LECTURES ON PHYSICS 1 - ffs.iuh.edu.vnffs.iuh.edu.vn/files/nguyenthingocnu/baigiang/Lectures on Physics 1.… · mathematics, the tool that provides a bridge between theory and experiment

INDUSTRIAL UNIVERSITY OF HO CHI MINH CITY

PH.D. NGUYEN THI NGOC NU

LECTURES ON PHYSICS 1

HO CHI MINH CITY – 2016

Lectures on Physics 1

2

Ph.D. Nguyen Thi Ngoc Nu

CONTENTS

Chapter 1. Physics and Measurement

1.1 Standards of Length, Mass, and Time

1.2 Matter and Model Building

1.3 Dimensional Analysis

1.4 Conversion of Units

1.5 Estimates and Order-of-Magnitude Calculations

1.6 Significant Figures

Chapter 2. Motion in One Dimension

2.1 Position, Average Velocity and Speed

2.2 Instantaneous Velocity and Speed

2.3 Particle Under Constant Velocity

2.4 Acceleration

2.5 One-Dimensional Motion with Constant Acceleration

2.6 Freely Falling Objects

Chapter 4. Motion in Two Dimensions

4.1 The Position, Velocity, and Acceleration Vectors

4.2 Two-Dimensional Motion with Constant Acceleration

4.3 Projectile Motion

4.4 Uniform Circular Motion

4.5 Tangential and Radial Acceleration

4.6 Relative Velocity and Relative Acceleration

Chapter 5. The Laws of Motion

5.1 The Concept of Force

5.2 Newton’s First Law and Inertial Frames

5.3 Mass

5.4 Newton’s Second Law


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5.5 The Gravitational Force and Weight

5.6 Newton’s Third Law

5.7 Analysis Models Using Newton’s Second Law

5.8 Forces of Friction

Chapter 6. Circular Motion and Other Applications of Newton’s Laws

6.1 Extending the Particle in Uniform Circular Motion Model

6.2 Nonuniform Circular Motion

6.3 Motion in Accelerated Frames


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Chapter 1

PHYSICS AND MEASUREMENT

Like all other sciences, physics is based on experimental observations and

quantitative measurements. The main objectives of physics are to identify a limited

number of fundamental laws that govern natural phenomena and use them to

develop theories that can predict the results of future experiments. The

fundamental laws used in developing theories are expressed in the language of

mathematics, the tool that provides a bridge between theory and experiment.

1.1 Standards of Length, Mass, and Time

In 1960, an international committee established a set of standards for the

fundamental quantities of science. It is called the SI (Système International), and its

fundamental units of length, mass, and time are the meter, kilogram, and second,

respectively. Other standards for SI fundamental units established by the committee

are those for temperature (the kelvin), electric current (the ampere), luminous

intensity (the candela), and the amount of substance (the mole).

The meter is defined as the distance traveled by light in vacuum during a

time of 1/299 792 458 second.

The SI fundamental unit of mass, the kilogram (kg), is defined as the mass

of a specific platinum–iridium alloy cylinder kept at the International Bureau of

Weights and Measures at Sèvres, France.

One second is defined as 9 192 631 770 times the period of vibration of

radiation from the cesium-133 atom.

1.2 Matter and Model Building

In 1897, J. J. Thomson identified the electron as a charged particle and as a

constituent of the atom. This led to the first atomic model that contained internal

structure. Following the discovery of the nucleus in 1911, an atomic model was

developed in which each atom is made up of electrons surrounding a central

nucleus. A nucleus of gold is shown in Figure 1.1. This model leads, however, to a

new question: Does the nucleus have structure? That is, is the nucleus a single

particle or a collection of particles? By the early 1930s, a model evolved that

described two basic entities in the nucleus: protons and neutrons. The proton carries

a positive electric charge, and a specific chemical element is identified by the

number of protons in its nucleus. This number is called the atomic number of the


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element. For instance, the nucleus of a hydrogen atom contains one proton (so the

atomic number of hydrogen is 1), the nucleus of a helium atom contains two protons

(atomic number 2), and the nucleus of a uranium atom contains 92 protons (atomic

number 92). In addition to atomic number, a second number—mass number,

defined as the number of protons plus neutrons in a nucleus—characterizes atoms.

Figure. 1.1.

Is that, however, where the process of breaking down stops? Protons,

neutrons, and a host of other exotic particles are now known to be composed of six

different varieties of particles called quarks, which have been given the names of

up, down, strange, charmed, bottom, and top.The up, charmed, and top quarks have

electric charges of +2/3 that of the proton, whereas the down, strange, and bottom

quarks have charges of -1/3 that of the proton. The proton consists of two up quarks

and one down quark as shown at the bottom of Figure 1.1 and labeled u and d. This

structure predicts the correct charge for the proton. Likewise, the neutron consists of

two down quarks and one up quark, giving a net charge of zero.


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1.3 Dimensional Analysis

In physics, the word dimension denotes the physical nature of a quantity. The

distance between two points, for example, can be measured in feet, meters, or

furlongs, which are all different ways of expressing the dimension of length.

The symbols we use to specify the dimensions of length, mass, and time are

L, M, and T, respectively. We shall often use brackets [ ] to denote the dimensions

of a physical quantity. For example, the symbol we use for speed in this book is v,

and in our notation, the dimensions of speed are written [v] =L/T.

In many situations, you may have to check a specific equation to see if it

matches your expectations. A useful procedure for doing that, called dimensional

analysis, can be used because dimensions can be treated as algebraic quantities. For

example, quantities can be added or subtracted only if they have the same

dimensions. Furthermore, the terms on both sides of an equation must have the

same dimensions. By following these simple rules, you can use dimensional

analysis to determine whether an expression has the correct form. Any relationship

can be correct only if the dimensions on both sides of the equation are the same.

Example 1.2.1

Show that the expression v = at is dimensionally correct, where v represents

speed, a acceleration, and t an instant of time.

Example 1.2.2

Suppose we are told that the acceleration a of a particle moving with uniform

speed v in a circle of radius r is proportional to some power of r, say rn, and some

power of v, say vm. Determine the values of n and m and write the simplest form of

an equation for the acceleration.

Quick Quiz 1.2.1

Newton’s second law of motion says that the mass of an object times its

acceleration is equal to the net force on the object. Which of the following gives the

correct units for force?

A. kg/m.s2

B. kg.m2/s

2

C. kg.m/s2

D. kg.m2/s


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Problem 1.2.1

Newton’s law of universal gravitation is represented by F=GMm/r2

, where

F is the magnitude of the gravitational force exerted by one small object on another,

M and m are the masses of the objects, and r is a distance. Force has the SI units

kg .m/s2. What are the SI units of the proportionality constant G?

1.4 Conversion of Units

Sometimes it is necessary to convert units from one measurement system to

another or convert within a system (for example, from kilometers to meters).

Conversion factors between SI and U.S. customary units of length are as follows:

1 mile = 1 609 m = 1.609 km

1 ft = 0.304 8 m = 30.48 cm

1 m = 39.37 in. = 3.281 ft

1 in. = 0.025 4 m = 2.54 cm

1 km/h = 5/18 m/s

1 m/s = 3.6 km/h

Quick Quiz 1.4.1

The distance between two cities is 100 mi. What is the number of kilometers

between the two cities?

A. smaller than 100

B. larger than 100

C. equal to 100

Problem 1.4.1

A solid piece of lead has a mass of 23.94g and a volume of 2.1 cm3. From

these data, calculate the density of lead in SI units.

Problem 1.4.2

The mean radius of the Earth is 6.37×106m, and that of the Moon is

1.74×108 cm. From these data calculate

a) the ratio of the Earth’s surface area to that of the Moon.


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b) the ratio of the Earth’s volume to that of the Moon.

1.5 Estimates and Order-of-Magnitude Calculations

It is often useful to compute an approximate answer to a given physical

problem even when little information is available. This answer can then be used to

determine whether or not a more precise calculation is necessary. Such an

approximation is usually based on certain assumptions, which must be modified if

greater precision is needed. We will sometimes refer to an order of magnitude of a

certain quantity as the power of ten of the number that describes that quantity.

Usually, when an order-of magnitude calculation is made, the results are reliable to

within about a factor of 10. If a quantity increases in value by three orders of

magnitude, this means that its value increases by a factor of about 103=1000. We

use the symbol ∼ for ―is on the order of‖. Thus,

0.008 6 ∼ 10-2

0.002 1 ∼ 10-3

720 ∼ 103

1.6 Significant Figures

When certain quantities are measured, the measured values are known only

to within the limits of the experimental uncertainty. The value of this uncertainty

can depend on various factors, such as the quality of the apparatus, the skill of the

experimenter, and the number of measurements performed. The number of

significant figures in a measurement can be used to express something about the

uncertainty.

