Module 1: Units and Significant Figures

1.1 The Speed of Light

When we observe and measure phenomena in the world, we try to assign numbers to the physical quantities with as much accuracy as we can possibly obtain from our measuring equipment. For example, we may want to determine the speed of light, which we can calculate by dividing the distance a known ray of light propagates over its travel time,

speed of light = distancetime

. (1.1.1)

In 1983 the General Conference on Weights and Measures defined the speed of light to be

c = 299, 792, 458 meters/second . (1.1.2)

This number was chosen to correspond to the most accurately measured value of the speed of light and is well within the experimental uncertainty.

1.2 International System of System of Units

The three quantities time, length, and the speed of light are directly intertwined. Which quantities should we consider as base and which ones as derived from the base quantities? For example, are length and time base quantities while speed is a derived quantity? This question is answered by convention. The basic system of units used throughout science and technology today is the internationally accepted Systme International (SI). It consists of seven base quantities and their corresponding base units:

Mechanics is based on just the first three of these quantities, the MKS or meter-kilogram-second system. An alternative metric system to this, still widely used, is the so-called CGS system (centimeter-gram-second). So far as distance and time measurements are concerned, there is also wide use of British Imperial units (especially in the USA) based on the foot (ft), the mile (mi), etc., as units of length, and also making use of the minute, hour, day and year as units of time.

Base Quantity Base Unit Length meter (m) Mass kilogram (kg) Time second (s) Electric Current ampere (A) Temperature Kelvin (K) Amount of Substance mole (mol) Luminous Intensity candela (cd)

We shall refer to the dimension of the base quantity by the quantity itself, for example

dim length length L, dim mass mass M, dim time time T. (1.2.1)

1.3 The Atomic Clock and the Definition of the Second

Isaac Newton, in the Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy), distinguished between time as duration and an absolute concept of time,

Absolute true and mathematical time, of itself and from its own nature, flows equably without relation to anything external, and by another name is called duration: relative, apparent, and common time, is some sensible and external (whether accurate or unequable) measure of duration by means of motion, which is commonly used instead of true time; such as an hour, a day, a month, a year. 1.

The development of clocks based on atomic oscillations allowed measures of timing with accuracy on the order of 1 part in

1014 , corresponding to errors of less than one microsecond (one millionth of a second) per year. Given the incredible accuracy of this measurement, and clear evidence that the best available timekeepers were atomic in nature, the second (s) was redefined in 1967 by the International Committee on Weights and Measures as a certain number of cycles of electromagnetic radiation emitted by cesium atoms as they make transitions between two designated quantum states:

The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium 133 atom.

1.4 The meter

The meter was originally defined as 1/10,000,000 of the arc from the Equator to the North Pole along the meridian passing through Paris. To aid in calibration and ease of comparison, the meter was redefined in terms of a length scale etched into a platinum bar

1 Isaac Newton. Mathematical Principles of Natural Philosophy. Translated by Andrew Motte (1729).

Revised by Florian Cajori. Berkeley: University of California Press, 1934. p. 6.

preserved near Paris. Once laser light was engineered, the meter was redefined by the 17th Confrence Gnrale des Poids et Msures (CGPM) in 1983 to be a certain number of wavelengths of a particular monochromatic laser beam.

The metre is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second.

Example 1: Light Year Astronomical distances are sometimes described in terms of light-years (ly). A light-year is the distance that light will travel in one year (yr). How far in meters does light travel in one year?

Solution: Using the relationship

distance = (speed of light) (time) , one light year corresponds to a distance. Since the speed of light is given in terms of meters per second, we need to know how many seconds are in a year. We can accomplish this by converting units. We know that

1 year = 365.25 days, 1 day = 24 hours, 1 hour = 60 minutes, 1 minute = 60 seconds

Putting this together we find that the number of seconds in a year is

1 year = 365.25 day( ) 24 hours1day

60 min1 hour

60 s

1 min

=31,557,600 s . (1.4.1)

So the distance that light travels in a one year is

1 ly = 299,792,458 m1s

31,557,600 s

1 yr

1 yr( )= 9.461 1015 m . (1.4.2)

The distance to the nearest star, Alpha Centauri, is three light years.

A standard astronomical unit is the parsec. One parsec is the distance at which there is one arcsecond = 1/3600 degree angular separation between two objects that are separated by the distance of one astronomical unit, 111AU 1.50 10 m= , which is the mean distance between the earth and sun. One astronomical unit is roughly equivalent to eight light minutes, 1AU 8.3l-min= One parsec is equal to 3.26 light years, where one light year is the distance that light travels in one earth year, 51pc 3.26ly 2.06 10 AU= = where 151ly 9.46 10 m= .

1.5 Mass

The unit of mass, the kilogram (kg), remains the only base unit in the International System of Units (SI) that is still defined in terms of a physical artifact, known as the International Prototype of the Standard Kilogram. The prototype was

made in 1879 by George Matthey (of Johnson Matthey) in the form of a cylinder, 39 mm high and 39 mm in diameter, consisting of an alloy of 90 % platinum and 10 % iridium. The international prototype is kept at the Bureau International des Poids et Mesures (BIPM) at Sevres, France under conditions specified by the 1st Confrence Gnrale des Poids et Msures (CGPM) in 1889 when it sanctioned the prototype and declared This prototype shall henceforth be considered to be the unit of mass. It is stored at atmospheric pressure in a specially designed triple bell-jar. The prototype is kept in a vault with six official copies.

Figure 1.1 International Prototype of the Standard Kilogram

The 3rd CGPM (1901), in a declaration intended to end the ambiguity in popular usage concerning the word weight confirmed that:

The kilogram is the unit of mass; it is equal to the mass of the international prototype of the kilogram.

There is a stainless steel one-kilogram standard that can travel for comparisons. In practice it is more common to quote a conventional mass value (or weight-in-air, as measured with the effect of buoyancy), than the standard mass. Standard mass is normally only used in specialized measurements wherever suitable copies of the prototype are stored.

Example 2: The International Prototype Kilogram Determine the type of shape and dimensions of the platinum-iridium prototype kilogram such that it has the smallest surface area for a given volume. The standard kilogram is an alloy of 90 % platinum and 10 % iridium. The density of the alloy is

= 21.56 g cm3 . You may want to consider the following questions:

1) Is there any reason that the surface area of the standard could be important?

2) What is the appropriate density to use?

3) What shape (that is, sphere, cube, right cylinder, parallelepiped, etc.) has the smallest surface area for a given volume?

4) Why was a right-circular cylinder chosen?

Solution: The standard kilogram is an alloy of 90 % platinum and 10 % iridium. The density of platinum is

21.45 g cm- 3 and the density of iridium is

22.55 g cm3 . Thus the density of the standard kilogram,

= 21.56 g cm3 , and its volume is

V = m / 1000 g / 22 g cm3 46.38 cm3 . (1.5.1)

Corrosion would affect the mass through chemical reaction; platinum and iridium were chosen for the standards composition as they resist corrosion.

To further minimize corrosion, the shape should be chosen to have the least surface area. Ideally, this would be a sphere, but as spheres roll easily they become impractical, whereas cylinders have flat surfaces which prevent this. The volume for a cylinder or radius

r and height

h is a constant and given by

V = pir 2h . (1.5.2)

The surface area can be expressed in terms of the radius

r as

A = 2pir 2 + 2pirh = 2pir 2 + 2Vr

. (1.5.3)

To find the smallest surface area, minimize the area with respect to the radius

dAdr

= 4pir 2Vr

2 = 0 . (1.5.4)

Solve for the radius

r3

=

V2pi

=

pir 2h2pi

. (1.5.5)

Thus the radius is one half the height,

r =h2

. (1.5.6)

For the standard mass, the radius is

r =V2pi

1 3

=

46.38 cm3

2pi

1 3

1.95 cm . (1.5.7)

Twice this radius is the diameter of the standard kilogram.

Alternative Definition of Mass

Since the prototype kilogram is an artifact, there are some intrinsic problems associated with its use as a standard. It may be damaged, or destroyed. The prototype gains atoms due to environment wear and cleaning, at a rate of change of mass corresponding to approximately

1 g / year (

1 g 1microgram 1 10-6 g ).

Several new approaches to defining the SI unit of mass (kg) are currently being explored. One possibility is to define the kilogram as a fixed number of atoms of a particular substance, thus relating the kilogram to an atomic mass. Silicon is a good candidate for this approach because it can be grown as a large single crystal, in a very pure form.

Example 3: Mass of a Silicon Crystal

A given standard unit cell of silicon has a volume

V0 and contains

N0 atoms. The number of molecules in a given mole of substance is given by Avogadros constant

N A = 6.0221415 1023

mole-1 . The molar mass of silicon is given by

Mmolar . Find the

mass

m of a volume

V in terms of

V0 ,

N0 , V ,

Mmolar , and

N A .

