classical dynamics and thermodynamics

Classical Dynamics and Thermodynamics

Shafiq R.∗Sarawak Matriculation College, Kuching, Sarawak

(Dated: March 18, 2021)

Here, I present short notes to supplement lectures on the topic of Classical Dynamics and Ther-modynamics found in the Malaysian Matriculation Programme Physics curriculum specification.The main reference text is the Cutnell’s 2015 10th edition textbook, Introduction to Physics.

Keywords: Newton’s Laws, Energy, Oscillations, Gravity

CONTENTS

I. Kinematics of Linear Motion 1

II. Momentum and Impulse 3

III. Forces 4

IV. Work, Energy and Power 5

V. Circular Motion 6

VI. Newtonian Gravitation 7

VII. Rotation of Rigid Bodies 8

VIII. Simple Harmonic Motion 9

IX. Waves 11

X. Solid Deformation 13

XI. Heat 14

XII. Kinetic Model of Gas 15

XIII. Thermodynamics 17

∗ Correspondence email address: [email protected]

Part One: Dynamics and SolidDeformation

I. KINEMATICS OF LINEAR MOTION

Time allocation:1.5h (Lecture) + 7h (Tutorial)Learning Outcomes:

1. Define

• Instantaneous Velocity/Acceleration• Average Velocity/Acceleration• Uniform Velocity/Acceleration

2. Discuss physical meaning of 3 graphs:

• Displacement-Time• Velocity-Time• Acceleration-Time

3. Apply equations of motion with uniform ac-celeration:

v = u+ at; s = ut+1

2at2; v2 = u2 + 2as

4. Describe projectile motion launched at an-gle θ

5. Solve problems related to projectile motion

LO1:

1. Instantaneous (velocity/acceleration)= (Velocity/Acceleration) at a specific point inspace and time.

2. Average (velocity/acceleration)= (Velocity/Acceleration) averaged over time inte-val

3. Uniform (velocity/acceleration)= constant rate of change of (displace-ment/velocity)

mailto:[email protected]

LO2:Assumption: Non-zero, constant acceleration

a=0

Time

DisplacementDisplacement -Graph Time

Equation: s = si + ut+ 12at

2

Description: In the case of a 6= 0, the graph plotted fol-lows a quadratic equation graph form with a y-interceptof si. The gradient of the plot at a certain point givesthe velocity.

Time

VelocityVelocity -Graph Time

Equation: v = u+ atDescription: The graph plotted is a linear plot whena = 0 with y-intercepts of the initial velocity, u. Thegradient of the plot at a certain point gives the accel-eration and the area under the graph represent thedisplacement.

Time

AccelerationAcceleration-Graph Time

Equation: dadt = 0

Description: The area under the graph representthe velocity.LO3 requires practice! But here we present a simplederivation that merely requires memorization of 1 equa-tion:

dv

dt= a→ v = u+ at

ds

dt= v = u+ at→ s = si + ut+

1

2at2

Projectile Motion

Description: A projectile launched at an angle θ ex-hibits a parabolic path.Problem Solving Strategy: Resolve ("Split") the vinto its components, vx and vy, with ax = 0 and ay = −g,where g is the gravitational acceleration.

Components x y

u0ux

cos(θ)uy

sin(θ)

Acceleration 0 -gVelocity vx = ux vy = uy − gt

Displacement sx = s0x + uxt sy = s0y + uyt− 12gt

2

Max height when vy = 0

II. MOMENTUM AND IMPULSE


1. Define momentum and impulse, ~J = ~F∆t

2. Solve impulse/ impulse-momentum prob-lems ~J = ∆~p = m(~vf − ~vi)

3. Determine impulse from F − t graph

4. State the principle of linear momentum con-servation, ∆(

∑~p) = 0 →

∑~pfinal =∑

~pinitial

5. Apply ddt (∑~p) = 0 in 1D and 2D elastic and

inelastic collisions.

6. Differentiate between elastic and inelasticcollision

LO1 - Definitions:

Momentum The vector product of mass of the bodywith it’s velocity, ~p = m~v.

Impulse The change in momentum of a body withmass m when it changes velocity from~vintial to ~vfinal, ~J = ∆~p = m(~vfinal −~vinitial) = ~F∆t

LO3: Determine impulse from F − t graphReferring to ~J = ~F∆t, we’d deduce that the area underthe F − t graph give us the impulse.

LO4: Conservation of Linear MomentumStatement: In a closed system, the total momentum isconserved regardless of the change in body mass of ve-locity. That is to say,

∆p = ~pfinal − ~pinitial = 0→ ~pfinal = ~pinitial

LO5 and LO6: Application of momentum conservationlaw and categorizing collisions.Begin with differentiating between elastic and inelasticcollision:

Elastic Obeys kinetic energy conservation,∆(∑Ekinetic) = 0.

Inelastic Does not obey kinetic energy conserva-tion, ∆(

∑Ekinetic) 6= 0

To solve problems, there are general key concepts tolook for in the questions such as:

• A body comes to a stop → vfinal = 0.

• A body started at rest → vinitial = 0

• sticks together → Inelastic equation,∑(∆Ekinetic) = 0 not applicable here.