As an example of significant figures, suppose we are asked to measure the

radius of a compact disc using a meterstick as a measuring instrument. Let us

assume the accuracy to which we can measure the radius of the disc is 0.1 cm.

Because of the uncertainty of 0.1 cm, if the radius is measured to be 6.0 cm, we

can claim only that its radius lies somewhere between 5.9 cm and 6.1 cm. In this

case, we say that the measured value of 6.0 cm has two significant figures. Note that

the significant figures include the first estimated digit. Therefore, we could write the

radius as (6.0 0.1) cm.

Zeros may or may not be significant figures. Those used to position the

decimal point in such numbers as 0.03 and 0.007 5 are not significant. Therefore,

there are one and two significant figures, respectively, in these two values. When

the zeros come after other digits, however, there is the possibility of


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misinterpretation. For example, suppose the mass of an object is given as 1 500 g.

In this case, we would express the mass as 1.5x103

g if there are two significant

figures in the measured value, 1.50x103

g if there are three significant figures, and

1.500x103

g if there are four. The same rule holds for numbers less than 1, so

2.3x10-4

has two significant figures (and therefore could be written 0.000 23) and

2.30x10-4

has three significant figures (also written as 0.000 230).

When multiplying several quantities, the number of significant figures in the

final answer is the same as the number of significant figures in the quantity having

the smallest number of significant figures. The same rule applies to division.

12.71 3.46 43.9766 → 12.71 3.46 44.0

When numbers are added or subtracted, the number of decimal places in the

result should equal the smallest number of decimal places of any term in the sum or

difference.

23.2 + 5.174 = 28.4 .

(15.47 0.1) cm (15.5 0.1) cm .

(15.42 0.1) cm (15.4 0.1) cm .

Quick Quiz 1.6.1

What is the sum of the measured values 21.4 s + 15 s +17.17 s + 4.00 3 s?

A. 57.573 s

B. 57.57 s

C. 57.6 s

D. 58 s

E. 60 s

Problem 16.1

Carry out the arithmetic operations

a) the sum of the measured values 756, 37.2, 0.83, and 2;

b) the product 0.0032 x 356.3;

Problem 1.6.2

A rectangular plate has a length of (21.3 0.2) cm and a width of (9.8

0.1) cm. Calculate the area of the plate, including its uncertainty.


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Chapter 2

MOTION IN ONE DIMENSION

As a first step in studying classical mechanics, we describe the motion of an

object while ignoring the interactions with external agents that might be affecting or

modifying that motion. This portion of classical mechanics is called kinematics. In

this chapter, we consider only motion in one dimension, that is, motion of an object

along a straight line.

In our study of translational motion, we use what is called the particle model

and describe the moving object as a particle regardless of its size. In general, a

particle is a point-like object, that is, an object that has mass but is of infinitesimal

size.

2.1 Position, Average Velocity and Speed

a) Position

A particle’s position x is the location of the particle with respect to a chosen

reference point that we can consider to be the origin of a coordinate system. The

motion of a particle is completely known if the particle’s position in space is known

at all times.

The displacement x of a particle is defined as its change in position in some

time interval. When a particle moves along the x axis from some initial position xi

to some final position xf, its displacement is:

f ix x x (2.1)

Distance is the length of a path followed by a particle.

Distance is always represented as a positive number, whereas displacement

can be either positive or negative.

Displacement is an example of a vector quantity. Many other physical

quantities, including position, velocity, and acceleration, also are vectors. In

general, a vector quantity requires the specification of both direction and magnitude.

By contrast, a scalar quantity has a numerical value and no direction.


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b) Average Velocity

The average velocity xv of a particle is defined as the particle’s

displacement x divided by the time interval t during which that displacement

occurs:

f ix

f i

x xxv

t t t

(2.2)

We can interpret average velocity geometrically by drawing a straight line

between any two points on the position–time graph in Figure 2.1. This line forms

the hypoten use of a right triangle of height x and base t.The slope of this line is

the ratio x/t.

Figure. 2.1.

c) Average Speed

The average speed of a particle is equal to the ratio of the total distance it

travels to the total time interval during which it travels that distance:

a

total distance sv

total time t

(2.3)

The SI unit of average speed is the same as the unit of average velocity:

meters per second. Unlike average velocity, however, average speed has no

direction and is always expressed as a positive number.


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Problem 2.1.1

The position versus time for a certain particle moving along the x axis is

shown in Figure P2.1.1. Find the displacement, distance, average velocity and

average speed in the time intervals:

a) 0 to 4s.

b) 2 to 8s.

Figure P2.1.1

Problem 2.1.2

A person walks first at a constant speed of 5.00 m/s along a straight line from point

A to point B and then back along the line from B to A at a constant speed of 3.00

m/s. What is:

a) Her average speed over the entire trip?

b) Her average velocity over the entire trip?

2.2 Instantaneous Velocity and Speed

The instantaneous velocity of a particle is defined as the limit of the ratio

x/t as t approaches zero. By definition, this limit equals the derivative of x with

respect to t, or the time rate of change of the position:

xt 0

x dxv lim x '

t dt

(2.4)


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The instantaneous velocity can be positive, negative, or zero. When the slope

of the position–time graph is positive, such as at any time during the first 10 s in

Figure 2.1, vx is positive and the car is moving toward larger values of x. After point

B, vx is negative because the slope is negative and the car is moving toward smaller

values of x. At point B, the slope and the instantaneous velocity are zero and the car

is momentarily at rest.

From here on, we use the word velocityto designate instantaneous velocity.

When we are interested in average velocity, we shall always use the adjective

average.

The instantaneous speed of a particle is equal to the magnitude of its

instantaneous velocity. As with average speed, instantaneous speed has no direction

associated with it.

Problem 2.21

The position of a particle moving along the x axis varies in time according to

the expression x=3t2, where x is in meters and t is in seconds. Evaluate

a) its position at t = 3.00 s

b) the velocity at t = 3.00 s.

Problem 2.2.2

Find the instantaneous velocity of the particle described in Figure P2.2.2 at

the following times:

a) t = 1.0 s

b) t = 3.0 s

c) t = 4.5 s

d) t = 7.5 s


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Figure P2.2.2

2.3 Particle Under Constant Velocity

If the velocity of a particle is constant, its instantaneous velocity at any

instant during a time interval is the same as the average velocity over the interval.

That is, xxv v const . Therefore, Equation 2.2 gives us an equation to be used in

the mathematical representation of this situation:

x

xv

t

Remembering that f ix x x , we see that x f iv (x x ) / t , or

f i xx x v t

This equation tells us that the position of the particle is given by the sum of

its original position xi at time t0 plus the displacement vxt that occurs during the

time interval t. In practice, we usually choose the time at the beginning of the

interval to be ti =0 and the time at the end of the interval to be tf = t, so our equation

becomes

f i xx x v t (2.5)

We can also use the defining equation 2.4 for velocity to derive our

kinematic equation 2.5:

t

x x f i x f i x

0

dxv dx v dt x x v dt x x v t

dt


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Figure 2.2 is a graphical representation of the particle under constant velocity.

On this position–time graph, the slope of the line representing the motion is

constant and equal to the magnitude of the velocity.

Figure 2.2

Problem 2.3. 1

A person takes a trip, driving with a constant speed of 89.5 km/h, except for

a 22.0-min rest stop. If the person’s average speed is 77.8 km/h,

a) how much time is spent on the trip?

b) how far does the person travel?

2.4 Acceleration

When the velocity of a particle changes with time, the particle is said to be

accelerating. For example, the magnitude of a car’s velocity increases when you

step on the gas and decreases when you apply the brakes. Let us see how to quantify

acceleration.

The average acceleration of a particle is defined as the ratio of the change in

its velocity vx divided by the time interval t during which that change occurs.

x xf xix

f i

Δv v - va =

Δt t - t (2.6)

The instantaneous acceleration is equal to the limit of the ratio vx/t as t

approaches 0. By definition, this limit equals the derivative of vx with respect to t, or

the time rate of change of the velocity:


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x xx x

0

Δv dva lim = v '(t)=x''(t)

Δt dtt (2.7)

The acceleration at any time is the slope of the velocity–time graph at that

time.

The SI unit of acceleration is meters per second squared (m/s2).

When the object’s velocity and acceleration are in the same direction

(ax.vx>0), the object is speeding up.