Solution: The mass

m0 of the unit cell is the density of silicon cell multiplied by the volume of the cell

V0 ,

m0 = V0 . (1.5.8)

The number of moles in the unit cell is the total mass,

m0 , of the cell, divided by the molar mass

Mmolar ,

n0 = m0 / Mmolar = V0 / Mmolar . (1.5.9)

The number of atoms in the unit cell is the number of moles

n0 times the Avogadro constant,

N A ,

N0 = n0N A =V0N A

Mmolar

(1.5.10)

The density of the crystal is related to the mass

m of the crystal divided by the volume

V of the crystal,

= m / V (1.5.11)

So the number of atoms in the unit cell can be expressed as

N0 =mV0N AVM

molar

(1.5.12)

So the mass of the crystal is

m =M

molar

N A

VV0

N0 (1.5.13)

The molar mass, unit cell volume and volume of the crystal can all be measured directly. Notice that

Mmolar / N A is the mass of a single atom, and

(V / V0 )N0 is the number of atoms in the volume. This approach is therefore reduced to the problem of measuring the Avogadro constant,

N A , with a relative uncertainty of 1 part in 108, which is equivalent

to the uncertainty in the present definition of the kilogram.

1.6 Radians and Steradians

Radians

Consider the triangle drawn in Figure 1.6.1

Figure 1.2 Trigonometric relations

You know the basic trigonometric functions of an angle in a right-angled triangle

ONB :

sin( ) = y / r , (1.6.1)

cos( ) = x / r , (1.6.2)

tan( ) = y / x (1.6.3)

It is very important to become familiar with using the measure of the angle itself as expressed in radians [rad]. Let be the angle between two straight lines

OX and

OP . If we draw a circle of any radius

r

centered at

O , the lines

OP and

OX cut the circle at the points

A and

B where

OA = OB = r . If the length of the arc

AB is

s ,

the radian measure of is given by

= s / r ,

and is the same for circles of all radii centered at

O -- just as the ratios

y / r

and

y / x

are the same for all right triangles with the angle at

O . As approaches 360o, s approaches the complete circumference

2pir of the circle, so that 360o = 2pi rad .

Lets compare the behavior of

sin( ) ,

tan( ) and itself for small angles. One can see from the diagram that

s / r > y / r . It is less obvious that

y / x > . It is very instructive to plot

sin( ) ,

tan( ) , and as functions of

[rad] between

0 and

pi / 2 on the same graph (see Figure 1.3).

Figure 1.3 Radians compared to trigonometric functions.

For small , the values of all three functions are almost equal. But how small is small? An acceptable condition is for

can be used almost interchangeably, within some small percentage error. This is the basis of many useful approximations in physics calculations.

Steradians

The steradian (sr) is the unit of solid angle that, having its vertex in the center of a sphere, cuts off an area of the surface of the sphere equal to that of a square with sides of length equal to the radius of the sphere. The conventional symbol for steradian measure is the uppercase greek Omega. The total solid angle sphere of a sphere is then found by dividing the surface area of the sphere by the square of the radius,

sphere = 4pir2 / r2 = 4pi (1.6.5)

Note that this result is independent of the radius of the sphere. Note also that it was implied that the solid angle was measured from the center of the sphere (the radius r is constant). It turns out that the above result does not depend on the position of the vertex as long as the vertex is inside the sphere.

The SI unit, candela, is the luminous intensity of a source that emits monochromatic radiation of frequency

540 1012 s-1 , in a given direction, and that has a radiant intensity in that direction of 1/683 watts per steradian. Note that "in a given direction" cannot be taken too literally. The intensity is measured per steradian of spread, so if the radiation has no spread of directions, the luminous intensity would be infinite.

1.7 Dimensions of Commonly Encountered Quantities

Many physical quantities are derived from the base quantities by set of algebraic relations defining the physical relation between these quantities. The dimension of the derived quantity is written as a power of the dimensions of the base quantitites,

For example velocity is a derived quantity and the dimension is given by the relationship

dim velocity = (length)/(time) = L T-1 . (1.6.6)

where

L length ,

T time .

Force is also a derived quantity and has dimension

dim force = (mass)(dim velocity)(time) . (1.6.7)

where

M mass .We could express force in terms of mass, length, and time by the relationship

dim force = (mass)(length)(time)2 = M L T-2

. (1.6.8)

The derived dimension of kinetic energy is

dim kineticenergy = (mass)(dim velocity)2 , (1.6.9)

which in terms of mass, length, and time is

dim kineticenergy = (mass)(length)2

(time)2 = M L2

T-2 (1.6.10)

The derived dimension of work is

dim work = (dim force)(length) , (1.6.11)

which in terms of our fundamental dimensions is

dim work = (mass)(length)2

(time)2 = M L2

T-2 (1.6.12)

So work and kinetic energy have the same dimensions.

Power is defined to be the rate of change in time of work so the dimensions are

dim power = dim worktime

=(dim force)(length)

time=

(mass)(length)2(time)3 = M L

2 T-3 (1.6.13)

In Table 1.1 we include the derived dimensions of some common mechanical quantities in terms of mass, length, and time.

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Table 1.1 Dimensions of Some Common Mechanical Quantities

M mass ,

L length ,

T time

Quantity Dimension MKS unit Angle dimensionless Dimensionless = radian Steradian dimensionless Dimensionless = radian2 Area

L2

m2 Volume

L3

m3 Frequency

T-1

s1 = hertz = Hz Velocity

L T-1

m s1

Acceleration

L T-2

m s2

Angular Velocity

T-1

rad s1 Angular Acceleration

T-2

rad s2 Density

M L-3

kg m3 Momentum

M L T-1

kg m s1 Angular Momentum

M L2 T-1

kg m2 s1 Force

M L T-2

kg m s- 2 = newton = N Work, Energy

M L2 T-2

kg m2 s2 = joule =J Torque

M L2 T-2

kg m2 s2 Power

M L2 T-3

kg m2 s3= watt = W Pressure

M L-1 T-2

kg m1 s2 = pascal= Pa

Dimensional Analysis

There are many phenomena in nature that can be explained by simple relationships between the observed phenomena.

Example 5: Period of a Pendulum

Consider a simple pendulum consisting of a massive bob suspended from a fixed point by a string. Let

Tperiod denote the time (period of the pendulum) that it takes the bob to complete one cycle of oscillation. How does the period of the simple pendulum depend on the quantities that define the pendulum and the quantities that determine the motion?

Solution:

What possible quantities are involved? The length of the pendulum

l , the mass of the pendulum bob

m , the gravitational acceleration

g , and the angular amplitude of the bob

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0 are all possible quantities that may enter into a relationship for the period of the swing. Have we included every possible quantity? We can never be sure but lets first work with this set and if we need more than we will have to think harder!

Our problem is then to find a function

f such that

Tperiod = f l,m, g,0( ) (1.6.14)

We first make a list of the dimensions of our quantities as shown in Table 1.2. Choose the set: mass, length, and time, to use as the base dimensions.

Table 1.2 Dimensions of quantities that may describe the period of pendulum

Name of Quantity Symbol Dimensional Formula Time of swing

t

T Length of pendulum

l

L Mass of pendulum

m

M Gravitational acceleration

g

L T-2 Angular amplitude of swing

0 No dimension

Our first observation is that the mass of the bob cannot enter into our relationship, as our final quantity has no dimensions of mass and no other quantity can remove the dimension of the pendulum mass. Lets focus on the length of the string and the gravitational acceleration. In order to eliminate length, these quantities must divide each other in the above expression for

Tperiod must divide each other. If we choose the combination

l / g , the dimensions are

dim[l / g] = lengthlength/(time)2 = (time)

2 (1.6.15)

It appears that the time of swing is proportional to the square root of this ratio. We have an argument that works for our choice of constants, which depend on the units we choose for our fundamental quantities. Thus we have a candidate formula

Tperiod :lg

1/ 2

(1.6.16)

(in the above expression, the symbol

: represents a proportionality, not an approximation).

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Since the angular amplitude

0 is dimensionless, it may or may not appear. We can account for this by introducing some function

y 0( ) into our relationship, which is beyond the limits of this type of analysis. Then the time of swing is

Tperiod = y ( ) lg

1/ 2

(1.6.17)

We shall discover later on that

y 0( ) is nearly independent of the angular amplitude

0 for very small amplitudes and is equal to

y 0( )= 2pi ,

Tperiod = 2pilg

1/ 2

(1.6.18)

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Module 2: Problem Solving Strategies and Estimation

2.1 Problem Solving

Solving problems is the most common task used to measure understanding in technical and scientific courses, and in many aspects of life as well. In general, problem solving requires factual and procedural knowledge in the area of the problem, plus knowledge of numerous schema, plus skill in overall problem solving. Schema is loosely defined as a specific type of problem such as principal, rate, and interest problems, one-dimensional kinematic problems with constant acceleration, etc. In most introductory university courses, improving problem solving relies on three things:

1. increasing domain knowledge, particularly definitions and procedures 2. learning schema for various types of problems and how to recognize that a

particular problem belongs to a known schema 3. becoming more conscious of and insightful about the process of problem solving.

To improve your problem solving ability in a course, the most essential change of attitude is to focus more on the process of solution rather than on obtaining the answer. For homework problems there is frequently a simple way to obtain the answer, often involving some specific insight. This will quickly get you the answer, but you will not build schema that will help solve related problems further down the road. Moreover, if you rely on insight, when you get stuck on a problem, youre stuck with no plan or fallback position. At MIT you will see very few exam problems that are exactly the same as problems you have seen before, but most will use the same schema.

