Px148 Notes

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University of Warwick

PX148

Classical Mechanics and Special Relativity

Mike Allen

Based heavily on the material of Prof. T. R. Marsh

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2

Foreword

These notes mostly show the essentials of the lectures, i.e. what I write on the visualizer and what students should know. Occasionally I include information that I may have said but not written during

lectures, when I think it would help you follow the notes. Really I may put little prompts in the margin too.

important equations and ideas appear in boxes like this and you should try to learn and remember them.

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). 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.

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Lecture 1

Introduction to PX148

Objectives

• To introduce the course.

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”: eventually pretty much everything falls to the ground and stops

mov-ing. “Common sense

is the collection of prejudices acquired by age eighteen.”, A.Einstein

Galileo’s experiments – real and “thought” – overturned these ideas. Newton, born shortly after Galileo died in 1642, put Galileo’s ideas into a rigorous mathematical framework in “Principia” (1687).

Newton’s work 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 stars, forces holding up a bridge – are termed Classical Mechanics.

Classical mechanics is the bedrock of physics and engineering, and is capable of great precision for motion at everyday speeds. One can’t ap-preciate the rest of physics without a firm grasp of classical mechanics, so this is where we start.

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LECTURE 1. INTRODUCTION TO PX148 4

1.2 Newton’s Laws

1.2.1 Newton’s First Law – N1

This comes straight from Galileo.

Every body continues in its state of rest or uniform motion unless acted on by a force.

More compactly:

A body moves at constant velocity unless acted upon by a force.

uniform motion: in a straight line at constant speed.

constant velocity: the same; constant direction and speed

at rest: a special case of constant velocity

Newton, of course, invented calculus. Or was it Leibniz?

Aristotle believed that the velocity of an object was proportional to the force acting on it. Newton realized that it is the acceleration (rate of change of velocity) that is proportional to force.

The property of objects that resists changes in velocity is known as “inertia”; we measure it as mass (sometimes “inertial mass”).

There is an implicit assumption here. N1 only holds in special frames of reference known as “inertial frames”: non-accelerating, non-rotating. Otherwise we would measure accelerations even in the absence of forces.

What defines an “inertial frame” (e.g. what do they not accelerate or rotate relative to?) is a deep and as-yet-unresolved question, but if you are in one, you feel no force acting upon you!

1.2.2 Newton’s Second Law – N2

The rate of change of momentum of a body is equal to the total force acting upon it.

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Force equals rate of change of momentum.

momentum: mass × velocity, i.e. p = mv

Really vectors!

According to N2 therefore

F = dp dt,

where t is the time. If the mass m is constant this can also be written

F = mdv

dt = ma,

where a is the acceleration of the body, its rate of change of velocity.

Example 1.1. An oil tanker has mass m = 500 000 t = 5 × 108kg (1 t is a metric tonne, 1000 kg). Its engine and propeller can exert a maximum force of F = 5 × 106N.

How long does it take to reach a speed of 5 m s−1, starting from rest? Starting from 5 m s−1, what is the tanker’s “stopping distance”? [Ignore water resistance.]

Answer. a = F m = 5 × 106 5 × 108 = 10 −2_{m s}−2_.

dv/dt = a, so it will take 500 s to reach v = 5 m s−1.

Using v2 = u2+ 2as, with initial speed u = 5 m s−1, final speed v = 0 m s−1, acceleration a = −0.01 m s−2, then the stopping distance s is given by s = v 2 _{− u}2 2a = −25 2 × −0.01 = 1250 m.

1.2.3 Newton’s Third Law – N3

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action: in more modern parlance, “force”

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Lecture 2

Forces

Keypoints from Lecture 1

• Classical mechanics is the bedrock of physics and engineering. • Newton’s Laws: physical principles, not mathematical axioms. • Newton’s Laws: consistent with experimental observations;

fur-ther experiments could prove them wrong.

Objectives

• Meet the Forces: reaction, friction, springs.

2.1 Types of Force

In an elementary analysis of a bridge or building, we are concerned with forces in balance with no resultant acceleration (the next step would be a dynamic analysis to understand a structure’s ability to withstand earthquakes perhaps). Here we look at some different forces.

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LECTURE 2. FORCES 8

2.1.1 Reaction forces

N

Reaction forces are communicated through inter-atomic forces. Try to squeeze nuclei closer together, electron “clouds” and nuclei repel each other through electrostatic forces. Often denoted N , or ~N , for “normal” to surface.

2.1.2 Static Friction

Almost no surface is completely smooth at the atomic level. For ex-ample, 0.1µm = 10−7m roughness is a very smooth surface, but still means peaks and valleys ∼ 1000 atoms in height or depth!

Small N

Large N

F

S

F

S

It requires a certain minimum sideways force to make two surfaces start to slide past each other. Experimentally it is found that this force required is linearly proportional to the normal reaction force N .

No slippage: FS ≤ µSN.

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LECTURE 2. FORCES 9

static frictional force: FS

coefficient of static friction: µS

A few values µS steel–steel 0.3 teflon–teflon 0.04 rubber–rubber 1.2 NB It is possible to have µS > 1.

The friction acts parallel to the surfaces. Consider a stationary block

on a slope: Draw diagrams

for all but the most trivial problems!

W=mg

_θ

F

_s

N

θ

1. Only the forces on the block are shown. See UP ch4 on

“free body” diagrams. I don’t personally like their use of wiggly lines to scrub forces that they also plot resolved into

components. I recommend drawing each force once only.

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. If static, the total force on the block is zero.

4. All forces are drawn acting through the same point because we are not interested in rotation (for now at least).

5. FS must act up the slope.

6. Force is a vector and so must balance separately along all inde-pendent directions or “degrees of freedom” (two in this case).

7. Must therefore resolve forces in this number (two) of directions.

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LECTURE 2. FORCES 10

Here, perpendicular and parallel to the inclined plane are good choices: N − W cos θ = 0,

FS − W sin θ = 0. Hence, FS = N tan θ ≤ µSN , for no slippage, i.e.

tan θ ≤ µS.

So, measuring the critical θ, at which slipping occurs, gives µS.

2.1.3 Kinetic friction

Once slippage occurs, friction still occurs, and is still found to be proportional to the normal force, but the coefficient changes. The force then is

Slippage: FK = µKN.

kinetic frictional force: FK

coefficient of kinetic or dynamic friction: µK

Note that the speed does not appear.

In general µK < µS, so once slippage starts, it continues, but µK is similar in size to µS.

2.1.4 Spring forces

Springs are resistant to being stretched or compressed. For small dis-placements from equilibrium, x, they offer a force that is in proportion to x:

F = −kx,

force constant: k (characteristic of the spring)

minus sign: force always opposes direction of displacement Example 2.1. A mass m is attached to a spring with constant k and dragged along a horizontal table with static coefficient of friction µS. How far must the spring stretch to get it moving?

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LECTURE 2. FORCES 11

Answer. Normal reaction supports weight so N = mg. Force needed to make mass move therefore equals F = µSmg. Therefore the spring stretches by

x = F

k =

µSmg

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Lecture 3

Equilibrium

Keypoints from Lecture 2

• At a microscopic level, reaction and frictional forces are a man-ifestation of the electromagnetic forces between atoms, in com-bination with quantum mechanics.

• Maximum static frictional force is proportional to the normal force N :

FS ≤ µSN.

• Kinetic frictional force is proportional to the normal force and (almost) independent of the sliding velocity:

FK = µKN.

• µK < µS; also, friction coefficients can exceed 1.