III. FORCES

Time allocation:1h (Lecture) + 3h (Tutorial)Learning Outcomes:

1. Identify forces (Weight, Tension, Normal,Friction, External) acting on a body in dif-ferent situations.

2. Sketch Free Body Diagram (FBD)

3. Determine static and kinetic friction:

fs ≤ µsN ; fk = µkN

4. State Newton’s Laws of Motion

5. Apply Newton’s Laws of Motion

LO1, 4 types of forces and its definition:

1. Weight = Force exerted upon a body interactingwith a gravitation field.

2. Tension force = Force transmitted axially througha massless one-dimensional continuous element.

3. Normal force = support force, perpendicular to thesurface, exerted upon a body in contact with a sta-ble object.

4. Frictional force = A force that stems from 2 roughsurfaces that in is contact and moving relative toeach other.

LO2, Sketching a free body diagram requires practice,however, herewith we present the directions of each typeof forces:

1. Weight = towards the Earth/gravitational source.

2. Tension force = along the cable/string/cord andaway from the body.

3. Normal force = Perpendicular to the surface thebody is in contact with.

4. Frictional force = Against the direction of motion.

LO3, Determination of static and kinetic friction.Static friction = Frictional force between two bodies

that are in contact and not moving relative to each other.Kinetic friction = Frictional force between two bodiesthat are in contact and moving relative to each other.

The equation for frictional force is

Ffriction = µN

where the µ is friction coefficient. The value of thisfriction coefficinet depends on the material as well as

whether or not the bodies are moving relative to eachother.

Static friction are generally higher than kinetic friction(special cases exists! e.g. Rubber friction), and the rea-son is the asperities (roughness) interaction between 2surfaces that are in contact. This interaction includes

• Asperities forms "cold welding" (intermolecularbonding).

• Asperities from both surfaces interlocks.

Both of which creates adhesion between the 2 surfaces.This adhesion must be overcome before the body canstart moving relative to each other.

The diagram above shows how frictional force variesas applied force is increased for 2 surfaces in contact.The peak indicates the motion threshold which is theamount of force needed to start moving the body. Thisoccurs at ~Ffriction = µstatic ~N . Once the body startsmoving (relative to the surface it is in contact with), thefrictional force decrease and becomes constant at somerange of velocity/speed.

LO. 4 and 5, Newton’s Laws of Motion:

1st Law An isolated body moves with constant ve-locity, ~F = 0⇒ ~v = constant

2nd Law The rate of change of linear momentum ofa body is equal to the force acting on thebody, ~F = d~p

dt .3rd Law Whenever two bodies interact, the force

~F21 which 1 exerts on 2 is equal and op-posite to the force ~F21 which 2 exerts on 1,~F12 = −~F21.

Application of these laws will be shown in tutori-als.

IV. WORK, ENERGY AND POWER


1. Define work done by a constant force, W =~F · ~s

2. Apply work done by a constant force andfrom a force-displacement graph

3. State the energy conservation principle

4. Apply energy conservation principle, involv-ing mechanical energy and heat energy dueto friction

5. State and apply the work-energy theorem,W = ∆K

6. Define and use:

• average power, Pav = ∆W∆t

• Instantaneous Power, P = ~F · ~v

LO1, LO3,LO6: Definitions and statements:

1. Work done,W , can be defined as the scalar productof force, F , and the displacement, ~s, :

W = ~F · ~s

2. Energy conservation statement - Total energy of anisolated system is conserved, that is ∆

∑E = 0.

3. Average power, Pav, can be defined as the averageamount of work done,∆W , in a time interval, ∆t, :

Pav =∆W

∆t

4. Work-energy theorem states that the change in ki-netic energy is equal to the work done by the force,that is:

Wi→f = ∆Ek =1

2m(v2

f − v2i )

5. Instantaneous power, Pinstant, is the scalar productof force, ~F , and velocity, ~v,:

Pinstant = ~F · ~v

6. Some useful equations:

• Ekinetic = 12mv

2

• Generally, F = −dEpot.

dx

• Egrav.pot. = mgh

• Espring pot. = 12kx

2

Note that for the definition of work done, W , and instan-taneous power, Pinstant is a scalar produce of the twovector, this is a reminder that we are only interested inthe components of force that is parallel to the displace-ment and that the angle between them must be takeninto account when it is not zero.LO2 - Force-displacement graph:

The area under the force-displacement graph representsthe work done. The "proof" can be demonstrated if onewere to consider the equation dW = F · dr and inte-grate both sides or sum up sections of the areas underthe graph.

V. CIRCULAR MOTION


1. Describe uniform circular motion

2. Convert units:

Degrees(o) and radians revolution/rotation

3. Define centripetal acceleration

4. Solve centripetal force for uniform circularmotion problems for the following cases:

(a) horizontol circular motion(b) vertical circular motion(c) conical pendulum

LO2- Unit conversion:

1 revolution/rotation = 360o = 2π radian

LO1- Uniform circular motion description:Uniform circular motion is the motion of an object travel-ling at a constant (uniform) speed on a circular paths.Anobject travelling in a circular path constantly changes it’sdirection. Constant changes in its direction indicates aconstant change in the velocity of the object. This con-stant change in object velocity is only allowed if thereexist an acceleration, which we call centripetal accelera-tion.

