Engineering Mechanics Dynamics 2/mohanad... · Engineering Mechanics Dynamics Dynamics is that branch of mechanics which deals with the motion of bodies under the action of forces

Engineering Mechanics

Dynamics

Dynamics is that branch of mechanics which deals with the motion of bodies

under the action of forces. The study of dynamics in engineering usually follows the

study of statics, which deals with the effects of forces on bodies at rest. Dynamics has

two distinct parts: kinematics, which is the study of motion without reference to the

forces which cause motion, and kinetics, which relates the action of forces on bodies to

their resulting motions.

Applications of Dynamics

Only since machines and structures have operated with high speeds and

appreciable accelerations has it been necessary to make calculations based on the

principles of dynamics rather than on the principles of statics. The rapid technological

developments of the present day require increasing application of the principles of

mechanics, particularly dynamics.

These principles are basic to the analysis and design of moving structures, to

fixed structures subject to shock loads, to robotic devices, to automatic control systems,

to rockets, missiles, and spacecraft, to ground and air transportation vehicles, to electron

ballistics of electrical devices, and to machinery of all types such as turbines, pumps,

reciprocating engines, hoists, machine tools, etc.

Definition Space is the geometric region occupied by bodies. Position in space is

determined relative to some geometric reference system by means of linear and angular

measurements.

Time Is a measure of the succession of events and is considered an absolute

quantity in Newtonian mechanics.

Mass Is the quantitative measure of the inertia or resistance to change in motion

of a body. Mass may also be considered as the quantity of matter in a body as well as

the property which gives rise to gravitational attraction.

Force A force has been defined as an action of one body on another in statics.

In dynamics we will see that a force is defined as an action which tends to cause

acceleration of a body. A force is a vector quantity, because its effect depends on the

direction as well as on the magnitude of the action.

Thus, forces may be combined according to the parallelogram law of vector

addition.

particle Is a body of negligible dimensions. When the dimensions of a body are

irrelevant to the description of its motion or the action of forces on it, the body may be

treated as a particle. An airplane, for example, may be treated as a particle for the

description of its flight path.

rigid body Is a body whose changes in shape are negligible compared with the

overall dimensions of the body or with the changes in position of the body as a whole.

As an example of the assumption of rigidity, the small flexural movement of the wing

tip of an airplane flying through turbulent air is clearly of no consequence to the

description of the motion of the airplane as a whole along its flight path.

Scalars: only magnitude is associated.

Ex: time, volume, density, speed, energy, mass

Vectors: possess direction as well as magnitude, and must obey the parallelogram law of addition (and the triangle law).

Ex: displacement, velocity, acceleration, force, moment, momentum

Equivalent Vector: V = V1 + V2 (Vector Sum) parallelogram law of vector

addition

Note: Speed is the magnitude of velocity.

Newton’s Laws

Law I. A particle remains at rest or continues to move with uniform velocity (in

a straight line with a constant speed) if there is no unbalanced force acting on it.

Law II. The acceleration of a particle is proportional to the resultant force acting

on it and is in the direction of this force.

Law III. The forces of action and reaction between interacting bodies are equal

in magnitude, opposite in direction, and collinear.

Newton’s second law forms the basis for most of the analysis in dynamics. For a

particle of mass m subjected to a resultant force F, the law may be stated as

F = ma (1/1)

where a is the resulting acceleration measured in a nonaccelerating frame of

reference.

Newton’s first law is a consequence of the second law since there is no

acceleration when the force is zero, and so the particle is either at rest or is moving with

constant velocity.

The third law constitutes the principle of action and reaction with which you

should be thoroughly familiar from your work in statics.

Chapter 12

Kinematics of a Particle

12.1 Introduction

Mechanics is a branch of the physical sciences that is concerned with the

state of rest or motion of bodies subjected to the forces action .

Engineering mechanics is divided into two areas of study namely statics and

dynamics. statics is concerned with the equilibrium of a body that is either at rest or

move with constant velocity. Here we will consider dynamics, which deals with the

accelerated motion of a body. The subject of dynamic will be presented in two part :

kinematics which treat only the geometric aspect of the motion and kinetics, which is

the analysis of the forces causing the motion. To develop these principles, the dynamics

of a particle will be discussed first , followed by topics in rigid-body dynamics in two

and then three dimensions.

12.2 Rectilinear Kinematics: Continuous Motion

We will begin our study of dynamic. by discussing the kinematic of a particle

that moves along a rectilinear or straight-line path. Recall that a particle has a mass but

negligible size and shape. Therefore we must limit application to those object that have

dimensions that are of no consequence in the analysis of the motion. In most problems,

we will be interested in bodies of finite size, such as rockets, projectiles or vehicles.

Each of these objects can be considered as a particle, as long as the motion is

characterized by the motion of its mass center and any rotation of the body is neglected.

Rectilinear Kinematics. The kinematics of a particle is characterized by

specifying at any given instant, the particle's position velocity, and acceleration.

Position. The straight-line path of a particle defined by using a single coordinate

axis s, Fig. 12-1a. The origin 0 on the path is a fixed point, and from this point the

position coordinates is used to specify the location of the particle at any given instant.

The magnitude of s is the distance from 0 to the particle, usually measured in meters

(m) or feet (ft) and the sense of direction is defined by the algebraic sign on s. Although

the choice is arbitrary, in this case s is positive since the coordinate axis is positive to

the right of the origin. Likewise it is negative if the particle is located to the left of 0 .

Realize that position is a vector quantity since it has both magnitude and direction. Here

however, it is being represented by the algebraic scalar s since the direction always

remains along the coordinate axis.

Displacement. The displacement of the particle is defined as the change in its

position. For example, if the particle moves from one point to another, Fig. 12-1 b, the

displacement is

∆푠 = 푠 − 푠

In this case ∆푠 is positive since the

particle's final position is to the right of its

initial position i.e., s' > s. Likewise if the

final position were to the left of it initial

position ∆푠 would be negative.