General Approach to Problem Solving

A great many physics textbook authors (e.g. Young and Freedman, Knight, Halliday, Resnick and Cartwright) recommend overall problem solving strategies. These are typically four-step procedures that descend from George Polyas influential book, How to Solve It, on problem solving2. Here are his four steps:

I. Understand get a conceptual grasp of the problem

What is the problem asking? What are the given conditions and assumptions? What domain of knowledge is involved? What is to be found and how is this determined or constrained by the given conditions?

What knowledge is relevant? E.g. in physics, does this problem involve kinematics, forces, energy, momentum, angular momentum, equilibrium? If the problem involves two different areas of knowledge, try to separate the problem into parts. Is there motion or is it static? If the problem involves vector quantities such as velocity or momentum, think of these geometrically

2 G. Polya, How to Solve It, 2nd ed., Princeton University Press, 1957.

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(as arrows that add vectorially). Get conceptual understanding: is some physical quantity (energy, momentum, angular momentum, etc.) conserved? Have you done problems that involve the same concepts in roughly the same way?

Model: Real life contains great complexity, so in physics (chemistry, economics) you actually solve a model problem that contains the essential elements of the real problem. The bike and rider become a point mass (unless angular momentum is involved), the ladders mass is regarded as being uniformly distributed along its length, the car is assumed to have constant acceleration or constant power (obviously not true when it shifts gears), etc. Become sensitive to information that is implicitly assumed (Presence of gravity? No friction? That the collision is of short duration relative to the timescale of the subsequent motion? ).

Advice: Write your own representation of the problems stated data; redraw the picture with your labeling and comments. Get the problem into your brain! Go systematically down the list of topics in the course or for that week if you are stuck.

II. Devise a Plan - set up a procedure to obtain the desired solution

General - Have you seen a problem like this i.e., does the problem fit in a schema you already know? Is a part of this problem a known schema? Could you simplify this problem so that it is? Can you find any useful results for the given problem and data even if it is not the solution (e.g. in the special case of motion on an incline when the plane is at

q = 0 )? Can you imagine a route to the solution if only you knew some apparently not given information? If your solution plan involves equations, count the unknowns and check that you have that many independent equations.

In Physics, exploit the freedoms you have: use a particular type of coordinate system (e.g. polar) to simplify the problem, pick the orientation of a coordinate system to get the unknowns in one equation only (e.g. only the

x -

direction), pick the position of the origin to eliminate torques from forces you dont know, pick a system so that an unknown force acts entirely within it and hence does not change the systems momentum Given that the problem involves some particular thing (constant acceleration, momentum) think over all the equations that involve this concept.

III. Carry our your plan solve the problem!

This generally involves mathematical manipulations. Try to keep them as simple as possible by not substituting in lengthy algebraic expressions until the end is in sight, make your work as neat as you can to ease checking and

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reduce careless mistakes. Keep a clear idea of where you are going and have been (label the equations and what you have now found), if possible, check each step as you proceed.

IV. Look Back check your solution and method of solution

Can you see that the answer is correct now that you have it often simply by retrospective inspection? Can you solve it a different way? Is the problem equivalent to one youve solved before if the variables have some specific values?

In physics: Check dimensions if analytic, units if numerical. Check special cases (for instance, for a problem involving two massive objects moving on an inclined plane, if

m1 = m2 or q = 0 does the solution reduce to a simple expression that you can easily derive by inspection or a simple argument?) Is the scaling what youd expect (an energy should vary as the velocity squared, or linearly with the height). Does it depend sensibly on the various quantities (e.g. is the acceleration less if the masses are larger, more if the spring has a larger

k )? Is the answer physically reasonable (especially if numbers are given or reasonable ones substituted).

Review the schema of your solution: Review and try to remember the outline of the solution what is the model, the physical approximations, the concepts needed, and any tricky math manipulation.

2.2 Significant Figures, Scientific Notation, and Rounding

Significant Figures

We shall define significant figures by the following rules.3

1. The leftmost nonzero digit is the most significant digit.

2. If there is no decimal place, the rightmost nonzero digit is the least significant digit.

3. If there is a decimal point then the right most digit is the least significant digit even if it is a zero.

4. All digits between the least and most significant digits are counted as significant digits.

3 Philip R Bevington and D. Keith Robinson, Data Reduction and Error Analysis for the Physical Sciences,

2nd Edition, McGraw-Hill, Inc., New York, 1992.

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When reporting the results of an experiment, the number of significant digits used in reporting the result is the number of digits needed to state the result of that measurement (or a calculation based on that measurement) without any loss of precision.

There are exceptions to these rules, so you may want to carry around one extra significant digit until you report your result. For example if you multiply

2 0.56 = 1.12 , not

1.1 .

There is some ambiguity about the number of significant figure when the rightmost digit is 0, for example 1050, with no terminal decimal point. This has only three significant digits. If all the digits are significant the number should be written as 1050., with a terminal decimal point. To avoid this ambiguity it is wiser to use scientific notation.

Scientific Notation

Careless use of significant digits can be easily avoided by the use of decimal notation times the appropriate power of ten for the number. Then all the significant digits are manifestly evident in the decimal number. So the number

1050 = 1.05 103 while the number

1050. = 1.050 103 .

Rounding

To round off a number by eliminating insignificant digits we have three rules. For practical purposes, rounding will be done automatically by a calculator or computer, and all we need do is set the desired number of significant figures for whichever tool is used.

1. If the fraction is greater than 1/2, increment the new least significant digit.

2. If the fraction is less than 1/2, do not increment.

3. If the fraction equals 1/2, increment the least significant digit only if it is odd.

The reason for Rule 3 is that a fractional value of 1/2 may result from a previous rounding up of a fraction that was slightly less than 1/2 or a rounding down of a fraction that was slightly greater than 1/2. For example, 1.249 and 1.251 both round to three significant digits 1.25. If we were to round again to two significant digits, both would yield the same value, either 1.2 or 1.3 depending on our convention in Rule 3. Choosing to round up if the resulting last digit is odd and to round down if the resulting last digit is even reduces the systematic errors that would otherwise be introduced into the average of a group of such numbers.

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2.3 Order of Magnitude Estimates - Fermi Problems

Counting is the first mathematical skill we learn. We came to use this skill by distinguishing elements into groups of similar objects, but we run into problems when our desired objects are not easily identified, or there are too many to count.

Rather than spending a huge amount of effort to attempt an exact count, we can try to estimate the number of objects in a collection. For example, we can try to estimate the total number of grains of sand contained in a bucket of sand. Since we can see individual grains of sand, we expect the number to be very large but finite. Sometimes we can try to estimate a number which we are fairly sure but not certain is finite, such as the number of particles in the universe (See Chapter 20).

We can also assign numbers to quantities that carry dimensions, such as mass, length, time, or charge, which may be difficult to measure exactly. We may be interested in estimating the mass of the air inside a room, or the length of telephone wire in the United States, or the amount of time that we have slept in our lives, or the number of electrons inside our body. So we choose some set of units, such as kilograms, miles, hours, and coulombs, and then we can attempt to estimate the number with respect to our standard quantity.

Often we are interested in estimating quantities such as speed, force, energy, or power. We may want to estimate our natural walking speed, or the force of wind acting against a bicycle rider, or the total energy consumption of a country, or the electrical power necessary to operate this institute. All of these quantities have no exact, well-defined value; they instead lie within some range of values.

When we make these types of estimates, we should be satisfied if our estimate is reasonably close to the middle of the range of possible values. But what does reasonably close mean? Once again, this depends on what quantities we are estimating. If we are describing a quantity that has a very large number associated with it, then an estimate within an order of magnitude should be satisfactory. The number of molecules in a breath of air is close to

1022 ; an estimate anywhere between

1021 and

1023 molecules is close enough. If we are trying to win a contest by estimating the number of marbles in a glass container, we cannot be so imprecise; we must hope that our estimate is within 1% of the real quantity.

These types of estimations are called Fermi Problems. The technique is named after the physicist Enrico Fermi, who was famous for making these sorts of back of the envelope calculations.

Methodology for Estimation Problems

Estimating is a skill that improves with practice. Here are two guiding principles that may help you get started.

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(1) You must identify a set of quantities that can be estimated or calculated.

(2) You must establish an approximate or exact relationship between these quantities and the quantity to be estimated in the problem.

Estimations may be characterized by a precise relationship between an estimated quantity and the quantity of interest in the problem. When we estimate, we are drawing upon what we know. But different people are more familiar with certain things than others. If you are basing your estimate on a fact that you already know, the accuracy of your estimate will depend on the accuracy of your previous knowledge. When there is no precise relationship between estimated quantities and the quantity to be estimated in the problem, then the accuracy of the result will depend on the type of relationships you decide upon. There are often many approaches to an estimation problem leading to a reasonably accurate estimate. So use your creativity and imagination!

Example: Lining Up Pennies

Suppose you want to line pennies up, diameter to diameter, until the total length is

1 kilometer . How many pennies will you need? How accurate is this estimation?