When the object’s velocity and acceleration are in opposite directions

(ax.vx<0), the object is slowing down.

Example 2.3.1

The velocity of a particle moving along the x axis varies in time according to

the expression vX = (40 – 5t2) m/s, where t is in seconds.

a) Find the average acceleration in the time interval t = 0 to t = 2.0 s.

b) Determine the acceleration at t = 2.0 s.

Problem 2.3.1

A particle moves along the x axis according to the equation

x = 2.00 + 3.00t - 1.00t2 , where x is in meters and t is in seconds. At t = 3.00 s,

find

a) the position of the particle,

b) its velocity,

c) its acceleration.

Problem 2.3.2

An object moves along the x axis according to the equation

x(t) = (3.00t2 - 2.00t + 3.00)m. Determine

a) the average speed between t=2.00s and t=3.00s,

b) the instantaneous speed at t = 2.00 s and at t=3.00 s,

c) the average acceleration between t=2.00s and t=3.00s,

d) the instantaneous acceleration at t =2.00 s and t=3.00 s.


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2.5 Particle Under Constant Acceleration

A very common and simple type of one-dimensional motion is that in which

the acceleration is constant. In such a case, the average acceleration xa over any

time interval is numerically equal to the instantaneous acceleration ax at any instant

within the interval, and the velocity changes at the same rate throughout the motion.

If we replace xa by ax in Equation 2.6 and take ti =0 and tf to be any later

time t, we find that

xf xix

v - va =

t - 0

or v vxf xi xa t (2.8)

This powerful expression enables us to determine an object’s velocity at

anytime t if we know the object’s initial velocity vxi and its (constant) acceleration

ax. A velocity–time graph for this constant-acceleration motion is shown in Figure

2.3. The graph is a straight line, the slope of which is the acceleration ax.

Because velocity at constant acceleration varies linearly in time according to

Equation 2.8, we can express the average velocity in any time interval as the

arithmetic mean of the initial velocity vxi and the final velocity vxf:

xi xfx

v vv

2

(2.9)

Figure 2.3.

We can now use Equations 2.2 and 2.9 to obtain the position of an object as a

function of time. Recognizing that t = tf - ti = t -0 = t, we find that

1(v v )

2f i xi xfx x t (2.10)

This equation provides the final position of the particle at time tin terms of

the initial and final velocities. We can obtain another useful expression for the


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position of a particle under constant acceleration by substituting Equation 2.8 into

Equation 2.10

21v

2f i xi xx x t a t (2.11)

This equation provides the final position of the particle at time tin terms of

the initial position, the initial velocity, and the constant acceleration.

We can also use the defining equations for acceleration and velocity to

derive two of our kinematic equations, Equations 2.8 and 2.11 (t

xf xi x

0

v v a dt and

t

f i x

0

x x v dt ).

Finally, we can obtain an expression for the final velocity that does not

contain time as a variable by substituting the value of t from Equation 2.8 into

Equation 2.10:

2 2v v = 2a (x x )xf xi x f i (2.12)

Quick Quiz 2.5.1

A racing car starts from rest at t = 0 and reaches a final speed v at time t. If

the acceleration of the car is constant during this time, which of the following

statements are true?

A. The car travels a distance vt.

B. The average speed of the car is v/2.

C. The magnitude of the acceleration of the car is v/t.

D. The velocity of the car remains constant.

E. None of statements (a) through (d) is true.

Example 2.5.1

A car traveling at a constant speed of 45.0m/s passes a trooper hidden behind

a billboard. One second after the speeding car passes the billboard, the trooper

sets out from the billboard to catch it, accelerating at a constant rate of 3.00m/s2.

How long does it take her to overtake the car?


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Problem 2.5.1

An object moving with uniform acceleration has a velocity of 12.0cm/s in the

positive x direction when its x coordinate is 3.00cm. If its x coordinate 2.00s later is

- 5.00 cm, what is its acceleration?

Problem 2.5.2

A truck covers 40.0 m in 8.50 s while smoothly slowing down to a final

speed of 2.80 m/s.

a) Find its original speed.

b) Find its acceleration.

Problem 2.5.3

A particle moves along the x axis. Its position is given by the equation:

x = 2 + 3t- 4t2 with x in meters and t in seconds. Determine:

a) its position when it changes direction.

b) its velocity when it returns to the position it had at t = 0.

2.6 Freely Falling Objects

A freely falling object is any object moving freely under the influence of

gravity alone, regardless of its initial motion.

Any freely falling object experiences an acceleration directed downward,

regardless of its initial motion. We shall denote the magnitude of the free-fall

acceleration, also called the acceleration due to gravity, by the symbol g. The value

of g decreases with increasing altitude above the Earth’s surface. Furthermore,


20


slight variations in goccur with changes in latitude. At the Earth’s surface, the value

of g is approximately 9.80 m/s2. Unless stated otherwise, we shall use this value for

g when performing calculations. For making quick estimates, use g =10 m/s2.

Quick Quiz 2.6.1

Consider the following choices: (a) increases, (b) decreases, (c) increases and

then decreases, (d) decreases and then increases, (e) remains the same. From these

choices, select what happens to (i) the acceleration and (ii) the speed of a ball after

it is thrown upward into the air.

Quick Quiz 2.6. 2

A student at the top of a building of height h throws one ball upward with a

speed of vi and then throws a second ball downward with the same initial speed vi.

Just before it reaches the ground, is the final speed of the ball thrown upward

A. larger

B. smaller,

C. the same in magnitude, compared with the final speed of the ball thrown

downward.

Quick Quiz 2.6. 3

A ball is thrown straight up in the air. For which situation are both the

instantaneous velocity and the acceleration zero?

A. on the way up.

B. at the top of its flight path.

C. on the way down.

D. halfway up and halfway down.

E. none of the above.

Problem 2.6.1

A ball is thrown directly downward with an initial speed of 8.00 m/s from a

height of 30.0 m. After what time interval does it strike the ground?

Problem 2.6. 2

A baseball is hit so that it travels straight upward after being struck by the

bat. A fan observes that it takes 3.00 s for the ball to reach its maximum height.

Find


21


a) the ball’s initial velocity.

b) the height it reaches.


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Chapter 4

MOTION IN TWO DIMENSIONS

In this chapter, we explore the kinematics of a particle moving in two

dimensions. We begin by studying in greater detail the vector nature of position,

velocity, and acceleration. We then treat projectile motion and uniform circular

motion as special cases of motion in two dimensions. We also discuss the concept

of relative motion, which shows why observers in different frames of reference

may measure different positions and velocities for a given particle.

4.1 The Position, Velocity, and Acceleration Vectors

In one dimension, a single numerical value describes a particle’s position,

but in two dimensions, we indicate its position by its position vector r , drawn from

the origin of some coordinate system to the location of the particle in the xy plane

as in Figure 4.1. The position vector for a particle moving in the xy plane can be

written:

r i j ( , )x y x y

Figure 4.1

At time ti, the particle is at point A, described by position vector ir . At some

later time tf, it is at point B, described by position vector fr . The path followed by

the particle from A to B is not necessarily a straight line (Figure 4.2).

x

y

M

O

r

x

y

i

j


23


Figure 4.2

The displacement vector r for a particle is the difference between its final

position vector and its initial position:

r r rf i

(4.1)

The direction of r is indicated in Figure 4.2. As we see from the figure, the

magnitude of r is less than the distance traveled along the curved path followed by

the particle.

We define the average velocity v

of a particle during the time interval t as

the displacement of the particle divided by the time interval:

rv

t

(4.2)

The average velocity is a vector quantity directed along r . The average

velocity between points is independent of the path taken.

The instantaneous velocity v is defined as the limit of the average velocity

as t approaches zero. That is, the instantaneous velocity equals the derivative of

the position vector with respect to time:

rv ( r ) '

d

dt

(4.3)

In the xy plane:

v v i v j (v , v ) (x', y')x y x y

(4.4)


24


The direction of the instantaneous velocity vector at any point in a particle’s

path is along a line tangent to the path at that point and in the direction of motion

(Figure 4.2).

Figure 4.3

The magnitude of the instantaneous velocity vector v v of a particle is

called the speed of the particle, which is a scalar quantity.

