PX148 Classical Mechanics and Special Relativity PDF

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Full notes for PX148 Warwick University, David Quigley...

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PX148 Classical Mechanics & Special Relativity

LECTURE NOTES Updated: December 4, 2018

Forward – please read: These notes mostly show the essentials of the lectures, i.e. what I write on the visualiser and what you should know. Occasionally I include information, highlighted in this manner , that I may have said but not written when I think it might help you follow the notes. Particularly important equations and ideas will appear in boxes like this . You should just know these. You be able to derive all other equations. The notes are terse, and brief to the point of grammatical inaccuracy. This is because they are notes and are not intended to replace the book (University Physics, 13th ed., by Young & Freedman – “UP” for short in the notes). I make them available in case you had to miss a lecture or find it difficult to make notes during lectures, but if you rely on these notes alone and do not read books as well, you will struggle. Reading multiple treatments of the same topic is one of the most effective ways to learn new physics.

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I put extra information off the main track in the margin too.

Chapter 1 Newton’s Laws & Forces 1.1 Classical Mechanics Until ∼ 1600, Aristotle’s view of the Universe, in which Earth defined a “state of rest”, was widely believed in Europe. Common sense: pretty much everything falls to the ground and stops moving doesn’t it? Galileo’s experiments – real and “thought” – overturned these ideas. Newton, born a year after Galileo died in 1642, put Galileo’s ideas into a rigorous mathematical framework in “Principia” (1687). Newton showed that, starting from a few experimentally-based principles, a huge range of physical phenomena could be understood quantitatively. The physics of these phenomena – e.g the motions of planets, forces holding up bridges and buildings - are termed “Classical Mechanics”.

Common sense is sometimes nonsense Principia’s full title in English: Mathematical Principles of Natural Philosophy

Classical mechanics is the bedrock of physics and engineering, and is capable of great precision for motion at everyday speeds. It is classical mechanics that has allowed us to send probes to other planets for example. Many techniques developed for classical mechanics extend to other areas of physics, and it will crop up repeatedly throughout other physics modules. One cannot appreciate physics without a firm grasp of classical mechanics.

1.2 Newton’s Laws 1.2.1 Newton’s First Law – N1 Every body continues in its state of rest or uniform motion unless acted on by an external force. Straight from Galileo.

More compactly: A body moves at constant velocity unless acted upon by a force. 3

PX148 Classical Mechanics & Special Relativity

Chapter 1.2

We know the property of an object that resists changes in velocity as its “mass” (also “inertia” or “inertial mass”). N1 only holds in special “frames of reference” known as “inertial frames” – nonaccelerating, non-rotating – otherwise we would measure accelerations even in the absence of forces. What exactly defines inertial frames (e.g. what do they not accelerate or rotate relative to?) is a deep and unresolved question.

1.2.2 Newton’s Second Law – N2 Defining a new quantity, momentum, as the product of mass times velocity: p = mv, N2 says: The rate of change of momentum of a body is equal to the total external force acting upon it. or, for short: Force equals rate of change of momentum

According to N2 therefore F =

dp , dt

where F is the (total) force acting and t is the time. If the mass m is constant this can be written F =m

dv = ma, dt

where a is the rate of change of velocity (acceleration) of the body. Example 1.1. An oil tanker has mass m = 500 000 t. Its engine and propeller can exert a maximum force of F = 5 × 106 N. Ignoring water resistance, how long does it take to reach a speed of 5 m s−1 , starting from rest?

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PX148 Classical Mechanics & Special Relativity

Answer. a=

Chapter 1.4

F 5 × 106 = 10−2 m s−2 . = 5 × 108 m

So it will take 500 s to reach v = 5 m s−1 .

1.2.3 Newton’s Third Law – N3 To every action there is an equal and opposite reaction.

For “action” and “reaction” read the more modern “force”. N3 says that if I push a car with a force of 200 N, then the car will push back on me with a force of 200 N.

1.3 Scalars and Vectors Physical quantities seen so far such as m, v, p, a, F , and t fall into two classes: scalars:

quantities like mass m and time t that have a size, scale or magnitude, but no direction. They can be positive or negative.

 that vectors: quantities like velocity v, momentum p, acceleration a or force F have magnitude and direction. I will denote vectors using underlined symbols in lectures. You may also see the use of arrows over symbols. In these notes I indicate vectors using bold face with arrows or hats (denoting unit vectors) on top. Written properly, N2 in the form “F = ma” becomes  = m F a. We will see many such vector relations. A scalar cannot equal a vector e.g. the following are both invalid:

r = 2v , ✗ m + v = p. ✗

1.4 Contact Forces When two objects touch, their atoms interact through electromagnetic and quantum effects giving rise to forces between them. 5

In 1D, the scalar form “F = ma” is fine, but try to think in terms of vectors.