LO2- Centripetal acceleration = a property in a bodytraversing in circular path, that is radially directed to thecircle centre and has a magnitude, |ac| = v2

r .

Using Newton’s 2nd Law of Motion, ~F = m~a, we canthen define centripetal force as follows:

Fc =mv2

r

with m = mass of body, v = linear velocity tangent tothe circle and r = circle radius.

LO4- 3 cases to be considered:

1. Horizontal Circular MotionWe can start by sketching out the path of theobject as well as its Free Body Diagram:

,From these 2 diagrams, we can deduce thatFc = mv2

r and N = mg.

2. Vertical Circular MotionAgain, we start with with sketching out the pathof the object, but to ease calculations, only thebottom half of the path is presented here as it willbe generalized to the rest of the pathway:

Here we see that Fc = mv2

r = T −mg cos(θ) for astring and Fc = mv2

r = N −mg cos(θ) for a bodyon a circular track, e.g. roller coaster ride.

3. Conical PendulumObject path:

Here we see that F cos(θ) = mg andF sin(θ) = mv2

r . Three equations that maybe obtained from this are:

(a) tan (θ) = v2

rg = rh , from dividing the two equa-

tions,(b) F = mg

√1 + ( rh )2, from the fact that

sin2(θ) + cos2(θ) = 1, and,

(c) Period, T = 2π√

hg , from v = ωr = 2π

T r and

tan (θ) = v2

rg

VI. NEWTONIAN GRAVITATION


1. State and use Newton’s Law of Gravitation,F = GMm

r2

2. State and use gravitational field strength,ag = GM

r2

3. Define and use gravitational potential en-ergy, U = −GMm

r

4. Derive and use escape velocity equation,vesc

√2GMR =

√2gR

5. Derive and use satellite motion equation:

• Velocity, v =√

GMr

• Period, T = 2π√

r3

GM

LO1, LO2- Newton’s Law of Gravitation (Fg), grav-itational field strength (ag):

Fg = m(ag) = m(GM

r2)

where m and M are masses of the interacting bodies,G is the Newtonian gravitational constant, and r is thedisplacement between the interacting bodies.

In words, gravitational field strength at a point is thegravitational force exerted experienced per unit mass atthat point.LO3-

Gravitational Potential Energy = energy of a body dueto its position within the gravitational field. Derivation:

Fg = −dUgdr

Ug = −∫Fg dr

= −GmM∫r−2dr

= −GMm

r

LO4 - Derivation of escape velocity:Idea: For a body overcome the gravitational attractionof a more massive body, the kinetic energy of that bodymust be equal to, or more than the gravitational potentialenergy, i.e. Ek > Ug.

The former describes a body escaping the gravitational

attraction with the final velocity of zero.

Ek = −Ug1

2mv2

esc =GmM

R2

vesc =

√2GM

R=√

2gR

LO5- Satellite motionAssumption: Satellite motion has a circular path, thissimplifies equations.Referring to the past chapter, we utilize centripetal forceand compare it with the gravitational force equation toobtain the equation for satellite velocity:

Fcentripetal = Fgrav

mv2

r=GMm

r2

v =

√GM

r

The period of the satellite motion can be derived by uti-lizing

v =2πr

T

to show that period of satellite is

T = 2π

√r3

GM

VII. ROTATION OF RIGID BODIES


1. Define and use:

• Angular displacement, θ.• Average angular velocity, ωave.• Instantaneous angular velocity, ω.• Average angular acceleration, αave.• Instantaneous angular acceleration, α.

2. Relate rotational parameters with their cor-responding linear parameters:

s = rθ; v = rω; atrans. = rα,

acentri. = rω2 =v2

r

3. Solve constant angular acceleration rota-tional motion problems:

ω = ωo + αt;

θ = ωot+1

2αt2

ω2 = ω2o + 2αθ

4. Define torque, ~τ = ~r × ~F .

5. Solve uniform rigid body equilibrium prob-lems.

6. Define and use the moment of inertia of auniform rigid body (sphere, cylinder, ring,disc and rod).

7. State and use torque, τ = Iα.

8. Define and use angular momentum, L = Iω

9. State and use the principle of angular mo-mentum conservation.

Note: For the most part of this chapter, the units areof radians, i.e. the units for angular displacement andangular velocity are radians (rad) and radians per second(rads−1) respectively.LO1, LO4, LO6, LO8 - Definitions:

1. LO1:

• Angular displacement, θ = the anglethrough which a rigid object rotates about afixed axis• Average angular (velocity, ωave / acceleration,αave) = angular (displacement / velocity) di-

vided by the elapsed time during which the(displacement / velocity change) occurs:

ω =∆θ

∆t; α =

∆ω

∆t

• Instantaneous angular (velocity, ω / accelera-tion, α) = the angular (velocity / acceleration)that exists at any given instant.