Solution: The first step is to consider what type of quantity is being estimated. In this example we are estimating a dimensionless scalar quantity, the number of pennies. We can now give a precise relationship for the number of pennies needed to mark off 1 kilometer

# of pennies = totaldistancediameter of penny

. (2.3.1)

We can estimate a penny to be approximately 2 centimeters wide. Therefore the number of pennies is

# of pennies= totaldistancelengthof a penny

=

(1 km)(2 cm)(1 km / 105 cm) = 5 10

4 pennies . (2.3.2)

When applying numbers to relationships we must be careful to convert units whenever necessary.

How accurate is this estimation? If you measure the size of a penny, you will find out that the width is

1.9 cm , so our estimate was accurate to within 5%. This accuracy was fortuitous. Suppose we estimated the length of a penny to be 1 cm. Then our estimate for the total number of pennies would be within a factor of 2, a margin of error we can live with for this type of problem.

1

Example: Estimate the total mass of all the water in the earth's oceans.

Solution: In this example we are estimating mass, a quantity that is a fundamental in SI units, and is measured in kg. Initially we will try to estimate two quantities: the density of water and the volume of water contained in the oceans. Then the relationship we want is

(mass)ocean

=(density)water (volume)ocean . (2.3.3)

One of the hardest aspects of estimation problems is to decide which relationship applies. One way to check your work is to check dimensions. Density has dimensions of mass/volume, so our relationship is

massocean( )= massvolume

volume

ocean( ). (2.3.4)

The density of fresh water is

water = 1.0 g cm

3 ; the density of seawater is slightly higher, but the difference wont matter for this estimate. You could estimate this density by envisioning how much mass is contained in a one-liter bottle of water. (The density of water is a point of reference for all density problems. Suppose we need to estimate the density of iron. If we compare iron to water, we estimate that iron is 5 to 10 times denser than water. The actual density of iron is

iron = 7.8 g cm-3 ).

Since there is no precise relationship, estimating the volume of water in the oceans is much harder. Lets model the volume occupied by the oceans as if they completely cover the earth, forming a spherical shell (Figure 1.5, which is decidedly not to scale). The volume of a spherical shell of radius

Rearth and thickness d is

( )2shell earthvolume 4piR d , (2.3.5)

where

Rearth is the radius of the earth and d is the average depth of the ocean.

Figure 1.5 A model for estimating the mass of the oceans.

2

We first estimate that the oceans cover about 75% of the surface of the earth. So the volume of the oceans is

volumeocean

0.75( ) 4pi Rearth2 d( ). (2.3.6)

We therefore have two more quantities to estimate, the average depth of the ocean, which we can estimate the order of magnitude as

d 1km , and the radius of the earth, which is approximately

Rearth 6 10

3km . (The quantity that you may remember is the circumference of the earth, about

25,000 miles . Historically the circumference of the earth was defined to be

4 107 m ). The radius

Rearth and the circumference s are exactly

related by

s = 2pi Rearth . (2.3.7)

Thus

Rearth =

s

2pi=

2.5 104 mi( )1.6 km mi-1( )2pi

= 6.4 103 km (2.3.8)

We will use

Rearth 6 10

3km ; additional accuracy is not necessary for this problem, since the ocean depth estimate is clearly less accurate. In fact, the factor of

75% is not needed, but included more or less from habit.

Altogether, our estimate for the mass of the oceans is

(mass)ocean

=(density)water (volume)ocean water 0.75( ) 4pi Rearth2 d( ), (2.3.9)

(mass)ocean

1g

cm3

1 kg103 g

(105 cm)3(1 km)3

(0.75)(4 )(6 103 km)2(1km) , (2.3.10)

(mass)ocean

3 1020 kg 1020 kg

3

Module 3: Cartesian Coordinates and Vectors

Philosophy is written in this grand book, the universe which stands continually open to our gaze. But the book cannot be understood unless one first learns to comprehend the language and read the letters in which it is composed. It is written in the language of mathematics, and its characters are triangles, circles and other geometric figures without which it is humanly impossible to understand a single word of it; without these, one wanders about in a dark labyrinth.

Galileo Galilee in Assayer

3.1 Introduction

Mathematics and physics have been historically interwoven since the time of the ancient Greeks.

The calculus had its origins in the logical difficulties encountered by the ancient Greek mathematicians in their attempt to express their intuitive ideas on the ratios and proportionalities of lines, which they vaguely recognized as continuous, in terms of numbers, which they regarded as discrete.4

The modern science of kinematics began in the 16th century, culminating with Galileos description of the motion of bodies, published in at the start of the 17th century. Galileo used geometric techniques derived from Euclids Elements to introduce the concepts of velocity and acceleration. The reliance on mathematics to describe nature is the foundation on which science is built.

With the introduction of analytic geometry, the systematic use of variables, and a use of the infinitesimal, Leibnitz and Newton developed algorithms for introducing and calculating the derivative and the integral, the core concepts of calculus. Although the calculus was not used by Newton in his development of the Principles of the Mechanics, its use was found to be indispensable in solving countless problems.

In physics, there are well-defined problem solving methodologies that provide step-by-step instructions on applying physics concepts but the key question is: How do these methodologies fit into the larger context of expert problem solving?

Although algorithmic steps can be articulated (and memorized), it is essential for students to understand the meaning of these methodologies. In particular how the methodologies are related to the core physics concepts. The process of understanding and mastering these methodologies is slow because it is built on practice. It is also difficult because it requires a simultaneous understanding of physical and mathematical concepts. The methodologies can bring out the subtleties of the physical concepts while the concepts illustrate the need for the mathematics. This is a non-algorithmic learning

4 Carl B. Boyer,The History of the Calculus and its Conceptual Development, Dover, New York, 1949, p. 4

4

process of synthesizing two difficult abstract knowledge systems simultaneously, mathematical physics.

The conceptual foundation of the methodologies provides the framework for thinking about the physics problem. Thinking about physics problems and designing problem solving strategies is the fusion that turns students into expert problem solvers. Learning to think about a physics problem using force diagrams is an example of this type of fusion.

An emphasis on modeling of physical systems is an effective way to mitigate against two of the main hazards of mathematics classes:

1. The students recognize certain formulas as relevant to other subjects, but view mathematical reasoning as an isolated game with fixed rules that they are forced to master, but with no connection to other subjects.

2. Because they can pass the course by memorizing some templated techniques (and because of point 1) the mathematics they learn is not truly available to them when solving problems in other domains.

Today, the first-year student at MIT is expected to understand single variable calculus (18.01), multiple variable calculus (18.02), along with Newtonian Mechanics (8.01), and Electricity and Magnetism (8.02). These General Institute Requirements are in fact a rite of passage for the modern MIT student into the world of science and engineering.

The understanding of physics will deepen because students can solve more mathematically challenging homework problems. This will reinforce the idea that calculus is an important part of physics. The understanding of mathematics will deepen because physics provides a rich source of problems

3.2 Cartesian Coordinate System

Physics involve the study of phenomena that we observe in the world. In order to connect the phenomena to mathematics we begin by introducing the concept of a coordinate system. A coordinate system consists of four basic elements:

(1) Choice of origin (2) Choice of axes (3) Choice of positive direction for each axis (4) Choice of unit vectors for each axis

There are three commonly used coordinate systems: Cartesian, cylindrical and spherical. What makes these systems extremely useful is the associated set of infinitesimal line, area, and volume elements that are key to making many integration

5

calculations in classical mechanics, such as finding the center of mass and moment of inertia.

Cartesian Coordinates

Cartesian coordinates consist of a set of mutually perpendicular axes, which intersect at a common point, the origin O . We live in a three-dimensional environment; for that reason, the most common system we will use has three axes, for which we choose the directions of the axes and position of the origin are.

(1) Choice of Origin

Choose an originO . If you are given an object, then your choice of origin may coincide with a special point in the body. For example, you may choose the mid-point of a straight piece of wire.

(2) Choice of Axis

Now we shall choose a set of axes. The simplest set of axes are known as the Cartesian axes, x -axis, y -axis, and the z -axis. Once again, we adapt our choices to the physical object. For example, we select the x -axis so that the wire lies on the x -axis, as shown in Figure 3.1

Figure 3.1 A segment of wire of length a lying along the x -axis of a Cartesian coordinate system.

Then each point P in space our S can be assigned a triplet of values ( , , )P P Px y z , the Cartesian coordinates of the point P . The ranges of these values are: Px < < + ,

Py < < + , Pz < < + .

The collection of points that have the same the coordinate Py is called a level surface. Suppose we ask what collection of points in our space S have the same value of Py y= . This is the set of points { }( , , ) such thatPy PS x y z S y y= = . This set PyS is a

6

plane, the x z

plane (Figure 3.2), called a level set for constant Py . Thus, the y -coordinate of any point actually describes a plane of points perpendicular to the y -axis.

Figure 3.2 Level surface set for constant value Py .