The displacement of a particle is

also a vector quantity and it should be

distinguished from the distance 풔푻 the

particle travels . Specifically the distance

traveled is a positive scalar that

represents the total length of path over

which the particle travels.

Velocity. If the particle move through a displacement ∆s during the time interval

∆t the average velocity of the particle during this time interval is

푣 =∆∆

If we take smaller and smaller values of ∆t, the magnitude of ∆s becomes

smaller and smaller. Consequently, the instantaneous velocity is a vector defined as

푣 = lim∆ →

∆푠∆푡 , or

→ 푣 = (12.1)

Since ∆t or dt is always positive

the sign used to define the sense of the

velocity is the same as that of ∆s or ds.

For example if the particle is moving to

the right, Fig. 12-1 c, the velocity is

positive; whereas if it is moving to the left, the velocity is negative.( This is emphasized

here by the arrow written at the left of Eq. 12- 1.) The magnitude of the velocity is

known as the speed and it is generally expressed in unit of m/s or ft/s .

Occasionally the term average speed" is used. The average speed is always a

positive scalar and is defined as the total distance traveled by a particle sT divided

by the elapsed time ∆t ; i.e.

풗풔풑 풂풗품=풔푻∆풕

For example, the particle in Fig.12-1d travel along the path of length sT in time

∆t, so its average speed is (vsp)avg = sT/∆t, but its average velocity is vavg = - ∆s / ∆t .

Acceleration. Provided the velocity of

the particle is known at two point , the average

acceleration of the particle during the time interval ∆t is defined as

푎 =∆푣∆푡

Here v represents the difference in the

velocity during the time interval ∆t, i.e. ∆v =

푣 − 푣 ,Fig. 12- 1e.

The instantaneous acceleration at time t is a vector that is found by taking

mailerand mailer values of ∆t and corresponding smaller and smaller value of ∆v , so

that

a = lim∆ →∆∆

or

→ 풂 = 풅풗풅풕

(12.2)

Substituting Eq.1 2-1 into this result we can also write

→ 풂 =풅ퟐ풔풅풕ퟐ

Both the average and instantaneous acceleration can be either positive or

negative. In particular when the particle is slowing down, or its speed is decreasing the

particle is said to be decelerating. In this case v' in Fig. 12-1f is less than v and so

∆v =v '- v will be negative. Consequently, a will also be negative, and therefore it will

act to the left in the opposite sense to v.

Also, notice that if the particle is originally at rest, then it can have an

acceleration if a moment later it has a velocity v'; and if the velocity is constant, then the

acceleration is zero since ∆v =v '- v = 0. Units commonly used to express the

magnitude of acceleration are m/s2 or ft/s2.

Finally, an important differential relation involving the displacement, velocity,

and acceleration along the path may be obtained by eliminating the time differential (dt)

between Eqs. 12-1 and12- 2 which gives

(→) ads = v dv (12-3)

Although we have now produced three important kinematic equations realize

that the above equation is not independent of Eqs. 12-1 and12- 2.

Constant Acceleration

IF a=ac When the acceleration is constant, each of the three kinematic

equation v = ds / dt , ac= dv/dt , and ac ds = v dv can be integrated to obtain formula

that relate ac , v , s, and t.

Velocity as a Function of Time. Integrate ac= dv/dt , assuming that initially

v = v0 when t = 0.

푑푣 = 푑푡

→ 풗 = 풗ퟎ +풂풄풕 (12-4)

Position as a Function of Time. Integrate v = ds/ dt = v0 + act ,

Assuming that initially s= s0 when t = 0.

푑푠 = (풗ퟎ +풂풄풕)푑푡

→ 푠 = 푠 + 푣 푡 + 푎 푡 (12-5)

푐푛푠푡푎푛푎푡푎푐푐푒푙푒푟푎푡푖표푛

Velocity as a Function of Position. Either solve for t in Eq.12-4 and substitute

into Eq.12- 5 or integrate v dv = ac d assuming that initially v = v0 at s = s0 .

푣푑푣 = 풂풄 푑푠

→ 푣 = 푣 + 2푎 (푠 −푠 ) (12-6)

푐푛푠푡푎푛푎푡푎푐푐푒푙푒푟푎푡푖표푛

The algebraic signs of s0 v0 , and ac, used in the above three equations, are

determined from the positive direction of the s axis as indicated by the arrow written at

the left of each equation. Remember that the equation are useful only when the

acceleration is constant and when t = 0, s = s0 , v = v0 . A typical example of constant

accelerated motion occurs when a body falls freely toward the earth. If air resistance

is neglected and the distance of fall is short then the downward

acceleration of the body when it is close to the earth is constant and

approximately 9. 1 m/ s2 or 32.2 ft/s2.

Important point

• Dynamics is concerned with bodies that have accelerated motion.

• Kinematic is a study of the geometry of the motion.

• Kinetics is a study of the forces that cause the motion.

• Rectilinear kinematics refers to straight-line motion.

• Average speed is the total distance traveled divided by the total

time. This is different from the average velocity, which is the

displacement divided by the time.

• A particle that is slowing down is decelerating.

• A particle can have an acceleration and yet have zero velocity.

• The relationship a ds = v dv is derived from a = dv / dt and

v = ds/dt , by eliminating dt.

Procedure for Analysis

Coordinate System.

• Establish a position coordinate s along the path and specify its fixed origin and

positive direction.

• Since motion is along a straight line, the vector quantities position, velocity, and

acceleration can be represented as algebraic scalars. For analytical work the sense of s,

v, and a is then defined by their algebraic signs.

• The positive sense for each of these scalars can be indicated by an arrow shown

alongside each kinematic equation as it is applied.

Kinematic Equations.

• If a relation is known between any two of the four variables a, v. s, and t. then a

third variable can be obtained by using one of the kinematic equations, a = dv/dt ,

v =ds/dt or ads= v dv since each equation relates all three variables.

• Whenever integration is performed, it is important that the position and velocity

be known at a given instant in order to evaluate either the constant of integration if an

indefinite integral is used, or the limits of integration if a definite integral is used.

