Introduction to Unit 1

1. Introduction to Unit 1

2. Mechanics Unit 1 Summary • Linear Kinematics • 1D & 2D kinematics • Relativistic Motion • Relativistic Mass • Relativistic Energy • Other Relativistic Effects • Angular Kinematics • Angular Notation • Kinematic Relationships • Tangential Speed • Rotational Dynamics • Centripetal acceleration / force • Torque • Newtons Laws • Angular Momentum • Angular Kinetic Energy • Gravity • Newtons Law of Gravitation • Gravitational force & fields • Gravitational Potential & Energy • Orbital Mechanics • Escape Velocity • Simple Harmonic Motion • Analysis of SHM • Energy in SHM • Damping • Wave particle duality • Theory • Examples of Duality • Quantum Mechanics • Atomic Models • Quantum Mechanics Theory

3. 1D Mechanics At higher we dealt exclusively with situations where the acceleration of an object was constant. This is not always the case. One common example is in friction where acceleration is dependant on velocity. If acceleration is a function of time (eg a=0.4t), then the standard equations of motion DO NOT APPLY. Instead we must use a calculus method. In general (and by definition)

4. 1 D Mechanics • Worked Example A 4kg object, initially at rest is subject to a force which varies with time according to the equation F=0.80t. • Calculate an expression for the acceleration of this object • Now, by calculus find expressions for v and s in terms of time • Find the displacement of the object after 12 seconds • Find the velocity of the object after 18 seconds

5. Do you get it – Try this. A 14kg object is initially moving at 6ms-1. From this moment it is subject to a force in Newtons according to the expression F=18-0.4t. • Formulate expressions for the acceleration, velocity and displacement of the object. • At what time is there no force on the object and what is its velocity at that time? • How long after the start of the experiment does the object come to rest? • How long after the start of the experiment does the object return to its initial position and what is its velocity at this time?

6. Here’s why its useful In an experiment we tracked the position of an object in one dimension and plotted the data in the graph below. Use your maths skills to get the equation of this graph, and use it to form an equation for the acceleration of the object. If we know the mass of the object 1.35kg, we can calculate the force on the object

7. Rigorous proof of higher formulae Using your knowledge of calculus: Considering an object with initial speed ‘u’ at t=0 and a constant acceleration of ‘a’, and knowing that a=dv/dt, prove that a=(v-u)/t. Now considering the same object and knowing that v=ds/dt, prove that s=ut+½at2 For both proofs use either • definite integrals, • or indefinite integrals and evaluate the constant of integration.

8. Relativity

9. A challenge to start with • How fast are we moving??? • Data Diameter of earth = 12000 km Radius of our orbit of Sun = 940 million kilometres Assume you are standing on the equator and that the Earth and the orbital path are perfectly circular. • Velocity due to earth rotation = 436 m/s • Velocity due to solar orbit = 187,284 m/s

10. Relativistic mass Although the concept of relativistic mass is odd, it is quite simple. As the velocity of an object relative to an observer increases, its mass will also increase. These effects are very small until we start to go faster than 0.1c (10% the speed of light). The relativistic mass of an object with rest mass mo moving at a speed v relative to a stationary observer is given by:

11. Relativistic Mass Questions • Find the relativistic mass of each of the following • An electron moving at 2.3x108 ms-1 • A neutron moving at 1.3x108 ms-1 • A 700kg space probe moving at 8x107 ms-1 • A 3kg meteorite accelerated to 9x106 ms-1 • If an object has a rest mass of 6kg, but a relativistic mass of 6.7kg, calculate its velocity • An object has a rest mass of 24x10-5 kg. Its relativistic mass is 12% greater. Calculate its velocity.

12. Change in mass as velocity changes • Using your knowledge of relativistic mass, plot a graph of how the mass of an object changes as its velocity increases from rest to 97%, if the object has a rest mass of 1kg. • (calculate m at 10% intervals, and then 95% and 97%)

13. Relativistic Energy The most important concept in relativity is that the mass of an object is a measure of its total energy. The Newtonian concept of kinetic energy (½mv2) begins to fail at relativistic velocities (>0.1c) as the mass is no longer constant as the object was accelerated. Instead we use Einstein’s equation as follows: The Rest Energy of an object is calculated using its rest mass (Eo=moc2), while the Relativistic Energy is calculated using the relativistic mass (E=mc2). The kinetic energy of the object is the difference between its relativistic energy and rest energy. E = Ek + Eo Ek = E – Eo Ek = mc2 - moc2

14. Relativistic Energy • The graph below shows how the Newtonian and Relativistic definitions of kinetic energy differ and how the difference between the theories increases dramatically as the speed of light is approached. Use Excel to plot a graph of the kinetic energy of a 1kg object as its velocity is Increased from rest to 0.85c. Plot both the kinetic energy calculated by Newtonian theory and the relativistic theory and note at what velocity Newtonian 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.

