Newtonian Mechanics: Introduction

8.01

W01D3 2006

Metric System of Units

Système International (SI)

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)

Speed of Light

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

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

c =299,792,458 meters/second

Atomic Clock and the Definition of the Second

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.

Adopted In 1967

Length as a Derived Quantity

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 preserved near Paris.

Once laser light was engineered, the meter was redefined to be a certain number of wavelengths of a particular monochromatic laser beam.

The meter (m) is now defined as the distance traveled by a light wave in vacuum in 1/299,792,458 seconds.

Inertial Mass• Inertial mass is a “quantity of matter”

• The SI standard of mass is a platinum-iridium

cylinder assigned a mass ms= 1 kg

• The mass of all other bodies will be determined relative to the mass of our standard body.

• By applying the same action (force) to the standard body and an unknown body, the unknown mass is obtained in terms of ratio of accelerations

u s

s u

m a

m a=

Group Problem: Proportions of the Standard Kilogram

The standard kilogram is a cylindrical alloy of 90 % platinum and 10 % iridium. The density of alloy is

Your job is to design a strategy for finding the optimal height and radius for the standard kilogram keeping in mind that the surface is occasionally cleaned of unwelcome atoms (dust). You don’t have to solve this.

ρ =21.56 g⋅cm-3

Proportions of Standard Kilogram

• Strategy; Since atoms collect on the surface, chose the radius and height to minimize surface area.

•Constant volume for cylinder:

• The surface area is

• Minimize the area with respect to radius:

• Radius is one half height:

•The volume determined from density

•The radius is

•The height is

V =πr2h

A =2πr2 + 2πrh=2πr2 + 2V / r

dA / dr =4πr −2V / r2 =0

30 02 / 2V r r hπ= ⇒ =

V =m/ ρ ≅1000 g /22 g⋅cm-3 ≅46.38cm3

r

0= V / 2π( )

1 3; 1.95 cm

h = 2r0 ; 3.90 cm

Problem Solving

• Measure understanding in technical and scientific courses

• Problem solving requires factual and procedural knowledge, 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..

Problem Solving

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.

General Approach to Problem Solving

George Polya’s How to Solve It, on problem solving.

I. Understand – get a conceptual grasp of the problem

II. Devise a Plan - set up a procedure to I.

III. Carry our your plan – solve the problem!

IV. Look Back – check your solution and method of solution

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?

• Model: Real life contains great complexity, so in physics you actually solve a model problem that contains the essential elements of the real problem.

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

desired solutionGeneral– Have you seen a problem like this

– 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

– 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.

III. Carry our your plan – solve the problem!

– mathematical manipulations.

– keep 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 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

solutionCheck work:– Check dimensions if analytic, units if numerical.– Check special cases – Is the scaling what you’d expect – Does it depend sensibly on the various quantities– Is the answer physically reasonable– consider special limits (zero, infinity)

Review the schema of your solution:– what is the model, – the physical approximations,– the concepts needed– any tricky math manipulation.

Estimation Problems

• Quantitative estimation of phenomena

• Order of magnitude estimations

Estimations Strategies

1) What type of quantity is being estimated?

2) How is that quantity related to other quantities, which can be estimated more accurately?

3) Establish an approximate or exact relationship between these quantities and the quantity to be estimated in the problem

Estimation Strategies

3) identify a set of quantities that can be estimated or calculated

4) establish an approximate or exact relationship between these quantities and the quantity to be estimated in the problem

Concept Question: Estimation Volume of Earth’s Atmosphere

What is your best estimate for the volume of the earth’s atmosphere?

1) between 101 and 105 cubic meters

2) between 105 and 1010 cubic meters

3) between 1010 and 1015 cubic meters

4) between 1015 and 1020 cubic meters

5) between 1020 and 1025 cubic meters

6) between 1025 and 1030 cubic meters

Example Estimation: Number of Molecules in Atmosphere

1. Estimation quantity is a scalar: N number of molecules and is related to volume of atmosphere.

2. Number of molecules = (number of molecules in a mole)(number of moles in atmosphere)

3. Number of moles in atmosphere = (volume of atmosphere)/(volume of mole)

• number of molecules in a mole• at STP (Standard Temperature and Pressure): one mole of an

ideal gas occupies 22.4 L = 22.4 x 10-3 m3.

Number of molecules in atmosphere:

N =(6 ×1023 molecules/mole)(4 ×1018 m3) / (22.4 ×10−3 m3 / mole)

=1×1044 molecules

N A≅6 ×1023 molecules/mole

Force Law

• Force Law: Discover experimental relation between force exerted on object and change in properties of object

• Induction: Extend force law from finite measurements to all cases within some range creating a model

Universal Law of Gravitation • Gravitational force between two bodies with masses m1 and m2

• Gravitational force is attractive

• The force is proportional to the product of the masses, and inversely proportional to square of the distance of separation, r1,2 .

rF

1,2=−G

m1m2

r1,22 r̂1,2

11 2 26.67 10 N m /kgG −= × ⋅

Models in Physics: Fundamental Forces of Nature

• Electromagnetism• Weak Force• Strong Force• Gravity• Fifth Force?