• Spring forces act to restore an equilibrium length and for small extension x, are given by

F = −kx.

Objectives

• How to set up and solve problems in statics

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LECTURE 3. EQUILIBRIUM 13

3.1 Forces in balance

When nothing moves, all forces must balance. This applies to individ-ual components and the whole system. In all cases forces must sum to zero. Consider:

2

θ 1

Cylindrical logs at equilibrium in a box.

Assuming that there is no friction then the forces acting appear as follows: W₁ W₂ N₁ N₁ N 2 N2 R F F R

Forces on each log and the box. (We take the box to be fixed!).

All contact forces have a normal reaction component only (no friction). N3 has been imposed on the action/reaction pairs. ~W1 = m1~g and

~

W2 = m2~g are the weights (assumed known). We also assume that the angle θ is known (fixed by the geometry). There are four unknown forces. The zero force condition allows us to derive relations between them, e.g. on log 1:

~

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This is really two equations in one.

horizontally: N1 − R cos θ = 0,

vertically: F − W1 − R sin θ = 0 .

Something similar applies to log 2. Equally we could consider the two logs together: ~ N1 + ~F + ~W1 + ~W2 + ~N2 = ~0, which becomes horizontally: N1 − N2 = 0, vertically: F − W1 − W2 = 0.

Four unknowns, four equations, giving answers in terms of W1, W2, and θ. F = W1 + W2; R = W2 sin θ; N1 = N2 = W2cos θ sin θ

Example 3.1. A 1 kg mass is placed at the mid-point of a near-horizontal, inextensible, light rope 2 m long. At equilibrium, the mid-point of the rope lies 1 cm below its two ends. What is the tension in the rope?

Answer. Translating the words into a figure:

mg T θ θ h l T where ` = 1 m, h = 1 cm and m = 1 kg

Resolving forces vertically: 2T sin θ = mg.

Geometry: sin θ = h `. Therefore T = mg` 2h = 1 × 9.81 × 1 2 × 0.01 = 490.5 N.

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3.2 Reflections on Newton’s Laws

Here are some left-overs from the first two lectures.

3.2.1 N1 is a special case of N2

If F = 0, N2 says dv/dt = 0, i.e. the velocity v remains constant in the absence of force, which is N1 .

3.2.2 N2 is not just a definition of force

N2 has not merely introduced a new abstract concept “force” to rep-resent the rate of change of momentum, dp/dt.

It is more than this: can devise ways to exert a constant force (e.g. a spring, mass on a string over a pulley) which can be applied to different masses so N2 is experimentally testable.

3.2.3 N3 is needed for consistency with N1 &

N2

Consider an object at rest. Imagine it to have two halves, which may exert forces on each other. Unless these forces are equal and opposite, the object would accelerate on its own, violating N1 . Hence N3 is required by N1 and N2 .

3.2.4 Scalars vs Vectors

Physical quantities seen so far such as m, v, p, a, F , and t fall into two classes.

Some (such as mass m and time t) have a size or “scale” or “magni-tude” but no direction. These are sometimes called “scalar” quantities, and can be represented mathematically by real numbers.

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Others (velocity v, momentum p, force F ) have a magnitude but also a direction. Mathematically we associate them with vectors.

Written properly, N2 in the form “F = ma” becomes

~

F = m~a.

We will see many such vector relations. A scalar cannot equal a vector; the following are therefore both invalid:

r = 2~v, ₇ m + ~v = ~p. ₇

For simple one-dimensional problems we can stick to scalars. In general try to think in terms of vectors. There wasn’t time in the lecture to do the following example, but you

might like to look at it anyway.

Example 3.2. Estimate the force exerted by a hammer on a nail.

Answer. Must guess reasonable values. Let’s take the mass of the hammer’s head to be m = 0.5 kg, its speed at contact with the nail v = 4 m s−1, and the nail to be driven d = 1 cm in to the wood by each blow.

Assume the force F to be constant, for t seconds, until the hammer comes to rest. Then from N2 the change in momentum ∆p is given by

∆p = −mv = −F t, so

F = mv

t .

The speed drops uniformly from v to 0, so the mean speed = v/2, and thus t = d v/2 = 2d v , therefore F = mv 2 2d = 0.5 × 42 2 × 0.01 = 400 N.

Both this example and the second part of the first lecture’s example could be worked out from KE = work done = force × distance.

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A suddenly acting force is sometimes called an “impulse”. The impulse is defined as force × time = F t, or equally well as the change in momentum ∆p.

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Lecture 4

Newton’s Law of

Gravity

Keypoints from Lecture 3

• Newton’s Laws are self-consistent; N3 is required for N1 and N2 to be true.

• Physical quantities require a variety of mathematical objects to represent them. We will use scalars and vectors (in later modules you will meet objects called “tensors” which extend this further). • It is important to combine these correctly in equations.

• N1 applies in situations of static equilibrium. All forces must then be in balance. This condition and geometry then allow one to work out the forces.

Objectives

• To introduce and discuss gravitational attraction.

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LECTURE 4. NEWTON’S LAW OF GRAVITY 19

4.1 Newton’s Law of Gravity

1

12

F

₁₂

2

^

r

The gravitational force between two point masses m1 and m2 a distance r apart is given by

F = Gm1m2

r2 ,

where G = 6.67 × 10−11N m2kg−2 is the gravitational constant.

It is an attractive force that acts along the line joining the two masses. These statements can be combined into a single vector expression:

~

F12 = −

Gm1m2 r2 ˆr12,

where ~F12 is the force exerted by 1 on 2 and ˆr12 is a unit vector pointing from 1 to 2. The minus sign ⇒ attraction.

Pedantic note: in the often-seen equation

F = −Gm1m2 r2 ,

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

Newton’s shell theorem: the same equation applies to two spheres if r is the distance between their centres.

4.2 Inertial vs Gravitational mass

For a moment, distinguish the inertial mass mi which appears in N2 from the gravitational mass mg which appears in the above equations.

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Why should they be the same? Applying N2 to mass 2:

F = mi₂a2 = Gmg₁mg₂ r2 , hence a2 = Gmg₁ r2 mg₂ mi₂ .

If mg₁ = ME, the mass of Earth, and r = RE, the radius of Earth, then this says g2 = GME R2_E mg₂ mi₂ ,

hence the acceleration due to gravity varies as mg/mi. Galileo’s fa- Unclear whether he did so in reality as opposed to thought.

mous experiment from the leaning tower of Pisa implies that mg/mi is constant. Modern experiments show mg/mi constant to < 1 part in 1013. Very odd coincidence in Newtonian mechanics; a key principle of General Relativity.

From now on we do not distinguish between “gravitational” and “in-ertial” mass, so we drop the superscripts. The acceleration due to gravity is the same for all objects:

g = GME

R_E2 = 9.81 m s −2_.

4.2.1 N3 and gravity.

What about the gravitational force ~F21 of particle 2 on particle 1? Swap the labels 1 and 2, and note that ˆr21 = −ˆr12 with r unchanged:

~ F21 = − Gm2m1 r2 rˆ21 = Gm1m2 r2 rˆ12 = − ~F12.

So Newton’s Law of gravity is consistent with N3 . Would this be

the case if F = Gm2

1m2/r2?

What about G(m1+ m2)/r2? Example 4.1. 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.

