Advanced Higher Physics Unit 1 Mechanics Saturday, 20 May 2006 Introduction to Unit 1

Advanced Higher Physics

Unit 1

MechanicsSaturday, 20 May 2006Saturday, 20 May 2006

Introduction to Unit 1Introduction to Unit 1

Thursday 20 April 2023Thursday 20 April 2023

Mechanics Unit 1 SummaryMechanics Unit 1 Summary► Linear KinematicsLinear Kinematics

1D & 2D kinematics1D & 2D kinematics► Relativistic MotionRelativistic Motion

Relativistic MassRelativistic Mass Relativistic EnergyRelativistic Energy Other Relativistic EffectsOther Relativistic Effects

► Angular KinematicsAngular Kinematics Angular NotationAngular Notation Kinematic RelationshipsKinematic Relationships Tangential SpeedTangential Speed

► Rotational DynamicsRotational Dynamics Centripetal acceleration / Centripetal acceleration /

forceforce TorqueTorque Newtons LawsNewtons Laws Angular MomentumAngular Momentum Angular Kinetic EnergyAngular Kinetic Energy

► GravityGravity Newtons Law of GravitationNewtons Law of Gravitation Gravitational force & fieldsGravitational force & fields Gravitational Potential & Gravitational Potential &

EnergyEnergy Orbital MechanicsOrbital Mechanics Escape VelocityEscape Velocity

► Simple Harmonic MotionSimple Harmonic Motion Analysis of SHMAnalysis of SHM Energy in SHMEnergy in SHM DampingDamping

► Wave particle dualityWave particle duality TheoryTheory Examples of DualityExamples of Duality

► Quantum MechanicsQuantum Mechanics Atomic ModelsAtomic Models Quantum Mechanics TheoryQuantum Mechanics Theory


1D Mechanics1D MechanicsAt higher we dealt exclusively with situations where the At higher we dealt exclusively with situations where the

acceleration of an object was constant. This is not always the acceleration of an object was constant. This is not always the case. One common example is in friction where acceleration case. One common example is in friction where acceleration is dependant on velocity.is dependant on velocity.

If acceleration is a function of time (eg a=0.4t), then the If acceleration is a function of time (eg a=0.4t), then the standard equations of motion DO NOT APPLY. Instead we must standard equations of motion DO NOT APPLY. Instead we must use a calculus method.use a calculus method.

In general (and by definition)In general (and by definition)

dt

dva dtav

dt

dsv dtvs


1 D Mechanics1 D Mechanics► Worked ExampleWorked Example

A 4kg object, initially at rest is subject to a force A 4kg object, initially at rest is subject to a force which varies with time according to the which varies with time according to the equation F=0.80t.equation F=0.80t.

1.1. Calculate an expression for the acceleration of this Calculate an expression for the acceleration of this objectobject

2.2. Now, by calculus find expressions for v and s in terms Now, by calculus find expressions for v and s in terms of timeof time

3.3. Find the displacement of the object after 12 secondsFind the displacement of the object after 12 seconds

4.4. Find the velocity of the object after 18 secondsFind the velocity of the object after 18 seconds


Do you get it – Try this.Do you get it – Try this.

A 14kg object is initially moving at 6msA 14kg object is initially moving at 6ms-1-1. From this . From this moment it is subject to a force in Newtons according moment it is subject to a force in Newtons according to the expression F=18-0.4t.to the expression F=18-0.4t.

1.1. Formulate expressions for the acceleration, velocity and Formulate expressions for the acceleration, velocity and displacement of the object.displacement of the object.

2.2. At what time is there no force on the object and what is its At what time is there no force on the object and what is its velocity at that time?velocity at that time?

3.3. How long after the start of the experiment does the object How long after the start of the experiment does the object come to rest?come to rest?

4.4. How long after the start of the experiment does the object How long after the start of the experiment does the object return to its initial position and what is its velocity at this return to its initial position and what is its velocity at this time?time?


Here’s why its usefulHere’s why its useful

In an experiment we tracked the position of an object In an experiment we tracked the position of an object in one dimension and plotted the data in the graph in one dimension and plotted the data in the graph below.below.

0

10

20

30

40

50

60

0 5 10 15 20t / s

d / mUse your maths Use your maths skills to get the skills to get the equation of this equation of this graph, and use graph, and use it to form an it to form an equation for the equation for the acceleration of acceleration of the object. If we the object. If we know the mass know the mass of the object of the object 1.35kg, we can 1.35kg, we can calculate the calculate the force on the force on the objectobject


Rigorous proof of higher formulaeRigorous proof of higher formulae

Using your knowledge of calculus:Using your knowledge of calculus:

Considering an object with initial speed ‘u’ at Considering an object with initial speed ‘u’ at t=0 and a constant acceleration of ‘a’, and t=0 and a constant acceleration of ‘a’, and knowing that a=dv/dt, prove that a=(v-u)/t.knowing that a=dv/dt, prove that a=(v-u)/t.

Now considering the same object and knowing Now considering the same object and knowing that v=ds/dt, prove that s=ut+½atthat v=ds/dt, prove that s=ut+½at22

For both proofs use either For both proofs use either ► definite integrals, definite integrals, ► or indefinite integrals and evaluate the or indefinite integrals and evaluate the

constant of integration.constant of integration.


Unit 1


RelativityRelativity


A challenge to start withA challenge to start with

►How fast are we moving???How fast are we moving???►DataData

Diameter of earth = 12000 kmDiameter of earth = 12000 km

Radius of our orbit of Sun = 940 million Radius of our orbit of Sun = 940 million kilometreskilometres

Assume you are standing on the equator and Assume you are standing on the equator and that the Earth and the orbital path are that the Earth and the orbital path are perfectly circular.perfectly circular.

► Velocity due to earth rotation = 436 m/sVelocity due to earth rotation = 436 m/s► Velocity due to solar orbit = 187,284 m/sVelocity due to solar orbit = 187,284 m/s


Relativistic massRelativistic massAlthough the concept of relativistic mass is odd, it is quite Although the concept of relativistic mass is odd, it is quite

simple. As the velocity of an object relative to an simple. As the velocity of an object relative to an observer increases, its mass will also increase. These observer increases, its mass will also increase. These effects are very small until we start to go faster than 0.1c effects are very small until we start to go faster than 0.1c (10% the speed of light).(10% the speed of light).

The relativistic mass of an object with rest mass The relativistic mass of an object with rest mass mmoo moving moving at a speed at a speed vv relative to a stationary observer is given by: relative to a stationary observer is given by:

2

21

cv

mm o


Relativistic Mass QuestionsRelativistic Mass Questions► Find the relativistic mass of each of the Find the relativistic mass of each of the

followingfollowing

1.1. An electron moving at 2.3x10An electron moving at 2.3x1088 ms ms-1-1

2.2. A neutron moving at 1.3x10A neutron moving at 1.3x1088 ms ms-1-1

3.3. A 700kg space probe moving at 8x10A 700kg space probe moving at 8x1077 ms ms-1-1

4.4. A 3kg meteorite accelerated to 9x10A 3kg meteorite accelerated to 9x1066 ms ms-1-1

5.5. If an object has a rest mass of 6kg, but a If an object has a rest mass of 6kg, but a relativistic mass of 6.7kg, calculate its velocityrelativistic mass of 6.7kg, calculate its velocity

6.6. An object has a rest mass of 24x10An object has a rest mass of 24x10-5-5 kg. Its kg. Its relativistic mass is 12% greater. Calculate its relativistic mass is 12% greater. Calculate its velocity.velocity.