The average acceleration of a particle is defined as the change in its

instantaneous velocity vector divided by the time interval t during which that

change occurs divided by the time interval t during which that change occurs:

v v va

f i

f it t t

(4.5)

The instantaneous acceleration is defined as the limiting value of the ratio

v

t

as t approaches zero. In other words, the instantaneous acceleration equals

the derivative of the velocity vector with respect to time.

va (v) ' ( r ) ''

d

dt

(4.6)

In the xy plane:

i j ( , ) (v ', v ') (x'', y'')x y x y x ya a a a a

(4.7)


25


Various changes can occur when a particle accelerates. First, the magnitude

of the velocity vector (the speed) may change with time as in straight-line (one

dimensional) motion. Second, the direction of the velocity vector may change with

time even if its magnitude (speed) remains constant as in two-dimensional motion

along a curved path. Finally, both the magnitude and the direction of the velocity

vector may change simultaneously.

Quick Quiz 4.1.1

Consider the following controls in an automobile: gas pedal, brake,

steering wheel. The controls in this list that cause an acceleration of the car are

A. all three controls.

B. the gas pedal and the brake.

C. only the brake.

D. only the gas pedal

Quick Quiz 4.1.2

Which of the following cannot possibly be acceleration?

A. An object moving with constant speed.

B. An object moving with constant velocity.

C. An object moving along a curve.

D. An object moving along a circular path.

4.2 Two-Dimensional Motion with Constant Acceleration

Motion in two dimensions can be modeled as two independent motions in

each of the two perpendicular directions associated with the x and y axes. That is,

any influence in the y direction does not affect the motion in the x direction and

vice versa.

Because the acceleration a of the particle is assumed constant in this

discussion, its components xa and ya also are constants. Therefore, we can model

the particle as a particle under constant acceleration independently in each of the

two directions and apply the equations of kinematics separately to the x and y

components of the velocity vector.

xf xi x

yf yi y

v v a t

v v a t

(4.8)

v v aif t


26


This result states that the velocity of a particle at some time tequals the

vector sum of its initial velocity v i

at time t = 0 and the additional velocity a t

at

acquired at time t as a result of constant acceleration.

The x and y coordinates of a particle under constant acceleration are:

2

2

1

2

1

2

f i xi x

f i yi y

x x v t a t

y y v t a t

(4.9)

21r r v a

2f i i t t

This equation tells us that the position vector r f

of a particle is the vector

sum of the original position r i

, a displacement v i t

arising from the initial velocity

of the particle, and a displacement 21a

2t

resulting from the constant acceleration of

the particle.

Problem 4.2.1

A particle starts from the origin at t=0 with an initial velocity having an x

component of 20 m/s and a y component of –15 m/s. The particle moves in the xy

plane with an x component of acceleration only, given by ax=4.0 m/s2.

a) Determine the components of the velocity vector at any time.

b) Determine the velocity of the particle at t = 2.0 s

c) Calculate the speed of the particle at t=5.0 s.

Problem 4.2.2

At t = 0, a particle moving in the xy plane with constant acceleration has a

velocity of v 3.00 2.00i i j m/s and is at the origin. At t = 3.00 s , the particle’s

velocity is v 9.00 7.00i j m/s. Find

a) The acceleration of the particle.

b) Its coordinates at any time t.


27


Problem 4.2.3

A particle initially located at the origin has an acceleration of 3.00a j

(m/s2) and an initial velocity of v 5.00i i (m/s).

a) Find the vector position of the particle at any time t.

b) Find the coordinates of the particle at t = 2.00 s.

c) Find the speed of the particle at t = 2.00 s.

4.3 Projectile Motion

Anyone who has observed a baseball in motion has observed projectile

motion. The ball moves in a curved path and returns to the ground. Projectile

motion of an object is simple to analyze if we make two assumptions:

(1) the free-fall acceleration is constant over the range of motion and is

directed downward,

(2) the effect of air resistance is negligible.

With these assumptions, we find that the path of a projectile, which we call

its trajectory, is alwaysa parabola as shown in Figure 4.4.

Figure 4.4.

The expression for the position vector of the projectile as a function of time

follows directly from Equation 4.9, with its acceleration being that due to gravity:

21r r v

2f i i t g t

(4.10)


28


where the initial x and y components of the velocity of the projectile are

cosθ

sin θ

xi i i

yi i i

v v

v v

In Section 4.2, we stated that two-dimensional motion with constant

acceleration can be analyzed as a combination of two independent motions in the x

and y directions, with accelerations ax and ay . Projectile motion can also be handled

in this way, with acceleration ax=0 in the x direction and a constant acceleration

ay=-g in the y direction. Therefore, when solving projectile motion problems, use

two analysis models: (1) the particle under constant velocity in the horizontal

direction and (2) the particle under constant acceleration in the vertical direction:

cosv

sin

xf xi i i

f

yf yi y i i

v v v

v v a t v gt

(4.11)

2

cos

r 1sin

2

f xi i i

f

f i i

x v t v t

y v t gt

(4.12)

Quick Quiz 4.3. 1

As a projectile thrown upward moves in its parabolic path, at what point

along its path are the velocity and acceleration vectors for the projectile

perpendicular to each other?

A. Nowhere. B. The highest point.

C. The launch. D. Everywhere.

Quick Quiz 4.3.2

As a projectile thrown upward moves in its parabolic path, at what point

along its path are the velocity and acceleration vectors for the projectile parallel to

each other?

A. Nowhere. B. The highest point.

C. The launch. D. Everywhere.

Maximum Height and Horizontal Range of a Projectile

Assume a projectile is launched from the origin at ti =0 with a positive vyi

component as shown in Figure 4.5 and returns to the same horizontal level. This


29


situation is common in sports, where baseballs, footballs, and golf balls often land

at the same level from which they were launched.

Two points in this motion are especially interesting to analyze: the peak

point A, which has coordinates (R/2, h), and the point B, which has coordinates

(R, 0). The distance R is called the horizontal range of the projectile, and the

distance h is its maximum height.

Figure 4.5

We can determine h by noting that at the peak vy= 0. Therefore, from the

particle under constant acceleration model, we can use the y direction version of

Equation 4.11 to determine the time tA at which the projectile reaches the peak:

sin 0yA i i Av v gt

sini iA

vt

g

Substituting this expression for tA into the ydirection version of Equation

4.12 and replacing yf = yA with h, we obtain an expression for h in terms of the

magnitude and direction of the initial velocity vector:

2 2sin

2

i ivh

g

(4.13)

The range R is the horizontal position of the projectile at a time that is twice

the time at which it reaches its peak, that is, at time tB = 2tA. Using the particle

under constant velocity we find that:

sincos cos 2 cos 2 i i

fB i i B i i A i i

vR x v t v t v

g

2 sin 2i ivR

g

(4.14)

The maximum value of R from Equation 4.14 is Rmax=vi2/g when i = 45°.


30


Quick Quiz 4.3.3

Rank the launch angles for the five paths in the Figure Q4.3.3 with

respect to time of flight, from the shortest time of flight to the longest.

Figure Q4.3.3

Quick Quiz 4.3.4

Suppose you are running at constant velocity and you wish to throw a ball

such that you will catch it as it comes back down. In what direction should you

throw the ball relative to you?

A. Straight up.

B. At an angle to the ground that depends on your running speed.

C. In the forward direction.

D. In the backward.

Problem 4.3.1

A projectile is fired in such a way that its horizontal range is equal to three

times its maximum height. What is the angle of projection?

Problem 4.3.2

A stone is thrown from the top of a building upward at an angle of 30.0° to

the horizontal with an initial speed of 20.0 m/s, as shown in Figure P4.3.2. If the

height of the building is 45.0 m,

a) how long does it take the stone to reach the ground?


31


b) What is the speed of the stone just before it strikes the ground?

Figure P4.3.2

Problem 4.3.3

A plane drops a package of supplies to a party of explorers. If the plane is

traveling horizontally at 40.0 m/s and is 100 m above the ground, where does the

package strike the ground relative to the point at which it is released?

Figure P4.3.3


32


4.4 Uniform Circular Motion

Figure 4.6 shows a car moving in a circular path; we describe this motion by

calling it circular motion. If the car is moving on this path with constant speed v, we

call it uniform circular motion.

Figure 4.6

It is often surprising to students to find that even though an object moves at a

constant speed in a circular path, it still has an acceleration. To see why, consider

the defining equation for acceleration, va d dt . Notice that the acceleration

depends on the change in the velocity. Because velocity is a vector quantity, an

acceleration can occur in two ways as mentioned in Section 4.1: by a change in the

magnitude of the velocity and by a change in the direction of the velocity. The latter

situation occurs for an object moving with constant speed in a circular path. The

constant-magnitude velocity vector is always tangent to the path of the object and

perpendicular to the radius of the circular path. Therefore, the direction of the

velocity vector is always changing (Figure 4.7).