• Remember that Eqs. 12-4 through 12-6 have only limited use. These equations

apply only when the acceleration is constant and the initial conditions are

s=s0, v=v0 when t =0.

*Some standard differentiation and integration formula are given in Appendix A.

EXAMPLE 12.1

The car on the left in the photo and in Fig. 12-2 moves in a straight line such that for a short time its velocity is defined by v = (312 + 21) ft/s, where t is in seconds. Determine its position and acceleration when / = 3 s. When / = 0, s = 0.

SOLUTION

Coordinate System. The position coordinate extends from the fixed origin O to the car, positive to the right.

Position. Since v = f(t), the car’s position can be determined from v = ds/dt, since this equation relates v, s, and t. Noting that s = 0 when t = 0, we have*

→ 푣 = (12.1)

푑푠 = (ퟑ풕ퟐ + ퟐ풕)푑푡

푠| = 푡 + 푡 | When t=3s S = 33 + 32 = 36 ft

Acceleration. Since v = f(t) the acceleration is determined from

a = dv/dt , since this equation relates a, v and c.

→ 푣 = = (3푡 + 2푡)

= 6t + 2

Where : t = 3

a = 6(3) + 2 = 20 ft/ s2 Ans.

NOTE: The formulas for constant acceleration cannot be used to solve

this problem, because the acceleration is a function of time.

*The same result can be obtained by evaluating a constant of integration rather than using definite limit on the integral. For example, integrating

ds = 3t2 + 2t)dt yield s = t3 + t2 C. Using the condition that at t = 0. s = 0. then C = 0.

Rectilinear Kinematics: Erratic Motion

When a particle has erratic or changing motion then its position, velocity,

and acceleration cannot be described by a single continuous mathematical function

along the entire path. Instead, a series of functions will be required to specify the motion

at different intervals. For this reason, it is convenient to represent the motion as a graph.

This method describes the geometric relationships between the acceleration, velocity,

and position diagrams of a particle undergoing rectilinear motion. These relationships

then are used to develop a simple numerical method for analyzing rectilinear motion

called the area method, which enables us to construct velocity and position diagrams

from a given acceleration diagram. The area method is particularly useful in cases

where the acceleration diagram is made up of straight lines.(what is that mean? And

why?

If a graph of the motion that relates any two of the variables s, v, a, t can be

drawn, then this graph can be used to construct subsequent graphs relating two other

variables since the variables are related by the differential relationships

v = ds/dt, a= dv/dt and ads = v dv.

The s-t, v-t, and a-t Graphs.

1. To construct the ( v-t ) graph given the( s-t )graph, Fig. 12-2, the equation

v = ds/dt should be used, since it relates the variables

s and t to v. This equation states that

slope of s-t graph = velocity

The slope of the position diagram at time ti is equal to

the velocity at that time, that is, (ds/dt)i = vi , as shown

in Fig. 12-2 a . For example, by measuring the slope

on the s-t graph when t = t1, the velocity is v1 , which

is plotted in Fig. 12-2 b. Then ,the v-t graph can be

constructed by plotting this and other values at each

instant.

NOW To construct the ( s-t ) graph given the(v-t )graph Fig. 12-3, it is possible to determine the s-t graph using v = ds/dt, is written as :

∆s = ∫v dt = area under the v-t graph

We begin with the particle's initial position

So and add (algebraically) to this small area

increments ∆s determined from the v-t graph. we

can write :

푠 − 푠 = ∫ 푣(푡)푑푡

Fig. 12-2a

Fig. 12-2b

FIG.12-3 (a)

Because the right-hand side of this equation is the area of the velocity diagram between t0 and tn, we arrive at

푠 − 푠 = 푎푟푒푎표푓푣 − 푡푑푖푎푔푟푎푚] (12-7)

2. The( a-t) graph can be constructed from the( v-t ) graph

In a similar manner as in Fig.12-4, the equation a = dv/dt should be used, since it

relates the variables v and t to a . This equation states that

slope of v-t graph = acceleration The slope of the velocity diagram at time ti

is equal to the acceleration at that time;

that is, (dv/dt)i = ai , as shown in Fig.

12.4(b). For example, by measuring the

slope on the v - t graph when t = t1, the

velocity is v1 , which is plotted in Fig.12-4.

Then ,the a-t graph can be constructed by

plotting this and other values at each

instant.

NOW To construct the (v-t) graph given

the(a-t)graph Fig. 12-5, it is possible to

determine the v-t graph using v = ds/dt, is

written as :

Fig.12-4(a)

∆v = ∫a dt = area under the a-t graph

Hence, to construct the v-t graph, we begin with the

particle's initial velocity vo and then add to this small

increments of area (∆v) determined from the a-t

graph. We can write :

푣 − 푣 = ∫ 푎(푡)푑푡

Because the right-hand side of this equation is the

area of the velocity diagram between t0 and tn,

we arrive at

푣 − 푣 = 푎푟푒푎표푓푎 − 푡푑푖푎푔푟푎푚] (12-8)

In this manner successive points, vi = vo +∆v, etc., for the v-t graph are determined,

Fig. 12-5b. Notice that an algebraic addition of the area increments of the a-t graph

is necessary, since areas lying above the( t axis )correspond to an increase in v

("positive" area), whereas those lying below the axis indicate a decrease in v

("negative" area).

If segments of the a-t graph can be described by a series of equations, then each of these

equations can be integrated to yield equations describing the corresponding segments of

Fig.12-4.(b)

Fig.12-5(a)

Fig.12-5(b)

Fig. 12.5 –(b)

the v-t graph. In a similar manner, the s-t graph can be obtained by integrating the

equations which describe the segments of the v-t graph. As a result, if the a-t graph

is linear (a first-degree curve), integration will yield a v-t graph that is parabolic (a

second-degree curve) and an s-t graph that is cubic (third degree curve).

The v-s and a-s Graphs.

