on the rim will rotate to occupy the position at point A. The
angular distance through which the pulley rotates is obviously
subtended by radii drawn to points A and B. the relationship
between the linear displacement of the weight and angular
displacement of the pulley is given by the equation: If we
differentiate Equation with respect to time
Note that r is the constant radius of the rotation. The term
defined as V, the linear velocity of the weight. The term
represents the time rate of the angular displacement and will be
called the angular velocity and be represented by the symbol . Thus
the angular velocity at any instant is defined by the
equation(1):
fe
e, representing the rate of change of displacement is
d
The common unit in radians per second but other units is used
such as degrees per second and revolution per minute. Rewriting the
equation: And differentiating with respect to time gives: The
equation in equation represents the time rate of change of the
magnitude of the velocity. It is preferable to denote this
acceleration by because it not only represents the linear
acceleration of the weight but is also tangential acceleration of a
point on the rim of a pulley. The expression represents the time
rate of change of angular velocity and will define as the angular
acceleration , according to the following equation(2):
The equation (1) and (2) are kinematic differential equation of
rotation, a third convenient relation may be found by eliminating .
Summarized differential equations of rotation: RECTILINEAR
MOTION:ROTATION
The common unit in radians per second, but other units may use.
Equation is written: t Since , the normal acceleration of any point
of the rim of the pulley is given by:
=
These relation differ only in the symbol used, they are
mathematical identical. They can be transformed into each other by
relations:
Kinetics of Translation
As an example, the flutter of an aircraft wing during the course
of a flight is clearly negligible relative to the motion of the
aircraft as a whole. Preliminaries: Motion Relative to Translating
Axes The positions, velocities and accelerations determined in this
way are referred to as absolute. Often it isn t possible or
convenient to use a fixed set of axes for the observation of
motion. Properties of Rigid Body Motion Translation, rectilinear
and curvilinear: Motion in which every line in the body remains
parallel to its original position. The motion of the body is
completely specified by the motion of any point in the body. All
points of the body have the same velocity and same acceleration.
Rotation about a fixed axis: All particles move in circular paths
about the axis of rotation. The motion of the body is completely
determined by the angular velocity of the rotation. General plane
motion: Any plane motion that is neither a pure rotation nor a
translation falls into this class. However, as we will see below, a
general plane motion can always be reduced to the sum of a
translation and a rotation. We proceed by demonstrating that every
motion of a planar rigid body is associated with a single angular
velocity and angular acceleration , describing the angular
displacement of an arbitrary line inscribed in the body relative to
a fixed direction. for an arbitrary line attached to the body.
Arguing analogously, the body can be associated with a unique
angular acceleration defined as,
Angular velocity and acceleration of a rigid body Sample
Problems 1. (Rotation about fixed axis) A cord is wrapped around a
wheel which is initially at rest (figure A). If a force is applied
to the cord and gives it an acceleration a = (4t)m/s2, where t is
in seconds, determine as a function of time (a) the angular
velocity of the wheel, and (b) the angular position of line OP in
radians.
Using this result, the wheel s angular velocity w can now be
determined from = d /dt,* since this equation relates , t and .
Integrating, with the initial condition that = 0 at t = 0,
yields
Part (b). Using this result, the angular position q of the
radial line OP can be computed from = d/dt, since this equation
relates ,and t. Integrating, with the initial condition = 0 at t =
0, we have
2. (Rotation about fixed axis) Disk A (figure A) starts from
rest and through the use of motor begins to rotate with a constant
angular acceleration of A = 2 rad/s2. If no slipping occurs between
the disks, determine the angular velocity and angular acceleration
of disk B just after A turns 10 revolutions.
First we will convert the 10 revolutions to radians. Since there
are 2 rad to one revolution, then9o
Since A is constant, the angular velocity of A is then
As shown in figure B the speed of the contacting point P on the
rim of A is (+ ) vP = rA = (15.9 rad/s)(0.6 m) = 9.54 m/s The
velocity is always tangent to the path of motion; and since no
slipping occurs between the disks, the speed of point P' on B is
the same as the speed of P on A*. The angular velocity of B is
therefore
The tangential components of acceleration of both disks are also
equal, since the disks are in contact with one another. Hence, from
figure C. (ap)t = (ap')t
Its is important notice that the normal components of
acceleration (ap)nand (ap')n act in opposite directions, since the
paths of motion for both points are different. Furthermore, (ap)n
(ap')n since the magnitudes of these components depend on both the
radius and angular velocity of each disk, i.e., (ap)n = rAand (ap)n
= rB. Consequently, apap,.