(3) Choice of Positive Direction

Our third choice is an assignment of positive direction for each coordinate axis. We shall denote this choice by the symbol + along the positive axis. Conventionally, Cartesian coordinates are drawn with the y z

plane corresponding to the plane of the paper. The horizontal direction from left to right is taken as the positive y -axis, and the vertical direction from bottom to top is taken as the positive z -axis. In physics problems we are free to choose our axes and positive directions any way that we decide best fits a given problem. Problems that are very difficult using the conventional choices may turn out to be much easier to solve by making a thoughtful choice of axes. The endpoints of the wire now have coordinates ( / 2,0,0)a and ( / 2,0,0)a .

(4) Choice of Unit Vectors

We now associate to each point P in space, a set of three unit directions vectors ( , , )P P Pi j k . A unit vector has magnitude one: 1P =i , 1P =j , and 1P =k . We assign the direction of Pi to point in the direction of the increasing x -coordinate at the point P . We define the directions for Pj and Pk in the direction of the increasing y -coordinate and z -coordinate respectively. (Figure 3.3).

7

Figure 3.3 Choice of unit vectors.

3.3 Vector Analysis

Introduction to Vectors

Certain physical quantities such as mass or the absolute temperature at some point only have magnitude. These quantities can be represented by numbers alone, with the

appropriate units, and they are called scalars. There are, however, other physical quantities which have both magnitude and direction; the magnitude can stretch or shrink, and the direction can reverse. These quantities can be added in such a way that takes into

account both direction and magnitude. Force is an example of a quantity that acts in a certain direction with some magnitude that we measure in newtons. When two forces act on an object, the sum of the forces depends on both the direction and magnitude of the two forces. Position, displacement, velocity, acceleration, force, momentum and torque are all physical quantities that can be represented mathematically by vectors. We shall

begin by defining precisely what we mean by a vector.

Properties of a Vector

A vector is a quantity that has both direction and magnitude. Let a vector be denoted by the symbol A

r. The magnitude of A

r is | AA |r . We can represent vectors as geometric

objects using arrows. The length of the arrow corresponds to the magnitude of the vector. The arrow points in the direction of the vector (Figure 3.4).

8

Figure 3.4 Vectors as arrows.

There are two defining operations for vectors:

(1) Vector Addition:

Vectors can be added. Let Ar

and Br

be two vectors. We define a new vector, = +C A Br r r

,

the vector addition of Ar

and Br

, by a geometric construction. Draw the arrow that represents A

r. Place the tail of the arrow that represents B

r at the tip of the arrow for A

r as

shown in Figure 3.5(a). The arrow that starts at the tail of Ar and goes to the tip of Br is defined to be the vector addition = +C A B

r r r. There is an equivalent construction for the

law of vector addition. The vectors Ar

and Br

can be drawn with their tails at the same point. The two vectors form the sides of a parallelogram. The diagonal of the parallelogram corresponds to the vector = +C A B

r r r, as shown in Figure 3.5(b).

Figure 3.5(a) Geometric sum of vectors. Figure 3.5 (b) Geometric sum of vectors.

Vector addition satisfies the following four properties:

(i) Commutivity:

The order of adding vectors does not matter;

9

+ = +A B B Ar rr r

. (3.3.1)

Our geometric definition for vector addition satisfies the commutivity property (i) since in the parallelogram representation for the addition of vectors, it doesnt matter which side you start with, as seen in Figure 3.6.

Figure 3.6 Commutative property of vector addition

(ii) Associativity:

When adding three vectors, it doesnt matter which two you start with

( ) ( )+ + = + +A B C A B Cr r r rr r (3.3.2)

In Figure 3.7, we add ( )+ +A B Cr rr , and ( )+ +A B Cr rr to arrive at the same vector sum in either case.

Figure 3.7 Associative law.

(iii) Identity Element for Vector Addition:

There is a unique vector, 0r

, that acts as an identity element for vector addition. For all vectors A

r,

10

+ = + =A 0 0 A Ar r r r r

(3.3.3)

(iv) Inverse Element for Vector Addition:

For every vector Ar

, there is a unique inverse vector

( 1) A Ar r (3.3.4)

such that ( )+ =A A 0r r r

The vector Ar

has the same magnitude as Ar

, | | | | A= =A Aur ur , but they point in opposite directions (Figure 3.8).

Figure 3.8 additive inverse.

(2) Scalar Multiplication of Vectors:

Vectors can be multiplied by real numbers. Let Ar

be a vector. Let c be a real positive number. Then the multiplication of A

r by c is a new vector, which we denote by the

symbol c Ar

. The magnitude of c Ar

is c times the magnitude of Ar

(Figure 3.9a),

c A Ac= (3.3.5)

Since 0c > , the direction of c Ar

is the same as the direction of Ar

. However, the direction of c A

r is opposite of A

r (Figure 3.9b).

11

Figure 3.9 Multiplication of vector Ar

by (a) 0c > , and (b) 0c < .

Scalar multiplication of vectors satisfies the following properties:

(i) Associative Law for Scalar Multiplication:

The order of multiplying numbers is doesnt matter. Let b and c be real numbers. Then

( ) ( ) ( ) ( )b c bc cb c b= = =A A A Ar r r r

(3.3.6)

(ii) Distributive Law for Vector Addition:

Vector addition satisfies a distributive law for multiplication by a number. Let c be a real number. Then

( )c c c+ = +A B A Br rr r

(3.3.7)

Figure 3.10 illustrates this property.

Figure 3.10 Distributive Law for vector addition.

(iii) Distributive Law for Scalar Addition:

12

The multiplication operation also satisfies a distributive law for the addition of numbers. Let b and c be real numbers. Then

( )b c b c+ = +A A Ar r r

(3.3.8)

Our geometric definition of vector addition satisfies this condition as seen in Figure 3.11.

Figure 3.11 Distributive law for scalar multiplication

(iv) Identity Element for Scalar Multiplication:

The number 1 acts as an identity element for multiplication,

1 =A Ar r

(3.3.9)

3.4 Application of Vectors

When we apply vectors to physical quantities its nice to keep in the back of our minds all these formal properties. However from the physicists point of view, we are interested in representing physical quantities such as displacement, velocity, acceleration, force, impulse, momentum, torque, and angular momentum as vectors. We cant add force to velocity or subtract momentum from torque. We must always understand the physical context for the vector quantity. Thus, instead of approaching vectors as formal mathematical objects we shall instead consider the following essential properties that enable us to represent physical quantities as vectors.

(1) Vectors can exist at any point P in space.

(2) Vectors have direction and magnitude.

(3) Vector Equality: Any two vectors that have the same direction and magnitude are equal no matter where in space they are located.

13

(4) Vector Decomposition: Choose a coordinate system with an origin and axes. We can decompose a vector into component vectors along each coordinate axis. In Figure 3.12 we choose Cartesian coordinates for the -x y plane (we ignore the z -direction for simplicity but we can extend our results when we need to). A vector A

r at P can be

decomposed into the vector sum,

x y= +A A Ar r r

, (3.4.1)

where xAr

is the x -component vector pointing in the positive or negative x -direction, and yA

r is the y -component vector pointing in the positive or negative y -direction

(Figure 3.12).

Figure 3.12 Vector decomposition

(5) Unit vectors: The idea of multiplication by real numbers allows us to define a set of unit vectors at each point in space. We associate to each point P in space, a set of three unit vectors ( , , )i j k . A unit vector means that the magnitude is one: | | 1=i , | | 1=j , and

| | 1=k . We assign the direction of i to point in the direction of the increasing x -coordinate at the point P . We call i the unit vector at P pointing in the + x -direction. Unit vectors j and k can be defined in a similar manner (Figure 3.13).

14

Figure 3.13 Choice of unit vectors in Cartesian coordinates.

(6) Vector Components: Once we have defined unit vectors, we can then define the x -component and y -component of a vector. Recall our vector decomposition,

x y= +A A Ar r r

. We can write the x-component vector, xAr

, as

x xA=A ir

(3.4.2)

In this expression the term xA , (without the arrow above) is called the x-component of the vector A

r. The x -component xA can be positive, zero, or negative. It is not the

magnitude of xAr

which is given by 2 1/ 2( )xA . Note the difference between the x -component, xA , and the x -component vector, xA

r.

In a similar fashion we define the y -component, yA , and the z -component, zA , of the vector A

r

,y y z zA A= =A j A k

r r (3.4.3)

A vector Ar

can be represented by its three components ( , , )x y zA A A=Ar

. We can also write the vector as

x y zA A A= + +A i j kr

(3.4.4)

(7) Magnitude: In Figure 3.13, we also show the vector components ( , , )x y zA A A=Ar

.

Using the Pythagorean theorem, the magnitude of Ar

is,

2 2 2x y zA A A A= + + (3.4.5)

(8) Direction: Lets consider a vector ( , ,0)x yA A=Ar

. Since the z -component is zero, the

vector Ar

lies in the -x y plane. Let denote the angle that the vector Ar

makes in the counterclockwise direction with the positive x -axis (Figure 3.14). Then the x -component and y -component are

), )x yA A A A = =cos( sin( (3.4.6)

15

Figure 3.14 Components of a vector in the x-y plane.