Change in mass as velocity Change in mass as velocity changeschanges

► Using your knowledge of relativistic mass, plot Using your knowledge of relativistic mass, plot a graph of how the mass of an object changes a graph of how the mass of an object changes as its velocity increases from rest to 97%, if as its velocity increases from rest to 97%, if the object has a rest mass of 1kg.the object has a rest mass of 1kg.

► (calculate m at 10% intervals, (calculate m at 10% intervals, and then 95% and 97%)and then 95% and 97%)


Relativistic EnergyRelativistic EnergyThe most important concept in relativity is that the mass of an The most important concept in relativity is that the mass of an

object is a measure of its total energy. The Newtonian concept object is a measure of its total energy. The Newtonian concept of kinetic energy (½mvof kinetic energy (½mv22) begins to fail at relativistic velocities ) begins to fail at relativistic velocities (>0.1c) as the mass is no longer constant as the object was (>0.1c) as the mass is no longer constant as the object was accelerated. Instead we use Einstein’s equation as follows:accelerated. Instead we use Einstein’s equation as follows:

The The Rest EnergyRest Energy of an object is calculated using its rest mass of an object is calculated using its rest mass (E(Eoo=m=moocc22), while the ), while the Relativistic EnergyRelativistic Energy is calculated using the is calculated using the relativistic mass (E=mcrelativistic mass (E=mc22). The kinetic energy of the object is the ). The kinetic energy of the object is the difference between its relativistic energy and rest energy.difference between its relativistic energy and rest energy.

E = EE = Ek k + E+ Eoo

EEkk = E – E = E – Eoo

EEkk = mc = mc22 - m - moocc22


Relativistic EnergyRelativistic Energy

► The graph below shows how the Newtonian The graph below shows how the Newtonian and Relativistic definitions of kinetic energy and Relativistic definitions of kinetic energy differ and how the difference between the differ and how the difference between the theories increases dramatically as the speed theories increases dramatically as the speed of light is approached.of light is approached.

Use Excel to plot a graph of the kinetic energy of a 1kg object as its velocity isIncreased from rest to 0.85c.

Plot both the kinetic energy calculated by Newtonian theory and the relativistic theory

and note at what velocityNewtonian mechanics begins to break down

Plot another graph showing thecorrelation between Newtonian

mechanics and relativistic mechanics at velocities <0.1c

Presentation is important here, so please play about with scales, etc to get

the graph looking appropriate.


Relativistic Energy ProblemsRelativistic Energy ProblemsAccelerating particles.Accelerating particles.Particle accelerators are used to accelerate protons, Particle accelerators are used to accelerate protons,

electrons and even small atomic nuclei to very high electrons and even small atomic nuclei to very high velocities for experimental work. However, as the velocities for experimental work. However, as the velocities reached are relativistic (>0.1c), the effects velocities reached are relativistic (>0.1c), the effects of relativity must be accounted for.of relativity must be accounted for.

We can assume that the charge of a particle does not We can assume that the charge of a particle does not change as it accelerates, not does the voltage of the change as it accelerates, not does the voltage of the field accelerating it.field accelerating it.

Equating the work done on the charge to the relativistic Equating the work done on the charge to the relativistic kinetic energy, we can calculate the velocity of the kinetic energy, we can calculate the velocity of the particle.particle.

► Combine the formulae for work done on a charge in Combine the formulae for work done on a charge in an electric field with the formula for relativistic kinetic an electric field with the formula for relativistic kinetic energy and rearrange this formula for voltage.energy and rearrange this formula for voltage.

► Also rearrange the formula for velocity (quite tricky).Also rearrange the formula for velocity (quite tricky).


Now you can work these out….Now you can work these out….1.1. Calculate the voltage of field required to Calculate the voltage of field required to

accelerate an accelerate an electronelectron to 8x10 to 8x107 7 msms-1-1. (19.2kV). (19.2kV)2.2. Calculate the velocity a Calculate the velocity a protonproton would reach in a would reach in a

field of 30kV. (2.40x10field of 30kV. (2.40x1066 ms ms-1-1))3.3. Calculate the velocity an Calculate the velocity an alpha particlealpha particle would would

reach in the same field. (1.69x10reach in the same field. (1.69x1066 ms ms-1-1) ) 4.4. Calculate the voltage of field required to Calculate the voltage of field required to

accelerate an accelerate an alpha particlealpha particle to 0.8c. (1.26x10 to 0.8c. (1.26x1099 V)V)

5.5. Calculate the voltage required to accelerate a Calculate the voltage required to accelerate a deuterium (deuterium (22

11H) nucleusH) nucleus to the same speed. to the same speed.6.6. From your knowledge of work done, calculate the From your knowledge of work done, calculate the

average force on the average force on the deuterium nucleusdeuterium nucleus in Q5 if it in Q5 if it accelerates over a distance of 800m. (2.58x10accelerates over a distance of 800m. (2.58x10--

1313N)N)


Other Relativistic Effects Other Relativistic Effects ► Time DilationTime DilationIn AH we do not need to analyse this effect In AH we do not need to analyse this effect

mathematically as it is quite complex and has effects mathematically as it is quite complex and has effects which are difficult to understand. One of the side which are difficult to understand. One of the side effects of relativity is that when two objects are in effects of relativity is that when two objects are in motion relative to each other, time does not run at a motion relative to each other, time does not run at a constant rate (it runs faster for the “stationary” constant rate (it runs faster for the “stationary” observer.observer.

If we set two clocks to run at identical times, then place If we set two clocks to run at identical times, then place one on a ship at a velocity close to the speed of light one on a ship at a velocity close to the speed of light and fly it in a circle, the clock which has been moving and fly it in a circle, the clock which has been moving will run slower than the stationary one.will run slower than the stationary one.

Essentially, as velocity increases, time slows down.Essentially, as velocity increases, time slows down.

► Length contractionLength contractionIn addition to this, at relativistic velocities, the length of In addition to this, at relativistic velocities, the length of

an object in the direction of travel will decrease as an object in the direction of travel will decrease as the object approaches the speed of light.the object approaches the speed of light.


Unit 1


Post Summer Brain Warm-Post Summer Brain Warm-upup


Mini testy type thingamajig….Mini testy type thingamajig….1.1. Write down the correct units for each of the followingWrite down the correct units for each of the following

(a) Momentum (a) Momentum (b) Charge (b) Charge (c) Pressure (c) Pressure (d) (d) AccelerationAcceleration(e) Capacitance(e) Capacitance (f) Weight(f) Weight (g) Relativistic Mass(g) Relativistic Mass

2.2. From the expression a = dv/dt , derive the equation of motion From the expression a = dv/dt , derive the equation of motion v = u + atv = u + at

3.3. Bobby (a 67kg ice skater) was skating along at 14msBobby (a 67kg ice skater) was skating along at 14ms -1-1, minding , minding his own business, when he collided with a stationary 39kg his own business, when he collided with a stationary 39kg kangaroo which was lost. The pair become entangled. What kangaroo which was lost. The pair become entangled. What was their velocity just after the collision?was their velocity just after the collision?

4.4. A small elephant (137kg) is accelerated in space to a velocity A small elephant (137kg) is accelerated in space to a velocity 1.5 times greater than “the velocity of light in a material with a 1.5 times greater than “the velocity of light in a material with a refractive index of 1.65”. Calculate the relativistic mass of the refractive index of 1.65”. Calculate the relativistic mass of the elephant at this velocity.elephant at this velocity.

5.5. If Mr Colquhoun had a rest mass of 87kg before the holidays, If Mr Colquhoun had a rest mass of 87kg before the holidays, but lost 7kg of mass between then and now, to what speed but lost 7kg of mass between then and now, to what speed would he have to be accelerated such that his relativistic mass would he have to be accelerated such that his relativistic mass equalled his rest mass before summer?equalled his rest mass before summer?