If the a-s graph can be constructed, then points on the v-s graph can be determined by

using v dv = a d s. Integrating this equation between the limits v = vo at s = so and

v = v1at s = s1 , we have, (푣 − 푣 ) = ∫ 푎푑푠 = 푎푟푒푎푢푛푑푒푟(푎 − 푠)푔푟푎푝ℎ

Therefore, if the black area in Fig. 12.6-(a) is determined, and the initial velocity vo at

So = 0 is known, then 푣 = 2∫ 푎푑푠 + 푣/

Fig. 12-6b. Successive points on the v-s graph can

be constructed in this manner. If the v-s graph is

known, the acceleration a at any position s can be

determined using a ds = v dv, written as

푎 = 푣푑푣푑푠

a=Acceleration = velocity times slope of v-s

graph

Fig. 12.6 –(a)

Fig. 12.6 –(b)

Curvilinear Motion: Rectangular Components

Occasionally the motion of a particle can best be described along a path

that can be expressed in terms of its x, y, z coordinates.

Position. If the particle is at point (x, y, z) on the curved path s

shown in Fig. 12-17a then its location i defined by the position vector

When the particle moves the x, y, z components of r will be functions of

time· i.e. x = x(t), y = y(t), z=z (t), so that r = r(t).

And the direction of r is specified by the unit vector Ur = r / r.

Velocity. . h fir t time derivative of r yield. the velocity of the

particle. Hence

When taking this derivative it is necessary to account for changes in both

the magnitude and direction of each of the vector component . For

example, the derivative of the i component of r is

The second term on the right side is zero, provided the x y, z reference

frame i fixed , and therefore the direction (and the magnitude) of i doe

not change with time. Differentiation of the j and k component may be

carried out in a similar manner, which yield the final result,

Motion of a Projectile

The free-flight motion of a projectile is often studied in term of its

rectangular component . To illustrate the kinematic analysis , consider a

projectile launched at point (x0 , y0) with an initial velocity of v0 , having

component (v0)x and (v0)_v Fig. 12- 20. When air resistance is neglected,

the only force acting on the projectile is it weight which cause the

projectile to have a constant downward acceleration

The fir t and last equation indicate that the horizontal component of

velocity always remains constant during the motion.

Vertical Motion. Since the positive y axis is directed upward, then

ay = - g.

Recall that the last equation can be formulated on the basis of eliminating

the time t from the first two equations, and therefore only two of the above

three equations are independent of one another.

Curvilinear Motion: Normal and Tangential Components

When the path along which a particle travels is known then it is often

convenient to describe the motion using n and t coordinate axe which act

normal and tangent to the path, respectively, and at the instant considered

have their origin located at the particle.

Planar Motion. Consider the particle shown in fig. 12- 24a, which

move in a plane along a fixed curve, such that at a given instant it i at

positions measured from point 0. We will now consider a coordinate

system that ha its origin on the curve and at the instant considered this

origin happens to coincide with the location of the particle. The taxis is

tangent to the curve at the point and is positive in the direction of

increasing. We will designate this position direction with the unit vector

ut . A unique choice for the normal axis can be made by noting that

geometrically the curve is constructed from a series of differential arc

segments ds fig. 12- 24b. each segment ds is formed from the arc of an

a associated circle having a radius of curvature p (rho) and center of

curvature O'. The normal axis n is perpendicular to the taxis with its

positive sense directed toward the center of curvature O' , Fig. 12-24a.

This positive direction which is always on the concave side of the curve

will be designated by the unit vector un • The plane which contains the n

and taxes is referred to as the embracing or osculating plane, and in this

case it is fixed in the plane of motion.

Velocity. Since the particle moves, s is a function of time. As indicated

in Sec. 12.4 the particle's velocity v has a direction that is always tangent

to the path, Fig. 12- 24c, and a magnitude that is determined by taking the

time derivative of the path function s = (t), i.e., v = ds/ dt ( q. 12- ).

Hence

To better understand the e results consider the following two special

cases of motion.

1. If the particle moves along a straight line, then ρ→∞and from

Eq. 12-20 an = 0. Thus a = at = 푣, and we can conclude that the

tangential component of acceleration represents the time rate of

change in the magnitude of the velocity.

2. If the particle move along a curve with a constant speed then

At = 푣 = 0 and a = an = v2 / ρ . Therefore, the normal component

of acceleration represents the time rate of change in the direction of

the velocity. Since an always acts towards the center of curvature

this component is sometimes referred to as the centripetal (or center

seeking) acceleration.

A a result of the e interpretations a particle moving along the cur cd

path in Fig. 25 will have accelerations directed a shown.

Acceleration. The acceleration of the particle is the time rate of

change of the velocity. Thus

In order to determine the time derivative 푢 , note that a the particle

move along the arc d in time dt , ut preserves its magnitude of unity;

however, it direction changes, and become u;, Fig. 24d. As shown in

Fig. 24e, we require u't = ut + dut • Here dut stretches between the

arrowhead of u, and u; which lie on an infinitesimal arc of radius u, = I.

Hence du, has a magnitude of dut = dɵ ut }, and its direction is defined by

un consequently, du, = dfJu,1 and therefore the time derivative become

Absolute Dependent Motion Analysis of Two Particles

In ome types of problem the motion of one particle will depend on the

corresponding motion of another particle. Thi dependency commonly

occurs if the particles, here represented by blocks are interconnected by

inextensible cord which are wrapped around pulleys. For example the

movement of block A downward along the inclined plane in Fig. 12-36

will cause a corresponding movement of block B up the other incline. We

can how thi mathematically by fir. t specifying the location of the blocks

u ing position coordinates A and s8 . Note that each of the coordinate axis (l) measured

from a fixed point (0) or fixed datum line, (2) measured

along each inclined plane in the direction of motion of each block, and

(3) has a positive en c from the fixed datum to A and to B. If the total

cord length is Lt, the two position coordinates are related by the equation

Here LCD is the length of the cord passing over arc CD. Taking

the time derivative of this expression, realizing that Leo and Lt remain constant while

sA and sb measure the segment of the cord that change in length, we have

The negative sign indicates that when block A has a velocity downward

i.e. in the direction of positive sA , it causes a corresponding upward

velocity of block B· i.e. B moves in the negative s8 direction.