We can now write a vector in the -x y plane as

cos( ) sin( )A A = +A i jr (3.4.7)

Once the components of a vector are known, the tangent of the angle can be determined by

) ))y

x

A AA A

= =

sin( tan(cos( , (3.4.8)

that yields

1tan y

x

AA

=

. (3.4.9)

Clearly, the direction of the vector depends on the sign of xA and yA . For example, if both 0xA > and 0yA > , then 0 / 2 pi< < , and the vector lies in the first quadrant. If, however, 0xA > and 0yA < , then / 2 0pi < < , and the vector lies in the fourth quadrant.

(9) Unit vector in the direction of Ar

: Let x y zA A A= + +A i j kr

. Let A denote a unit vector in the direction of A

r. Then

2 2 2 1/2

( )x y z

x y z

A A AA A A

+ += =

+ +

i j kAAA

r

r (3.4.10)

16

(10) Vector Addition: Let Ar

and Br

be two vectors in the x-y plane. Let A and B denote the angles that the vectors A

r and B

r make (in the counterclockwise direction)

with the positive x-axis. Then

) )A AA A = +A cos( i sin( jr

(3.4.11)

) )B BB B = +B cos( i sin( jr

(3.4.12)

In Figure 3.15, the vector addition = +C A Br r r

is shown. Let C denote the angle that the vector C

r makes with the positive x-axis.

Figure 3.15 Vector addition with components

Then the components of Cr

are

,x x x y y yC A B C A B= + = + (3.4.13)

In terms of magnitudes and angles, we have

) ) )) ) )

x C A B

y C A B

C C A BC C A B

= = +

= = +

cos( cos( cos(sin( sin( sin( (3.4.14)

We can write the vector Cr

as

( ) ( ) ) )x x y y C CA B A B C C = + + + = +C i j cos( i sin( jr

(3.4.15)

17

Example 1: Given two vectors, 2 3 7= + +A i j kr and 5 2= + +B i j kr , find: (a) Ar ; (b) Br

; (c) +A Br r

; (d) A Br r

; (e) a unit vector A pointing in the direction of Ar

; (f) a unit vector B pointing in the direction of B

r.

(a) ( )1/ 22 2 22 ( 3) 7 62 7.87= + + = =Ar

(b) ( )1/ 22 2 25 1 2 30 5.48= + + = =Br

(c)

( ) ( ) ( ) (2 5) ( 3 1) (7 2)

7 2 9

x x y y z zA B A B A B+ = + + + + +

= + + + + +

= +

A B i j ki j k

i j k

r r

(d)

( ) ( ) ( ) (2 5) ( 3 1) (7 2)

3 4 5

x x y y z zA B A B A B = + +

= + +

= +

A B i j ki j k

i j k

r r

(e) A unit vector A in the direction of Ar

can be found by dividing the vector A

r by the magnitude of A

r. Therefore

( ) / 2 3 7 / 62= = + +A A A i j kr r

(f) In a similar fashion, ( ) / 5 2 / 30= = + +B B B i j kr r .

Example 2: Consider two points located at 1rr

and 2rr

, separated by distance 12 1 2r = r rr r

.

Find a vector Ar

from the origin to the point on the line between 1rr

and 2rr

at a distance

12xr from the point at 1rr

, where x is some number.

Solution: Consider the unit vector pointing from 1rr

and 2rr

given by

12 1 2 1 2 1 2 12 / / r= = r r r r r r rr r r r r r

. The vector r in the figure connects Ar

to the point at 1rr

,

therefore we can write ( ) ( )12 12 12 1 2 12 1 2 /xr xr r x= = = r r r r rr r r r r . The vector 1 = +r A rr r

.

Therefore ( )1 1 1 2 1 2(1 )x x x= = = +A r r r r r rr r r r r r r r .

3.5 Dot Product

18

We shall now introduce a new vector operation, called the dot product or scalar product that takes any two vectors and generates a scalar quantity (a number). We shall see that the physical concept of work can be mathematically described by the dot product between the force and the displacement vectors.

Let Ar

and Br

be two vectors. Since any two non-collinear vectors form a plane, we define the angle to be the angle between the vectors A

r and B

r as shown in Figure

3.16. Note that can vary from 0 to pi .

Figure 3.16 Dot product geometry.

Definition: Dot Product

The dot product rA

rB of the vectors A

r and B

r is defined to be product of the

magnitude of the vectors Ar

and Br

with the cosine of the angle between the two vectors:

Aur

Bur

= ABcos() (3.5.1)

Where | |A = Ar and | |B = Br represent the magnitude of Ar and Br respectively. The dot product can be positive, zero, or negative, depending on the value of cos . The dot product is always a scalar quantity.

The angle formed by two vectors is therefore

= cos1rA

rB

rA

rB

(3.5.2)

The magnitude of a vector rA is given by the square root of the dot product of the vector

rA with itself.

19

rA = ( rA rA)1/ 2 (3.5.3)

We can give a geometric interpretation to the dot product by writing the definition as

rA

rB = ( Acos( )) B (3.5.4)

In this formulation, the term cosA is the projection of the vector Br in the direction of the vector B

r. This projection is shown in Figure 3.17(a). So the dot product is the

product of the projection of the length of Ar in the direction of Br with the length of Br . Note that we could also write the dot product as

rA

rB = A(Bcos( )) (3.5.5)

Now the term

Bcos( ) is the projection of the vector Br in the direction of the vector Ar as shown in Figure 3.17(b). From this perspective, the dot product is the product of the projection of the length of Br in the direction of Ar with the length of Ar .

Figure 3.17(a) and 2.17(b) Projection of vectors and the dot product.

From our definition of the dot product we see that the dot product of two vectors that are perpendicular to each other is zero since the angle between the vectors is / 2pi and

cos(pi / 2) = 0 .

Algebraic Properties of Dot Product

The first property involves the dot product between a vector c Ar

where c is a scalar and a vector B

r,

(1a) c

rA

rB = c( rA rB) (3.5.6)

20

The second involves the dot product between the sum of two vectors Ar

and Br

with a vector C

r,

(2a) ( rA + rB) rC = rA rC + rB rC (3.5.7)

Since the dot product is a commutative operation,

rA

rB =

rB

rA , (3.5.8)

similar definitions hold;

(1b)

rA c

rB = c( rA rB) (3.5.9)

(2b)

rC ( rA + rB) = rC rA + rC rB . (3.5.10)

Vector Decomposition and the Dot Product

With these properties in mind we can now develop an algebraic expression for the dot product in terms of components. Lets choose a Cartesian coordinate system with the vector B

r pointing along the positive x -axis with positive x -component xB , i.e.,

Bur

= Bxi . The vector A

r can be written as

Aur

= Ax

i + Ay j + Az k (3.5.11)

We first calculate that the dot product of the unit vector i with itself is unity:

i i =| i || i | cos(0) = 1 (3.5.12)

since the unit vector has magnitude

| i |= 1 and cos(0) 1= . We note that the same rule applies for the unit vectors in the y and z directions:

j j = k k = 1 (3.5.13)

The dot product of the unit vector i with the unit vector j is zero because the two unit vectors are perpendicular to each other:

i j =| i || j | cos(pi / 2) = 0 (3.5.14)

21

Similarly, the dot product of the unit vector i with the unit vector k , and the unit vector j with the unit vector k are also zero:

i k = j k = 0 (3.5.15)

The dot product of the two vectors now becomes

rA

rB = ( A

xi + Ay j + Az k) Bx i

= Ax

i Bx

i + Ay j Bx i + Az k Bx i property (2a)= A

xB

x(i i) + Ay Bx (j i) + Az Bx ( k i) property (1a) and (1b)

= AxB

x

(3.5.16)

This third step is the crucial one because it shows that it is only the unit vectors that undergo the dot product operation.

Since we assumed that the vector Br

points along the positive x -axis with positive x -component xB , our answer can be zero, positive, or negative depending on the x -component of the vector A

r. In Figure 3.18, we show the three different cases.

Figure 3.18 Dot product that is (a) positive, (b) zero or (c) negative.

The result for the dot product can be generalized easily for arbitrary vectors

rA = A

xi + Ay j + Az k (3.5.17)

and

rB = B

xi + By j + Bz k (3.5.18)

to yield

22

rA

rB = A

xB

x+ Ay By + Az Bz (3.5.19)

Example 3: Given two vectors, 2 3 7= + +A i j kr and 5 2= + +B i j kr , find A Br r .

(2)(5) ( 3)(1) (7)(2) 21x x y y z zA B A B A B = + +

= + + =

A Br r

Example 4: Find the cosine of the angle between the vectors 3= + +A i j kr and 2 3= B i j kr .

Solution: The dot product of two vectors is equal to cos =A B A Br rr r

where is the angle between the two vectors. Therefore

2 2 2 1/ 2 2 2 2 1/ 2

2 2 2 1/ 2 2 2 2 1/ 2

1/ 2 1/ 2

cos /

( ) /( ) ( )((3)( 2) (1)( 3) (1)( 1)) /((3) (1) (1) ) ( ( 2) ( 3) ( 1) )( 10) /(11) (14) 0.806

x x y y z z x y z x y zA B A B A B A A A B B B

=

= + + + + + +

= + + + + + +

= =

A B A Br rr r

Example 5: Show that if = +A B A Br rr r

, then Ar

is perpendicular to Br

.