Force laws are mathematical models of physical processes

Fundamental UnitsDimensions and values of the fundamental constants

1. Planck constant

2. Speed of light

3. Universal Gravitation constant

4. Boltzmann constant

kB=1.3806505(24) ×10−23 J ⋅K -1

dim h⎡⎣ ⎤⎦= energy( ) time( ) =mass( ) length( )

2

time( ) h =6.6260693(11) ×10−34kg⋅m2 ⋅s-1

dim c⎡⎣ ⎤⎦=length( )time( )

c =299792458 m⋅s-1

dim G⎡⎣ ⎤⎦=length( )

3

mass( ) time( )2 G =6.6742(10) ×10−11N⋅m2 ⋅kg-2

dim kB

⎡⎣ ⎤⎦=energy( )

temperature( )=

mass( )(length)2

temperature( )(time)2

Example: Planck Mass

Let’s construct a combination that depends only on mass. The dimension of the product of Planck’s constant and the speed of light is

• Thus the following quantity has the dimensions of mass

• Value

dim hc⎡⎣ ⎤⎦=mass( ) length( )

3

time( )2

dimhc

G

⎛

⎝⎜⎞

⎠⎟

1 2⎡

⎣⎢⎢

⎤

⎦⎥⎥=

mass( ) length( )3

time( )2

/length( )

3

mass( ) time( )2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

1 2

= mass( )

hc

G

⎛

⎝⎜⎞

⎠⎟

1 2

=(6.6260693×10−34kg⋅m2 ⋅s-1)(2.99792458 ×108m⋅s-1)

(6.6742 ×10−11N⋅m2 ⋅kg-2 )

⎛

⎝⎜⎞

⎠⎟

1 2

=5.3053×10−8kg

Example: Mass to Length Conversion

What are the dimensions and magnitude of the Gravitational constant divided by the speed of light squared?

• This ratio acts as a conversion factor between mass to length in the same manner that speed of light can be used as a conversion factor between length and time. The magnitude of this conversion factor is

dimG

c2=

force( ) length( )2/ mass( )

2

length( )2/ time( )

2=

force( ) time( )2

mass( )2

=mass( )(length)(time)2

(time)2 mass( )2

=(length)mass( )

G

c2=

6.6742 ×10−11N ⋅m2 ⋅kg-2

(2.99792458 ×108m⋅s-1)2=7.4260 ×10−28m⋅kg-1

Example: Planck Length

fundamental length unit, we use the conversion

• Thus the following quantity has the dimensions of length

• Value

dimG

c2

⎡

⎣⎢

⎤

⎦⎥=

(length)mass( )

dimGh

c3

⎛

⎝⎜⎞

⎠⎟

1 2⎡

⎣⎢⎢

⎤

⎦⎥⎥=dim

Gc2

⎡

⎣⎢

⎤

⎦⎥dim

hcG

⎛

⎝⎜⎞

⎠⎟

1 2⎡

⎣⎢⎢

⎤

⎦⎥⎥= mass( )

(length)mass( )

=length

hG

c3

⎛

⎝⎜⎞

⎠⎟

1 2

=(6.6260693×10−34kg⋅m2 ⋅s-1)(6.6742 ×10−11N ⋅m2 ⋅kg-2 )

(2.99792458 ×108m⋅s-1)3⎛

⎝⎜⎞

⎠⎟

1 2

=4.0513×10−35m

Example: Mass to Time Conversion

The dimensions and magnitude of the speed of light cubed divided by Gravitational constant

• This ratio acts as a conversion factor between mass to time

dim (c2 / G)⎡⎣ ⎤⎦dim c⎡⎣ ⎤⎦=mass( )

(length)(length)

time( )=

mass( )time( )

c3

G=

(2.99792458 ×108m⋅s-1)3

6.6742 ×10−11N⋅m2 ⋅kg-2=4.0370 ×1035kg⋅s-1

Example: Planck Time

Use the conversion

• Thus the following quantity has the dimensions of time

• Value

dimG

c3

⎡

⎣⎢

⎤

⎦⎥=

time( )mass( )

dimhG

c5

⎛

⎝⎜⎞

⎠⎟

1 2⎡

⎣⎢⎢

⎤

⎦⎥⎥=dim

Gc3

⎡

⎣⎢

⎤

⎦⎥dim

Gc3

hcG

⎛

⎝⎜⎞

⎠⎟

1 2⎡

⎣⎢⎢

⎤

⎦⎥⎥=

time( )mass( )

mass( ) = time( )

hG

c5

⎛

⎝⎜⎞

⎠⎟

1 2

=(6.6260693×10−34kg⋅m2 ⋅s-1)(6.6742 ×10−11N⋅m2 ⋅kg-2 )

(2.99792458 ×108m⋅s-1)5⎛

⎝⎜⎞

⎠⎟

1 2

=1.3514 ×10−43s