Figure 4.7

Let us first argue that the acceleration vector in uniform circular motion is

always perpendicular to the path and always points toward the center of the circle. If

that were not true, there would be a component of the acceleration parallel to the

path and therefore parallel to the velocity vector. Such an acceleration component

would lead to a change in the speed of the particle along the path. This situation,


33


however, is inconsistent with our setup of the situation: the particle moves with

constant speed along the path. Therefore, for uniform circular motion, the

acceleration vector can only have a component perpendicular to the path, which is

toward the center of the circle (Figure 4.8). It is called a centripetal acceleration

(centripetal means center-seeking). The magnitude of the centripetal acceleration is:

2

c

va

r (4.15)

(Figure 4.8)

In many situations, it is convenient to describe the motion of a particle

moving with constant speed in a circle of radius r in terms of the period T, which is

defined as the time interval required for one complete revolution of the particle:

2 rT

v

(4.16)

The angular speed the particle, measured in radians/s or s-1

:

2

T

v r

2

ca r

Quick Quiz 4.4.1

Which of the following correctly describes the centripetal acceleration vector

for a particle moving in a circular path?

A. constant and always perpendicular to the velocity vector for the particle.

B. constant and always parallel to the velocity vector for the particle.

C. of constant magnitude and always perpendicular to the velocity vector for

the particle.


34


D. of constant magnitude and always parallel to the velocity vector for the

particle.

Quick Quiz 4.4.2

A particle moves in a circular path of radius r with speed v. It then increases

its speed to 2v while traveling along the same circular path. The centripetal

acceleration of the particle has changed by a factor of

A. 0.25 B. 0.5 C. 2 D. 4

Quick Quiz 4.4.3

A particle moves along a path and its speed increases with time. In which of

the following cases are its acceleration and velocity vectors perpendicular

everywhere along the path?

A. the path is circular. B. the path is straight.

C. the path is a parabola. D. never.

Quick Quiz 4.4.4

A particle moves along a path and its speed is constant. In which of the

following cases are its acceleration and velocity vectors perpendicular everywhere

along the path?



4.5 Tangential and Radial Acceleration

As the particle moves along the curved path in Figure 4.9, the direction of

the total acceleration vector a changes from point to point. At any instant, this

vector can be resolved into two components based on an origin at the center of the

dashed circle corresponding to that instant: a radial component ra along the radius

of the circle and a tangential component ta perpendicular to this radius. The total

acceleration vector a can be written as the vector sum of the component vectors:

t ra a a


35


Figure 4.9

The tangential acceleration component causes a change in the speed v of the

particle. This component is parallel to the instantaneous velocity, and its magnitude

is given by:

't

dva v

dt (4.17)

The radial acceleration component arises from a change in direction of the

velocity vector and is given by:

2

r

va

r (4.18)

where r is the radius of curvature of the path at the point in question. Because ra

and ta are perpendicular component vectors of a , it follows that the magnitude of

a :

2 2

t ra a a (4.19)

At a given speed, ar is large when the radius of curvature is small (as at

points A and B in Fig. 4.9) and small when r is large (as at point C). The direction

of ta is either in the same direction as v (if v is increasing) or opposite v (if v is

decreasing).

In uniform circular motion, where v is constant, at = 0 and the acceleration is

always completely radial as described in Section 4.4. In other words, uniform

circular motion is a special case of motion along a general curved path.

Furthermore, if the direction of v does not change, there is no radial acceleration

and the motion is one dimensional (in this case, ar = 0, but at may not be zero).


36


Quick Quiz 4.5.1

A particle moves along a path and its speed increases with time. In which of

the following cases are its acceleration and velocity vectors parallel?



Problem 4.5.1

An automobile whose speed is increasing at a rate of 0.60 m/s2 travels along

a circular road of radius 20.0 m. When the instantaneous speed of the automobile

is 4.00 m/s, find:

a) the tangential acceleration component.

b) the centripetal acceleration component.

c) the magnitude of the total acceleration.

Problem 4.5.2

Figure P4.4 represents the total acceleration of a particle moving

clockwise in a circle of radius 2.50 m at a certain instant of time. At this instant,

find:

Figure P4.5.2

a) the radial acceleration,

b) the speed of the particle,

c) its tangential acceleration.

4.6 Relative Velocity and Relative Acceleration

In this section, we describe how observations made by different observers in

different frames of reference are related to one another. A frame of reference can be


37


described by a coordinate system for which an observer is at rest with respect to the

origin.

Consider two observers watching a man walking on a moving beltway at an

airport in Figure 4.10. The woman standing on the moving beltway sees the man

moving at a normal walking speed. The woman observing from the stationary floor

sees the man moving with a higher speed because the beltway speed combines with

his walking speed. Both observers look at the same man and arrive at different

values for his speed. Both are correct; the difference in their measurements results

from the relative velocity of their frames of reference.

Figure 4.10

In a more general situation, consider a particle located at point Pin Figure

4.11. Imagine that the motion of this particle is being described by two observers,

observer A in a reference frame SA fixed relative to the Earth and a second observer

B in a reference frame SBmoving to the right relative to SA (and therefore relative

to the Earth) with a constant velocity BAv . In this discussion of relative velocity, we

use a double-subscript notation; the first subscript represents what is being

observed, and the second represents who is doing the observing. Therefore, the

notation BAv means the velocity of observer B (and the attached frame SB) as

measured by observer A. With this notation, observer B measures A to be moving

to the left with a velocity AB BAv v . For purposes of this discussion, let us place

each observer at her or his respective origin.

We define the time t=0 as the instant at which the origins of the two

reference frames coincide in space. Therefore, at time t, the origins of the reference


38


frames will be separated by a distance vBAt. We label the position P of the particle

relative to observer A with the position vector PAr and that relative to observer B

with the position vector PBr , both at time t. From Figure 4.11, we see that the

vectors PAr and PBr are related to each other through the expression:

r r vPA PB BA t

(4.20)

By differentiating Equation 4.22 with respect to time, noting that BAv is

constant, we obtain:

v v vPA PB BA

(4.21)

Equations 4.20 and 4.21 are known as Galilean transformation equations.

Although observers in two frames measure different velocities for the particle, they

measure the same accelerationwhen BAv is constant. We can verify that by taking

the time derivative of Equation 4.21:

a aPA PB

(4.22)

Quick Quiz 4.6.1

Two projectiles are inflight at the same time. The acceleration of one relative

to the other:

A. is always 9.8 m/s2.

B. can be as large as 19.8 m/s2.

C. can be horizontal.

D. is zero.

Quick Quiz 4.6.2

A student throws a heavy red ball horizontally from a balcony of a tall

building with an initial speed vi. At the same time, a second student drops a lighter

blue ball from the balcony. Neglecting air resistance, which statement is true?

A. The blue ball reaches the ground first.

B. The balls reach the ground at the same instant.

C. The red ball reaches the ground first.

D. Both balls hit the ground with the same speed.


39


Chapter 5

THE LAWS OF MOTION

In Chapters 2 and 4, we describedthe motion of an object in terms of its

position, velocity, and acceleration without considering what might influence that

motion. Now we consider that influence: Why does the motion of an object change?

What might cause one object to remain at rest and another object to accelerate?

Why is it generally easier to move a small object than a large object? The two main

factors we need to consider are the forces acting on an object and the mass of the

object. In this chapter, we begin our study of dynamics by discussing the three basic

laws of motion, which deal with forces and masses and were formulated more than

three centuries ago by Isaac Newton.

5.1 The Concept of Force

Everyone has a basic understanding of the concept of force from everyday

experience. When you push your empty dinner plate away, you exert a force on it.

Similarly, you exert a force on a ball when you throw or kick it. In these examples,

the word forcer efers to an interaction with an object by means of muscular activity

and some change in the object’s velocity. Forces do not always cause motion,

however. For example, when you are sitting, a gravitational force acts on your body

and yet you remain stationary. As a second example, you can push (in other words,

exert a force) on a large boulder and not be able to move it.

When a coiled spring is pulled, as in Figure 5.1a, the spring stretches. When

a stationary cart is pulled, as in Figure 5.1b, the cart moves. When a football is

kicked, as in Figure 5.1c, it is both deformed and set in motion. These situations are

all examples of a class of forces called contact forces.That is, they involve physical

contact between two objects. Other examples of contact forces are the force exerted

by gas molecules on the walls of a container and the force exerted by your feet on

the floor.