Mini testy type thingymajig….Mini testy type thingymajig….1.1. Write down the correct units for each of the followingWrite down the correct units for each of the following

(a) kgms(a) kgms-1-1 or Ns or Ns (b) C (b) C (c) Pa or Nm(c) Pa or Nm-2-2 (d) ms(d) ms--

22

(e) F(e) F (f) N(f) N (g) kg(g) kg

2.2. DerivationDerivation

3.3. 8.85 ms8.85 ms-1-1

4.4. 329 kg329 kg

5.5. 1.18 x 101.18 x 1088 ms ms-1-1

atuv

uvat

vat

dvdta

dt

dva

vu

t

0

0

v

u

t

0


Some Quick Revision…post Some Quick Revision…post summersummer

1.1. The motion of an object is the be analysed and a The motion of an object is the be analysed and a student measures its displacement. If its motion is student measures its displacement. If its motion is described by the equation described by the equation s=0.3ts=0.3t33-0.6t-0.6t22+1.3t+1.3t, , calculate:calculate:(a) Its velocity after 5 seconds (17.8m/s)(a) Its velocity after 5 seconds (17.8m/s)(b) Its acceleration after 8 seconds (13.2 m/s(b) Its acceleration after 8 seconds (13.2 m/s22))

2.2. Find the relativistic mass of a 1.6 tonne African Find the relativistic mass of a 1.6 tonne African elephant moving with a velocity of 1.3x10elephant moving with a velocity of 1.3x108 8 msms-1-1..

3.3. By calculating the difference between the rest By calculating the difference between the rest energy and the relativistic energy, determine the energy and the relativistic energy, determine the kinetic energy of an 87kg human being moving with kinetic energy of an 87kg human being moving with a velocity of 60,000 kmsa velocity of 60,000 kms-1-1..


Unit 1


Rotational MotionRotational Motion


Angular TerminologyAngular Terminology

Rotational motion works in much the same way as Rotational motion works in much the same way as motion in a straight line if we analyse it in the motion in a straight line if we analyse it in the correct way. The following symbols will be used correct way. The following symbols will be used throughout this section.throughout this section.

θθ - Angular displacement- Angular displacement - rad- rad

ωω00 - Initial angular velocity- Initial angular velocity - rad s- rad s-1-1

ωω - Final angular velocity- Final angular velocity - rad s- rad s-1-1

- Angular acceleration- Angular acceleration - rad s- rad s-2-2

Terminology aside, the basic equations of motion Terminology aside, the basic equations of motion for angular kinematics can be derived in exactly for angular kinematics can be derived in exactly the same way as they were for linear the same way as they were for linear kinematics.kinematics.


Deriving the angular equations of Deriving the angular equations of motionmotion

► Starting from Starting from

► Derive the angular equations of motion for an object Derive the angular equations of motion for an object accelerating from ‘accelerating from ‘ωω00’ to ‘’ to ‘ωω’ with a constant angular ’ with a constant angular acceleration ‘acceleration ‘’ in time ‘t’. In performing this ’ in time ‘t’. In performing this motion, the object passes through angular motion, the object passes through angular displacement displacement θθ..

dt

d


Working with angular velocitiesWorking with angular velocities

When using angular velocities in calculations, When using angular velocities in calculations, they must always be measured in rad sthey must always be measured in rad s-1-1. . Figures for angular velocity are often given Figures for angular velocity are often given in different ways. Convert each of the in different ways. Convert each of the following to radians per second.following to radians per second.

1.1. 3000 rpm3000 rpm

2.2. 24 rpm24 rpm

3.3. 1 revolution per day1 revolution per day

4.4. 1 revolution in 28 days1 revolution in 28 days

5.5. 50 revolutions per second50 revolutions per second

1 . 314 rad/s2. 2.51 rad/s3.7.3x10-5 rad/s4.2.6x10-6 rad/s5.314 rad/s


Centripetal acceleration Centripetal acceleration

In order for an object to move in anything other In order for an object to move in anything other than a straight line, it must accelerate. In the than a straight line, it must accelerate. In the case of pure circular motion, this acceleration case of pure circular motion, this acceleration is always towards the centre of the circle. is always towards the centre of the circle. This is called centripetal acceleration and is This is called centripetal acceleration and is entirely different from angular acceleration. entirely different from angular acceleration.

Two formulae for centripetal acceleration can Two formulae for centripetal acceleration can be derived from your knowledge of circular be derived from your knowledge of circular geometry and vector addition and geometry and vector addition and subtraction.subtraction.

This derivation can be found in your SCHOLAR This derivation can be found in your SCHOLAR textbooks.textbooks.


Centripetal acceleration Centripetal acceleration We consider a point moving in a circle with radius We consider a point moving in a circle with radius rr and with and with

constant angular velocity constant angular velocity ωω moving though an angle moving though an angle ΔθΔθ from from point A to point B as shown here.point A to point B as shown here.

vb

vaA

B

Δθr

ω

In order to find the centripetal In order to find the centripetal acceleration we must acceleration we must determine the change in determine the change in velocity of the point in velocity of the point in travelling from A to B. Its travelling from A to B. Its angular velocity will not angular velocity will not have changed, nor will the have changed, nor will the magnitude of its linear magnitude of its linear velocity, but the direction velocity, but the direction of the tangential velocity of the tangential velocity has changed. This change has changed. This change in velocity can be in velocity can be expressed asexpressed as

ΔΔvv == vvbb -- vvaa



vb

vaA

B

Δθr

ω

This is a vector sum, and is carried out as followsThis is a vector sum, and is carried out as followsΔΔv = vv = vbb - v - vaa

-va

vb

Δv

Now, by some clever maths, we Now, by some clever maths, we can show that as we make can show that as we make ΔθΔθ very small, the change in very small, the change in velocity becomes velocity becomes perpendicular to vperpendicular to vaa. In other . In other words the acceleration of the words the acceleration of the point is towards the centre of point is towards the centre of the circle.the circle.

Centripetal acceleration is Centripetal acceleration is written aswritten as

a



vb

-va

Δv

With this right angled triangle we can say that With this right angled triangle we can say that ΔΔv = v v = v ΔθΔθ , , which is true for small angles measured in radians (which is true for small angles measured in radians (θθ = tan = tan θθ for for small angles)small angles)

Δθ

r

vva

vadt

dva

t

va

t

va

2

We can define the centripetal We can define the centripetal acceleration asacceleration as

Then substitute in for Then substitute in for ΔΔv = v v = v ΔθΔθ

Then in the limit as Then in the limit as ΔΔt tends to zero we t tends to zero we get the differential get the differential

And finally since v=rAnd finally since v=rωω, , ωω=v/r giving =v/r giving


Centripetal ForceCentripetal Force► Firstly, a myth to dispel. Firstly, a myth to dispel. ► You are in a car driving round a circular track You are in a car driving round a circular track

anticlockwise. There is a force acting on you, but in anticlockwise. There is a force acting on you, but in which direction does it act?which direction does it act? Forward?Forward? Backward?Backward? Left?Left? Right?Right?

► Centrifugal force does not exist, it is a reaction force Centrifugal force does not exist, it is a reaction force experienced as a result of centripetal force which acts experienced as a result of centripetal force which acts towards the centre of the circle.towards the centre of the circle.

► In the same way that when standing on the floor you In the same way that when standing on the floor you are pushed up by the floor, when travelling in a circle are pushed up by the floor, when travelling in a circle you feel like you are being pushed out from the you feel like you are being pushed out from the centre – this is not the case!!centre – this is not the case!!