PX148 Classical Mechanics & Special Relativity

Chapter 1.4

On our large “macroscopic” scale, we describe the contact force as a combination of a normal reaction and a frictional force. “normal” here means acting perpendicular to the contact.

1.4.1 Static Friction Consider applying a steadily increasing horizontal force F to a stationary object of weight W at rest on a horizontal surface:

N

F FS

Advice: draw diagrams for all but the most trivial problems! They are a key visualisation aid.

W Figure: Forces on a block. W and N are offset for clarity. They must in fact act in line with each other since no rotation occurs. Reaction forces to W and FS that act on the surface that the block is on are not shown for clarity. N is the normal reaction, FS is the static frictional force which opposes F . While the block is at rest, N1 shows us that zero total force acts, hence N = W and FS = F . There is a maximum to the force that static friction can provide. If F exceeds it, the block will accelerate. Experimentally the maximum static friction is found to be proportional to N : max(FS ) = µS N,

Amontons’ Law

where µS is the coefficient of static friction. Thus in general FS ≤ µS N. e.g. steel/steel µS = 0.3; teflon/teflon 0.04; rubber/rubber 1.2.

NB It is possible to have µS > 1.

1.4.2 Kinetic friction Once motion occurs, friction is still roughly proportional to the normal force, but the coefficient changes: FK = µK N, where µK is the coefficient of kinetic or dynamic friction. Note that the speed does not appear.

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PX148 Classical Mechanics & Special Relativity

Chapter 1.6

In general µK < µS , so once slippage starts, it continues, but µK is similar in size to µS . The equations of friction do not have the same fundamental basis as Newton’s Laws, but are approximations to the complex phenomena that occur between surfaces. They allow us to make progress with problems of great practical importance but numerical results should always be treated with caution in critical cases. The simple addition of oil can dramatically alter the coefficient of friction for instance. Engineers standardly build in safety factors into their designs.

1.5 Example: block on a slope See UP ch4 on “free body” diagrams. I don’t personally much like their use of wiggly lines to scrub forces that they also plot resolved into components. I recommend drawing the resolved forces only.

Consider a stationary block on a slope:

N

Fs

θ W=mg

θ

1. Only the forces on the block are shown. 2. Equal and opposite forces to FS and N act on the slope, but we are not interested in them here, so we don’t show them. 3. Stationary, so by N1 the total force on the block is zero. 4. Force is a vector and so must balance separately along all independent directions or “degrees of freedom” (two in this case). Must therefore resolve forces in two directions. 5. The choice of directions is up to you: try to make things easy. Here, perpendicular and parallel to the inclined plane are good choices: N − W cos θ = 0, FS − W sin θ = 0.

In rushing to finish lecture 2 I wrote tan θ ≥ µS on my pad. Corrected in lecture 3.

Hence, for no slippage: FS = W sin θ ≤ µS N = µS W cos θ, 7

PX148 Classical Mechanics & Special Relativity

Chapter 1.6

and so, dividing through by W cos θ , tan θ ≤ µS . This provides an easy way to measure µS : gradually increase θ until the block starts to move at a critical angle θ = θC , then µS = tan θC . How might you measure µK ?

1.6 Gravitational Force

r

1

~ F 12

ˆr

2

The gravitational force between two point masses m1 and m2 a distance r apart is given by Newton’s Law of Gravity: F =

Gm1 m2 , r2

where G = 6.674 × 10−11 N m2 kg−2 is the gravitational constant. [N.B In the lecture I meant to write G with these units, but misremembered and wrote something else. Some helpful chap suggested m3 kg−1 s−2 instead which is equivalent to the above as 1 N = 1 kg m s−2 . In exams you will be given G (and its units) if you need it]. It is an attractive force that acts along the line joining the two masses. These statements are contained in the single vector expression:  12 = − Gm1 m2 rˆ , F r2  12 is the force exerted by mass 1 on mass 2 and r where F ˆ is a unit vector pointing from 1 to 2. The -ve sign ⇒ attraction. In this notation, the distance r is the magnitude of the vector r 12 = r 2 − r1 which connects from 1 → 2, i.e. r = |r 12| and ˆr = r 12/r. [Pedantic note: in the often-seen equation F =−

Gm1 m2 , r2

the minus sign has no clear meaning because it is not vectorial.]