In a similar manner time differentiation of the velocities yields the

relation between the accelerations i.e.,

A more complicated example is shown in Fig. 12- 37a. In this case the

position of block A is specified by sA, and the position of the end of the

cord from which block B is suspended is defined by sB. As above we have

chosen position coordinates which (1) have their origin at fixed points

or datum's (2) are measured in the direction of motion of each block, and

(3) from the fixed datum are positive to the right for sA and positive

downward for 8 . During the motion , the length of the red color d

segment of the cord in Fig. 37a remains constant. If l represent the

total length of cord minus these segment , then the position coordinates

can be related by the equation

2s8 + h + sA = I

since I and h arc constant during the motion the two time derivative yield

Hence, when B move downward ( + 8) , A moves to the left (- A) with

twice the motion.

This example can also be worked by defining the position of block B

from the center of the bottom pulley (a fixed point) , Fig. 12- 37b. In

this case

2(h - s8) + h + sA = I

ime differentiation yield.

2vb = vA

Here the sign arc the same.

7-1

Chapter 2 Newton’s Laws of Motion

I have not as yet been able to discover the reason for these properties of gravity from phenomena, and I do not feign hypotheses. For whatever is not deduced from the phenomena must be called a hypothesis; and hypotheses, whether metaphysical or physical, or based on occult qualities, or mechanical, have no place in experimental philosophy. In this philosophy particular propositions are inferred from the phenomena, and afterwards rendered general by induction. 1

2.1 Force and Quantity of Matter

Isaac Newton

In our daily experience, we can cause a body to move by either pushing or pulling that body. Ordinary language use describes this action as the effect of a person’s strength or force. However, bodies placed on inclined planes, or when released at rest and undergo free fall, will move without any push or pull. Galileo referred to a force acting on these bodies, a description of which he published in 1623 in his Mechanics. In 1687, Isaac Newton published his three laws of motion in the Philosophiae Naturalis Principia Mathematica (“Mathematical Principles of Natural Philosophy”), which extended Galileo’s observations. The First Law expresses the idea that when no force acts on a body, it will remain at rest or maintain uniform motion; when a force is applied to a body, it will change its state of motion.

Many scientists, especially Galileo, recognized the idea that force produces motion before Newton but Newton extended the concept of force to any circumstance that produces acceleration. When a body is initially at rest, the direction of our push or pull corresponds to the direction of motion of the body. If the body is moving, the direction of the applied force may change both the direction of motion of the body and how fast it is moving. Newton defined the force acting on an object as proportional to the acceleration of the object.

An impressed force is an action exerted upon a body, in order to change its state, either of rest, or of uniform motion in a right line.2

In order to define the magnitude of the force, he introduced a constant of

proportionality, the inertial mass, which Newton called “quantity of matter”.

7-2

a

s

F F

a a

F as1

The quantity of matter is the measure of the same, arising from its density and bulk conjointly.

Thus air of double density, in a double space, is quadruple in quantity; in a triple space, sextuple in quantity. The same thing is to be understood of snow, and fine dust or powders, that are condensed by compression or liquefaction, and of all bodies that are by any causes whatever differently condensed. I have no regard in this place to a medium, if any such there is, that freely pervades the interstices between the parts of bodies. It is this quantity that I mean hereafter everywhere under the name of body or mass. And the same is known by the weight of each body, for it is proportional to the weight, as I have found by experiment on pendulums, very accurately made, which shall be shown hereafter.3

Suppose we apply an action to a body (which we refer to as the standard body) that will induce the body to accelerate with a magnitude

that can be measured by an

accelerometer (any device that measures acceleration). The magnitude of the force F

acting on the object is the product of the mass m with the magnitude of the acceleration a . Force is a vector quantity. The direction of the force on the standard body is defined

to be the direction of the acceleration of the body. Thus F ms a (7.1.1)

In order to justify the statement that force is a vector quantity, we need to apply two

forces

and simultaneously to our body and show that the resultant force FT is the

vector sum of the two forces when they are applied one at a time.

Figure 7.1 Acceleration add as vectors Figure 7.2 Force adds as vectors.

We apply each force separately and measure the accelerations and , noting that

m

(7.1.2)

3 Ibid. p. 1.

1 2

1 2.

1

7-3

F as2

a

s

F F1 2

m

. (7.1.3)

When we apply the two forces simultaneously, we measure the acceleration . The force by definition is now

FT m a . (7.1.4) We then compare the accelerations. The results of these three measurements, and for that matter any similar experiment, confirms that the accelerations add as vectors (Figure 7.1)

a a1 a2 . (7.1.5)

Therefore the forces add as vectors as well (Figure 7.2),

FT F F . (7.1.6)

1 2

This last statement is not a definition but a consequence of the experimental result described by Equation (7.1.5) and our definition of force.

Example 7.1 Vector Decomposition Solution Two horizontal ropes are attached to a post that is stuck in the ground. The ropes pull the

post producing the vector forces 70 N i 20 N j and

30 N i 40 N j as

shown in Figure 7.1. Find the direction and magnitude of the horizontal component of a third force on the post that will make the vector sum of forces on the post equal to zero.

Figure 7.3 Example 7.1

Figure 7.4 Vector sum of forces

Solution: Since the ropes are pulling the post horizontally, the third force must also have a horizontal component that is equal to the negative of the sum of the two horizontal forces exerted by the rope on the post Figure 7.4. Since there are additional vertical

2

7-4

3

F 1 y

(40 N)2 (60 N)2 F

in

s

in in

am a

forces acting on the post due to its contact with the ground and the gravitational force exerted on the post by the earth, we will restrict our attention to the horizontal component of the third force. Let F denote the sum of the forces due to the ropes. Then we can write

3

the vector F as

3 ( F1x F2x ) i ( F F2 y ) j (70 N + 30 N) i (20 N + 40 N) j

(40 N) i (60 N) j Therefore the horizontal component of the third force of the post must be equal to

F F (F F ) = (40 N) i (60 N) j .

hor 3 1 2

The magnitude is

hor 72 N . The horizontal component of the force makes an angle

tan1 60 N 56.3

40 N

as shown in the figure above.