Centripetal ForceCentripetal Force► The basic concept of centripetal force is very simple to The basic concept of centripetal force is very simple to

understand. We have proved the expression for understand. We have proved the expression for centripetal acceleration.centripetal acceleration.

► If we consider the case of a point mass following a If we consider the case of a point mass following a circular path, a force must exist to give it this circular path, a force must exist to give it this centripetal acceleration. This force must act towards centripetal acceleration. This force must act towards the centre of the circle and must obey Newton’s 2the centre of the circle and must obey Newton’s 2ndnd law.law.

r

mvmrF

maF2

2


Applications of centripetal forceApplications of centripetal force

► Confirming the formula experimentallyConfirming the formula experimentally

►Motion in a horizontal circleMotion in a horizontal circle

►Motion in a vertical circleMotion in a vertical circle

► Conical pendulumConical pendulum

► Cars corneringCars cornering

► Banked cornersBanked corners


Confirming the formulaConfirming the formulaMass experiencing centripetal force

Mass used to counter centripetal force

In this experiment we slowly increase the angular velocity of the turntable until the centripetal force on m1 exceeds the weight of m2. When this occurs m1 will move. When m1 moves, maintain the rotational velocity and measure it (time a number of rotations). Since the mass and initial radius of m1 , the weight of m2 and the angular velocity are known, the formula can be checked.

Perform the experiment twice at each of two radial positions. (4 times in total)

m1

m2

2mrF


Motion in a Horizontal CircleMotion in a Horizontal Circle► Try the following questionsTry the following questions

1.1. A young girl of mass 37kg is on a roundabout at a A young girl of mass 37kg is on a roundabout at a radius of 1.6m. Her angular velocity is constant at 2.4 radius of 1.6m. Her angular velocity is constant at 2.4 rad srad s-1-1. Calculate her tangential velocity, her . Calculate her tangential velocity, her centripetal acceleration and the centripetal force centripetal acceleration and the centripetal force acting on her.acting on her.

2.2. In a high g simulator a 300kg pod is rotated at high In a high g simulator a 300kg pod is rotated at high speed around a central pivot. The radius of the arc is speed around a central pivot. The radius of the arc is 6m. If the tension force safety limit of the arm 6m. If the tension force safety limit of the arm connecting the pod to the pivot is 20kN, Calculate the connecting the pod to the pivot is 20kN, Calculate the maximum angular velocity at which the simulator can maximum angular velocity at which the simulator can safely be used. Calculate the corresponding tangential safely be used. Calculate the corresponding tangential velocity at maximum safe limits.velocity at maximum safe limits.

3.3. A 850kg car drives round a corner at 40.0 km hA 850kg car drives round a corner at 40.0 km h-1-1 . If the . If the

corner has a radius of 8.00m, calculate the centripetal corner has a radius of 8.00m, calculate the centripetal force required to negotiate the corner safely.force required to negotiate the corner safely.


Motion in a Vertical CircleMotion in a Vertical Circle► Consider an object being whirled vertically Consider an object being whirled vertically

on a piece of string. The forces acting on it on a piece of string. The forces acting on it will change depending on the position of the will change depending on the position of the object as shown below.object as shown below.

mg

mg

mg

mgT

T

TT

a⊥

a⊥

a⊥

a⊥

In each position, we can state In each position, we can state that the sum of the forces that the sum of the forces (T and mg) must equal the (T and mg) must equal the centripetal force (mrcentripetal force (mrωω22).).

Form an equation for the Form an equation for the tension in the string when tension in the string when the mass m is atthe mass m is at(a) the top(a) the top(b) the bottom(b) the bottom(c) the horizontal position(c) the horizontal position


Conical PendulumConical PendulumThe period of a conical pendulum, similarly to a The period of a conical pendulum, similarly to a

standard pendulum, can be proven to be independent standard pendulum, can be proven to be independent of the mass on the end as long as the connection (eg of the mass on the end as long as the connection (eg string) is of negligible mass.string) is of negligible mass.

mg

T

a⊥

ΦΦ

l

ω

mg

T cos Φ

a⊥

T sin Φ

Use the above free body diagrams Use the above free body diagrams and with the aid of your scholar and with the aid of your scholar books derive expressions forbooks derive expressions for(a) the tension in the string(a) the tension in the string(b) the angular velocity of the bob(b) the angular velocity of the bob(c) the period of the bob(c) the period of the bob


Cars cornering on the flatCars cornering on the flat

We can assume that a car, or other object driving round a We can assume that a car, or other object driving round a corner is travelling in a horizontal circle with a constant corner is travelling in a horizontal circle with a constant angular velocity. As a result we know the only angular velocity. As a result we know the only unbalanced force on the car is the centripetal force. unbalanced force on the car is the centripetal force.

On a flat road, this force must be provided by the friction On a flat road, this force must be provided by the friction of the tyres. The weight of the car is exactly balanced of the tyres. The weight of the car is exactly balanced by the reaction force from the road.by the reaction force from the road.

This means that at a given velocity, there is a limit to how This means that at a given velocity, there is a limit to how tight a corner a car can go round before skidding.tight a corner a car can go round before skidding.

Reaction Force

Friction from tyres

Weight

ExampleExample

A 700 kg car travelling at 50kmhA 700 kg car travelling at 50kmh-1-1 is fitted with tyres providing a is fitted with tyres providing a maximum frictional force of maximum frictional force of 4500N. What is the smallest 4500N. What is the smallest radius of corner it can go round radius of corner it can go round at this velocity?at this velocity?


Banked cornersBanked cornersOn a banked corner, similar to a conical pendulum, a On a banked corner, similar to a conical pendulum, a

component of the reaction force from the road component of the reaction force from the road provides part of the centripetal force required.provides part of the centripetal force required.

The car is still assumed to be travelling in a horizontal The car is still assumed to be travelling in a horizontal circle, so the weight and the vertical component of circle, so the weight and the vertical component of the reaction force must be balanced.the reaction force must be balanced.

Reaction Force

Weight

ExampleExample

For a 600kg car, calculate the For a 600kg car, calculate the angle of banking required such angle of banking required such that a car travelling at 50 kmhthat a car travelling at 50 kmh-1-1 need produce no friction from need produce no friction from the tyres (ie, horizontal the tyres (ie, horizontal component of the reaction component of the reaction force equals the centripetal force equals the centripetal force required), for a corner force required), for a corner with a radius of 36m.with a radius of 36m.

Φ


Unit 1


Rotational DynamicsRotational DynamicsTorque & MomentTorque & Moment


Torque & MomentTorque & MomentLike the equations of motion, there are commonalities Like the equations of motion, there are commonalities

between linear and angular dynamics (the study of between linear and angular dynamics (the study of forces and motion).forces and motion).

Torque and moment are in essence the same thing. They Torque and moment are in essence the same thing. They are a force acting on a mass which causes it to rotate or are a force acting on a mass which causes it to rotate or turn. Technically, a moment is defined as:turn. Technically, a moment is defined as:

The The tangential componenttangential component of a force multiplied by its of a force multiplied by its radiusradius from the point of rotation. It is measured in Nm. from the point of rotation. It is measured in Nm.

If the force is If the force is perpendicular to the perpendicular to the radiusradius

F

T = Frr

If the force is not If the force is not perpendicular to the perpendicular to the radiusradius

F

T = Fr sin Φ

r

Φ

F sin Φ

F cos Φ


Balancing torquesBalancing torquesBalanced torques work in exactly the same way as balanced forces. For an Balanced torques work in exactly the same way as balanced forces. For an

object which is stationary or rotating, the torques must be balanced.object which is stationary or rotating, the torques must be balanced.