15. Relativistic Energy Problems Accelerating particles. Particle accelerators are used to accelerate protons, electrons and even small atomic nuclei to very high velocities for experimental work. However, as the velocities reached are relativistic (>0.1c), the effects of relativity must be accounted for. We can assume that the charge of a particle does not change as it accelerates, not does the voltage of the field accelerating it. Equating the work done on the charge to the relativistic kinetic energy, we can calculate the velocity of the particle. • Combine the formulae for work done on a charge in an electric field with the formula for relativistic kinetic energy and rearrange this formula for voltage. • Also rearrange the formula for velocity (quite tricky).

16. Now you can work these out…. • Calculate the voltage of field required to accelerate an electron to 8x107 ms-1. (19.2kV) • Calculate the velocity a proton would reach in a field of 30kV. (2.40x106 ms-1) • Calculate the velocity an alpha particle would reach in the same field. (1.69x106 ms-1) • Calculate the voltage of field required to accelerate an alpha particle to 0.8c. (1.26x109 V) • Calculate the voltage required to accelerate a deuterium (21H) nucleus to the same speed. • From your knowledge of work done, calculate the average force on the deuterium nucleus in Q5 if it accelerates over a distance of 800m. (2.58x10-13N)

17. Other Relativistic Effects • Time Dilation In AH we do not need to analyse this effect mathematically as it is quite complex and has effects which are difficult to understand. One of the side effects of relativity is that when two objects are in motion relative to each other, time does not run at a constant rate (it runs faster for the “stationary” observer. 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 and fly it in a circle, the clock which has been moving will run slower than the stationary one. Essentially, as velocity increases, time slows down. • Length contraction In addition to this, at relativistic velocities, the length of an object in the direction of travel will decrease as the object approaches the speed of light.

18. Post Summer Brain Warm-up

19. Mini testy type thingamajig…. • Write down the correct units for each of the following(a) Momentum (b) Charge (c) Pressure (d) Acceleration(e) Capacitance (f) Weight (g) Relativistic Mass • From the expression a = dv/dt , derive the equation of motion v = u + at • Bobby (a 67kg ice skater) was skating along at 14ms-1, minding his own business, when he collided with a stationary 39kg kangaroo which was lost. The pair become entangled. What was their velocity just after the collision? • 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 refractive index of 1.65”. Calculate the relativistic mass of the elephant at this velocity. • If Mr Colquhoun had a rest mass of 87kg before the holidays, but lost 7kg of mass between then and now, to what speed would he have to be accelerated such that his relativistic mass equalled his rest mass before summer?

20. Mini testy type thingymajig…. • Write down the correct units for each of the following(a) kgms-1 or Ns (b) C (c) Pa or Nm-2 (d) ms-2(e) F (f) N (g) kg • Derivation • 8.85 ms-1 • 329 kg • 1.18 x 108 ms-1

21. Some Quick Revision…post summer • The motion of an object is the be analysed and a student measures its displacement. If its motion is described by the equation s=0.3t3-0.6t2+1.3t, calculate:(a) Its velocity after 5 seconds (17.8m/s)(b) Its acceleration after 8 seconds (13.2 m/s2) • Find the relativistic mass of a 1.6 tonne African elephant moving with a velocity of 1.3x108 ms-1. • By calculating the difference between the rest energy and the relativistic energy, determine the kinetic energy of an 87kg human being moving with a velocity of 60,000 kms-1.

22. Rotational Motion

23. Angular Terminology Rotational motion works in much the same way as motion in a straight line if we analyse it in the correct way. The following symbols will be used throughout this section. θ - Angular displacement - rad ω0- Initial angular velocity - rad s-1 ω- Final angular velocity - rad s-1  - Angular acceleration - rad s-2 Terminology aside, the basic equations of motion for angular kinematics can be derived in exactly the same way as they were for linear kinematics.

24. Deriving the angular equations of motion • Starting from • Derive the angular equations of motion for an object accelerating from ‘ω0’ to ‘ω’ with a constant angular acceleration ‘’ in time ‘t’. In performing this motion, the object passes through angular displacement θ.