2.1.1 Mass Calibration

So far, we have only used the standard body to measure force. Instead of performing experiments on the standard body, we can calibrate the masses of all other bodies in terms of the standard mass by the following experimental procedure. We shall refer to the mass measured in this way as the inertial mass and denote it by m .

We apply a force of magnitude F to the standard body and measure the

magnitude of the acceleration a . Then we apply the same force to a second body of

unknown mass m and measure the magnitude of the acceleration a . Since the same force is applied to both bodies,

F min ain ms as , (1.7)

Therefore the ratio of the inertial mass to the standard mass is equal to the inverse ratio of the magnitudes of the accelerations,

m in s

s in

. (1.8)

Therefore the second body has inertial mass equal to

7-5

aa

m m s . (1.9) in s

in

This method is justified by the fact that we can repeat the experiment using a different force and still find that the ratios of the acceleration are the same. For simplicity we shall denote the inertial mass by m .

2.2Newton’s First Law

The First Law of Motion, commonly called the “Principle of Inertia,” was first realized by Galileo. (Newton did not acknowledge Galileo’s contribution.) Newton was particularly concerned with how to phrase the First Law in Latin, but after many rewrites Newton perfected the following expression for the First Law (in English translation):

Law 1: Every body continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed upon it.

Projectiles continue in their motions, so far as they are not retarded by the resistance of air, or impelled downwards by the force of gravity. A top, whose parts by their cohesion are continually drawn aside from rectilinear motions, does not cease its rotation, otherwise than as it is retarded by air. The greater bodies of planets and comets, meeting with less resistance in freer spaces, preserve their motions both progressive and circular for a much longer time.4

The first law is an experimental statement about the motions of bodies. When a

body moves with constant velocity, there are either no forces present or there are forces acting in opposite directions that cancel out. If the body changes its velocity, then there must be an acceleration, and hence a total non-zero force must be present. We note that velocity can change in two ways. The first way is to change the magnitude of the velocity; the second way is to change its direction.

After a bus or train starts, the acceleration is often so small we can barely perceive

it. We are often startled because it seems as if the station is moving in the opposite direction while we seem to be still. Newton’s First Law states that there is no physical way to distinguish between whether we are moving or the station is, because there is essentially no total force present to change the state of motion. Once we reach a constant velocity, our minds dismiss the idea that the ground is moving backwards because we think it is impossible, but there is no actual way for us to distinguish whether the train is moving or the ground is moving.

7-6

p

7.2 Momentum, Newton’s Second Law and Third Law

Newton began his analysis of the cause of motion by introducing the quantity of motion:

Definition: Quantity of Motion:

The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjointly.

The motion of the whole is the sum of the motion of all its parts; and therefore in a body double in quantity, with equal velocity, the motion is double, with twice the velocity, it is quadruple.5

Our modern term for quantity of motion is momentum and it is a vector quantity

p mv . (7.3.1)

where m is the inertial mass and v is the velocity of the body (velocity is a vector quantity). Newton’s Second Law is the most important experimental statement about motion in physics.

Law II: The change of motion is proportional to the motive force impressed, and is made in the direction of the right line in which that force is impressed.

If any force generates a motion, a double force will generate double the motion, a triple force triple the motion, whether that force is impressed altogether and at once or gradually and successively. And this motion (being always directed the same way with the generating force), if the body moved before, is added or subtracted from the former motion, according as they directly conspire with or are directly contrary to each other; or obliquely joined, when they are oblique, so as to produce a new motion compounded from the determination of both.6

Suppose that a force is applied to a body for a time interval t . The impressed force or impulse (a vector quantity I ) that we denote by produces a change in the momentum of the body,

I Ft . (7.3.2)

5 Ibid. p. 1. 6 Ibid. p. 13.

7-7

F m . (7.3.5)

From the commentary to the second law, Newton also considered forces that were applied continually to a body instead of impulsively. The instantaneous action of the total force acting on a body at a time t is defined by taking the mathematical limit as the time interval t becomes smaller and smaller,

F lim p dp . (7.3.3)

t0 t dt

When the mass remains constant in time, the Second Law can be recast in its more familiar form,

F m dv

. (7.3.4) dt

Because the derivative of velocity is the acceleration, the force is the product of mass and acceleration,

a

Because we defined force in terms of change in motion, the Second Law appears to be a restatement of this definition, and devoid of predictive power since force is only determined by measuring acceleration. What transforms the Second Law from just merely a definition is the additional input that comes from force laws that are based on experimental observations on the interactions between bodies. Throughout this book, we shall investigate these force laws and learn to use them in order to determine the left-hand side of Newton’s Second Law.

The right-hand-side of Newton’s Second Law is the product of mass with

acceleration. Acceleration is a mathematical description of how the velocity of a body changes. If we know the acceleration of a body we can in principle, predict the velocity and position of that body at all future times by integration techniques.

7.3 Newton’s Third Law: Action-Reaction Pairs

Newton realized that when two bodies interact via a force, then the force on one body is equal in magnitude and opposite in direction to the force acting on the other body.

Law III: To every action there is always opposed an equal reaction: or, the mutual action of two bodies upon each other are always equal, and directed to contrary parts.

Whatever draws or presses another is as much drawn or pressed by that other. If you press on a stone with your finger, the finger is also pressed by the stone.7

7-8

1, 2 F

F

The Third Law, commonly known as the “action-reaction” law, is the most surprising of the three laws. Newton’s great discovery was that when two objects interact, they each exert the same magnitude of force on each other. We shall refer to objects that interact as an interaction pair.