For each of the following, draw out the diagram, and find the value For each of the following, draw out the diagram, and find the value of the missing force or radius.of the missing force or radius.

6N F

0.3m0.6m

3N F

0.9m1.5m

96N

160N

r7m

20N

F

5m4m

90N

3m

85N

40Nr

0.9m

12N

F1.9m

1.3m

60°


Basic Structural MechanicsBasic Structural Mechanics

This principle can be applied to any object or This principle can be applied to any object or structure, or section of a structurestructure, or section of a structure

ΦMass=200kg

Mass=300kg

3m5m

Φ

3m

2.5mmg

mg


Torque & AccelerationTorque & Acceleration

The aim of this experiment is to prove the linear The aim of this experiment is to prove the linear relationship between torque and angular relationship between torque and angular acceleration.acceleration.

Mass used to create force

m2

r


Torque & Angular AccelerationTorque & Angular AccelerationTo derive the angular equivalent of Newton’s 2To derive the angular equivalent of Newton’s 2ndnd Law Law

(F=ma) we analyse a point mass ‘m’ moving in a (F=ma) we analyse a point mass ‘m’ moving in a circle with an angular acceleration ‘circle with an angular acceleration ‘’. We already ’. We already know that:know that:

F = maF = ma T = FrT = Fr a = ra = r

And all of these laws must hold for our object, so:And all of these laws must hold for our object, so:F=maF=maF=m r F=m r T/r = m r T/r = m r T = mrT = mr2 2 The quantity ‘mr2’ is a property of the object which is rotating and is called the moment of inertia.

For a rotating point mass the moment of inertia is given by: I = mr2

T = I

I is the moment of inertia measured in kgm2


Finding Compound Moments of Finding Compound Moments of InertiaInertia

► When calculating the moment When calculating the moment of inertia of an object, the of inertia of an object, the back of the data book back of the data book provides the following provides the following formulae for different shapes formulae for different shapes of object.of object.

► Note that it is their Note that it is their distribution of mass around distribution of mass around the axis of rotation that the axis of rotation that matters, so a doughnut shape matters, so a doughnut shape uses the same formula as a uses the same formula as a point mass as all the mass is point mass as all the mass is at a single radius.at a single radius.

► We can also add or subtract We can also add or subtract shapes from each other to shapes from each other to allow us to analyse more allow us to analyse more complex shapes.complex shapes.

2

2

2

2

2

52I centreabout Sphere

21I centreabout Disc

31I endabout Rod

121I centreabout Rod

mrI massPoint

mr

mr

ml

ml


Finding Compound Moments of Finding Compound Moments of InertiaInertia

► Find the moment of inertia for each shapeFind the moment of inertia for each shape

0.28m

1.9kg

3.9kgSolid sphere

0.36m

(a)(b)

(c) Metre stick about 1 end(d) A bouncy ball (rolling)(e) Plastic pipe

i. Around 1 endii. Around centreiii. Around lengthways central axis

0.83m

5.2kg

0.08m


Angular MomentumAngular MomentumAgain, angular momentum works exactly the same as Again, angular momentum works exactly the same as

linear momentum in that momentum is always linear momentum in that momentum is always conserved. This gives rise to the formula:conserved. This gives rise to the formula:

The same principles apply to the angular system in that The same principles apply to the angular system in that momentum is always conserved, and that in an momentum is always conserved, and that in an elastic collision, the rotational kinetic energy is also elastic collision, the rotational kinetic energy is also conserved. conserved.

The only additional consideration is that changes in The only additional consideration is that changes in angular velocity can be brought about by changes in angular velocity can be brought about by changes in moment of inertia.moment of inertia.

IL Angular Momentum (kgm2s-1)

Moment of Inertia (kg m2)

Angular velocity (rad s-1)

221 IE

rotk


Try theseTry these

1.1. Calculate the angular momentum of Calculate the angular momentum of a)a) A 12kg disc of radius 0.85m rotating at 6.5rad sA 12kg disc of radius 0.85m rotating at 6.5rad s-1-1

b)b) A 2.6m, 4.3kg rod rotating about the middle at A 2.6m, 4.3kg rod rotating about the middle at 8.2rad s8.2rad s-1-1

2.2. A ballet dancer is trying to achieve a faster A ballet dancer is trying to achieve a faster spinning speed. She starts spinning with her spinning speed. She starts spinning with her arms out, giving her an angular velocity of arms out, giving her an angular velocity of 12.3 rad s12.3 rad s-1-1. In this position her moment of . In this position her moment of inertia 15kgminertia 15kgm22. She then draws her arms in . She then draws her arms in which reduces her moment of inertia by 20%. which reduces her moment of inertia by 20%. Calculate her new angular velocity. Calculate Calculate her new angular velocity. Calculate whether her kinetic energy has increased, whether her kinetic energy has increased, decreased or stayed the same.decreased or stayed the same.


Outcome 3 – Measuring IOutcome 3 – Measuring IUsing the air table fitted with a central pulley, determine Using the air table fitted with a central pulley, determine

the moment of inertia of the disc and any associated the moment of inertia of the disc and any associated masses.masses.

Determine the torque applied to the disc and its angular Determine the torque applied to the disc and its angular acceleration. Repeat the experiment for a number of acceleration. Repeat the experiment for a number of different applied torques. Use this data to determine the different applied torques. Use this data to determine the moment of inertia of the disc by graphical method.moment of inertia of the disc by graphical method.

It is recommended that the measurement of time and It is recommended that the measurement of time and angular displacement is used to determine the angular angular displacement is used to determine the angular acceleration of the disc, rather than an attempt to acceleration of the disc, rather than an attempt to measure the instantaneous velocity of the disc.measure the instantaneous velocity of the disc.

A full report of the experiment should be produced using A full report of the experiment should be produced using the guidelines including an analysis of errors and a full the guidelines including an analysis of errors and a full evaluation of the experiment.evaluation of the experiment.


Unit 1


GravitationGravitationThe Laws of AttractionThe Laws of Attraction


Newtons Law of GravitationNewtons Law of GravitationWe are used to thinking of gravity as the source of weight We are used to thinking of gravity as the source of weight

always acting downwards. In fact there is a force of always acting downwards. In fact there is a force of gravitational attraction between any two objects in the gravitational attraction between any two objects in the universe. Usually, for these forces to be significant, the universe. Usually, for these forces to be significant, the masses must be large and the distances reasonable.masses must be large and the distances reasonable.

The force of attraction (which acts on both objects) is The force of attraction (which acts on both objects) is given by the formula:given by the formula:

221

r

mGmF

Where:

F = Force of attraction (N)

m1= mass of object 1 (kg)

m2= mass of object 2 (kg)

r = the distance between the objects (m)

G = Gravitational constant = 6.67x10-11 N m2 kg-2


Interesting ideasInteresting ideas

Zero gravity pointsZero gravity pointsIt stands to reason that there is some location It stands to reason that there is some location between the moon and earth where the moon’s between the moon and earth where the moon’s gravity cancels out the earths leading to a point of gravity cancels out the earths leading to a point of “zero gravity”.“zero gravity”.

Find this point between the earth and the moonFind this point between the earth and the moon


Orbital MechanicsOrbital MechanicsBy equating the gravitational force to the formula for By equating the gravitational force to the formula for

centripetal force we can perform detailed analysis on centripetal force we can perform detailed analysis on objects in circular orbits of planets or moons.objects in circular orbits of planets or moons.

The orbital period of an object The orbital period of an object is only dependant on the mass of is only dependant on the mass of the planet / star around which it is the planet / star around which it is orbiting, and the radius of the orbit.orbiting, and the radius of the orbit.