Solution: Recall that ( ) ( )1/ 2 1/ 2( ) ( ) ( 2 ) = = + A B A B A B A A A B B Br r r r r rr r r r r r . Similarly ( ) ( )1/ 2 1/ 2( ) ( ) ( 2 )+ = + + = + + A B A B A B A A A B B Br r r r r rr r r r r r . So if

= +A B A Br rr r

, then 0 =A Br r

hence Ar

is perpendicular to Br

.

Work:

23

A typical physics application of the dot product between two vectors involves the calculation of the work done by a force F

r on an object that undergoes a displacement

rr . If the force is uniform in space and constant in time during the entire displacement,

then the work done by the force on the object is given by

WF =

rF rr (3.5.20)

3.6 Cross Product

We shall now introduce our second vector operation, called the cross product that takes any two vectors and generates a new vector. The cross product is a type of multiplication law that turns our vector space (law for addition of vectors) into a vector algebra (a vector algebra is a vector space with an additional rule for multiplication of vectors). The first application of the cross product will be the physical concept of torque about a point P that can be described mathematically by the cross product between two vectors: a vector from P to where the force acts, and the force vector.

Definition: Cross Product

Let Ar

and Br

be two vectors. Since any two non-parallel vectors form a plane, we define the angle to be the angle between the vectors A

r and B

r as shown in

Figure 3.19. The magnitude of the cross product rA

rB of the vectors A

r and B

r

is defined to be product of the magnitude of the vectors Ar

and Br

with the sine of the angle between the two vectors,

rA

rB = ABsin( ) , (3.6.1)

where A and B denote the magnitudes of Ar

and Br

, respectively. The angle between the vectors is limited to the values

0 pi insuring that

sin( ) 0 .

Figure 3.19 Cross product geometry.

The direction of the cross product is defined as follows. The vectors Ar

and Br

form a plane. Consider the direction perpendicular to this plane. There are two possibilities: we

24

shall choose one of these two (shown in Figure 3.19) for the direction of the cross product

rA

rB using a convention that is commonly called the right-hand rule.

Right-hand Rule for the Direction of Cross Product

The first step is to redraw the vectors Ar

and Br

so that their tails are touching. Then draw an arc starting from the vector A

r and finishing on the vector B

r. Curl your right

fingers the same way as the arc. Your right thumb points in the direction of the cross product

rA

rB (Figure 3.20).

Figure 3.20 Right-Hand Rule.

You should remember that the direction of the cross product rA

rB is perpendicular to

the plane formed by Ar

and Br

.

We can give a geometric interpretation to the magnitude of the cross product by writing the magnitude as

rA

rB = A(Bsin ) (3.6.2)

The vectors Ar

and Br

form a parallelogram. The area of the parallelogram is equal to the height times the base, which is the magnitude of the cross product. In Figure 3.22, two different representations of the height and base of a parallelogram are illustrated. As depicted in Figure 3.21(a), the term

Bsin( ) is the projection of the vector Br in the direction perpendicular to the vector B

r. We could also write the magnitude of the cross

product as

rA

rB = ( Asin( )) B (3.6.3)

Now the term

Asin( ) is the projection of the vector Ar in the direction perpendicular to the vector B

r as shown in Figure 3.21(b).

25

Figure 3.21(a) and 3.21(b) Projection of vectors and the cross product

The cross product of two vectors that are parallel (or anti-parallel) to each other is zero since the angle between the vectors is 0 (or pi ) and

sin(0) = 0 (or

sin(pi ) = 0 ). Geometrically, two parallel vectors do not have a unique component perpendicular to their common direction.

Properties of the Cross Product

(1) The cross product is anti-commutative since changing the order of the vectors cross product changes the direction of the cross product vector by the right hand rule:

rA

rB =

rB

rA (3.6.4)

(2) The cross product between a vector c Ar where c is a scalar and a vector Br is

c

rA

rB = c( rA rB) (3.6.5)

Similarly,

rA c

rB = c( rA rB) (3.6.6)

(3) The cross product between the sum of two vectors Ar and Br with a vector Cr

is

( rA + rB) rC = rA rC + rB rC (3.6.7)

Similarly,

rA ( rB + rC) = rA rB + rA rC (3.6.8)

Vector Decomposition and the Cross Product

We first calculate that the magnitude of cross product of the unit vector i with j :

26

| i j |=| i || j |sin(pi / 2) = 1 (3.6.9)

since the unit vector has magnitude

| i |=| j |= 1 and

sin(pi / 2) = 1. By the right hand rule, the direction of

i j is in the +k as shown in Figure 3.22. Thus

i j = k .

Figure 3.22 Cross product of

i j

We note that the same rule applies for the unit vectors in the y and z directions,

j k = i, k i = j (3.6.10)

Note that by the anti-commutatively property (1) of the cross product,

j i = k, i k = j (3.6.11)

The cross product of the unit vector i with itself is zero because the two unit vectors are parallel to each other, (

sin(0) = 0 ),

| i i |=| i || i | sin(0) = 0 (3.6.12)

The cross product of the unit vector j with itself and the unit vector k with itself are also zero for the same reason.

j j = 0, k k = 0 (3.6.13)

With these properties in mind we can now develop an algebraic expression for the cross product in terms of components. Lets choose a Cartesian coordinate system with the vector B

r pointing along the positive x-axis with positive x-component xB . Then the

vectors Ar

and Br

can be written as

rA = A

xi + Ay j + Az k (3.6.14)

and

27

rB = B

xi (3.6.15)

respectively. The cross product in vector components is

rA

rB = ( A

xi + Ay j + Az k) Bx i (3.6.16)

This becomes, using properties (3) and (2),

rA

rB = ( A

xi B

xi) + ( Ay j Bx i) + ( Az k Bx i)

= AxB

x(i i) + Ay Bx (j i) + Az Bx ( k i)

= Ay Bx k + Az Bx j (3.6.17)

The vector component expression for the cross product easily generalizes for arbitrary vectors

rA = A

xi + Ay j + Az k (3.6.18)

and

rB = B

xi + By j + Bz k

(3.6.19)

to yield

rA

rB = ( Ay Bz Az By ) i + ( Az Bx Ax Bz ) j + ( Ax By Ay Bx ) k . (3.6.20)

Example 6: Given two vectors, 2 3 7= + +A i j kr and 5 2= + +B i j kr , find ABr r .

Solution:

( ) ( ) ( ) (( 3)(2) (7)(1)) ((7)(5) (2)(2)) ((2)(1) ( 3)(5))

13 31 17

y z z y z x x z x y y xA B A B A B A B A B A B = + +

= + +

= + +

A B i j ki j k

i j k

r r

Example 6: Law of Sines: Prove that / sin / sin =B Arr and / sin / sin =B Crr using the cross product. (Hint: Consider the area of a triangle formed by three vectors Ar , Br

, and Cr

, where 0+ + =A B Cr rr

.)

Solution: Since 0+ + =A B Cr rr

, we have that 0 ( )= + + = + A A B C A B A Cr r r r r rr r

. Thus

28

= A B A Cr r rr

or = A B A Cr r rr

.

From the figure we see that sin =A B A Br rr r

and sin =A C A Cr r r r . Therefore sin sin =A B A Cr r rr , and hence / sin / sin =B Crr . A similar argument shows that

/ sin / sin =B Arr proving the law of sines.

Example 8: Show that the volume of a parallelpiped with edges formed by the vectors Ar

, Br

, and Cr

is given by ( ) A B Cr rr

.

Solution: The volume of a parallelpiped is given by area of the base times height. If the base is formed by the vectors B

rand C

r, then the area of the base is given by the

magnitude of B Crr

. The vector = B C B C nr rr r

where n is a unit vector perpendicular

to the base. The projection of the vector Ar along the direction n gives the height of the parallelpiped. This projection is given by taking the dot product of Ar with a unit vector and is equal to height =A n

r. Therefore

rA ( rB rC) = rA ( rB rC ) n = ( rB rC ) rA n = (area)(height) = (volume) .

29

Module 4: One-Dimensional Kinematics

4.1 Introduction

Kinematics is the mathematical description of motion. The term is derived from the Greek word kinema, meaning movement. In order to quantify motion, a mathematical coordinate system, called a reference frame, is used to describe space and time. Once a reference frame has been chosen, we can introduce the physical concepts of position, velocity and acceleration in a mathematically precise manner. Figure 4.1 shows a Cartesian coordinate system in one dimension with unit vector i pointing in the direction of increasing

x -coordinate.

Figure 4.1 A one-dimensional Cartesian coordinate system.

4.2 Position, Time Interval, Displacement

Position

Consider an object moving in one dimension. We denote the position coordinate of the center of mass of the object with respect to the choice of origin by ( )x t . The position coordinate is a function of time and can be positive, zero, or negative, depending on the location of the object. The position has both direction and magnitude, and hence is a vector (Figure 4.2),

( ) ( )t x t=x ir . (4.2.1)

We denote the position coordinate of the center of the mass at 0t = by the symbol 0 ( 0)x x t = . The SI unit for position is the meter [m] (see Section 1.3).

Figure 4.2 The position vector, with reference to a chosen origin.

30

Time Interval

Consider a closed interval of time 1 2[ , ]t t . We characterize this time interval by the difference in endpoints of the interval such that

2 1t t t = . (4.2.2)

The SI units for time intervals are seconds [s].