Consider two bodies engaged in a mutual interaction. Label the bodies 1 and 2

respectively. Let F be the force on body 2 due to the interaction with body 1, and

2,1

be the force on body 1 due to the interaction with body 2. These forces are depicted in Figure 7.5.

Figure 7.5 Action-reaction pair of forces These two vector forces are equal in magnitude and opposite in direction,

1, 2

F2,1

. (7.4.1)

With these definitions, the three laws and force laws constitute Newtonian

Mechanics, which will be able to explain a vast range of phenomena. Newtonian mechanics has important limits. It does not satisfactorily explain systems of objects moving at speeds comparable to the speed of light ( v 0.1 c ) where we need the theory of special relativity, nor does it adequately explain the motion of electrons in atoms, where we need quantum mechanics. We also need general relativity and cosmology to explain the large-scale structure of the universe.

Kinetics of Particles: Work and Energy

ME101 - Division III Kaustubh Dasgupta 1

Total work done is given by: Modifying this eqn to account for the potential energy terms:

' U’1-2 + (-ΔVg) + (-ΔVe) = ΔT U12 T

V

U’1-2 is work of all external forces other than the gravitational and spring forces

ΔT is the change in kinetic energy of the particle ΔV is the change in total potential energy

•More convenient form because only the end point positions of the particle and end point lengths of elastic spring are of significance.

'

T1 V1 U12 T2 V2

If the only forces acting are gravitational, elastic, and nonworking constraint forces

E = T+V is the total mechanical energy of the particle and its attached spring

U12 T2 T1 T

T1 V1 T2 V2 or E1 E2



Conservation of Energy •During the motion, only transformation of KE into PE occurs and it can be vice versa

A ball of weight W is dropped from a height h above the ground (datum) •PE of the ball is maximum before it is dropped, at which time its KE is zero.

Total mechanical energy of the ball in its initial position is:

•When the ball has fallen a distance h/2, its speed is: Energy of the ball at mid-height position:



Potential Energy Example: A 3 kg slider is released from rest at position 1 and slides with negligible friction in vertical plane along the circular rod. Determine the velocity of the slider as it passes position 2. The spring has an unstretched length of 0.6 m.

Solution:

•Reaction of rod on slider is normal to the motion does no work U’1-2 = 0 Defining the datum to be at the level of position 1 Kinetic Energy: T1 = 0 and T2 = ½ (3)(v2)2

Gravitational Potential Energies: V1 = 0 and V2 = -mgh = -3(9.81)(0.6) = -17.66 J Initial and final elastic potential energies: V1 = ½kx1

2 =0.5(350)(0.6)2 = 63J and V2 = ½kx22=0.5(350)(0.6√2 – 0.6)2= 10.81J

'

T1 V1 U12 T2 V2

0 + (0+63) + 0 = ½ (3)(v2)2 + (-17.66 + 10.81) v2 = 6.82 m/s


Kinetics of Particles :: Impulse and Momentum

Third approach to solution of Kinetics problems

•Integrate the equation of motion with respect to time (rather than disp.)

•Cases where the applied forces act for a very short period of time (e.g., Impact loads) or over specified intervals of time

Linear Impulse and Linear Momentum

Resultant of all forces acting on a particle equals its time rate of change of linear momentum

Invariability of mass with time!!! Fixed Origin


Kinetics of Particles Linear Impulse and Linear Momentum

Three scalar components of the eqn:

Linear Impulse-Momentum Principle •Describes the effect of resultant force on linear momentum of the particle over a finite period of time

Multiplying the eqn by dt ∑F dt = dG and integrating from t1 to t2

G1 = linear momentum at t1 = mv1

G2 = linear momentum at t1 = mv2

The product of force and time is defined as Linear Impulse of the Force.

Alternatively:

Initial linear momentum of the body plus the linear impulse applied to it equals its final linear momentum

Impulse integral is a vector!!


Kinetics of Particles Linear Impulse and Momentum Impulse-Momentum Equation

It is necessary to write this eqn in component form and then combine the integrated components:

The three scalar impulse- momentum eqns are completely independent

Impulse-Momentum Diagram

In the middle drawing linear impulses due to all external forces should be included (except for those forces whose magnitudes

are negligible) Impulse-Momentum diagrams can also show the components


Kinetics of Particles Linear Impulse and Linear Momentum Impulsive Forces: Large forces of short duration (e.g., hammer impact) •In some cases Impulsive forces constant over time they can be brought outside the linear impulse integral. Non-impulsive Forces: can be neglected in comparison with the impulsive forces (e.g., weight of small bodies)

In few cases, graphical or numerical integration is required to be performed.

The impulse of this force from t1 to t2

is the shaded area under the curve

Conservation of Linear Momentum If resultant force acting on a particle is zero during an interval of time, the impulse momentum equation requires that its linear momentum G remains constant. The linear momentum of the particle is said to be conserved (in any or all dirn).

This principle is also applicable for motion of two interacting particles with equal and opposite interactive forces


Kinetics of Particles: Linear Impulse and Linear Momentum

Example

Solution: Construct the impulse-momentum diagram



Example Solution:

Using the impulse-momentum eqns:

The velocity v2:



Example

Solution: The force of impact is internal to the system composed of the block and the bullet. Further, no other external force acts on the system in the plane of the motion. Linear momentum of the system is conserved G1 = G2

Final velocity and direction:

Kinetics of Particles

Angular Impulse and Angular Momentum Velocity of the particle is Momentum of the particle:

Moment of the linear momentum vector mv about the origin O is defined as Angular Momentum HO of P about O and is given by:

Scalar components of angular momentum:

11



Angular Impulse and Angular Momentum A 2-D representation of vectors in plane A is shown:

Magnitude of the moment of mv @ O = linear momentum mv times the moment arm HO = mvr sinθ This is the magnitude of the cross product HO = r x mv

Units of Angular Momentum: kg.(m/s).m = kg.m2/s or N.m.s



Angular Impulse and Angular Momentum

Rate of Change of Angular Momentum - To relate moment of forces and angular momentum

Moment of resultant of all forces acting on P @ origin:

Differentiating with time:

v and mv are parallel vectors v x mv = 0

Using this vector equation, moment of forces and angular momentum are related

Moment of all forces @ O = time rate of change of angular momentum

Scalar components:



Angular Impulse and Angular Momentum

Angular Impulse-Momentum Principle

-This eqn gives the instantaneous relation between moment and time rate of change of angular momentum

Integrating:

The product of moment and time is defined as the angular impulse. The total angular impulse on m about the fixed point O equals the corresponding change in the angular momentum of m about O. Alternatively:

Initial angular momentum of the particle plus the angular impulse applied to it equals the final angular momentum.