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“Big G”

PX148 Classical Mechanics & Special Relativity

Chapter 1.6

1.6.1 Newton’s shell theorem the same equation applies to two uniform spheres of masses m1 and m2 if r is the distance between their centres. Big G is difficult to measure: need large, dense objects to get significant force. Cavendish (1798) used lead spheres to “weigh Earth”. (Read up on Cavendish’s experiment.) Example 1.2. Three identical iron spheres of radius 5 m are placed in contact as shown below. Calculate the gravitational force on sphere A due to B and C. B A R

C

Answer. The force on A due to B alone has magnitude FBA =

GmB mA . r2

Setting r = 2R, and mA = mB = then FBA =

4π 3 R ρ, 3

16Gπ 2 R6 ρ2 4Gπ 2 4 2 = R ρ. 9 × 4R2 9

Putting R = 5 m and ρ = 7874 kg m−3 , the density of iron, FBA =

( )2 4 × 6.674 × 10−11 × π 2 × 54 × 7.874 × 103 = 11.33 N. 9

The total force is not twice this since the forces from B and C upon A are not parallel. Instead only the component towards the centre of mass of the three spheres matters, giving √ Ftot = 2FBA cos(30◦ ) = FBA 3 = 19.6 N.

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I assume the shell theorem without proof; it took Newton years to prove (he had to develop calculus). It is not hard to prove, but would cost a full lecture. I encourage you to look it up.

PX148 Classical Mechanics & Special Relativity

Chapter 1.6

[A quick note on significant figures here since this is the first problem we’ve encountered which needed a calculator. I’m using R exactly equal to 5 m since the question specifies that parameter. In these notes I’m aiming for 3 s.f. (a sensible default) in my final result so have worked with ρ and G accurate to 4 s.f., and also kept the intermediate FBA to 4 s.f. In lectures I’m usually just looking to get a feeling for the scale of a number (e.g. could we measure the above force in a lab) so will probably be rather sloppy with significant figures. For the problem sheets you should make sensible choices on how accurate to be. Defaulting to 3 s.f. will usually be fine.] Example 1.3. How much less would a person of mass m = 70 kg weigh at the top of the Eiffel tower compared to the ground? Answer. Weight on ground WG =

GME m , 2 RE

where ME and RE are the mass and radius of Earth. Weight at top WT =

GME m , (RE + h)2

where h = 320 m is the height of the Eiffel tower. Difference WG − WT = = = ≈ = =

(

) 1 1 GME m − , 2 (RE + h)2 RE ( ) (RE + h)2 − R2E GME m , RE2 (RE + h)2 ( ) 2RE h + h2 GME m , 2 (R + h)2 RE E 2GME mh , 3 RE ) ( ) ( GME h , 2 m 2 RE RE ) ( h mg = 0.07 N. 2 RE

where the approximations of the 4th line follow because h ≪ RE so h2 ≪ 2RE h, and the expression for the acceleration due to gravity has been extracted.

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Chapter 2 Systems of particles, Accelerations 2.1 Systems of particles How do Newton’s Laws apply to finite-size objects (as opposed to particles) for which different points can have different accelerations? Break down object into a set of N particles with both inter-particle (internal) and external forces acting:

1

F1

3

F31 F13

4 F2

5

2 6

 ji , and the external force on Let the force of particle j acting on i be denoted by F  i. particle i, F Then the total force acting upon particle i is given by ∑  ji = mi a i + F  i. F j=i

Summing over all particles, the total force acting on the N particles is ∑ ∑ ∑ ∑  ji = i + F F mi ai . i

i

j=i

i

N3 ⇒ the second term = 0 since for every Fji , say, there is a cancelling Fij . The first 11

PX148 Classical Mechanics & Special Relativity

Chapter 2.2

 =∑ F  i , so term is the total external “applied” force F i ∑  = F mia i. i

∑  = m We can write this as F acm, where m = i mi is the total mass, if we define ∑ mia i . acm = ∑i i mi

Remembering that

d2r i , dt2 where r i is the position vector of particle i, then (∑ ) d2r cm d2 i m ir i ∑ , = acm = 2 dt2 dt i mi ai =

where

r cm

∑ mir i = ∑i . i mi

This is a mass-weighted average of the position vectors of the particles. It is known as the “centre of mass”. Hence we have the important result: For finite-size objects, Newton’s second law applies unchanged (F = ma) if “a” is the acceleration of the centre of mass.