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A B

C

R

Answer. The force on A just due to B has magnitude

FBA = GmBmA r2 . Setting r = 2R, and mA = mB = 4π 3 R 3 ρ,

where R = 5 m is the radius of the spheres and ρ = 7870 kg m−3 is the density of iron, then

FBA = 16Gπ2R6ρ2 9 × 4R2 = 4Gπ2 9 R 4_ρ2_. This gives FBA = 4 × 6.67 × 10−11× π2 9 × 5 4 _{× 7.87 × 10}32 _{= 11.33 N.}

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 = 2FBAcos(30◦) = FBA √

3 = 19.62 N.

Example 4.2. A person of mass m = 71.4 kg weighs 700 N standing on the ground. How much less would she weigh at the top of the Eiffel tower (height h = 320 m)? [Radius of Earth RE = 6370 km. Hint: use h/RE 1].

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Answer. Weight on ground

W0 = GMEm R_E2 = mg. Weight at height h Wh = GMEm (RE + h)2 = mg R2_E (RE + h)2 = mg (1 + h/RE)2,

Since h/RE 1 (the symbol means “much less than”) we can expand using (1 + x)−2 = 1 − 2x + · · · Wh ≈ mg 1 − 2h RE Wh− W0 ≈ − 2h RE mg = −0.07 N,

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Lecture 5

Systems of Particles

Keypoints from Lecture 4

• Newton’s Law of Gravity: ~

F12 = −

Gm1m2

r2 rˆ12.

• Note: both sides are vectors, as they must be.

• By experiment it is found that the masses that appear in New-ton’s Law of Gravity are the same as the masses required for

N2 .

• Newton’s Law of Gravity is consistent with N3 .

Objectives

• Discuss how to handle composite objects

• Define centre of mass and mention centre of gravity

5.1 Many Particles

How do Newton’s Laws apply to composite objects, with multiple ac-celerations and forces?

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LECTURE 5. SYSTEMS OF PARTICLES 24

Consider a set of N particles with both inter-particle and external forces acting:

F

₃₁

F

₁₃

1

2

3 F

₂

F

1

6

4 ₅

Let the force of particle j acting on i be called ~Fji and the external force on particle i be simply ~Fi.

Then the total force acting upon particle i is given by

~ Fi+ X j6=i ~ Fji = mi~ai.

Summing over all particles, the total force acting on the N particles is

X i ~ Fi + X i X j6=i ~ Fji = X i mi~ai.

N3 ⇒ the second term = ~0 since for every Fji, say, there is a cancelling Fij. The first term is the total external “applied” force ~F =

P iF~i, so ~ F =X i mi~ai.

We can write this as ~F = m~acm, where m = P

imi is the total mass, if we define ~ acm = P imi~ai P imi . Remembering that ~ ai = d2~ri dt2 ,

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LECTURE 5. SYSTEMS OF PARTICLES 25 then ~ acm = d2 dt2 P imi~ri P imi = d 2_~_r cm dt2 , where ~rcm = P imi~ri P imi .

This is a mass-weighted average of the position vectors of the particles. This is the definition of the “centre of mass”.

Hence we have the important result:

Newton’s second law applies to the total mass and the acceler-ation of the centre of mass of composite bodies.

5.2 Centre of Mass of Continuous Bodies

If N → ∞ while mi → 0, then the centre-of-mass sums become inte-grals:

~rcm =

R ~r dm R dm .

This applies to each component one-by-one, e.g.

xcm =

R x dm R dm . “dm” is an infinitesimal element of mass.

Example 5.1. A bar of length L has an increasing density 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. By definition, dm = ρdx. The total mass of the bar is

M = Z dm = Z L 0 ρ dx = Z L 0 αx dx = 1₂αL2. Similarly, Z x dm = Z L 0 xρ dx = Z L 0 αx2dx = 1₃αL3.

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LECTURE 5. SYSTEMS OF PARTICLES 26 Hence xcm = R x dm R dm = 2 3L.

Example 5.2. Find the centre of mass of a two-dimensional uniformly filled semi-circle of radius R.

Answer. Orient the semi-circle with its straight edge on the y-axis:

x y

dx

By symmetry, ycm = 0. We seek xcm. Assume a density per unit area of σ. Semicircle area is 1₂πR2.

M = Z

dm = 1₂πσR2.

Slice the semi-circle up vertically, so the area of each slice is 2ydx, so dm = 2yσ dx, and x2 + y2 = R2, so Z x dm = Z R 0 x 2yσ dx = 2σ Z R 0 x R2 − x21/2 dx = 2σ h −1 3 R 2 _{− x}23/2iR 0 = 2 3σR 3 . Therefore, finally xcm = 2σR3/3 πσR2_/2 = 4 3πR ≈ 0.424R.

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LECTURE 5. SYSTEMS OF PARTICLES 27

5.3 Centre of Gravity

In the lecture I demonstrated the location of the centre of gravity but did not explicitly write down the equation.

The total external gravitational force on a set of particles

~ FG =

X i

mi~g(~ri),

can be reduced to a single force acting through one point, the centre of gravity.

If ~g is constant then the centre of gravity and the centre of mass are one and the same.

I hope to return and give a proper definition of the centre of gravity later.

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Lecture 6

Force and acceleration

Keypoints from Lecture 5

• N2 applies to composite objects with ~F as the total externally-applied force, m as the total mass and ~a as the acceleration of the centre of mass.

• The centre-of-mass is defined by

~rcm = P imi~ri P imi .

• In the “continuum limit”

~rcm =

R ~r dm R dm .

• Gravity acts through the centre of gravity. It coincides with the centre of mass, if ~g is constant, but otherwise in general it will not be the same.

Objectives

• How to set up and solve dynamical, non-equilibrium, problems.

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LECTURE 6. FORCE AND ACCELERATION 29

6.1 Forces out of balance

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

W=mg

_θ

F

_K

N

θ

_a

Again, only forces on block are shown; the friction force is now the kinetic friction FK = µKN . The acceleration is indicated with a double arrow: it is not a force, but the result of the forces.

Resolving perpendicular to and parallel to (down) the slope: N − mg cos θ = 0,

mg sin θ − FK = ma.

Therefore, as before, N = mg cos θ, and so FK = µKmg cos θ, and so

a = g(sin θ − µKcos θ).

Example 6.1. 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?

Answer. Draw diagrams with forces on each component sepa-rately: θ θ 2 a₂ N θ M a M m 1 N N’ W’ Forces on wedge Forces on block mg m a

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relative to the wedge. The true acceleration of the block (i.e. rel-ative to an inertial frame) is ~aB = ~a1 + ~a2, and this is what we must use in N2 . We resolve forces and accelerations.

Wedge, horizontal:

N sin θ = M a2. (6.1)

Wedge, vertical: not needed, as we are not interested in N0. Block, perpendicular to slope:

N − mg cos θ = −ma2sin θ. (6.2)

N.B. ~a1 has no component perpendicular to the slope, so the ac-celeration on the right is all from ~a2.

Block, parallel to slope:

mg sin θ = m(a1 − a2cos θ). (6.3)

Equally well we could have resolved horizontally and vertically for the block, which would give two equations different from, but ex-actly equivalent to, Eqs. (6.2), (6.3).