25. Working with angular velocities When using angular velocities in calculations, they must always be measured in rad s-1. Figures for angular velocity are often given in different ways. Convert each of the following to radians per second. • 3000 rpm • 24 rpm • 1 revolution per day • 1 revolution in 28 days • 50 revolutions per second • 1 . 314 rad/s • 2. 2.51 rad/s • 7.3x10-5rad/s • 2.6x10-6rad/s • 314 rad/s

26. Centripetal acceleration In order for an object to move in anything other than a straight line, it must accelerate. In the case of pure circular motion, this acceleration is always towards the centre of the circle. This is called centripetal acceleration and is entirely different from angular acceleration. Two formulae for centripetal acceleration can be derived from your knowledge of circular geometry and vector addition and subtraction. This derivation can be found in your SCHOLAR textbooks.

27. Centripetal acceleration We consider a point moving in a circle with radius r and with constant angular velocity ω moving though an angle Δθ from point A to point B as shown here. In order to find the centripetal acceleration we must determine the change in velocity of the point in travelling from A to B. Its angular velocity will not have changed, nor will the magnitude of its linear velocity, but the direction of the tangential velocity has changed. This change in velocity can be expressed as ω vb Δθ r B A va Δv=vb-va

28. Centripetal acceleration Now, by some clever maths, we can show that as we make Δθ very small, the change in velocity becomes perpendicular to va. In other words the acceleration of the point is towards the centre of the circle. Centripetal acceleration is written as Δv = vb - va This is a vector sum, and is carried out as follows -va Δv ω vb vb Δθ r B A va

29. Centripetal acceleration -va Δθ Δv vb With this right angled triangle we can say that Δv = v Δθ , which is true for small angles measured in radians (θ = tan θ for small angles) We can define the centripetal acceleration as Then substitute in for Δv = v Δθ Then in the limit as Δt tends to zero we get the differential And finally since v=rω, ω=v/r giving

30. Centripetal Force • Firstly, a myth to dispel. • You are in a car driving round a circular track anticlockwise. There is a force acting on you, but in which direction does it act? • Forward? • Backward? • Left? • Right? • Centrifugal force does not exist, it is a reaction force experienced as a result of centripetal force which acts towards the centre of the circle. • In the same way that when standing on the floor you are pushed up by the floor, when travelling in a circle you feel like you are being pushed out from the centre – this is not the case!!

32. Centripetal Force • The basic concept of centripetal force is very simple to understand. We have proved the expression for centripetal acceleration. • If we consider the case of a point mass following a circular path, a force must exist to give it this centripetal acceleration. This force must act towards the centre of the circle and must obey Newton’s 2nd law.

33. Applications of centripetal force • Confirming the formula experimentally • Motion in a horizontal circle • Motion in a vertical circle • Conical pendulum • Cars cornering • Banked corners

34. Confirming the formula Mass experiencing centripetal force m1 Mass used to counter centripetal force m2 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)

35. Motion in a Horizontal Circle • Try the following questions • A young girl of mass 37kg is on a roundabout at a radius of 1.6m. Her angular velocity is constant at 2.4 rad s-1. Calculate her tangential velocity, her centripetal acceleration and the centripetal force acting on her. • In a high g simulator a 300kg pod is rotated at high speed around a central pivot. The radius of the arc is 6m. If the tension force safety limit of the arm connecting the pod to the pivot is 20kN, Calculate the maximum angular velocity at which the simulator can safely be used. Calculate the corresponding tangential velocity at maximum safe limits. • A 850kg car drives round a corner at 40.0 km h-1. If the corner has a radius of 8.00m, calculate the centripetal force required to negotiate the corner safely.

36. Motion in a Vertical Circle a⊥ T mg a⊥ a⊥ T T mg mg T a⊥ mg • Consider an object being whirled vertically on a piece of string. The forces acting on it will change depending on the position of the object as shown below. In each position, we can state that the sum of the forces (T and mg) must equal the centripetal force (mrω2). Form an equation for the tension in the string when the mass m is at(a) the top(b) the bottom(c) the horizontal position

37. Conical Pendulum The period of a conical pendulum, similarly to a standard pendulum, can be proven to be independent of the mass on the end as long as the connection (eg string) is of negligible mass. T cos Φ T Φ Φ l T sin Φ mg mg a⊥ a⊥ Use the above free body diagrams and with the aid of your scholar books derive expressions for(a) the tension in the string(b) the angular velocity of the bob(c) the period of the bob ω

38. Cars cornering on the flat Reaction Force Friction from tyres Weight We can assume that a car, or other object driving round a corner is travelling in a horizontal circle with a constant angular velocity. As a result we know the only unbalanced force on the car is the centripetal force. On a flat road, this force must be provided by the friction of the tyres. The weight of the car is exactly balanced by the reaction force from the road. This means that at a given velocity, there is a limit to how tight a corner a car can go round before skidding. Example A 700 kg car travelling at 50kmh-1 is fitted with tyres providing a maximum frictional force of 4500N. What is the smallest radius of corner it can go round at this velocity?