2.2 Centre of Mass of Continuous Bodies For continuous bodies, we let N → ∞ and mi → 0. In this limit, the centre-of-mass sums become integrals: ∫

r dm r cm = ∫ . dm

The integral in the numerator implies one over each ordinate, i.e. (∫ ) ∫ ∫ ∫ r dm = x dm, y dm, z dm , thus, for example, the x-component of the C-of-M is ∫ x dm xcm = ∫ . dm “dm” is an infinitesimal element of mass.

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PX148 Classical Mechanics & Special Relativity

Chapter 2.2

Example 2.1. A bar of length L has an increasing mass per unit length of ρ = αx where α is a constant and x is the distance from one end. What is the position of its centre of mass? Answer. The total mass of the bar is ∫ ∫ L ∫ M = dm = ρ dx = 0

Similarly,

Hence

∫

x dm =

∫

L

xρ dx = 0

xcm

∫

L 0

L 0

The limits 0 and L shouldn’t appear until after writing the integral over x. I was sloppy with this in the lecture.

1 αx dx = αL2 . 2

1 αx2 dx = αL3 . 3

∫ x dm 2 = ∫ = L. 3 dm

Example 2.2. Find the centre of mass of a filled semi-circle of radius R. Answer. Orient the semi-circle with its straight edge on the y-axis so that the C-of-M is evidently on the x-axis:

y

x

dx Assume a density per unit area of σ. Therefore ∫ 1 M = dm = πσR 2 . 2

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PX148 Classical Mechanics & Special Relativity

Chapter 2.3

Slicing the semi-circle up vertically then ∫

x dm =

∫

R

x(2yσ dx), 0

∫

R

( )1/2 x R2 − x2 dx, 0 [ ]R ) 1( 2 2 3/2 = 2σ − R − x , 3 0 2 = σR3 . 3 = 2σ

Therefore, finally xcm =

4 2σR3 /3 = R ≈ 0.424R. 2 πσR /2 3π

2.3 Forces out of balance When forces are out of balance, accelerations occur. Consider the block on a slope again:

N

FK

θ

a W=mg

θ

Again, only forces on block are shown; the friction force is now the kinetic friction FK = µK N . The acceleration is indicated with a double arrow: it is not a force, but the result of the forces. Resolving perpendicular and parallel (down) to the slope: N − mg cos θ = 0,

mg sin θ − FK = ma. Therefore, as before, N = mg cos θ, and so FK = µK mg cos θ, and so a = g(sin θ − µK cos θ). Example 2.3. Consider the block and slope again, but let’s assume that the slope is a wedge that can move, and that there is no friction. What will the accelerations of block and wedge be? 14

I repeat: “ma” should never be drawn as a force.

PX148 Classical Mechanics & Special Relativity

Chapter 2.4

Answer. With multiple components, draw force / acceleration diagrams for each component separately: N a2

a2 θ

N

a1 mB g

W’

θ

Forces on block

N’

θ

Forces on wedge

 = The acceleration of the block a  1 is defined relative to the wedge. To apply “ F m a” to the block we must add the acceleration of the wedge so we are judging  2. relative to an inertial frame, i.e. aB = a1 + a Wedge, horizontal: N sin θ = mW a2 .

(2.1)

Wedge, vertical not needed – not interested in N ′ . Block, perpendicular to slope, outwards: N − mB g cos θ = −mB a2 sin θ.

(2.2)

NB The acceleration on the right is the component of the block’s acceleration perpendicular to the slope which is all from a2 . Block, parallel, down slope: mB g sin θ = mB (a1 − a2 cos θ).

(2.3)

Use Eq. 2.2 to eliminate N from Eq. 2.1: mB (g cos θ − a2 sin θ) sin θ = mW a2 , so, rearranging, a2 =

mB g cos θ sin θ . mW + mB sin2 θ

(2.4)

Eq. 2.3 and Eq. 2.4 then tell us a1 : a1 = g sin θ + which reduces to a1 =

mB g cos2 θ sin θ , mW + mB sin2 θ

(mW + mB )g sin θ . mW + mB sin2 θ

Check: dimensions ✓; intuition: a2 = 0 at θ = 0 and 90◦ ✓; a1 = 0 at θ = 0 ✓; a1 = g at θ = 90◦ ✓; a2 → 0 as mB /mW → 0 ✓. 15

PX148 Classical Mechanics & Special Relativity

Chapter 2.5

2.4 Physics jargon Physics problems can seem rather artificial, but the point is to try to isolate the physics of interest without the clutter of multiple effects. To this end one needs to know the meaning of a few “code” words, some of which we have had already, but now seems a good point to list them all. smooth: zero friction. light:

e.g. “a light pulley” – zero mass, no force needed to accelerate it.

rigid:

does not bend at all

flexible: totally bendy. Strings are almost alw...