Kinetics of Particles Angular Impulse and Angular Momentum Angular Impulse-Momentum Principle In the component form: The x-component of this eqn: Similarly other components can be written

Plane Motion Applications - In most applications, plane motions are encountered instead of 3-D motion. - Simplifying the eqns Using the scalar form of the principle betn 1 and 2:


Kinetics of Particles Angular Impulse and Angular Momentum

Conservation of Angular Momentum If the resulting moment @ a fixed point O of all forces acting on a particle is zero during an interval of time, the angular momentum of the particle about that point remain constant.

Principle of Conservation of Angular Momentum Also valid for motion of two interacting particles with equal and opposite

interacting forces


Kinetics of Particles Angular Impulse and Angular Momentum

Example

Solution



Impact • Collision between two bodies during a very short period of time. • Generation of large contact forces (impulsive) acting over a very short interval

of time. • Complex phenomenon (material deformation and recovery, generation of heat

and sound) • Line of Impact is the common normal to the surfaces in contact during impact.

Impact primarily classified as two types: Central Impact: Mass centers of two colliding bodies are located on line of

impact Current chapter deals with central impact of two particles. Eccentric Impact: Mass centers are not located on line of impact.

Central Impact Eccentric Impact



Central Impact - Can be classified into two types Direct Central Impact (or Direct Impact)

• Velocities of the two particles are directed along the line of impact • Direction of motion of the particles will also be along the line of impact

Oblique Central Impact (or Oblique Impact) • Velocity and motion of one or both particles is at an angle with the line of

impact. • Initial and final velocities are not parallel.

Direct Central Impact Oblique Central


Kinetics of Particles: Impact

Direct Central Impact Collinear motion of two spheres (v1 > v2) • Collision occurs with contact forces directed

along the line of impact (line of centers) • Deformation of spheres increases until contact

area ceases to increase. Both spheres move with the same velocity.

• Period of restoration during which the contact area decreases to zero

• After the impact, spheres will have different velocities (v’1 & v’2) with v’1 < v’2

During impact, contact forces are equal and opposite. Further, there are no impulsive external forces linear momentum of the system remains unchanged. Applying the law of conservation of linear momentum:

Assumptions • Particles are perfectly smooth and frictionless. • Impulses created by all forces (other than the internal forces of contact) are negligible

compared to the impulse created by the internal impact force. • No appreciable change in position of mass centers during the impact.



Coefficient of Restitution The momentum eqn contains 2 unknowns v’1 & v’2 (assuming that v1 & v2 are known)

Another equation is required Coefficient of Restitution (e)

Using the definition of impulse:

Fr and Fd = magnitudes of the contact forces during the restoration and deformation periods.

t0 = time for the deformation t = total time of contact

Further using Linear Impulse-Momentum Principle, we can write this equation in terms of change in momentum


Kinetics of Particles: Impact Coefficient of Restitution

For Particle 1:

For Particle 2:

Eliminating v0 between the two expressions:

Change in momentum (and thus the velocity) is expressed in the same direction as impulse (and thus the force)

If v1, v2, and e are known, the final velocities v’1 & v’2 can be obtained using the two eqns eqn for e and momentum eqn


Kinetics of Particles: Impact Coefficient of Restitution

Energy Loss during impact • Impact always associated with energy loss (heat, inelastic deformation, etc) • Energy loss may be determined by finding the change in the KE of the system

before and after the impact.

Classical theory of impact: e = 1 Elastic Impact (no energy loss) • Put e = 1 in the eqn v’2 - v’1 = v1 - v2

• Relative velocities before & after impact are equal • Particles move away after impact with the same velocity

with which they approached each other before impact. e = 0 Inelastic or Plastic Impact (max energy loss) • Put e = 0 in the eqn v’1 = v’2

• The particles stick together after collision and move with a common velocity

Real conditions lie somewhere betn these extremes e varies with impact velocity, and size and shape of the colliding bodies. e is considered constant for given geometries and a given combination of contacting

materials. e approaches unity as the impact velocity approaches zero.



Oblique Impact • In-plane initial and final velocities are not parallel • Choosing the n-axis along the line of impact, and the

t-axis along the common tangent. • Directions of velocity vectors measured from t-axis. Initial velocity components are: (v1)n = -v1sinθ1, (v1)t = v1cosθ1

(v2)n = v2sinθ2, (v2)t = v2cosθ2

Four unknowns (v’1)n, (v’1)t, (v’2)n, (v’2)t Four equations are required.

Particles are assumed to be perfectly smooth and frictionless the only impulses exerted on the particles during the impact are due to the internal forces directed along the line of impact (n-axis).

Impact forces acting on the two particles are equal and opposite: F and –F (variation during impact is shown). Since there are no impulsive external forces, total momentum of the system (both particles) is conserved along the n-axis

Line of



Oblique Impact: • total momentum of the system is conserved along

the n-axis (1)

• Component along t-axis of the momentum of each particle, considered separately, is conserved (since there is no impulse on particles along t-direction)

(2 and 3)

t-component of velocity of each particle remains unchanged. • Coefficient of the Restitution is the positive ratio of the recovery impulse to the

deformation impulse the eqn will be applied to the velocity components along n-direction.

(4)

Once the four final velocities are determined, angles θ1’ and θ2’ can be easily determine


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