For example, using the following data, calculate the For example, using the following data, calculate the radius and therefore, the altitude of the orbit of the radius and therefore, the altitude of the orbit of the moon above the earth (HINT: first calculate moon above the earth (HINT: first calculate ωω from from orbital period)orbital period)::Mass of EarthMass of Earth = 6.0x10= 6.0x102424kgkgMass of MoonMass of Moon = 7.3x10= 7.3x102222kgkgRadius of EarthRadius of Earth = 6.4x10= 6.4x1066mmLunar Orbital PeriodLunar Orbital Period = 28 days= 28 days

232

212

21

rGm

rmr

mGmF

Answer = 3.84x108m


Orbital MechanicsOrbital Mechanics

When an object is in a circular orbit around a planet, the When an object is in a circular orbit around a planet, the only force acting on it is the gravitational pull of the only force acting on it is the gravitational pull of the planet. This force must therefore be equal to the planet. This force must therefore be equal to the centripetal force on the object.centripetal force on the object.

By equating these two formulae, derive expressions for:By equating these two formulae, derive expressions for:► The tangential velocity of the satellite at a known The tangential velocity of the satellite at a known

orbital radius around a given planet.orbital radius around a given planet.► The period of a satellite given the radius of orbit The period of a satellite given the radius of orbit

around a known planetaround a known planet► The required radius of orbit to achieve a given period The required radius of orbit to achieve a given period

of rotation.of rotation.

► Use the last formula to calculate the radius of orbit and Use the last formula to calculate the radius of orbit and therefore the altitude above earth of a geostationary therefore the altitude above earth of a geostationary satellitesatellite


Design a solar SystemDesign a solar System► You have been tasked with You have been tasked with

the design of a single planet the design of a single planet within a solar system, based within a solar system, based around a star slightly larger around a star slightly larger than our sun. The star has than our sun. The star has a mass of 2.5x10a mass of 2.5x103030kg and a kg and a radius of 7.1x10radius of 7.1x1088m.m.

► You must form a “Fact You must form a “Fact sheet” for your planet in the sheet” for your planet in the following format:following format:

► You should also supply a You should also supply a calculations sheet stating calculations sheet stating what assumptions you what assumptions you made and showing all your made and showing all your calculations.calculations.

► Fact sheet should be a Fact sheet should be a single A4 sheet.single A4 sheet.

Planet Name

Radius of planet

Mass of Planet

Radius of orbit

Length of 1 “year”

Length of one “day”

Gravity at Surface

Height of Geostationary Orbit

Moons (and associated geometrical and orbital details)

Any other information


Solar System Design TipsSolar System Design Tips► All the calculations you’ll All the calculations you’ll

be using are based on a be using are based on a few formulae which can few formulae which can be rearranged to be rearranged to calculate the calculate the information you need.information you need.

► When trying to describe When trying to describe time, specify whether time, specify whether you are referring to a you are referring to a day on earth, or a day day on earth, or a day on your planet/moon.on your planet/moon.

221

r

mGmF

2mrF

T

2


Weight on planetsWeight on planets

When dealing with a When dealing with a relatively small object relatively small object on a planet we consider on a planet we consider the mass and radius of the mass and radius of the planet to work out the planet to work out what we have previously what we have previously used as ‘g’.used as ‘g’.

So for an object on earth So for an object on earth (mass=5.97x10(mass=5.97x102424 kg) kg) (Radius=6.38x10(Radius=6.38x106 6 m), m), we can say the we can say the gravitational field gravitational field strength(g) is:strength(g) is:

2

2

2

221

saycan weSo

planet

planet

object

planet

planet

planet

objectplanet

r

Gmg

mr

GmWeight

r

mGmWeight

r

mGmF


Cavendish-Boys ExperimentCavendish-Boys ExperimentFirst carried out in the 18First carried out in the 18thth century, this experiment can be used to century, this experiment can be used to

determine a reasonably accurate value for G. It uses a torsion determine a reasonably accurate value for G. It uses a torsion balance which measures the torque due to the gravitational balance which measures the torque due to the gravitational attraction between the large and smaller masses. A mirror attraction between the large and smaller masses. A mirror mounted on the torsion wire can be used to determine the mounted on the torsion wire can be used to determine the angular displacement of the wire and hence the torque. (T=kangular displacement of the wire and hence the torque. (T=kθθ, , where k is a constant)where k is a constant)

m1

m1

m2

m2

Torsion wire

Torsion wire

m2

m2

m1

m1


Cavendish-Boys ExperimentCavendish-Boys Experiment

Torsion wire

m2

m2

m1

m1

kr

lmmG

kFl

kFr

221

2

balance from torqueforces grav from Torque

Length Since all values except G can be measured to a reasonable degree of accuracy in this experiment, it can be used to determine a value for G.


Gravitational PotentialGravitational PotentialAt SG and higher EAt SG and higher Epp=mgh, however this assumes that g is constant which =mgh, however this assumes that g is constant which

for an object gaining significant height, we know it is not true. for an object gaining significant height, we know it is not true.

The The gravitational potentialgravitational potential (V measured in J kg (V measured in J kg-1-1) at a point is defined ) at a point is defined as:as:

the work done by external forces in moving a unit mass from infinity to the work done by external forces in moving a unit mass from infinity to that point.that point.

This will This will always be a negative valuealways be a negative value and is calculated by integrating the and is calculated by integrating the work done formula (since F varies with radius) from infinity to r.work done formula (since F varies with radius) from infinity to r.

r

r

drr

GmV

drF

V

m

21

2

Perform this integration to get an expression for V

r

GmV


Gravitational Potential EnergyGravitational Potential Energy

The actual potential energy of a specific object (mThe actual potential energy of a specific object (m22) ) due to its position in the gravitational field of another due to its position in the gravitational field of another object(mobject(m11) is simply the gravitational potential at that ) is simply the gravitational potential at that point multiplied by the mass of the object in the point multiplied by the mass of the object in the field(mfield(m22).).

r

mGmVm 21

2Energy Potential


Drawing Gravitational fieldsDrawing Gravitational fields

The gravitational field strength (g) can be The gravitational field strength (g) can be shown diagrammatically by drawing lines shown diagrammatically by drawing lines around a picture of the object. around a picture of the object.

► The lines show the direction of the force The lines show the direction of the force acting on a body placed in the fieldacting on a body placed in the field

► The closer together the lines are the stronger The closer together the lines are the stronger the gravitational field.the gravitational field.

► Use your SCHOLAR books to sketch the Use your SCHOLAR books to sketch the gravitational field aroundgravitational field around A single point massA single point mass Two identical point masses in proximityTwo identical point masses in proximity


Now try theseNow try these► On page 92(ish) of SCHOLAR books try Quiz 1 and On page 92(ish) of SCHOLAR books try Quiz 1 and

Quiz 2 (Page 78 & 83) Quiz 2 (Page 78 & 83) ► Extension (use planetary data from scholar book)Extension (use planetary data from scholar book)

1.1. A 1200kg satellite moves from a circular orbit A 1200kg satellite moves from a circular orbit around the earth at an altitude of 13,000km above around the earth at an altitude of 13,000km above the surface into a higher circular orbit. If its the surface into a higher circular orbit. If its gravitational potential energy is increased by 5GJ, gravitational potential energy is increased by 5GJ, determine the new altitude of the satellite.determine the new altitude of the satellite.(17.9x10(17.9x1066m)m)

2.2. How much energy would need to be given to a 5kg How much energy would need to be given to a 5kg object on the surface of the earth in order for it to object on the surface of the earth in order for it to reach infinity (ie never stop). What velocity would it reach infinity (ie never stop). What velocity would it need to be moving at to have this much kinetic need to be moving at to have this much kinetic energy (non-relativistic).energy (non-relativistic).(-312MJ , 11.17x10(-312MJ , 11.17x1033m/s)m/s)