Another class of forces, known as field forces, does not involve physical

contact between two objects. These forces act through empty space. The

gravitational force of attraction between two objects with mass, illustrated in Figure

5.1d, is an example of this class of force. The gravitational force keeps objects

bound to the Earth and the planets in orbit around the Sun. Another common field

force is the electric force that one electric charge exerts on another (Fig. 5.1e), such


40


as the attractive electric force between an electron and a proton that form a

hydrogen atom. A third example of a field force is the force a bar magnet exerts on

a piece of iron (Fig. 5.1f).

Figure 5.1

The only known fundamentalforces in nature are all field forces: (1)

gravitational forces between objects, (2) electromagnetic forces between electric

charges, (3) strong forces between subatomic particles, and (4) weak forcesthat arise

in certain radioactive decay processes. In classical physics, we are concerned only

with gravitational and electromagnetic forces.

It is possible to use the deformation of a spring to measure force (Figure 5.2).

The net force acting on an object is defined as the vector sum of all forces acting on

the object.


41


Figure 5.2

5.2 Newton’s First Law and Inertial Frames

If an object does not interact with other objects, it is possible to identify a

reference frame in which the object has zero acceleration. Such a reference frame is

called an inertial frame of reference.

Any reference frame that moves with constant velocity relative to an inertial

frame is itself an inertial frame. When you and the train accelerate, however, you

are observing the puck from a noninertial reference frame because the train is

accelerating relative to the inertial reference frame of the Earth’s surface.

A reference frame that moves with constant velocity relative to the distant

stars is the best approximation of an inertial frame, and for our purposes we can

consider the Earth as being such a frame. The Earth is not really an inertial frame

because of its orbital motion around the Sun and its rotational motion about its own

axis, both of which involve centripetal accelerations. These accelerations are small

compared with g, however, and can often be neglected. For this reason, we model

the Earth as an inertial frame, along with any other frame attached to it.

Newton’s first law of motion:

In the absence of external forces and when viewed from an inertial reference

frame, an object at rest remains at rest and an object in motion continues in motion

with a constant velocity (that is, with a constant speed in a straight line).

In other words, when no force acts on an object, the acceleration of the

object is zero.


42


Quick Quiz 5.2.1

An example of an inertial reference frame is:

A. any reference frame that is not accelerating.

B. a frame attached to a particle on which there are no forces.

C. any reference frame that is at rest.

D. a reference frame attached to Earth.

Quick Quiz 5.2.2

An object moving at constant velocity in an inertial frame must:

A. have a net force on it.

B. eventually stop due to gravity.

C. have zero net force on it.

D. have no frictional force on it.

Quick Quiz 5.2.3

If an object is in equilibrium, which of the following statements is not true?

A. The speed of the object remains constant.

B. The net force acting on the object is zero.

C. The object must be at rest.

D. There are at least two forces acting on the object.

5.3 Mass

Mass is that property of an object that specifies how much resistance an

object exhibits to changes in its velocity, and as we learned in Section 1.1, the SI

unit of mass is the kilogram. Experiments show that the greater the mass of an

object, the less that object accelerates under the action of a given applied force.

Mass is an inherent property of an object and is independent of the object’s

surroundings and of the method used to measure it. Also, mass is a scalar quantity

and thus obeys the rules of ordinary arithmetic. For example, if you combine a 3-kg

mass with a 5-kg mass, the total mass is 8 kg.

Mass should not be confused with weight. Mass and weight are two different

quantities. The weight of an object is equal to the magnitude of the gravitational

force exerted on the object and varies with location. For example, a person

weighing 720 N on the Earth weighs only about 120 N on the Moon. On the other


43


hand, the mass of an object is the same everywhere: an object having a mass of 2 kg

on the Earth also has a mass of 2 kg on the Moon.

5.4 Newton’s Second Law

Newton’s first law explains what happens to an object when no forces act on

it. Newton’s second law answers the question of what happens to an object when

one or more forces act on it.

When viewed from an inertial reference frame, the acceleration of an object

is directly proportional to the net force acting on it and inversely proportional to its

mass:

F

am

F am

(5.1)

Equation 5.1 is a vector expression and hence is equivalent to three

component equation:

x x

y y

z z

F ma

F ma

F ma

(5.2)

The SI unit of force is the newton (N). A force of 1 N is the force that, when acting

on an object of mass 1 kg, produces an acceleration of 1 m/s2. From this definition

and Newton’s second law, we see that the newton can be expressed in terms of the

following fundamental units of mass, length, and time:

21 1 .N kg m s

Quick Quiz 5.4.1

The term ―mass‖ refers to the same physical concept as:

A. weight.

B. inertia.

C. force.

D. acceleration.


44


Quick Quiz 5.4.2

In SI units a force is numerically equal to the __________ , when the force is

applied to it.

A. velocity of the standard kilogram

B. speed of the standard kilogram

C. acceleration of the standard kilogram

D. acceleration of any object

Quick Quiz 5.4.3

Acceleration is always in the direction:

A. of the displacement.

B. of the velocity.

C. of the net force.

D. opposite to the frictional force.

Problem 5.4.1

When a certain force is applied to the standard kilogram, its acceleration

is 5.0 m/s2. When the same force is applied to another object, its acceleration is

one-fifth as much. Determine the mass of the object.

5.5 The Gravitational Force and Weight

All objects are attracted to the Earth. The attractive force exerted by the

Earth on an object is called the gravitational force Fg

. This force is directed toward

the center of the Earth, and its magnitude is called the weight of the object.

Gravitational force: F gg m

(5.3)

Weight of the object: gF mg (5.4)

Because it depends on g, weight varies with geographic location. Because g

decreases with increasing distance from the center of the Earth, objects weigh less

at higher altitudes than at sea level. For example, a 1 000-kg pallet of bricks used in

the construction of the Empire State Building in New York City weighed 9 800 N at

street level, but weighed about 1 N less by the time it was lifted from sidewalk level

to the top of the building. As another example, suppose a student has a mass of 70.0

kg. The student’s weight in a location where g =9.80 m/s2 is 686 N. At the top of a


45


mountain, however, where g = 9.77 m/s2, the student’s weight is only 684 N.

Therefore, if you want to lose weight without going on a diet, climb a mountain or

weigh yourself at 30 000 ft during an airplane flight!

The mass min Equation 5.4 determines the strength of the gravitational

attraction between the object and the Earth. This role is completely different from

that previously described for mass, that of measuring the resistance to changes in

motion in response to an external force. In that role, mass is also called inertial

mass. We call m in Equation 5.4 the gravitational mass. Even though this quantity

is different in behavior from inertial mass, it is one of the experimental conclusions

in Newtonian dynamics that gravitational mass and inertial mass have the same

value.

Quick Quiz 5.5.1

Which of the following quantities is NOT a vector?

A. Displacement.

B. Weight.

C. Acceleration.

D. Force

Quick Quiz 5.5.2

A feather and a lead ball are dropped from rest in vacuum on the Moon. The

acceleration of the feather is:

A. more than that of the lead ball.

B. the same as that of the lead ball.

C. less than that of the lead ball.

D. 9.8 m/s2

5.6 Newton’s Third Law

When your finger pushes on the book, the book pushes back on your finger.

This important principle is known as Newton’s third law:

If two objects interact, the force 12F exerted by object 1 on object 2 is equal

in magnitude and opposite in direction to the force 21F exerted by object 2 on

object 1:

12 21F F

(5.5)


46


Figure 5.3

Consider a computer monitor at rest on a table as in Figure 5.4a. The

gravitational force on the monitor is g EmF F . The reaction to this force is the force

mE EmF F emexerted by the monitor on the Earth. The monitor does not

accelerate because it is held up by the table. The table exerts on the monitor an

upward force tmN F , called the normal force. (Normal in this context means

perpendicular.) In general, whenever an object is in contact with a surface, the

surface exerts a normal force on the object. The normal force on the monitor can

have any value needed, up to the point of breaking the table. Because the monitor

has zero acceleration, Newton’s second law applied to the monitor gives us

0F N mg , so N = mg. The normal force balances the gravitational force on

the monitor, so the net force on the monitor is zero. The reaction force to N is the

force exerted by the monitor downward on the table, mt tmF F N .

Figure 5.4.


47


Notice that the forces acting on the monitor are gF and n as shown in

Figure 5.4b. The two forces mEF and mtF are exerted on objects other than the

monitor.

Figure 5.6b, by contrast, shows only the forces acting on one object, the

monitor, and is called a force diagramor a diagram showing the forces on the

object. The important pictorial representation in Figure 5.6c is called a free-body

diagram.

Quick Quiz 5.6.1

(i) If a fly collides with the windshield of a fast-moving bus, which

experiences an impact force with a larger magnitude?