Use Eq. (6.2) to eliminate N from Eq. (6.1):

m(g cos θ − a2sin θ) sin θ = M a2,

so, rearranging,

a2 =

mg cos θ sin θ

M + m sin2θ. (6.4)

Eq. (6.3) and Eq. (6.4) then tell us a1:

a1 = g sin θ + mg cos2θ sin θ M + m sin2θ , which reduces to a1 = (M + m)g sin θ M + m sin2θ . Check: dimensions _4;

horizontal slope for θ = 0, a1 = 0, a2 = 0 4; vertical slope for θ = 90◦, a1 = g, a2 = 0 4;

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6.2 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: as in “light pulley” – zero mass, no force needed to accelerate it.

rigid: does not bend at all

flexible: totally bendy. Strings are almost always “flexible”.

ideal: a string is ideal if it is light, flexible and does not stretch at all; an “ideal pulley” is light, frictionless and rigid.

uniform: e.g. “uniform sphere” – constant density throughout.

elastic: “elastic collision” – conserves kinetic energy (“inelastic” kinetic energy lost to heat, sound etc)

6.3 Strings and pulleys

Illustrate with an example.

Example 6.2. Consider the following arrangement. The pulleys are light and frictionless; the string is ideal; the table is smooth. Table and wall are fixed. Find the accelerations of the two blocks and the tension in the string.

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LECTURE 6. FORCE AND ACCELERATION 32 m₁ m 2 Table Wall

Answer. No force needed to turn the pulleys or accelerate the pulleys or the string ⇒ tension in string is constant:

m₂ _a 2 T T T m₁ m g₁ a₁

Therefore F = ma on block 1 (downwards):

m1g − T = m1a1, (6.5)

while block 2 (leftwards):

2T = m2a2. (6.6)

Two conditions, but three unknowns: need one more condition. String does not stretch ⇒ m1 moves twice as fast as m2:

a1 = 2a2. (6.7)

Eqs (6.5), (6.6) and (6.7) then imply:

a1 = 4m1g m2 + 4m1 , a2 = 2m1g m2 + 4m1 , T = m1m2g m2 + 4m1 .

Added after the lecture: the diagram below should make it clear why the bottom of the string moves twice as fast as the moving pulley. The

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numbers are just the lengths of the straight sections of string.

2 6 4 5 3 4

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Lecture 7

Equations of motion

Keypoints from Lecture 6

• If forces on an object don’t balance, it accelerates.

• Always be clear about what quantities you do and don’t know in a problem: you will need one condition for every “unknown”. • Code words: smooth, light, rigid, flexible, ideal, uniform, elastic,

inelastic.

• Tension in ideal strings is constant. With pulleys can double, triple, etc, forces.

Objectives

• How to solve equations of motion

7.1 Constant acceleration – “SUVAT”

“Equations of motion” are expressions for ~r, ~v and ~a as functions of time.

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LECTURE 7. EQUATIONS OF MOTION 35

Consider motion under constant acceleration, a, in 1D. We have

dv dt = a. Integrating: Z v u dv = Z t 0 a dt = a Z t 0 dt, so v = dx dt = u + at. (7.1)

Here u is the initial speed at t = 0 and v is the speed after t seconds. Integrating again Z x x0 dx = Z t 0 (u + at) dt, so s = x − x0 = ut + 1₂at2, (7.2)

where s is the distance travelled in t seconds.

If we square both sides of Eq. (7.1) we can simplify using Eq. (7.2):

v2 = u2 + 2u at + a2t2 = u2 + 2a ut + 1₂at2 | {z } s so finally v2 = u2 + 2as. (7.3)

Example 7.1. In the UK’s driving theory test, the “braking distance” at 70 miles per hour is said to be 75 m. What is the acceleration?

Answer. One mile = 1600 m, so

u = 1600 × 70

3600 = 31 m s −1

The final speed v = 0, so using Eq. (7.3),

a = v 2 _{− u}2 2s = 0 − 312 2 × 75 = −6.4 m s −2_, or −0.65g.

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7.2 Time-dependent acceleration

Suppose now that the acceleration is a function of time, a = a(t), then we must keep a inside the integral:

Z v u dv = v − u = Z t 0 a(t) dt.

7.3 Position-dependent acceleration

Suppose a = a(x), i.e. it is a function of position. Then

dv dt = a(x). Integrating Z dv = Z a(x) dt,

but we don’t know x = x(t) – that’s what we are trying to determine! Use the chain rule

dv dt = dv dx dx dt = dv dxv = a(x), so we can write Z v u v dv = Z x 0 a(x) dx, or 1 2(v 2 _{− u}2 ) = Z x 0 a(x) dx. RHS now calculable.

If we are lucky, we might be able to work out the integral, and integrate again to get x. Incidentally we can (again) relate this to change in energy equals work done, if we multiply both sides by m. See later.

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7.4 Velocity-dependent acceleration

Now suppose a = a(v), then

dv

dt = a(v), which can be integrated as follows:

Z v u dv a(v) = Z t 0 dt = t,

to relate v and t. Re-arrange to give v = v(t), integrate to get x = x(t).

Example 7.2. Starting from rest at t = 0, an object of mass m falls through air under gravity, experiencing a drag force

F = −mv τ ,

where τ is a constant with dimensions of time.

Calculate its speed, and the distance it has dropped, by time t.

Answer. N2 becomes:

mg − mv

τ = m

dv dt. Divide out m, rearrange, and integrate

Z t 0 dt = t = Z v 0 dv g − v/τ = h −τ ln(g − v/τ )iv 0, = −τ ln(g − v/τ ) − ln g, = −τ ln g − v/τ g . Re-arranging g − v τ = ge −t/τ_, _or _{v = gτ 1 − e}−t/τ_.

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Tends to a maximum value of gτ , the “terminal velocity”. Setting v = dx/dt, we can integrate again

x = Z t 0 gτ 1 − e−t/τ dt = gτ ht + τ e−t/τ it 0 , = gτ ht − τ (1 − e−t/τ)i.

x ∝ t2 at short times, and x ∝ t at long times.

0 1 2 3

t / τ

v gτ

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Lecture 8

Work and Energy

Keypoints from Lecture 7

• Starting from an expression for the acceleration ~a, and given initial values of position ~r and ~v, it is possible, in principle, to determine ~r(t), ~v(t) and ~a(t), the “equations of motion”.

• “SUVAT” is the simple case of constant a.

• More generally there are three cases a = a(t), a = a(x), a = a(v), each of which requires its own handling.

• (More generally still ⇒ differential equations.)

Objectives

• Work, kinetic energy, conservative forces, potential energy

8.1 Work and Kinetic Energy

Consider a 1D, position-dependent force, F (x) acting on a mass m:

F (x) = mdv dt.

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LECTURE 8. WORK AND ENERGY 40

As last lecture, we can use the chain rule

dv dt = dv dx dx dt = dv dxv.

Now let the mass move from position A to B, and integrate:

Z B A F (x) dx = Z B A mv dv = 1 2mv 2 B A . (8.1)

Defining the kinetic energy T by

T = 1₂mv2,

and the work done ∆W by

∆W = Z B

A

F (x) dx,

Eq. (8.1) says that

The work done equals the change in kinetic energy.

If F (x) = F , a constant, then

∆W = F Z B

A

dx = F (xB − xA),

i.e. “work done = force × distance”, but not true in general.

Work ≈ force × distance if the step is so small that F does not vary much. As distance → 0, an infinitesimal step dx, it becomes exact:

dW = F dx.

8.2 3D

Now consider a general position dependent force acting on a mass m:

~

F (~r) = md~v dt.

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Cannot simply extend previous approach because in

d~v dt = d~v d~r d~r dt,

the first derivative is undefined. Instead we generalise dW = F dx. A natural extension of this to 3D is

dW = Fxdx + Fydy + Fzdz = ~F · d~r. Using N2 dW = md~v dt · d~r = m d~v dt · ~vdt, = m~v · d~v, = d 1₂m~v · ~v .