2. LO4 - Torque, τ , = a measure of force acting on abody rotating about an axis:

~τ = ~r × ~F = rFsinθ

τ = Iα =dL

dt

3. LO6- Moment of Inertia = a measure of a body’sresistance to rotation, quantified by ratio of the netangular momentum of a system to its angular ve-locity along a principal axis, Lω .

I =

∫ M

0

r2dm

Common moments of Inertia:

Shape Rotation Axis EquationSphere (solid) centre 2

5MR2

Sphere (solid) ⊥ surface 75MR2

Cylinder (solid)/disk

symmetry axis 12MR2

Cylinder (solid) Central diame-ter

14MR2 + 1

12ML2

Ring symmetry axis MR2

Rod center 112ML2

Rod End 13ML2

4. LO8 - Angular Momentum, L = the product ofthe body’s moment of inertia, I, and its angularvelocity, ω, with respect to that axis:

L = Iω

LO2- Relate rotational parameters with their corre-sponding linear parameters.

Rotational parameters Linear counterparts EquationAngular displacement,θ[rad]

Linear displacement,s[m]

s = rθ

Angular velocity, ω linear velocity, v v = rω

In this chapter, when speaking of acceleration, thereare 2 types of acceleration - translational accelera-tion, αtrans., and centripetal acceleration, αcentri.

The latter is the same as the case for circular motion,αcentri. = v2

r = rω2. The former, refers to the tangentialcomponent of acceleration. If αtrans. = 0, the the motionis a uniform circular motion.LO3- Constant angular accceleration rotational motionproblems

The following equations describes rotational motion ofwhich the angular acceleration is constant:

ω = ωo + αt;

θ = ωot+1

2αt2

ω2 = ω2o + 2αθ

They resemble equations for constant acceleration, justwith parameters redefined in the context of rotationalmotion.LO5- Uniform Rigid Body Equilibrium:

The concept of the uniform rigid body Equilibrium isbased on a few key ideas:

1. Uniform refers to a special case in which mass dis-tribution is a constant.

2. Rigid body refers to the condition of the body to notshape its shape under external force, consequentlyneglecting strain/stress in the calculations.

3. Equilibrium refers to condition in which the sumof externally applied torques is 0, that is

∑τ = 0

(and therefore∑Fx = 0and

∑Fy = 0).

Guidelines to solving uniform rigid body equilibriumproblems:

1. Select the object you’ll be applying the equationsto.

2. Draw free body diagram to show the external forcesthat acts upon the object chosen.

3. Make your choice of axes and resolve the externalforces to its components on the chosen axes.

4. Apply∑Fx = 0and

∑Fy = 0.

5. Select a rotation axis.

6. Apply∑τ = 0

7. Rinse and repeat until goal achieved.

LO9- Principle of angular momentum conservation:Statement: When τexternal = 0, Lfinal = Linitial.

VIII. SIMPLE HARMONIC MOTION


1. Explain SHM

2. Solve problems related to SHM displace-ment equation, x = A sin (ωt)

3. Derive equations :

• Velocity, v = dxdt = ±ω

√A2 − x2.

• Acceleration, a = dvdt = −ω2x.

• Kinetic energy, K = 12mω

2(A2 − x2).

• Potential energy, U = 12mω

2x2

4. Emphasise the relationship between totalSHM energy and the amplitude

5. Apply equations for v, a,Kand U for SHM

6. Discuss the following graphs:

• Displacement-time• Velocity-time• Acceleration-time• Energy-displacement

7. Use expression for SHM period for simplependulum and single spring

LO 1 - Simple Harmonic Motion (SHM)A SHM is a periodic motion in which the restoring forceis directly proportional to its displacement and acts theopposite direction to the its displacement. It has thefollowing general equation:

d2x

dt2= −ω2x

where the ω is the angular frequency.Note that the negative sign in the general equation

tells us that the force is acting on the opposite directionto the displacement. The solution to the SHM generalequation is

x(t) = Asin(ωt± φ)

where A is the amplitude of oscillation and φ is the phaseshift (which we will not be concerned of).LO7- Single pendulum and single springOne example of this type of motion is seen in the Hooke’sLaw describing the amount of reaction force of an ex-tended/compressed spring, aptly giving the equation:

F = ma = md2x

dt2= −kx

From this, we will obtain the equation

d2x

dt2= − k

mx

and obtain ω2 = km

Keeping in mind that the period-angular frequencyequation, T = 2π

ω gives us

T = 2π

√m

k

Another example noted here is the simple pendulumcase, in which it has the equation F = −mg( xL ) describ-ing the restoring force, where L is the length of the pen-dulum string. (The small angle approximation,sin (θ) ≈θ = x

L is used here).We then obtain the following equations:

d2x

dt2= − g

Lx;

ω2 =g

L

T = 2π

√L

g

LO2- Deriving equations:Equations for velocity, kinetic energy and potential en-ergy can be done by knowing these equations:

• Velocity, v = dxdt ⇒ v = ±ω

√A2 − x2

• Kinetic Energy, K = 12mv

2 ⇒ K = 12mω

2(A2−x2)

• Potential Energy, U = 12kx

2 ⇒ U = 12mω

2x2

LO4- Conservation of energy states that E[total] =K + U , utilizing this and applying the equations ob-tained above, gives Etotal = 1

2kA2. We can see that the

Etotal ∝ A2 LO6- Graphs:

Displacement, velocity and acceleration against time.v → 0 twice, once at xmax and the second time whenamax.