A. The fly.

B. The bus.

C. The same force is experienced by both.

(ii) Which experiences the greater acceleration?

A. The fly.

B. The bus.

C. The same acceleration is experienced by both.

5.7 Analysis Models Using Newton’s Second Law

Remember that when Newton’s laws are applied to an object, we are

interested only in external forces that act on the object. If the objects are modeled as

particles, we need not worry about rotational motion. For now, we also neglect the

effects of friction in those problems involving motion, which is equivalent to stating

that the surfaces are frictionless.

a) The Particle in Equilibrium

If the acceleration of an object modeled as a particle is zero, the object is

treated with the particle in equilibrium model. In this model, the net force on the

object is zero:

F 0

(5.6)

Consider a lamp suspended from a light chain fastened to the ceiling as in

Figure 5.5a. The force diagram for the lamp (Fig. 5.5b) shows that the forces acting


48


on the lamp are the downward gravitational force gF and the upward force T

exerted by the chain:

00

0

x

y g g

y

FF T F T F

F

Again, notice that T and gF are notan action–reaction pair because they act

on the same object, the lamp. The reaction force to T is a downward force

exerted by the lamp on the chain

.

Figure 5.5

b) The Particle Under a Net Force

If an object experiences an acceleration, its motion can be analyzed with the

particle under a net forcemodel. The appropriate equation for this model is

Newton’s second law:

F am

Consider a crate being pulled to the right on a frictionless, horizontal floor as

in Figure 5.6a. Of course, the floor directly under the boy must have friction;

otherwise, his feet would simply slip when he tries to pull on the crate! Suppose you

wish to find the acceleration of the crate and the force the floor exerts on it. The

forces acting on the crate are illustrated in the free-body diagram in Figure 5.6b. We

can now apply Newton’s second law in component form to the crate:

0

x x x

gy y

F ma T ma

n FF ma


49


Figure 5.6

In the situation just described, the magnitude of the normal force N is equal

to the magnitude of gF , but that is not always the case. For example, suppose a

book is lying on a table and you push down on the book with a force F as in Figure

5.7. In this situation, the normal force is greaterthan the gravitational force:

gn F F .

Figure 5.7

Example 5.7.1

A traffic light weighing 122 N hangs from a cable tied to two other cables

fastened to a support, as in Figure E5.1. The upper cables make angles of =37.0°


50


and =53.0° with the horizontal. These upper cables are not as strong as the

vertical cable, and will break if the tension in them exceeds 100 N. Will the traffic

light remain hanging in this situation, or will one of the cables break?

Figure E5.7.1

Example 5.7.2

A car of mass m is on an icy driveway inclined at an angle q, as in Figure

E5.2.

a) Find the acceleration of the car, assuming that the driveway is

frictionless.

b) Suppose the car is released from rest at the top of the incline, and the

distance from the car’s front bumper to the bottom of the incline is d. How long

does it take the front bumper to reach the bottom, and what is the car’s speed as it

arrives there?

Figure E5.7.2


51


Example 5.7.3

When two objects of unequal mass are hung vertically over a frictionless

pulley of negligible mass as in Figure E5.3a, the arrangement is called an Atwood

machine. The device is sometimes used in the laboratory to determine the value of

g. Determine the magnitude of the acceleration of the two objects and the tension in

the lightweight string.

Figure E5.7.3

Example 5.7.4

A person weighs a fish of mass m on a spring scale attached to the ceiling of

an elevator, as illustrated in Figure E5.4.

a) Show that if the elevator accelerates either upward or downward, the spring

scale gives a reading that is different from the weight of the fish.

b) Suppose the elevator cable breaks, so that the elevator and its contents are in

free-fall. What happens to the reading on the scale?


52


Figure E5.7.4

Problem 5.7.1

An object of mass m1 = 5.00 kg placed on a frictionless, horizontal table is

connected to a string that passes over a pulley and then is fastened to a hanging

object of mass m2 = 9.00 kg as shown in Figure P5.1.

a) Draw free-body diagrams of both objects.

b) Find the magnitude of the acceleration of the objects.

c) Find the tension in the string.

Figure P5.7.1

5.8 Forces of Friction

When an object is in motion either on a surface or in a viscous medium such

as air or water, there is resistance to the motion because the object interacts with its

surroundings. We call such resistance a force of friction.Forces of friction are very

important in our everyday lives. They allow us to walk or run and are necessary for

the motion of wheeled vehicles.


53


The magnitude of the force of static friction between any two surfaces in

contact can have the values:

fs ≤ sn (5.7)

where the dimensionless constant s is called the coefficient of static friction and nis

the magnitude of the normal force exerted by one surface on the other.

The magnitude of the force of kinetic friction acting between two surfaces is:

fk = kn (5.8)

where k is the coefficient of kinetic friction.Although the coefficient of

kinetic friction can vary with speed, we shall usually neglect any such variations in

this text.

The values of k and s depend on the nature of the surfaces, but k is

generally less than s. Typical values range from around 0.03 to 1.0.

The direction of the friction force on an object is parallel to the surface with

which the object is in contact and opposite to the actual motion (kinetic friction) or

the impending motion (static friction) of the object relative to the surface.

The coefficients of friction are nearly independent of the area of contact

between the surfaces. We might expect that placing an object on the side having the

most area might increase the friction force. Although this method provides more

points in contact, the weight of the object is spread out over a larger area and the

individual points are not pressed together as tightly. Because these effects

approximately compensate for each other, the friction force is independent of the

area.

Quick Quiz 5.8.1

The manager of a department store is pushing horizontally with a force of

magnitude 200 N on a box of shirts. The box is sliding across the horizontal

floor with a forward acceleration. Nothing else touches the box. What must be

true about the magnitude of the force of kinetic friction acting on the box?

A. It is greater than 200 N.

B. It is less than 200 N.

C It is equal to 200 N.

D. None of those statements is necessarily true.


54


Quick Quiz 5.8.2

You are playing with your daughter in the snow . She sits on a sled and ask

you to slide her across a flat, horizontal field. You have a choice of (a) pushing her

from behind, by applying a force downward on her shoulders at 30o below the

horizontal, or (b) attaching a rope to the front of the sled and pulling with a force at

30o above the horizontal (Fig. Q5.2). Which would be easier for you and why?

Fig. Q5.8.2

Quick Quiz 5.8.3

A large crate of mass m is place on the flatbed of a truck but not tied down.

As the truck accelerates forward with acceleration a, the crate remains at rest

relative to the truck. What force causes the crate to accelerate?

A. The normal force. B. The gravitational force.

C. The friction force. D. No force is required.

Quick Quiz 5.8.4

A crate remains stationary after it has been placed on a ramp inclined at an

angle with the horizontal. Which of the following statements is correct about the

magnitude of the friction force that acts on the crate?

A. It is larger than the weight of the crate.

B. It is equal to sn.

C. It is equal to the component of the gravitational force acting down the ramp.

D. It is less than the component of the gravitational force acting down the

ramp.


55


Quick Quiz 5.8.5

The block shown moves with constant velocity on a horizontal surface. Two

of the forces on it are shown. A frictional force exerted by the surface is the

only other horizontal force on the block. The frictional force is

A. 2 N, leftward.

B. 2 N, rightward.

C. slightly more than 2 N, leftward.

D. slightly less than 2 N, leftward.

Problem 5.8.1

A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the

puck always remains on the ice and slides 115 m before coming to rest, determine

the coefficient of kinetic friction between the puck and ice.

Problem 5.8.2

A woman at an airport is towing her 20.0-kg suitcase at constant speed by

pulling on a strap at an angel q above the horizontal (Fig. P5.2). She pulls on the

strap with a 35.0-N force, and the friction force on the suitcase is 20.0 N.

a) Draw a free-body diagram of the suitcase.

b) What angle does the strap make with the horizontal?

c) What is the magnitude of the normal force that the ground exerts on the

suitcase?


56


Fig. P 5.8.2


57


Chapter 6

CIRCULAR MOTION AND OTHER APPLICATIONS OF

NEWTON’S LAWS

In the preceding chapter, we introduced Newton’s laws of motion and

incorporated them into two analysis models involving linear motion. Now we

discuss motion that is slightly more complicated. For example, we shall apply

Newton’s laws to objects traveling in circular paths. We shall also discuss motion

observed from an accelerating frame of reference and motion of an object through a

viscous medium. For the most part, this chapter consists of a series of examples

selected to illustrate the application of Newton’s laws to a variety of new

circumstances.