Therefore generalising the definition of kinetic energy to

T = 1₂m~v · ~v = 1₂mv2 = 1₂m v_x2 + v_y2 + v_z2, we recover work done = change in kinetic energy, but now

∆W = Z B A ~ F · d~r. Physical interpretation: θ A _B F r d

Defining θ as the angle between ~F and d~r:

dW = ~F · d~r = F cos θ dr,

so dW = the component of force in the direction of movement times the (infinitesimal) distance moved. As a result the term on the right in ∆W = Z B A ~ F · d~r.

is called a line integral. Simply think of this as the sum of the in-finitesimal contributions from each step d~r.

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8.3 Conservative forces, potential energy

The relation between work and kinetic energy is most useful in

con-servative force fields. “conservative”

as in energy conserving.

A conservative force is one for which the work done moving from one point to another is independent of the path.

B A P P’ Z B A ~ F · d~r P = Z B A ~ F · d~r P0 , or Note reversed limits on second integral. Circle on final integral indicates that the path is a closed loop. Z B A ~ F · d~r P + Z A B ~ F · d~r P0 = I ~ F · d~r = 0.

A conservative force ~F (~r) allows one to define a potential energy U (~r) because the work done by it in moving a particle from A to B is only a function of the end points, not the path taken:

∆W = Z B

A ~

F · d~r = − U (B) − U (A) = −∆U,

where U is a function to be determined. The minus sign is a convention to ensure that ∆W > 0 when moving from a high to a low potential. Since ∆T = ∆W , for conservative forces ∆T = −∆U , and so if we define the total energy E = T + U , ∆E = 0 and thus

E = T + U = constant.

This is the conservation of mechanical energy, which was later gen-eralised to the conservation of energy when heat was recognised as another form of energy.

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8.4 Deriving a force from a potential

If a potential can be defined then the corresponding force can be ob-tained as follows: F = −dU dx, in 1D ~ F = −∂U ∂xˆı − ∂U ∂yˆ − ∂U ∂z ˆ k = − ~∇U, in 3D where ~∇ = ∂ ∂x, ∂ ∂y, ∂

∂z is the gradient operator.

Not every force field ~F (~r) is conservative, i.e. not every force field is derived from a potential. For example, frictional forces are not conservative.

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Lecture 9

More energy

Keypoints from Lecture 8

• A force acting on a moving particle performs “work”, measured in Joules [J]. The work done moving from A to B is defined as

∆W = Z B

A

~ F · d~r.

• ∆W = ∆T , the change in kinetic energy, where

T = 1₂mv2.

• If the work done from A to B is independent of the path taken, ~

F is said to be conservative and can be represented in the form of a (scalar) potential function U , where ~F = − ~∇U .

• In this case T + U = E, the total mechanical energy, is constant.

Objectives

• Energy conservation, power.

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LECTURE 9. MORE ENERGY 45

9.1 The work you do

Last lecture discussed work done by a conservative force, which ends up in kinetic energy of a particle.

If you move a particle slowly from A to B in the same force field, by N3 you apply an equal but opposite force, so where last time we had

∆U = −∆W,

when expressed in terms of the work you do, ∆WY,

∆U = ∆WY.

i.e. you must do positive work to move the particle to a place of higher potential; the force does work moving the particle to a lower potential.

9.2 Gravitational potential energy

d dr r

s

M

If a mass m moves d~s under gravitational attraction from a mass M , I use d~s because d~r causes confusion here because of the use of dr to represent the change in distance between the masses; i.e. dr 6= |d~s|. then dW = ~F · d~s = −GM m r2 r · d~s = −ˆ GM m r2 dr.

As usual, infinitesimal terms of higher order are dropped.

Integrating: ∆W = −GM m Z B A dr r2 = GM m 1 r(B) − 1 r(A) .

This is only a function of the value of r at the end points, hence gravity is conservative with potential energy

U (r) = −GM m

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The constant is not uniquely defined, but usually we take U (∞) = 0, so the constant = 0 and

U (r) = −GM m

r .

Often we use the “gravitational potential” to mean the potential en-ergy per unit mass

u(r) = U (r)

m = −

GM r . Sometimes this is written φ(r).

9.3 Using energy conservation

Energy conservation is useful when one needs to know the speed at a particular place but not time.

Example 9.1. An object is released from rest at a height h = 3000 km above the surface of Earth. Ignoring air resistance, at what speed will it hit the ground? Compare your answer with assuming potential = mgh.

Answer. Gravity is conservative so

T + U = E = constant,

with T = 1

2mv

2 _and _{U = −}GMEm

r ,

where r is the distance of the object from the centre of the Earth. Conservation of energy means:

−GMEm RE + h = 1 2mv 2 ₋ GMEm RE ,

(T = 0 at the start). Therefore

v2 = 2GME 1 RE − 1 RE + h , = 2GMEh RE(RE + h) .

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Putting G = 6.67 × 10−11N m2kg−2, ME = 5.97 × 1024kg, RE = 6.37 × 106m and h = 3 × 106m, find v = 6.3 km s−1.

Using 1₂mv2 = mgh (inaccurate, since h is not small compared with RE) v2 = 2gh = 2GME R2 E h

one finds v = 7.7 km s−1. mgh over-estimates the force and hence the kinetic energy.

Example 9.2. Escape velocity: What speed is needed to escape the gravitational pull of an object of mass M and radius R?

Answer. The speed must be enough for T > 0 at r = ∞ where U = 0, i.e. E > 0. Therefore, by conservation, E = T + U > 0 at the start as well.

1 2mv 2 _> GM m R , or v > r 2GM R .

Escape velocity from Earth = 11.2 km s−1, from the Sun = 618 km s−1.

This is a fundamental constraint upon planetary atmospheres.

9.4 Power

The amount of work performed per unit time is the power, and is measured in Joules per second or Watts.

Since dW = ~F · d~r, then P = dW dt = ~F · d~r dt = ~F · ~v. Power is a scalar

Example 9.3. The “Lorentz force” is the force that acts on a charged particle in an electromagnetic field, and is given by

~

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where q is the particle’s electric charge, ~v is its velocity and ~E and ~B are the electric and magnetic fields (and × is the vector, or “cross”, product).

What is the rate of work on the particle?

Answer.

~

F · ~v = q( ~E · ~v + ~v × ~B · ~v) = q ~E · ~v,

i.e. only the electric field performs work on the particle.

There was not enough time in the lecture to go through the next example, but you may still find it interesting.

Example 9.4. The Dinorwig pumped storage power station in Wales stores water in one lake 500 m higher than another in order to provide a source of rapidly accessible power. When running at its full power of 1.7 GW, what is the flow rate of water?

Answer. If the water flows at a rate of ˙m = dm/dt, measured in mass per unit time, then the energy released per unit time (= power) falling through a height h is given by

P = ˙mgh. Therefore ˙ m = P gh = 1.7 × 109 9.81 × 500 = 3.47 × 10 5_{kg s}−1_, or 347 tonnes/sec.

In reality it is higher still – 390 tonnes/second because it is not 100% efficient.

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Lecture 10

Rockets and circular

motion

Keypoints from Lecture 9

• Gravity is a conservative force. The gravitational potential en-ergy of two masses m1 and m2 separated by a distance r is given

by

U = −Gm1m2 r , a scalar equation.

• If position but not time is defined in a problem, then energy conservation might be useful.

• The speed needed to give an object a positive total energy E = T + U > 0 in a gravitational field is known as its “escape veloc-ity”.

• Power, measured in Watts, is the rate of performing work.

Objectives

• The mechanics of rockets (changing mass!) • Making a start on circular motion.