Work and Energy Introduction The term Work and Kinetic Energyare
used to define certain mathematical expressions. A precedent for
this was found when we define the moment of inertia of an area
equivalent to the mathematical equation . Fundamental Work-Energy
Principle The mathematical expression defining work kinetic energy
as applied to translation are easily obtained by considering the
following equation:
(a)
Note that the first member if eq.(a) aquates the resultant force
acting at any instant to the corresponding acceleration.the second
value of eq. (a) express the instantaneous value of the
acceleration in terms of the instantaneous velocity Work Consider a
body subjected to the constant forces shown which move the body up
the incline.
Selecting the X-axis as positive in the direction of motion, the
resultant of the unbalanced force system is:
Whence, multiplying both sides of Eq. (a) by S, we find the
resultant work afer moving a displacement S is expressed by : Eq.
(b)
The first expression I n the term(b) is called the accelerating
work or Positive Work. The other expression are known as retarding
work or negative work.the algebraic sum of positive and negative
work is called resultant work.
Application of the Work-Energy Methodif a body is subjected to
different sets of forces during the different phases of motion, the
resultant work summed up for all this phases may be equated
directly to the total change in Kinetic Energy.
General Plan for Applying the Work-Energy Method 1) Determine
the direction of the motion. Confirmation is obtained by noting
that the resultant work must be positive to speed up a system, and
vice versa. 2) Determine the kinematic relation between the bodies
composing the system. 3) Apply the Work-Energy equation to the
entire system. 4) If the internal force in a connecting member is
desired, apply the work-energy equation to the free body diagram of
that part of the system on which this force then acts as an
internal force. If the internal force then acts as an external
force. If the internal force is not constant, however this step
will determine only its average value. The instantaneuos value of a
variable force must be found by the force-inertia method.
ExampleWhat force P will give the system of bodies shown in figure
a velocity of 30 after moving 20 ft.from rest
/s
= 89.88 + 200cos45(0.2) + 200sin45
=
Find the velocity of body A in fig. after it has moved 10 ft
from rest. Assume the pulleys to be weightless and frictionless
=2
1&2 eq. 4 eq. 4 & 3
2(200-12.42 400 24.84 100 = 34.165
= 300 + 9.32 = 300 + 9.32
= 7.65 ft/s
Example A body starts from rest in the position shown in the
figure. Determine its velocity after it has moved 15ft long the
frictionless surface.
Substitute VB to 1,
Impulse and Momentum Introduction: We have seen that the
work-energy method eliminates consideration of acceleration in
problems relating force ,displacement and velocity. This is
particularly convenient when forces act for very small time
intervals during which the force may vary, as in an impact or
sudden blow. We shall also discuss the technique, use, and
advantages of the impulse and momentum method applied to the
several motions. Fundamentals Impulse and Momentum Equation for
Translation. Proceeding as in work energy, let us eliminate the
acceleration from the equation for the motion of the motion of the
center of gravity in terms of velocity. These equation are:
Eliminating a leaves:
Assuming that the velocity has the value the impulse-moment
equation. =
when tis zero and the value and the value v at any form of
The symbol indicates vector subtraction of the terms on the
right side of the equation. We shall in the show following articles
that each term of this equation represents a vector quantity; hence
the necessity for vector instead of algebraic subtraction.
Resultant Linear Impulse . the expression is known as the resultant
linear impulse. Obviously this
expression can be evaluated only if R is constant or can be
expression dimensionally shows the unit of the linear impulse to be
lb-sec. Linear Momentum. The quantity symbolized by U. the quantity
is the linear momentum of the body at any instant and may be may be
represented by dU. Subtituting the dimentional equivalent in
the expression for the linear momentum gives.
Linear Impulse Since the condition of translation means that all
the particles have the same displacement, velocity, and
acceleration, there will be no need to use the bar sign to
distinguish the motion of the center of gravity of the body.
When the applied forces are constant, this equation becomes
Ex. 1) A 1500 N block is in contact with a level plane whose
coefficient of kinetic friction is 0.10. If the block is acted upon
a horizontal force of 250 N, what time will collapse before reaches
a velocity of 14.5 m/sec., starting from rest? If the 250 N force
is then removed, how much longer will the block continue to
move?
a) F = uN F= 0.10(1500) =150N
150)t= t = 22.2sec b) 8 sec. Ex.2) A horizontal force of 1500 N
pushes a 100 N block up an incline whose slope is 3 to 4
horizontal. If k=0.20, determine the time required to increase the
velocity of the block from 3 to 15m/sec.
N = 1000( = 800N F = 0.20(800)= 160N
t =1.65sec Ex. 3) The 100kg crate shown is originally at rest on
a smooth surface. If the force of 200 N acting at an angle 45 is
applied at the crate for 10secs, determine the final velocity of
the crate during the time interval.