6.1 Extending the Particle in Uniform Circular Motion Model

Imagine a moving object that can be modeled as a particle. If it moves in a

circular path of radius rat a constant speed v, it experiences a centripetal

acceleration. Because the particle is accelerating, there must be a net force acting on

the particle. That force is directed toward the center of the circular path and is given

by:

2vcF ma m

r (6.1)

Examples:

• the tension in a string of constant length acting on a rock twirled in a circle.

(Fig. 6.1)

• the gravitational force acting on a planet traveling around the Sun in a

perfectly circular orbit.

• the magnetic force acting on a charged particle moving in a uniform

magnetic field.

• the electric force acting on an electron in orbit around a nucleus in the Bohr

odel of the hydrogen atom.


58


Fig. 6.1

Quick Quiz 6.1.1

You are riding on a Ferris wheel (Fig. Q6.1.1) that is rotating with constant

speed. The car in which you are riding always maintains its correct upward

orientation—it does not invert. What is the direction of your centripetal acceleration

when you are at the top of the wheel? (a) upward (b) downward (c) impossible to

determine. What is the direction of your centripetal acceleration when you are at the

bottom of the wheel?

Fig. Q6.1.1

Example 6.1.1

A small ball of mass m is suspended from a string of length L. The ball

revolves with constant speed v in a horizontal circle of radius r as shown in Figure

E6.1.1 (Because the string sweeps out the surface of a cone, the system is known as


59


a conical pendulum.). Find an expression for v in terms of the geometry in Figure

E6.1.1.

Figure E6.1.1

Example 6.1.2

A 1500-kg car moving on a flat, horizontal road negotiates a curve as shown

in Figure E6.1.2a. If the radius of the curve is 35.0 m and the coefficient of static

friction between the tires and dry pavement is 0.523, find the maximum speed the

car can have and still make the turn successfully.

Figure E6.1.2

Example 6.1.3

A child of mass m rides on a Ferris wheel as shown in Figure E6.1.3. The

child moves in a vertical circle of radius 10.0 m at a constant speed of 3.00 m/s.

a) Determine the force exerted by the seat on the child at the bottom of the ride.

Express your answer in terms of the weight of the child, mg.

b) Determine the force exerted by the seat on the child at the top of the ride.


60


Figure E6.1.3

6.2 Nonuniform Circular Motion

Because the total acceleration is r ta a a

, the total force exerted on the

particle is r t

F F F as shown in Figure 6.2. (We express the radial and

tangential forces as net forces with the summation notation because each force

could consist of multiple forces that combine.) The vector r

F is directed toward

the center of the circle and is responsible for the centripetal acceleration. The vector

tF tangent to the circle is responsible for the tangential acceleration, which

represents a change in the particle’s speed with time.

Figure 6.2


61


Quick Quiz 6.2.1

A bead slides freely along a horizontal, curved wire at constant speed, as

shown in Figure Q6.2.1. Draw the vectors representing the force exerted by the wire

on the bead at points A, B, and C.

Figure Q6.2.1

Quick Quiz 6.2.2

In Figure Q6.2.1, the bead speeds up with constant tangential acceleration as

it moves toward the right. Draw the vectors representing the force on the bead at

points A, B, and C.

Example 6.2.1

A small sphere of mass m is attached to the end of a cord of length R and set

into motion in a vertical circle about a fixed point O, as illustrated in Figure E6.2.1.

Determine the tension in the cord at any instant when the speed of the sphere is v

and the cord makes an angle with the vertical.

Figure E6.2.1

Problem 6.2.1

A pail of water is rotated in a vertical circle of radius 1.00 m. What is the

minimum speed of the pail at the top of the circle if no water is to spill out?


62


Figure P6.2.1

6.3 Motion in Accelerated Frames

Newton’s laws of motion, which we introduced in Chapter 5, describe

observations that are made in an inertial frame of reference. In this section, we

analyze how Newton’s laws are applied by an observer in a noninertial frame of

reference, that is, one that is accelerating.

On the accelerating train, as you watch the puck accelerating toward the back

of the train, you might conclude based on your belief in Newton’s second law that a

force has acted on the puck to cause it to accelerate. We call an apparent force such

as this one a fictitious force because it is not a real force and is due only to

observations made in an accelerated reference frame. A fictitious force appears to

act on an object in the same way as a real force. Real forces are always interactions

between two objects, however, and you cannot identify a second object for a

fictitious force. (What second object is interacting with the puck to cause it to

accelerate?). In general, simple fictitious forces appear to act in the direction

opposite that of the acceleration of the noninertial frame. For example, the train

accelerates forward and there appears to be a fictitious force causing the puck to

slide toward the back of the train.

The train example describes a fictitious force due to a change in the train’s

speed. Another fictitious force is due to the change in the directionof the velocity

vector. To understand the motion of a system that is noninertial because of a change

in direction, consider a car traveling along a highway at a high speed and

approaching a curved exit ramp on the left as shown in Figure 6.3a. As the car takes

the sharp left turn on the ramp, a person sitting in the passenger seat leans or slides

to the right and hits the door. At that point the force exerted by the door on the

passenger keeps her from being ejected from the car. What causes her to move


63


toward the door? A popular but incorrect explanation is that a force acting toward

the right in Figure 6.3b pushes the passenger outward from the center of the circular

path. Although often called the ―centrifugal force,‖ it is a fictitious force. The car

represents a noninertial reference frame that has a centripetal acceleration toward

the center of its circular path. As a result, the passenger feels an apparent force

which is outward from the center of the circular path, or to the right in Figure 6.3b,

in the direction opposite that of the acceleration.

Let us address this phenomenon in terms of Newton’s laws. Before the car

enters the ramp, the passenger is moving in a straight-line path. As the car enters the

ramp and travels a curved path, the passenger tends to move along the original

straight-line path, which is in accordance with Newton’s first law: the natural

tendency of an object is to continue moving in a straight line. If a sufficiently large

force (toward the center of curvature) acts on the passenger as in Figure 6.3c,

however, she moves in a curved path along with the car. This force is the force of

friction between her and the car seat. If this friction force is not large enough, the

seat follows a curved path while the passenger tends to continue in the straight-line

path of the car before the car began the turn. Therefore, from the point of view of an

observer in the car, the passenger leans or slides to the right relative to the seat.

Eventually, she encounters the door, which provides a force large enough to enable

her to follow the same curved path as the car.

Figure 6.3


64


Another interesting fictitious force is the ―Coriolis force.‖ It is an apparent

force caused by changing the radial position of an object in a rotating coordinate

system.

For example, suppose you and a friend are on opposite sides of a rotating

circular platform and you decide to throw a baseball to your friend. Figure 6.4a

represents what an observer would see if the ball is viewed while the observer is

hovering at rest above the rotating platform. According to this observer, who is in

an inertial frame, the ball follows a straight line as it must according to Newton’s

first law. At t =0 you throw the ball toward your friend, but by the time tf when the

ball has crossed the platform, your friend has moved to a new position and can’t

catch the ball. Now, however, consider the situation from your friend’s viewpoint.

Your friend is in a noninertial reference frame because he is undergoing a

centripetal acceleration relative to the inertial frame of the Earth’s surface. He starts

off seeing the baseball coming toward him, but as it crosses the platform, it veers to

one side as shown in Figure 6.4b. Therefore, your friend on the rotating platform

states that the ball does not obey Newton’s first law and claims that a sideways

force is causing the ball to follow a curved path. This fictitious force is called the

Coriolis force.

Figure 6.4

Example 6.3.1

A small sphere of mass m hangs by a cord from the ceiling of a boxcar that is

accelerating to the right as shown in Figure 6.5. Both the inertial observer on the

ground in Figure 6.5a and the noninertial observer on the train in Figure 6.5b agree

that the cord makes an angle uwith respect to the vertical. The noninertial observer


65


claims that a force, which we know to be fictitious, causes the observed deviation of

the cord from the vertical. How is the magnitude of this force related to the boxcar’s

acceleration measured by the inertial observer in Figure 6.5a?

Figure 6.5

Example 6.3.2

Suppose a block of mass m lying on a horizontal, frictionless turntable is

connected to a string attached to the center of the turntable, as shown in Figure 6.6.

How would each of the observers write Newton’s second law for the block?


66


Figure 6.6


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REFERENCES

1. Physics for Scientists and Engineers with Modern Physics, Ninth

Edition. Raymond A. Serway and John W. Jewett, Jr.

2. Vật lí đại cương - tập 1, Đỗ Quốc Huy, NXBĐHCN.

3. Cơ sở vật lí tập 1, 2 - D.Hallday, R.Resnick, J.Walker, NXBGD.

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