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LECTURE 10. ROCKETS AND CIRCULAR MOTION 50

10.1 Rockets

Rockets burn fuel which is ejected at high speed in one direction while the rocket is pushed in the opposite direction.

10.1.1 Rockets in space

Assuming that the rocket and its exhaust form an isolated system, then there is no overall change in momentum. Consider a time t when the mass of a rocket is m and its speed v, followed by t + dt when its mass is m + dm, its speed v + dv having expelled a mass −dm at speed −ve relative to the rocket. ve is the exhaust speed, determined by the

combustion energy of the fuel. Convention! dm

will, of course, be a negative quantity.

Conserving momentum:

mv = (m + dm)(v + dv) + (−dm)(v − ve). Expanding the terms in brackets

mv = mv + m dv + v dm + dm dv − v dm + vedm.

Cancelling terms and dropping the dm dv second-order term we have Infinitesimals again: dm → 0 and dv → 0. m dv = −vedm. Integrating: Z v v0 dv = −ve Z m m0 dm m ,

and thus This is a classic

equation of rocketry. v − v0 = ∆v = veln m₀ m ,

where m0 is the initial mass.

10.1.2 Rockets at launch

When launched from Earth, the rocket and its exhaust are acted upon by gravity. During time dt, the force of gravity mg changes the overall momentum by −mg dt and thus

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Integrating as before (taking v0 = 0 and assuming g is constant)

v = veln m₀

m

− gt.

The rocket’s acceleration at launch is

a = dv dt = − ve m dm dt − g. The total thrust is

F = ma + mg = −ve dm

dt .

Example 10.1. At launch the Saturn V rocket of the Apollo missions had a mass of 3000 t. Burning 15 t of fuel per second, the thrust was 35 MN.

Calculate the exhaust speed and, assuming that both it and fuel burn rate remain constant, the “G-force” on the astronauts as a function of time, (a + g)/g, and the rocket speed, at t = 100 s.

Answer. We know that

m0 = 3.0 × 106kg, dm dt = −1.5 × 10 4_{kg s}−1_, F = 3.5 × 107N. Therefore ve = F −dm/dt = 3.5 × 107 1.5 × 104 = 2333 m s −1_.

The G-force experienced by the astronauts is

= a + g g = F mg = 3.5 × 107/9.81 3.0 × 106 _{− (1.5 × 10}4_t) = 1.19 1 − t/200,

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LECTURE 10. ROCKETS AND CIRCULAR MOTION 52 0 50 100 150 t / sec 0 1 2 3 4 5 G-force

The speed after 100 s is

v = veln m₀ m − gt, = 2333 ln 3000 3000 − 15 × 100 − 9.81 × 100, = 636 m s−1. The central engine was cut at 150 secs to prevent too high an acceleration. Almost twice the speed of sound.

10.2 Circular motion

Motion at constant speed in a circle is the simplest case of acceleration as the result of a change in direction rather than speed. Consider a particle at position ~r, rotating about an axis perpendicular to ~r.

t) δθ O r δ r r (t+ (t) δ

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In time δt, vector changes by δ~r, while staying constant in magnitude:

r = |~r(t)| = |~r(t + δt)|.

The velocity is given by

~v = lim δt→0

δ~r δt.

The magnitude of δ~r is NB In this case

|δ~r| 6= δr because the magnitude of ~r is constant. |δ~r| = 2r sin δθ 2 . For small x 1, sin x = x − x 3 3! + x5 5! − . . . ≈ x, so |δ~r| ≈ r δθ, and therefore v = |~v| = rdθ dt = rω,

where ω = dθ/dt is the angular velocity measured in radians per sec-ond. For circular motion, ~v is always perpendicular to ~r.

NB The above argument can be applied in a similar way to any vector ~

A rotating about a perpendicular axis. For example, for the rate of change of velocity ~v, we have a = vω, where the acceleration vector ~

a is directed perpendicular to ~v, inwards towards the centre of the circle.

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Lecture 11

More circular motion

Keypoints from Lecture 10

• If a rocket ejects mass −dm at exhaust speed ve, conservation of

momentum gives a change in velocity dv

m dv = −vedm ⇒ v − v0 = ∆v = veln

m₀ m

• A rotating vector ~A of constant magnitude A = ~ A , and per-pendicular to the axis of rotation, changes with magnitude

d ~A dt = ωA, ω = dθ

dt is the “angular frequency” or “angular velocity”. • The rate of change d ~A/dt is perpendicular to ~A itself.

• For the position vector ~r we get the important relations v = ωr for the tangential speed and a = ωv for the acceleration directed in towards the centre.

Objectives

• Centripetal acceleration, circular orbits

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LECTURE 11. MORE CIRCULAR MOTION 55

11.1 Centripetal acceleration

Just like the radius vector, the velocity vector of a particle moving in a circle at constant speed sweeps out a circle:

δθ 2 1 1 2 3 r δθ O δ v v(t+δt) v(t) 3

Thus just as v = ωr, so a = ωv, and ~a is perpendicular to ~v. It points towards O, the centre of the circle. We have three equivalent formulae:

a = ωv = ω2r = v 2 r .

These give the centripetal acceleration of an object moving in a circle of radius r, at speed v or, equivalently, with angular velocity ω.

Example 11.1. Draw the forces and acceleration on a car driving at constant speed in a circle. Obtain an expression for the maximum speed with no skidding.

Answer. The car moves in a circle because of the centripetal force provided by friction between tyre and track:

mg Rear view Top view a F v F a F = ma, so F = mv 2 r < µSmg,

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where µS = coefficient of static friction, so

v < √µSgr. e.g. µS = 0.8, R = 50 m, then v < 19.9 m s−1 ≈ 45 mph. NB No “centrifugal” outwards-pointing force!

Example 11.2. Calculate the centripetal acceleration of a point on Earth’s equator.

Answer. Here a = ω2r is the easiest to use because we know Earth’s spin period.Using ω = 2π/P , P = 1 day, and r = 6370 km we have a = ω2r = 2π 24 × 3600 2 × 6.37 × 106 = 0.0337 m s−2 = 0.0034 g.

11.2 Angular velocity vector

The equation v = ωr can be made vectorial by defining ~ω.

The angular velocity vector ~ω is a vector of length ω pointing along the axis of rotation.

The direction of ~ω is set by the right-hand rule: if the fingers of your right-hand point in the direction of rotation, your thumb points in the

direction of ~ω. Q: what is the

direction of ω for the Earth?

With this definition

~

v = ~ω × ~r,

where ~r is the position vector measured from a point on the axis.

r

sin

φ

P

ω

_r

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Let’s check. The point P spins round the axis marked by ~ω at speed ωr sin φ into the page, so as claimed

~

v = ~ω × ~r.

11.3 Circular orbits

Centripetal acceleration is key to understanding orbits. Consider a satellite in orbit around Earth:

V

F

a

Earth

R

The force acting is gravity which has magnitude

F = GMEm R2 ,

which causes centripetal acceleration of the satellite of a = V2/R so

F = GMEm

R2 = m

V2 R . Therefore the speed is given by

V2 = GME

R .

Using V = ωR, we also have

ω2 = 4π 2 P2 =

GME R3 ,

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where P is the orbital period. So, P2 ∝ R3_{: we will see later that this} is Kepler’s third law.