Escape velocityEscape velocity

Theoretically, ignoring the effect of air friction, if an Theoretically, ignoring the effect of air friction, if an object is launched vertically with a high enough object is launched vertically with a high enough velocity, it will continue on indefinitely rather than velocity, it will continue on indefinitely rather than fall back to the surface. This is when the kinetic fall back to the surface. This is when the kinetic energy it has at launch plus its gravitational energy it has at launch plus its gravitational potential energy (negative remember!) equals potential energy (negative remember!) equals zero.zero.

esc

2122

for vequation an for thisrearrange now

02

1

0

planet a of surface on the velocity escapeFor

planetesc

pk

r

mGmvm

EE

Using data in your scholar books find the escape velocity from the moon and the earth


Black HolesBlack Holes

► A black hole is essentially a massive star A black hole is essentially a massive star which is extremely dense. The result is that it which is extremely dense. The result is that it has a relatively small diameter and a mass has a relatively small diameter and a mass many times greater than our sun.many times greater than our sun.

► If the escape velocity from the surface of the If the escape velocity from the surface of the star is greater than the speed of light, then no star is greater than the speed of light, then no light will be emitted by the star and so to an light will be emitted by the star and so to an external observer, it will appear black.external observer, it will appear black.

► Black holes also have the tendency to “suck Black holes also have the tendency to “suck in” other objects. This increases their mass in” other objects. This increases their mass further.further.


Planetary DataPlanetary Data

ObjectObject NameName Mass (kg)Mass (kg) Radius Radius (m)(m)

Orbital Orbital Radius Radius (m)(m)

PlanetPlanet MercuryMercury 3.30x103.30x102323 2.44x102.44x1066 57.9x1057.9x1099

PlanetPlanet VenusVenus 4.87x104.87x102424 6.05x106.05x1066 108x10108x1099

PlanetPlanet MarsMars 6.42x106.42x102323 3.40x103.40x1066 228x10228x1099

PlanetPlanet JupiterJupiter 1.90x101.90x102727 71.5x1071.5x1066 778x10778x1099

Jupiter Jupiter MoonMoon

IoIo 8.94x108.94x102222 1.82x101.82x1066 422x10422x1077

StarStar SolSol 1.99x101.99x103030 695x10695x1066 N/AN/A


Unit 1


Simple Harmonic MotionSimple Harmonic Motion


Springs and spring constantsSprings and spring constantsAimAim

To determine the spring constant of a To determine the spring constant of a given spring.given spring.

MethodMethod

The change in displacement of a spring The change in displacement of a spring is measured with a ruler as known is measured with a ruler as known masses are hung from it.masses are hung from it.

ResultsResults

Mass (kg)Mass (kg) Weight (N)Weight (N) ∆ ∆ Displacement Displacement (m)(m)


Springs and spring constantsSprings and spring constants

AnalysisAnalysis

Plot a graph of force against Plot a graph of force against displacement. Find the displacement. Find the gradient of this graph (dF/ds). gradient of this graph (dF/ds). This is the spring constant.This is the spring constant.

Mass (kg)Mass (kg) Weight (N)Weight (N) ∆ ∆ Displacement Displacement (m)(m)

F

∆s


Wobbly thingsWobbly things

Anything wobbling or vibrating is usually undergoing Anything wobbling or vibrating is usually undergoing simple harmonic motion. Take this mass attached at simple harmonic motion. Take this mass attached at both sides to rubber bands.both sides to rubber bands.

Think about the forces acting on the box.Think about the forces acting on the box.

Where would the following occur:Where would the following occur:(a) Maximum positive acceleration?(a) Maximum positive acceleration?

(b) Maximum negative acceleration?(b) Maximum negative acceleration?

(c) Maximum positive velocity?(c) Maximum positive velocity?

(d) Maximum negative velocity?(d) Maximum negative velocity?

positive


Intro to simple harmonic motionIntro to simple harmonic motionMany objects exhibit simple harmonic motion, which is Many objects exhibit simple harmonic motion, which is

simply an oscillation of an object which is disturbed simply an oscillation of an object which is disturbed from its rest position. If we examine a spring and from its rest position. If we examine a spring and mass in a horizontal frictionless arrangement as mass in a horizontal frictionless arrangement as shown we can analyse this type of motion.shown we can analyse this type of motion.

Equilibrium positionPosition y=0Force = 0

Spring CompressedPosition y=negativeForce = positive

Spring ExtendedPosition y=positiveForce = negative


SHM MathsSHM Maths

m

k

m

ky

dt

ydm

kya

kyma

kyF

22

y

This differential equation is specific to the This differential equation is specific to the spring scenario, but represents a more spring scenario, but represents a more general equation for SHM.general equation for SHM.

The constant The constant ωω is called the angular is called the angular frequency, but is simply a constant for frequency, but is simply a constant for any given situation relating to the any given situation relating to the frequency of the oscillation.frequency of the oscillation.

The mathematical solution of this The mathematical solution of this differential equation is beyond the scope differential equation is beyond the scope of advanced higher.of advanced higher.

If we assume the object is moved to a If we assume the object is moved to a positive y position and then released we positive y position and then released we can say the force acting on it and causing can say the force acting on it and causing it to accelerate is:it to accelerate is:

ydt

yd 22

2


Solutions to the SHM EquationSolutions to the SHM Equation

There are two solutions (really only one) to this There are two solutions (really only one) to this equation. Both satisfy the equation, but are only in equation. Both satisfy the equation, but are only in fact different in their initial conditions as we will see fact different in their initial conditions as we will see later.later.

ydt

yd

tadt

yd

tadt

dy

tay

22

2

22

2 sin

cos

sin

ydt

yd

tadt

yd

tadt

dy

tay

22

2

22

2 cos

sin

cos


Velocity in SHMVelocity in SHM► Assuming the initial conditions for an object given an Assuming the initial conditions for an object given an

initial velocity at zero displacement, we can say:initial velocity at zero displacement, we can say:

► Starting with the mathematical relationshipStarting with the mathematical relationship

► Show that the velocity of the object is given by the Show that the velocity of the object is given by the expressionexpression

a

yt

tay

sin

sin

a

vt

tav

cos

cos

1)(cos)(sin

1cossin22

22

tt

22 yav


Energy in SHMEnergy in SHM

So far we know that a mass oscillating on a So far we know that a mass oscillating on a horizontal spring on a frictionless surface will horizontal spring on a frictionless surface will have a displacement of the form y = a sin have a displacement of the form y = a sin ωωt. t. We also know that the velocity of the mass is We also know that the velocity of the mass is given by v=given by v=±±ωω√(a√(a2 2 - y- y22).).

First on a scrap bit of paper, then in you jotter….

Substitute the velocity equation above into the kinetic energy equation

Look at the Energy in SHM section of your scholar book and note the proof for the potential energy stored in the spring.

Finally show that the total energy is given by Ep+Ek= ½mω2a2


Scholar graph sketching exerciseScholar graph sketching exercise

At the the end of the energy in SHM section At the the end of the energy in SHM section there is a scholar graphing exercise. Sketch there is a scholar graphing exercise. Sketch both these graphs by hand and stick in your both these graphs by hand and stick in your jotter.jotter.

You may draw both graphs on the same set of You may draw both graphs on the same set of axes assuming you clearly identify which is axes assuming you clearly identify which is which.which.