Energy variation with displacement.K decreases and U increases to keep Emax constant, asx increases.

IX. WAVES


1. Define wavelength, λ, and wave number, k.

2. Solve problems related to the equation ofprogressive wave, y(x, t) = Asin(ωt± kx)

3. Discuss and use the particle vibrational ve-locity and wave propagation velocity (groupvelocity)

4. Discuss the graphs of:

• displacement-time, y − t• displacement-distance, y − x.

5. State the principle of superposition of wavesfor constructive and destructive interfer-ences.

6. Use the standing wave equation, y =Acos(kx)sin(ωt)

7. Discuss progressive and standing wave.

8. Define and use sound intensity

9. Discuss the dependence of intensity on am-plitude and distance from a point source byusing graphical illustrations.

10. Solve problems related to the fundamentaland overtone frequencies for:

• stretched string• air columns (open and closed)

11. Use wave speed in a stretched string, v =√Tµ .

12. State Doppler effect for sound waves.

13. Apply Doppler effect equation, fapparent =

f(v±vov∓vs

), for relative motion between

source and observer. Limit to stationary ob-server and moving source, and vice versa.

LO1, LO8- Definitions:

1. LO1: Wavelength, λ, is the horizontal length of onewave cycle - the distance over which wave’s shaperepeats.

2. LO1: Wave number, k, is the number of waves perunit distance, k = 2π

λ .

3. LO8: Sound intensity, I, is defined to be the ac-coustic power that passes perpendicularly througha surface divided by the area of that surface, I = P

A

LO2, LO6, LO7- Waves:

1. Progressive wave:Sinusoidal equation: y(x, t) = Asin(ωt± kx)

2. Standing wave:y(x, t) = Acos(kx)sin(ωt)

where y(x, t) is the displacement on an element at po-sition x at time t, A is the amplitude, ω is the angularfrequency, k is the wavenumber.

LO3- Velocities:The aim here is to distinguish between particle vibra-tional velocities and wave propagation velocity. Picture,in your head, water waves. The water molecule, moves"up and down". This motion of "up and down" thencauses the next water molecule to move "up and down".The velocity of moving "up and down" is called par-ticle vibrational velocity whereas how fast the nextwater molecule reacts to the "up and down motion" ofthe water molecule before it is called wave propagationvelocity .Essentialy,

• particle vibrational velocity = how fast the particlevibrate.

• wave propagation velocity = how fast the wavepropagates.

LO4- Graphs:In the displacement-time graph, the distance between

peaks(troughs) is the period of the wave. In thedisplacement-distance graph, the distance between peaks(troughs) is the wavelength of the wave.

LO5- Principle of wave superposition:The principle of (linear) wave superposition simply statesthat when multiple waves are present simultaneously atthe same place (=interfere), the resultant disturbance isthe sum of the disturbances from the individual waves.Constructive (/Destructive) interference =

when the interference results in a wave with amplitudehigher(/lower) than the amplitude of the individualwaves.

LO9- Sound Intensity Dependence:Sound intensity is directly proportional to the square ofthe amplitude of the sound wave, I ∝ A2.For a point source, one may assume the sound wavepropagates uniformly in all directions and therefore maytake the area of a surface of the sphere for the area, thisgive I = P

4πr2 . This shows the relation I ∝ 1r2 .

LO10, LO11- Standing wave tones:

Nodes = points on a waveform where that does notvibrate at all.Antinodes = points on a waveform where maximumvibration occurs.Fundamental frequency, fo = the lowest frequency of aperiodic waveform.Overtone = harmonic frequencies above the fundamen-tal, f = nfo where n ∈ Z.

Here, it is crucial one particular aspect, which is thenaming of certain terms, that is worth mentioning,Overtones HarmonicsFundamental 1st Harmonic1st overtone 2nd Harmonicand so on and so forth. We can see that the nth overtoneis the (n+ 1)th harmonic.

In a stretched string, the boundary condition im-posed is that the ends of the string must be a node, thatis y(0, L) = 0. The fundamental and overtones are asfollows:

As we can see, the string length is in the integer valueof half a wavelength, that is L = nλ2 . We can then writedown frequencies as a function of length using f = v

λ ,where v is the wave speed to obtain:

fn =nv

2Lwhere n ∈ Z

The wave speed on a string is given by:

v2 =T

µ

where T is the tension, µ(= mL ) is the linear mass density.

For air columns, there are 2 situations to be considered:

1. Tube with one open end and one closed end("Closed Tube")Boundary condition- a node on the closed endand an antinode on the open end, which gives thefundamental tone:

We can see that these boundary conditions requiresL = nλ4 which gives us:

fn = (2n+ 1)v

4Lwhere n ∈ Z

2. Tube with 2 open ends ("Open Tube")Boundary condition- Antinodes on the open ends,which gives the fundamental tone:

We can see that these boundary conditions requiresL = nλ2 which gives us:

fn =nv

2Lwhere n ∈ Z

LO12, LO13- Doppler Effect:Doppler Effect = change in frequency detected by an ob-server due to difference in velocity of the sound sourceand observer with respect to the medium of sound prop-agation.