39. Banked corners On a banked corner, similar to a conical pendulum, a component of the reaction force from the road provides part of the centripetal force required. The car is still assumed to be travelling in a horizontal circle, so the weight and the vertical component of the reaction force must be balanced. Reaction Force Example For a 600kg car, calculate the angle of banking required such that a car travelling at 50 kmh-1 need produce no friction from the tyres (ie, horizontal component of the reaction force equals the centripetal force required), for a corner with a radius of 36m. Φ Weight

40. Rotational DynamicsTorque & Moment

41. Torque & Moment Like the equations of motion, there are commonalities between linear and angular dynamics (the study of forces and motion). Torque and moment are in essence the same thing. They are a force acting on a mass which causes it to rotate or turn. Technically, a moment is defined as: The tangential component of a force multiplied by its radius from the point of rotation. It is measured in Nm. If the force is perpendicular to the radius If the force is not perpendicular to the radius F sinΦ F F Φ F cos Φ r r T = Fr T = Fr sin Φ

42. Balancing torques Balanced torques work in exactly the same way as balanced forces. For an object which is stationary or rotating, the torques must be balanced. For each of the following, draw out the diagram, and find the value of the missing force or radius. F F 6N 3N 0.6m 0.3m 1.5m 0.9m 160N 90N F 96N 20N 7m r 4m 3m 5m 85N 12N 60° r 1.9m 40N 0.9m 1.3m F

43. Basic Structural Mechanics Mass=200kg Φ Mass=300kg 3m 5m This principle can be applied to any object or structure, or section of a structure Φ 2.5m mg mg 3m

44. Torque & Acceleration m2 Mass used to create force The aim of this experiment is to prove the linear relationship between torque and angular acceleration. r

45. Torque & Angular Acceleration To derive the angular equivalent of Newton’s 2nd Law (F=ma) we analyse a point mass ‘m’ moving in a circle with an angular acceleration ‘’. We already know that: F = ma T = Fr a = r And all of these laws must hold for our object, so: F=ma F=m r  T/r = m r  T = mr2  I is the moment of inertia measured in kgm2 T = I 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

46. Finding Compound Moments of Inertia • When calculating the moment of inertia of an object, the back of the data book provides the following formulae for different shapes of object. • Note that it is their distribution of mass around the axis of rotation that matters, so a doughnut shape uses the same formula as a point mass as all the mass is at a single radius. • We can also add or subtract shapes from each other to allow us to analyse more complex shapes.

47. Finding Compound Moments of Inertia • Find the moment of inertia for each shape (b) (a) 3.9kg Solid sphere 0.36m 1.9kg 0.28m 5.2kg • Metre stick about 1 end • A bouncy ball (rolling) • Plastic pipe • Around 1 end • Around centre • Around lengthways central axis 0.83m 0.08m

48. Angular Momentum Again, angular momentum works exactly the same as linear momentum in that momentum is always conserved. This gives rise to the formula: The same principles apply to the angular system in that momentum is always conserved, and that in an elastic collision, the rotational kinetic energy is also conserved. The only additional consideration is that changes in angular velocity can be brought about by changes in moment of inertia. Angular velocity (rad s-1) Angular Momentum (kgm2s-1) Moment of Inertia (kg m2)

49. Try these • Calculate the angular momentum of • A 12kg disc of radius 0.85m rotating at 6.5rad s-1 • A 2.6m, 4.3kg rod rotating about the middle at 8.2rad s-1 • A ballet dancer is trying to achieve a faster spinning speed. She starts spinning with her arms out, giving her an angular velocity of 12.3 rad s-1. In this position her moment of inertia 15kgm2. She then draws her arms in which reduces her moment of inertia by 20%. Calculate her new angular velocity. Calculate whether her kinetic energy has increased, decreased or stayed the same.

50. Outcome 3 – Measuring I Using the air table fitted with a central pulley, determine the moment of inertia of the disc and any associated masses. Determine the torque applied to the disc and its angular acceleration. Repeat the experiment for a number of different applied torques. Use this data to determine the moment of inertia of the disc by graphical method. It is recommended that the measurement of time and angular displacement is used to determine the angular acceleration of the disc, rather than an attempt to measure the instantaneous velocity of the disc. A full report of the experiment should be produced using the guidelines including an analysis of errors and a full evaluation of the experiment.