For low Earth orbit,

R = RE + 200 km = 6570 km ME = 5.97 × 1024kg

G = 6.67 × 10−11N m2kg−2 ⇒ V = 7.8 km s−1

⇒ P = 88.4 min

Example 11.3. In a classical model of a hydrogen atom, the electron We should mention that this model is incorrect! Quantum physics is needed

(mass me = 9.109 × 10−31kg) orbits the nucleus (a proton) in a circle of radius r = 5.29 × 10−11m, attracted by the electric “Coulomb” force which is given in this case by

F = e

2 4π0r2,

where e = 1.60 × 10−19C is the magnitude of the charge on the elec-tron, and 0 = 8.85 × 10−12F m−1 is the “permittivity of free space”. Calculate the orbital speed of the electron.

Answer. Using N2 and a = v2/r, we have

F = me v2 r = e2 4π0r2 , so v = s e2 4π0mer .

Putting in the numbers gives v = 2.19 × 106m s−1 < 0.01 c.

This is a non-relativistic speed. However for heavier atoms, when the nuclear charge is of order 100, e2 → 100e2_{, relativistic effects start to}

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Lecture 12

Torque and Angular

Momentum

Keypoints from Lecture 11

• Just like the position vector ~r, the velocity vector ~v of a particle moving at constant speed in a circle sweeps around at angular velocity ω.

• The resultant centripetal acceleration a is given by

a = ωv = ω2r = v

2

r .

• Circular motion can be described vectorially using the angular velocity vector ~ω. Any vector ~A which rotates around an axis in this way satisfies

d ~A

dt = ~ω × ~A.

• Gravitational force causes the centripetal acceleration of satel-lites and planets in circular orbits whereby

V2 = GM R .

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LECTURE 12. TORQUE AND ANGULAR MOMENTUM 60

Objectives

• Torque and angular momentum

12.1 Moments

Consider a simple (light) lever in balance:

+ 1 r2 y x P r W₁ W2 W₁ W₂

As Archimedes proposed, to find the correct ratio of W1 and W2 we need to balance moments:

r1W1 = r2W2.

Is this a new principle in addition to N1 , N2 and N3 ?

In fact no. To see this, first generalise “moments” by defining the torque ~τ exerted by a force ~F acting at position vector ~r to be

~ τ = ~r × ~F . Physical interpretation:

θ

O

F

r sin

θ

r

θ

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The moment of F around O has magnitude = F r sin θ. (The clos-est approach distance r sin θ is sometimes known as the “lever arm”.) The only sensible direction to associate with ~r and ~F is the direction perpendicular to both. All properties are satisfied by ~τ = ~r × ~F . NB Torques must always be measured with respect to a specific point – choose a different point, get a different torque.

From N2 ~ F = d~p dt = m d~v dt, so ~ τ = m~r × d~v dt. Now Since d~r/dt = ~v, and the cross product of a vector with itself is zero. d dt(~r × ~v) = d~r dt × ~v | {z } zero +~r × d~v dt = ~r × d~v dt.

Therefore we can write

~ τ = m~r × d~v dt = m d dt(~r × ~v) = d dt(~r × ~p) .

The quantity ~r × ~p is called the “moment of momentum”, or more usually the angular momentum. Defining

~ L = ~r × ~p, we have ~ τ = d~L dt .

Torque equals the rate of change of angular momentum.

12.2 Systems of particles

As before, consider a system of particles with internal and external forces. The force on particle i was given by

~ Fi + X j6=i ~ Fji = d~p_i dt .

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The torque around a point O is

~ri× ~Fi + X j6=i ~ri× ~Fji = d~Li dt .

The total torque on the particles is therefore

~ τ = X i ~ri× ~Fi + X i X j6=i ~ri × ~Fji = X i d~Li dt .

For the internal forces, the double sum contains both ij and ji terms:

~

ri× ~Fji + ~rj × ~Fij = ~ri − ~rj × ~Fji because of N3 .

For “central” forces ~Fji, which act along the vector between particles

~ri − ~rj, this is zero. So this just leaves the external forces. Again, the vectors are parallel. See diagram 21

F

₁₂

₂

O

1 F

so ~ τ = X i ~ri× ~Fi = d dt X i ~ Li ! = d~L dt. Thus

The total torque from external forces on a composite body equals the rate of change of its total angular momentum.

12.3 Back to levers

A body in equilibrium has zero torque about any point: It is your

choice: make things as simple as possible. In this case P looks good.

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LECTURE 12. TORQUE AND ANGULAR MOMENTUM 63 + 1 r2 y x P r W₁ W2 W₁ W₂ In vectors ~r1 = −r1ˆı, ~r2 = +r2ˆı, ~ F1 = −W1ˆ, ~ F2 = −W2ˆ,

so the total torque is

~

τ = X

i

~ri × ~Fi = (r1W1 − r2W2) ˆı × ˆ = ~0.

Now ˆı × ˆ = ˆk is not zero, and therefore

r1W1 = r2W2,

the standard equation for balancing a lever, obtained by “taking mo-ments about P ”.

Conclude: Archimedes’ lever equation is contained within Newton’s Laws, and in ~τ = ~r × ~F , ~L = ~r × ~p, and ~τ = d~L/dt, we have a much more powerful version of it.

12.4 Centre of Gravity

We return to discuss centre of gravity. Consider a rigid body composed of several masses mi at positions ~ri under the influence of gravity. The force on each mass will be ~f_i = wig, where the unit vector ˆˆ g points “down” and wi = mig is the weight. Consider the torque due to all these weights about some arbitrary point (which, for convenience, we choose to be the origin here). The position of the centre of gravity,

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~rcg, is defined such that, if the individual forces were replaced by the total force ~f = P

i~fi = P

iwi ˆg acting at ~rcg, then the torque on the body would be the same. Thus

~ rcg× ~f = X i ~ri × ~f_i X i wi ~rcg× ˆg = X i wi~ri × ˆg .

This equation should apply no matter what the orientation of the body, or equivalently, whatever the orientation of ˆg, so

X i wi ~ rcg = X i wi~ri ⇒ ~rcg = P iwi~ri P iwi .

If g is the same everywhere, so wi = mig, the factors of g cancel and this is the same equation as the one defining the centre of mass ~rcm. If the magnitude of g varies from place to place, ~rcg and ~rcm will not

coincide. For example, g varies slightly with height above the earth, as Actually, if this is the case, we can’t simply drop the dependence on ˆ g; things get a little more complicated.

a result of which (“University Physics”, p347) the centre of gravity of the Petronas towers in Kuala Lumpur lies about 2 cm below the centre of mass. In the general case of variable gravity, the problem has no consistent solution.

It is relatively easy to show that the standard method for locating the centre of gravity of a body by suspending it from several points in turn, is consistent with the above definition.

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Lecture 13

Moments of Inertia

Keypoints from Lecture 12

• The moment of a force ~F around a point O is called the torque ~

τ = ~r × ~F ,

where ~r is the position vector of the point at which the force acts relative to O.

• The moment of a particle’s momentum ~p about O is called its angular momentum ~L and is given by

~

L = ~r × ~p.

• Rate of change of angular momentum equals applied torque

~

τ = d~L dt.

This applies to particles and composite bodies.

• If the weight of a body is concentrated at the centre of gravity, the torque will be unchanged.

Objectives

• Moments of inertia

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LECTURE 13. MOMENTS OF INERTIA 66

13.1 Angular momentum of a particle

mov-ing in a circle

The angular momentum about a point O of a particle of momentum ~

p and position vector ~r relative to O is given by ~

L = ~r × ~p.

This is general. Now consider the specific case of circular motion around O at angular velocity ω:

v

m

O

r

With ~p = m~v, the angular momentum ~L around O is a vector parallel

to ~ω (i.e. along the axis of rotation, out of the page) with magnitude NB This holds only about O.