The Doppler effect can be quantitatively described us-ing the equation:

fapparent = fsource

(v ± vobserverv ∓ vsource

)Where v is the sound wave speed (≈ 343ms−1 in air of20oC).

In the numerator, the (plus/minus) sign is used whenthe observer moves (towards the / away from) source. Inthe denominator, the (minus/plus) sign is used if sourcemoves (towards the / away from) the observer.

X. SOLID DEFORMATION


1. Distinguish between stress and strain fortensile and compression force.

2. Discuss the graph of stress-strain for a metalunder tension.

3. Discuss elastic and plastic deformation.

4. Discuss graph of force-elongation for brittleand ductile materials.

5. Define Young’s modulus.

6. Discuss strain energy from the force-elongation graph.

7. Discuss strain energy per unit volume fromstress-strain graph.

8. Solve Young’s modulus problems.

LO1, LO5- Definitions:

1. Stress, δ = the measure of internal forces actingon neighbouring particles in a continuous media,quantified by the ratio of the magnitude of the forceto the area, δ = F

A .

2. Strain, ε = the measure of the deformation of thematerial, quantified by the ratio of the change in aquantity to the quantity, ε = ∆L

L .

3. Young’s modulus, Y = a measure of the stiffness ofsolid material quantified by the the ratio of stressto strain, Y = δ

ε .

LO2, LO3, LO4, LO6, LO7- Graphs:

1. LO2, LO3 & LO7 - Stress-Strain Graph:

Discussion:Young’s modulus is applicable between the begin-ning of the graph to the proportional limit (yieldstrength). In this part of the graph, it shows a lin-ear relation between stress and strain, by which one

can calculate the Young’s modulus from its gradi-ent. It is in this region an object goes throughelastic deformation.Elastic deformation = deformation in which theshape change due to external force is reversible,temporary shape changeNon-linearity beyond the proportional limit repre-sent the failure of Hooke’s Law and the beginningof plastic deformationPlastic deformation = deformation in which thethe shape change due to the external force is irre-versible, permanent shape change.One can observe from the graph that the stress con-tinues to increase with strain until it reaches a max-imum, a point aptly named the Ultimate TensileStrength(UTS). Between the yield strength andthe UTS, the strain accumulates to contribute tostrain hardening, which is caused by dislocationof crystal planes.Beyond the UTS, necking occurs until a fracturetakes place. Necking is simply when the cross-sectional area begins to become significantly lowerthan the average.

2. LO4 & LO6 - Force-Elongation Graph:

Discussion:The graph above shows that difference in theforce-elongation curve for a brittle and ductilematerial. The brittle material fracture point liesclose to the elastic limit where as for the ductilematerial, fracture point is quite an extent fromthe elastic limit. This means ductile goes throughsignificant plastic deformation before fracturing.This is because ductile materials are able tocontain much more strain energy than brittlematerials. Necking also does not take place priorto fracturing in a brittle material.From the difference in gradient, we can deduce thatit takes a lot more force to extend brittle materialthan the ductile material, that is Yductile > Ybrittle.This allows us to say that under the same stress,the strain energy is lower in the brittle material,

Yductile − Ybrittle > 0 ⇒ δ(

1εductile

− 1εbrittle

)> 0

which is obeyed iff εbrittle < εductile.

Part Two: Heat and Thermodynamics

XI. HEAT


1. Define heat conduction

2. Solve problems related to rate of heat trans-fer, dQdt = −kAdT

dx , through a cross-sectionalarea. (Maximum 2 insulated objects in se-ries)

3. Discuss temperature-distance, T − x graphfor for heat conduction through insulatedand noninsulated rods. (Maximum two rodsin series)

4. Define coefficient of linear, area and volumethermal expansion.

5. Solve problems related to thermal expansionof linear, area and volume (include expan-sion of liquid in a container):

∆L = αLo∆T ; β = 2α; γ = 3α

LO1, LO4 & LO5- Definitions:

1. LO1 - Heat Conduction = Conduction is the pro-cess whereby heat is transferred directly througha material, with any bulk motion of the materialplaying no role in the transfer.

2. LO5 - Coefficient of (linear/area/volume) expan-sion = a measure of (linear/area/volumetric) frac-tional change per degree change in temperature ata constant temperature.

∆L = αLo∆T,

∆A = βAo∆T,

∆V = γVo∆T,

β = 2α; γ = 3α

where L = length, A = Area, V = volume, T =temperature, {α, β, γ} = coefficient of {linear, area,volumetric} expansion.