Vertically oscillating massesVertically oscillating masses► Design and write up an experiment to determine the Design and write up an experiment to determine the

spring constant of a spring using a vertically suspended spring constant of a spring using a vertically suspended mass and measuring the angular frequency and mass mass and measuring the angular frequency and mass to determine k. Carry out the experiment, determine a to determine k. Carry out the experiment, determine a value for k and then check this value for your spring as value for k and then check this value for your spring as we did in the first experiment by calculating the ratio of we did in the first experiment by calculating the ratio of the weight to the extension of the spring.the weight to the extension of the spring.

► In order to do this you will need to derive the In order to do this you will need to derive the expression for angular frequency of a vertically expression for angular frequency of a vertically suspended oscillating mass on a spring.suspended oscillating mass on a spring.

► Take five values for time for 10 oscillations. Find Take five values for time for 10 oscillations. Find approximate random error, then find k and the error in approximate random error, then find k and the error in k.k.

► Good luckGood luck


Build a clockBuild a clock

► Read your scholar textbook section on simple Read your scholar textbook section on simple pendulums in SHM. Use the information to pendulums in SHM. Use the information to find the length of pendulum on earth find the length of pendulum on earth (g=9.8Nkg(g=9.8Nkg-1-1) which would pass the central ) which would pass the central point once per second (period=2s).point once per second (period=2s).

► Build this pendulum and test it for accuracy.Build this pendulum and test it for accuracy.


DampingDamping

When most objects undergo simple harmonic When most objects undergo simple harmonic motion there is some friction in the system. This motion there is some friction in the system. This causes the oscillation to decrease in amplitude, causes the oscillation to decrease in amplitude, but usually not to change in frequency.but usually not to change in frequency.

Sometimes damping is to be minimised when an Sometimes damping is to be minimised when an oscillation is desired, however often vibrations oscillation is desired, however often vibrations are to be minimised and so additional damping is are to be minimised and so additional damping is required such as in shock absorbers and required such as in shock absorbers and suspension systems.suspension systems.

Critically damped systemsCritically damped systems are damped such are damped such that when disturbed, they return to their that when disturbed, they return to their equilibrium position in the equilibrium position in the minimum possible minimum possible timetime..


Unit 1


Wave Particle DualityWave Particle Duality


Schizophrenic WavesSchizophrenic Waves

DiscussDiscuss

What is light????What is light????

►Think about:Think about: DiffractionDiffraction Photoelectric effectPhotoelectric effect Absorption and emission Absorption and emission

spectraspectra ReflectionReflection RefractionRefraction PhotonsPhotons WavesWaves

What do you think????What do you think????

OR

Wave

Particle


Wave Particle DualityWave Particle DualityUnder certain conditions waves can behave like Under certain conditions waves can behave like

particles and particles can behave like waves.particles and particles can behave like waves.

We know from higher that:We know from higher that:f=v/f=v/λλ for light for lightE=hf for a photonE=hf for a photonE=mcE=mc22 for a particle for a particle

The theory of wave-particle duality combines these The theory of wave-particle duality combines these three ideas and applies them all to waves and three ideas and applies them all to waves and particles.particles.

For this you must imagine light as a particle moving at For this you must imagine light as a particle moving at the speed of light.the speed of light.

Form an equation for the momentum (mv) of the photon Form an equation for the momentum (mv) of the photon as a function of its wavelength.as a function of its wavelength.


de Broglie Wavelengthde Broglie Wavelength► The de Broglie wavelength is given by the equationThe de Broglie wavelength is given by the equation

► Find the wavelength ofFind the wavelength of

1.1. A neutron moving at 10% of the speed of lightA neutron moving at 10% of the speed of light

2.2. An electron moving at 6x10An electron moving at 6x1055 ms ms-1-1

3.3. A golf ball moving at 5 msA golf ball moving at 5 ms-1-1

4.4. A person at full sprint (estimate values)A person at full sprint (estimate values)

p

hWavelength (m)

Plank’s Constant6.63x10-34 Js

Momentum (kgms-1)


de Broglie Wavelengthde Broglie Wavelength

The wavelengths of larger objects are so small The wavelengths of larger objects are so small that no wave phenomena are observable.that no wave phenomena are observable.

As a result of this, the concept of de Broglie As a result of this, the concept of de Broglie wavelength is only applied to particles of sub wavelength is only applied to particles of sub atomic or at most, atomic scales.atomic or at most, atomic scales.


Compton ScatteringCompton ScatteringIn 1923, Arthur Compton provided more evidence for In 1923, Arthur Compton provided more evidence for

wave particle duality by showing that X-rays (high wave particle duality by showing that X-rays (high frequency EM waves) can behave like particles.frequency EM waves) can behave like particles.

thin graphite sheet


Compton ScatteringCompton ScatteringIf he thought of the X-rays as particles (photons), If he thought of the X-rays as particles (photons),

which collided with the electrons in the graphite, which collided with the electrons in the graphite, and analysed the momentum of the collision.and analysed the momentum of the collision.

thin graphite sheet

electron

Less momentum so lower frequency

Some momentum transferred to electron


Unit 1


Wave Particle DualityWave Particle DualityAtomic ModelsAtomic Models


In the beginningIn the beginning

► EarthEarth

►WaterWater

► AirAir

► FireFire


Then came the chemists…..(in Then came the chemists…..(in 1869)1869)


But what about atomsBut what about atomsThe concepts of positive and negatives were known about The concepts of positive and negatives were known about

first. So Scientists made an assumptions. Large first. So Scientists made an assumptions. Large positive lump with little negative bits stuck in it. The positive lump with little negative bits stuck in it. The plum pudding model. Originally suggested by J.J plum pudding model. Originally suggested by J.J Thomson (the bloke who discovered electrons) in 1904Thomson (the bloke who discovered electrons) in 1904


Then came Rutherford (in 1911)Then came Rutherford (in 1911)

Zinc sulphide screen and eyepiece used to detect alpha particles

Experiment carried out in a vacuum

Thin gold foil

source

large deflection by nucleus

Small deflection by electrons

Rutherford scattering suggested a different model, Rutherford scattering suggested a different model, mostly made up of empty space with a small dense mostly made up of empty space with a small dense nucleus which was positively charged, orbited by nucleus which was positively charged, orbited by much smaller electrons, which were negatively much smaller electrons, which were negatively charged.charged.


But there were still questionsBut there were still questions

► The Rutherford model didn’t explain all the The Rutherford model didn’t explain all the interesting work being done on absorption interesting work being done on absorption and emission spectra at the time. It was all and emission spectra at the time. It was all very clever, but nobody knew how it all very clever, but nobody knew how it all worked. There was a missing piece of the worked. There was a missing piece of the puzzle.puzzle.


Neils Bohr (1913) - DenmarkNeils Bohr (1913) - Denmark

Mr Bohr cracked the basis for quantum theory in 1913 Mr Bohr cracked the basis for quantum theory in 1913 when he modified Rutherford’s model to explain when he modified Rutherford’s model to explain absorption and emission spectra and opening that absorption and emission spectra and opening that whole can of worms that is “quantum mechanics”.whole can of worms that is “quantum mechanics”.

Bohr proposed that there were Bohr proposed that there were distinct values at which the distinct values at which the amount of energy represented a amount of energy represented a wavelength that would exactly fit wavelength that would exactly fit into the orbit. After messing with into the orbit. After messing with some complicated maths he some complicated maths he discovered that for a stable orbit discovered that for a stable orbit the angular momentum was a the angular momentum was a multiple of h/2multiple of h/2ππ

ie. L = nh/2ie. L = nh/2ππ

mvr = nh/2mvr = nh/2ππ

Documents

Advanced Higher Physics Unit 1 Mechanics Saturday, 20 May 2006 Introduction to Unit 1