Dynamics

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


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 . 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 .


#

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


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


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.

Documents

Dynamics