LO2- Rate of heat transfer equation:The equation for the rate of heat transfer is given by:

dQ

dt= −kAdT

dx

where dQdt = heat change in a given body with respect to

time, A = area of contact, dTdx = temperature gradient

and k = heat transfer coefficient.For a metal of length L and cross sectional area A, theheat conducted in time t is

Q = kAt∆TendsL

where ∆Tends = temperature difference between ends ofthe metal.This could be extended for 2 insulated objects in series(assuming steady state conduction) to give

Q = tA(∆Tends)

(L1

k1+L2

k2

)−1

where the subscripts represents the objects involved.This is possible because the assumption of steady state

tells us that

dQ1

dt=dQ2

dt

dQdt may be thought of as "heat currents" and functions

analogously to an electrical system with resistances inseries.So it can be said that the "heat current" goingthrough both the material must be equal.LO3- Temperature-distance graph:

For the insulated rod, the heat loss through the sides ofthe rod is negligible relative to the non-insulated rod.This means that the heat flows only through the crosssectional area from the hot end to the cooler end. Whatthis means is that whilst the insulated rod temperaturedrops linearly with distance, the temperature dropsexponentially for the non-insulated rod.

XII. KINETIC MODEL OF GAS


1. Solve problems related to ideal gas equation,pV = nRT .

2. Discuss the following graphs of an ideal gas:

• p-V graph at constant temperature• V-T graph at constant pressure• p-T graph at constant volume

3. State the assumptions of kinetic theory ofgases.

4. Discuss root mean square (rms) speed of gasmolecules

5. Solve problems related to root mean square(rms) speed of gas molecules.

6. Solve problems related to the equations:

pV =1

3Nmv2

rms; p =1

3ρv2rms

7. Discuss translational kinetic energy of amolecule, Ktr = 3

2

(RNA

)T = 3

2kBT

8. Discuss degrees of freedom, f formonoatomic, diatomic and polyatomicgas molecules.

9. State the principle of equipartition of en-ergy.

10. Discuss internal energy of gas.

11. Solve problems related to internal energy,U = 1

2fNkBT

LO1 & LO2- Ideal Gas Equations:The ideal gas equation is given by

pV = nRT

where p = absolute pressure of the gas , V = volume,T = temperature in Kelvin, R = universal gas constant,n = number of moles.Standard temperature and pressure refers to 273Kand 1atm.Graphs:

1. p-V graph

2. V-T graph

3. p-T graph

Comments: The relationships between variables shownby the graphs above are fairly obvious. That is thevolume-temperature and pressure-temperature relationsare linear in nature, assuming pressure is constant inthe former and volume is constant for the latter. Thepressure-volume graph, on the other hand, has a inverselyproportional relation.LO3- Assumptions of the kinetic theory of gases:

Gases are composed of large number of non-null masspoint-like particles that obeys Newton’s laws of mo-tion and elastically collides with each other, of whichtheir average kinetic energy solely depends on thesystem’s absolute temperature.

It’s called the Kinetic Theory of gases because thenotion of kinetic energy is applied through Newton’slaws of motion in collisional force between particles.The speed distribution curves follows that of Maxwell-Boltzmann’s distribution.LO4 & LO5- Root mean square (rms) speed of gas

molecules:The idea of rms speed of gas molecules stems from a no-tion of the same name in the study of statistics. It stemsfrom the assumption for large number of particles, not all

gas particles has the same speed, but its distribution fol-lows that of what is called as the Maxwell-Boltzmanndistribution. These gas particles all moves in randomdirection. Speaking of the particle velocities is then dif-ficult due to directional component. It is easier then toconsider the translational kinetic energy of the particlesto equal to the average translational kinetic energy, suchthat 1

2mv2 = 1

2mv2. This is achievable if one were to

consider the root-mean-square speed of particles, vrmswhich is equal to

√v2. Note that v2 6= v2.

LO6:Post-introduction of the rms speed, it might be useful (&necessary!) for one to cast the ideal gas equation withrms speed as one of its variable. To do so, one beginswith pV = nRT , and then with consideration of notionsof force (ipso facto utilization of Newton’s Law of Motion) onto a wall by a cloud of gas molecules, one comes tothe conclusion of the following equation:

pV =1

3Nmv2

rms and p =1

3ρv2rms

where the mass density, ρ = NmV and number of particles,

N = n×NA.LO7 & LO9- Translational kinetic energy of a

molecule and the equipartition theorem:Moleculate move, and they move until they hit stuff(wall, another molecule etc.) but how much energy dothey have when they move? Well, we know that kineticenergy,Ktr = 1

2mv2, so we know that Ktr = 1

2mv2rms.

We can then have the following equation

pV =2

3NKtr

which when equated to the ideal gas equation gives

Ktr =3

2

(R

NA

)T =

3

2kBT

where kb = RNA

. This equation is important becauseit tells you that the kinetic energy of the molecules issolely dependent on the temperature (in Kelvin). In fact,this fact, 1

2mv2rms = 3

2kBT , is so crucial because it isthe direct consequence of the classical equipartitiontheorem .

This theorem states that for every degree of freedomthat a gas molecule has, it contributes to 1

2kBT to theaverage energy. Since the velocity components are foundin 3 directions, then the kinetic energy is 3× 1

2kBT .LO8, LO10 & LO11- Degrees of freedom and inter-

nal energy:In the discussion of degrees of freedom, it is importantfor one to understand what it means to have N degreesof freedom. For gas molecules, degrees of freedom can bedefined to be the number of physical variables that givesa particular system its characteristics. Mathematically, adegree of freedom is is any dynamical variable that con-tributes a squared term to the expression for the total en-ergy of the molecule (e.g. Ek = 1

2mv2, Erotation = 1

2Iω2,

Ehooke = 12kx

2).