L = mrv = mr2ω. The kinetic energy of the particle

T = 1 2mv 2 ₌ 1 2m(rω) 2 ₌ 1 2mr 2_ω2_.

Defining a new quantity I = mr2, the “moment of inertia”, we have I is the angular counterpart of mass m, but unlike m, I is usually direction dependent. L = Iω, T = 1 2Iω 2_.

13.2 Rigid-body rotation

The equation for kinetic energy is easily generalised to the rotation of rigid bodies. However, everything must be defined relative to the axis of rotation: Here r (or ri) is the perpendicular distance of each mass element (or particle) from the axis.

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ω

dm r

Summing over all particles we then have

T = 1 2 X i miv2i = 1 2 X i miri2 ! ω2,

so we still have T = 1₂Iω2 if

I = X i miri2 = Z r2dm = Z r2ρ dV,

where we take the continuum limit on the right (ρ is the density per unit volume).

This comes with a warning: if the axis is chosen in a different direction, in general the value of I will be different. This complicates things, and

we will not go into those complications in this module. I is not really a scalar: it is a second-rank tensor. This will become clear in later years. For these reasons, the connection between ~L and ~ω is more

compli-cated, in general, than the analogous link between ~p and ~v. In general ~

L and ~ω are not parallel. In this module we will only discuss cases for which ~L and ~ω are parallel:

~

L = I ~ω,

with I given as before. This is true for all directions through spheres or cubes, symmetry axes of cuboids, cylinders and discs. In the cases we shall encounter ~ τ = d~L dt = I d~ω dt ,

where d~ω/dt is the angular acceleration. This is the angular analogue of ~F = m~a.

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13.3 Example moments of inertia

Example 13.1. Calculate the moment of inertia of a thin ring of radius R, mass M about an axis through its centre, perpendicular to the plane of the ring

Answer. Looking along the rotation axis:

θ δθ R

With all elements of mass at the same distance R, the answer is obviously I = M R2, but let’s do it more formally as well.

Assume a mass per unit length along the ring of σ, then a small section dθ has mass dm = σR dθ, and we have

I = Z r2dm = σR3 Z 2π 0 dθ = 2πσR3.

Then, since M = 2πRσ, we have I = M R2.

Example 13.2. Calculate the moment of inertia of a uniform circular disc of radius R, mass M about an axis through its centre,

perpendic-ular to its face. e.g. a coin.

Answer. Split into thin rings, integrate.

r+dr r R

Assume mass per unit area σ, then mass of a thin ring radius r, radial thickness dr is

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It is r from the axis at all points, hence

I = Z r2dm = Z R 0 2πσr3dr = 1 42πσR 4 .

The mass of the disc M = πR2σ, and therefore

I = 1 2M R

2_.

Can calculate the following moments of inertia:

• uniform sphere, radius R, mass M , about an axis through its centre: I = 2₅M R2;

• thin rod, length L, mass M , about a perpendicular axis through its centre: I = ₁₂1 M L2;

• the same rod, but about a perpendicular axis through one end: I = 1₃M L2.

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Lecture 14

Angular momentum in

use

Keypoints from Lecture 13

• The kinetic energy of an object in rigid-body rotation is

T = 1 2Iω

2

,

where I, the moment of inertia, is given by

I = Z

r2dm,

where r is the perpendicular distance of element dm from the rotation axis.

• The angular momentum of an object in rigid-body rotation can (often, and always in this module) be written as

~

L = I ~ω.

• Some example moments of inertia were given.

Objectives

• Examples of angular momentum

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LECTURE 14. ANGULAR MOMENTUM IN USE 71

14.1 Cylinder rolling down a slope

F_S N mg θ ω v a

A cylinder of radius r, mass m and moment of inertia I rolls without slipping down a slope which makes an angle θ to the horizontal. What is its acceleration a?

Because of torques, must take care over where the forces act. The weight acts through the centre of gravity, equivalent to the centre of mass in a uniform gravitational field.

Unknowns: N , FS, a, dω/dt, [v, ω]. First, “ ~F =m~a”:

perpendicular to slope: N − mg cos θ = 0

It soon becomes obvious that we don’t need this equation, since N doesn’t appear anywhere else.

parallel to slope: mg sin θ − FS = ma.

Next, “~τ = d~L/dt”, around the axis of the cylinder with L = Iω:

rFS = I dω

dt.

[NB Both N and mg act through axis of cylinder and add no torque.] The final relation comes from the no-slip condition v = rω, or

a = rdω dt.

These last two equations combine to give

FS = I r2a.

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Inserting into the equation for acceleration down the slope:

mg sin θ − I

r2a = ma, so

a = g sin θ 1 + I/mr2. Clearly, the ratio I/mr2 plays a major role.

mass concentrated on axis: I = 0, a = g sin θ, like frictionless mass.

uniform cylinder: I = 1₂mr2, a = 2₃g sin θ, not so fast.

hollow tube: I = mr2, a = 1₂g sin θ, slower still.

14.2 Conservation of angular momentum

In some circumstances, no external torques act and then

d~L dt = 0,

so ~L is constant and

In the absence of external torques, angular momentum is con-served.

Example 14.1. During a Type II supernova the Earth-sized core (R = 6370 km) of a star collapses to a neutron star of radius R = 10 km in a few seconds. If the initial spin period of the core was P = 1000 s, calculate the final spin period.

Answer. Assuming uniform spheres, L = 2₅M R2ω. If L is con-served during the collapse, then since ω = 2π/P , we must have

R2/P = constant, so P ∝ R2. Hence Pf = 10 6370 2 × 1000 = 2.5 × 10−3s.

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LECTURE 14. ANGULAR MOMENTUM IN USE 73 This is why “pulsars” spin so rapidly – the current record-holder spins 716 times/second!

Conservation of angular momentum is important in formation of the solar system, the Galaxy, ice skaters in a spin, divers, cats falling, spacecraft, helicopters, . . .

14.3 Gyroscopic Precession (approx.)

θ dt Ω top view Ω side view θ Ω dL=(Ω dt)L sin θ mg ω O Ω r sin r θ L sin θ L L You need to imagine the tip of ~L tracing a circle about the vertical axis. Since it lies at angle θ to the vertical, the radius of the circle is L sin θ.

Consider a tilted gyroscope, spinning rapidly about its axis at angular velocity ω. The centre of the wheel is at ~r from the base. It is subject to a torque from gravity. The torque ~r × (m~g) points into the page, so d~L/dt also points into the page. This is perpendicular to ~L, so the direction (not the magnitude) of ~L changes.

d~L dt = ~τ = m~r × ~g ⇒ d~L dt = mgr sin θ

Given the direction of ~L, the gyroscope “precesses” around a vertical axis, with ~L sweeping out a cone. Only the horizontal component of ~L is changing, the vertical component remains constant. If it precesses at an angular velocity ~Ω then

d~L dt = ~Ω × ~L ⇒ d~L dt = ΩL sin θ

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If the gyroscope has moment of inertia I, then L = Iω where ω is its spin rate. Hence

mgr sin θ = Ω Iω sin θ. Therefore

Precession rate: Ω = mgr Iω .

NB Independent of θ. The derivation becomes inaccurate at lower spin rates, when we can no longer take ~L to point precisely along the axis of the gyroscope.

The Earth’s axis precesses once every 26,000 years because of torques from the Sun and Moon. Called “precession of the equinoxes” and visible through a slow alteration in the path the Sun traces out through the constellations during the year.