Definition: Displacement

The change in position coordinate of the mass between the times 1t and 2t is

2 1 ( ( ) ( )) ( )x t x t x t x i ir . (4.2.3)

This is called the displacement between the times 1t and 2t (Figure 4.3). Displacement is a vector quantity.

Figure 4.3 The displacement vector of an object over a time interval is the vector difference between the two position vectors

4.3 Velocity

When describing the motion of objects, words like speed and velocity are used in common language; however when introducing a mathematical description of motion, we need to define these terms precisely. Our procedure will be to define average quantities for finite intervals of time and then examine what happens in the limit as the time interval becomes infinitesimally small. This will lead us to the mathematical concept that velocity at an instant in time is the derivative of the position with respect to time.

Definition: Average Velocity

The component of the average velocity, xv , for a time interval t is defined to be the displacement x divided by the time interval t ,

31

x

xv

t

. (4.3.1)

The average velocity vector is then

( ) ( )xx

t v tt

=

v i ir . (4.3.2)

The SI units for average velocity are meters per second 1m s .

Instantaneous Velocity

Consider a body moving in one direction. We denote the position coordinate of the body by ( )x t , with initial position 0x at time 0t = . Consider the time interval [ , ]t t t+ . The average velocity for the interval t is the slope of the line connecting the points ( , ( ))t x t and ( , ( ))t x t t+ . The slope, the rise over the run, is the change in position over the change in time, and is given by

rise ( ) ( )run

x

x x t t x tv

t t

+ = =

. (4.3.3)

Lets see what happens to the average velocity as we shrink the size of the time interval. The slope of the line connecting the points ( , ( ))t x t and ( , ( ))t x t t+ approaches the slope of the tangent line to the curve ( )x t at the time t (Figure 4.4).

Figure 4.4 Graph of position vs. time showing the tangent line at time t .

In order to define the limiting value for the slope at any time, we choose a time interval [ , ]t t t+ . For each value of t , we calculate the average velocity. As 0t ,

32

we generate a sequence of average velocities. The limiting value of this sequence is defined to be the x -component of the instantaneous velocity at the time t .

Definition: Instantaneous Velocity

The x -component of instantaneous velocity at time t is given by the slope of the tangent line to the curve of position vs. time curve at time t :

0 0 0

( ) ( )( ) lim lim limx xt t tx x t t x t dx

v t vt t dt

+ = =

. (4.3.4)

The instantaneous velocity vector is then

( ) ( )xt v t=v ir

. (4.3.5)

Example 1: Determining Velocity from Position

Consider an object that is moving along the x -coordinate axis represented by the equation

20

1( )2

x t x bt= + (4.3.6)

where 0x is the initial position of the object at 0t = .

We can explicitly calculate the x -component of instantaneous velocity from Equation (4.3.4) by first calculating the displacement in the x -direction, ( ) ( )x x t t x t = + . We need to calculate the position at time t t+ ,

( )2 2 20 01 1( ) ( ) 22 2x t t x b t t x b t t t t+ = + + = + + + . (4.3.7)

Then the instantaneous velocity is

2 2 20 0

0 0

1 1( 2 )( ) ( ) 2 2( ) lim limx t tx b t t t t x bt

x t t x tv t

t t

+ + + + + = =

. (4.3.8)

This expression reduces to

0

1( ) lim2x t

v t bt b t

= +

. (4.3.9)

33

The first term is independent of the interval t and the second term vanishes because the limit as 0t of t is zero. Thus the instantaneous velocity at time t is

( )xv t bt= . (4.3.10)

In Figure 4.5 we graph the instantaneous velocity, ( )xv t , as a function of time t .

Figure 4.5 A graph of instantaneous velocity as a function of time.

4.4 Acceleration

We shall apply the same physical and mathematical procedure for defining acceleration, the rate of change of velocity. We first consider how the instantaneous velocity changes over an interval of time and then take the limit as the time interval approaches zero.

Average Acceleration

Acceleration is the quantity that measures a change in velocity over a particular time interval. Suppose during a time interval t a body undergoes a change in velocity

( ) ( )t t t = + v v vr r r . (4.4.1)

The change in the x -component of the velocity, xv , for the time interval [ , ]t t t+ is then

( ) ( )x x xv v t t v t = + . (4.4.2)

Definition: Average Acceleration

The x -component of the average acceleration for the time interval t is defined to be

( ( ) ( )) x x x x

x

v v t t v t va

t t t

+ = = =

a i i i ir . (4.4.3)

The SI units for average acceleration are meters per second squared, 2[m s ] .

34

Instantaneous Acceleration

On a graph of the x -component of velocity vs. time, the average acceleration for a time interval t is the slope of the straight line connecting the two points ( , ( ))xt v t and ( , ( ))xt t v t t+ + . In order to define the x -component of the instantaneous acceleration at time t , we employ the same limiting argument as we did when we defined the instantaneous velocity in terms of the slope of the tangent line.

Definition: Instantaneous Acceleration.

The x -component of the instantaneous acceleration at time t is the limit of the slope of the tangent line at time t of the graph of the x -component of the velocity as a function of time,

0 0 0

( ( ) ( ))( ) lim lim limx x x xx xt t tv t t v t v dv

a t at t dt

+ = =

. (4.4.4)

The instantaneous acceleration vector is then

( ) ( )xt a t=a ir

. (4.4.5)

In Figure 4.6 we illustrate this geometrical construction.

Figure 4.6 Graph of velocity vs. time showing the tangent line at time t .

Since velocity is the derivative of position with respect to time, the x -component of the acceleration is the second derivative of the position function,

35

2

2x

x

dv d xa

dt dt= = . (4.4.6)

Example 2: Determining Acceleration from Velocity

Lets continue Example 1, in which the position function for the body is given by 2

0 (1/ 2)x x bt= + , and the x -component of the velocity is xv bt= . The x -component of the instantaneous acceleration at time t is the limit of the slope of the tangent line at time t of the graph of the x -component of the velocity as a function of time (Figure 4.5)

0 0

( ) ( )lim limx x xx t tdv v t t v t bt b t bt

a bdt t t

+ + = = = =

. (4.4.7)

Note that in Equation (4.4.7), the ratio /v t is independent of t , consistent with the constant slope of the graph in Figure 4.4.

4.5 Constant Acceleration

Lets consider a body undergoing constant acceleration for a time interval [0, ]t t = . When the acceleration xa is a constant, the average acceleration is equal to the instantaneous acceleration. Denote the x -component of the velocity at time 0t = by

,0 ( 0)x xv v t = . Therefore the x -component of the acceleration is given by

,0( )x xxx x

v t vva a

t t

= = =

. (4.5.1)

Thus the velocity as a function of time is given by

,0( )x x xv t v a t= + . (4.5.2)

When the acceleration is constant, the velocity is a linear function of time.

Velocity: Area Under the Acceleration vs. Time Graph

In Figure 4.7, the x -component of the acceleration is graphed as a function of time.

36

Figure 4.7 Graph of the x -component of the acceleration for xa constant as a function of time.

The area under the acceleration vs. time graph, for the time interval 0t t t = = , is

Area( , )x xa t a t . (4.5.3)

Using the definition of average acceleration given above,

,0Area( , ) ( )x x x x xa t a t v v t v = = . (4.5.4)

Displacement: Area Under the Velocity vs. Time Graph

In Figure 4.8, we graph the x -component of the velocity vs. time curve.

Figure 4.8 Graph of velocity as a function of time for xa constant.

The region under the velocity vs. time curve is a trapezoid, formed from a rectangle and a triangle and the area of the trapezoid is given by

( ),0 ,01Area( , ) ( )2x x x xv t v t v t v t= + . (4.5.5)

Substituting for the velocity (Equation (4.5.2)) yields

37

2,0 ,0 ,0 ,0

1 1Area( , ) ( )2 2x x x x x x x

v t v t v a t v t v t a t= + + = + . (4.5.6)

Figure 4.9 The average velocity over a time interval.

We can then determine the average velocity by adding the initial and final velocities and dividing by a factor of two (see Figure 4.9).

,01 ( ( ) )2x x x

v v t v= + . (4.5.7)

The above method for determining the average velocity differs from the definition of average velocity in Equation (4.3.1). When the acceleration is constant over a time interval, the two methods will give identical results. Substitute into Equation (4.5.7) the x -component of the velocity, Equation (4.5.2), to yield

vx

=

12

(vx(t) + v

x ,0 ) =12

((vx ,0 + ax t) + vx ,0 ) = vx ,0 +

12

axt . (4.5.8)

Recall Equation (4.3.1); the average velocity is the displacement divided by the time interval (note we are now using the definition of average velocity that always holds, for non-constant as well as constant acceleration). The displacement is equal to

0( ) xx x t x v t = . (4.5.9)

Substituting Equation (4.5.8) into Equation (4.5.9) shows that displacement is given by

20 ,0

1( )2x x x

x x t x v t v t a t = = + . (4.5.10)

Now compare Equation (4.5.10) to Equation (4.5.6) to conclude that the displacement is equal to the area under the graph of the x -compone