Classically, the calculation of internal energies is basedon the number of degrees of freedom of the gas molecules,obeying

U =1

2fNAkBT

. To show its relationship with internal energy of a givensystem, let us consider 3 cases of the following:

1. Monatomic gas molecules:For Monatomic gas molecules (e.g. Ne gas), thecalculations are fairly straight forward. The as-sumption here is that there are no molecularrotation or molecular vibration. This meansthat the only contributor to the internal energyis the translational kinetic energy. The gasmolecules here has 3 directions in which it cantranslates in - (x, y, z) direction. Then the totalinternal energy for n moles of monoatomicgas molecules is simply

U = Ktr = n× 3

2kBT =

3

2nRT

2. Diatomic gas molecules:For the diatomic gas molecules (e.g. H2 gas), thereare 6 degrees of freedom - namely 3 translationalmotion and 2 corresponds to the rotational motion(about the x- and y-axis). Therefore, the totalinternal energy for n moles of diatomic gasmolecules is given by

U = Ktr + Erotation = n5

2RT

*This author would like to point out that this calcu-lation is only limited to classical treatment of theinternal energy of gases, which completely disre-gards the quantization of energy and leads tothe consideration of vibrational energy. This con-sideration has been experimentally showed to takeeffect for Cl2 gas at around 1500K, where the in-ternal energy follows 7

2RT . But this considerationis ignored at the current level. This represents thefailure of classical physics.

3. Polyatomic gas molecules:Essentially, one has to ask how many degrees offreedom (f) does the gas molecules have, break itdown to its individual components (Ktr, Erotation,etc...) and use the equipartition theorem. But clas-sically, U = 1

2fNkBT is sufficient.

XIII. THERMODYNAMICS


1. State the first law of thermodynamics.

2. Solve problem related to first law of ther-modynamics.

3. Define the following thermodynamics pro-cesses:

• Isothermal• Isochoric• Isobaric• Adiabatic

4. Discuss p-V graph for all the thermody-namic processes.

5. Discuss work done in isothermal, isochoricand isobaric processes.

6. Solve problem related to work done in

• isothermal process,W = nRT ln V2

V1= nRT ln p1

p2;

• isobaric process,W =

∫p dV = p (V2 − V1);

• isochoric process,W =

∫p dV = 0.

LO1, LO2 & LO4- First law of thermodynamics:The first law of thermodynamics states that:Changes between states thermal equilibrium leads to in-ternal energy change in the system that obeys

∆U = Q+W

in which Q = heat transferred to the system and W =total work done on the system.*Disclaimer: Some reference texts may show −W in-stead of +W and this is due to defining W as the totalwork done by the system. But we choose this form as itclearly shows the conservation of energy principle. Thatis when heat is added to the system, the internal energyincrease or when work is done on the system, the internalenergy also increases.*

Solving problems related to the first law of thermo-dynamics requires certain strategies. Irregardless of thestrategies involved, it might be useful to know that

1. for heat energies stemming from temperaturechange, Q = mc∆T

2. for heat energy stemming from phase change, Q =ml

3. Calculation of work done from volume changes,W =

∫pdV

The third equation, W =∫pdV , gives us a piece of in-

formation on interpreting calculation of work done fromthe p-V graphs, that is the work done is the area underthe graph, no matter how the graph looks like (i.e. whatthe thermodynamical process is involved).

LO3, LO4, LO5 & LO6-In the matriculation syllabus, we consider 4 type of ther-modynamical processes:

1. Isothermal (∆T = ∆U = 0⇒ Q = −W )In this case, the volume of the gas containerchanges. So,

W =

∫ Vf

Vi

pdV = nRT

∫ Vf

Vi

dV

V= nRTln

V2

V1

2. Isochoric (∆V = 0⇒ ∆U = Q)This thermodynamical process is also known as iso-volumetric process, which means the volume doesnot change and since work done that we considerdepends on the change in volume, no work is done,W = 0.

3. Isobaric (∆p = 0⇒ ∆U = Q+W )For the isobaric case, pressure is kept constant.

work done is then

W =

∫ Vf

Vi

pdV = p(V2 − V1)

4. Adiabatic (∆Q = 0⇒ ∆U = W )Adiabatic processes is a peculiar one, however thecalculation of the work done is not required atthe matriculation level. However, for those bravesouls willing to give it a go, one can derive the equa-tion of it. This can be done with the considerationthat whilst ∆Q = 0, ∆T 6= 0 and ∆V 6= 0, fromthe first law of thermodynamic, we can then writedU = dW . The ratio of specific heat capacitiesneeds to be considered in this case to show that∆(PV γ) = 0 where gamma is known as the ra-tio of specific heat capacities. This is the keyto deriving the work done equation for the case ofadiabatic processes.

The following shows p-V graph for all the processes:

End of notes

classical dynamics and thermodynamics

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