01-02-00 - basic physics - mechanics
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UCT EEE4106Z Introductory Nuclear Physics 2015 20150220.2349 UNIVERSITY OF CAPE TOWN EEE4106Z Introductory Nuclear Physics 01-02-00 Mechanics February 2015 Emeritus Professor David Aschman Room 4T7, Physics Department University of Cape Town mailto:david.aschman@uct.ac.za http://www.phy.uct.ac.za/courses/eee4106z/current/ . * 20150220.2349 uct-physics-dga 1 UCT EEE4106Z Introductory Nuclear Physics 2015 CT Gaunt EEE4106Z Syllabus UCT EEE4106Z 2015 Introductory nuclear physics and radiation for power supply 16 credits, NQF level 08 Course aims: To develop strong concepts of nuclear physics and radiation in the context of nuclear power reactors. Course outcomes: An understanding of nuclear fission process and ability to calculate reaction energy. Course outline: Nuclear physics and radiation in the context of nuclear power reactors: atomic nature of matter; binding energy; radioactive decay; nuclear fission; neutron efficiency; ionizing radiation; radiation detection and measurement; effects of radiation on matter and biological systems. 20150220.2349 uct-physics-dga 2 UCT EEE4106Z Introductory Nuclear Physics 2015 * 01-02-00 Mechanics Kinematics - describes where a particle is and how it is moving Kinetics - describes why a particle moves - Newton’s Laws; F = ma linear motion; 2-d motion (vectors); circular motion acceleration, force work done by a force. systems. matter and interaction energy: kinetic energy, potential energy. Conservative forces ( ∇ × F = 0) Potential F = −∇U ... 20150220.2349 uct-physics-dga 3 UCT EEE4106Z Introductory Nuclear Physics 2015 * Reading and references EEE4106Z Reading list: Reading list on course website any good first year physics textbook . . Cummings, Law, Redish and Cooney, Understanding Physics, Wiley. Recommended: Halliday, Resnick and Walker, Fundamentals of Physics, Wiley ¿= 6th ed. Young and Freedman, University Physics, Addison-Wesley, 10th ed. 20150220.2349 uct-physics-dga 4 UCT EEE4106Z Introductory Nuclear Physics 2015 * Serway, Physics for Scientists and Engineers, Thomson. Other: Halliday, Resnick and Krane, Physics, Wiley. 20150220.2349 uct-physics-dga 5 UCT EEE4106Z Introductory Nuclear Physics 2015 * Book for these mechanics notes Cummings, Law, Redish & Cooney Understanding Physics (Wiley) 20150220.2349 uct-physics-dga . 6 UCT EEE4106Z Introductory Nuclear Physics 2015 Cummings, Law, Redish and Cooney Chapters in CLRC Part 1 1) Measurement 2) Motion Along a Straight Line 3) Forces and Motion Along a Line 4) Vectors 5) Forces and Motion in Two Dimensions 6) Combined Forces 7) Translational Momentum 8) Extended Systems 9) Kinetic Energy and Work 10) Potential Energy and the Conservation of Energy 11) Rotation 12) Complex Rotations 20150220.2349 uct-physics-dga 7 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 Ch 1: Measurement . . . it doesn’t matter how beautiful your theory is, it doesn’t matter how smart you are – if it doesn’t agree with experiment, it’s wrong. - R.P. Feynman Physics is based on measurement and experiment. 20150220.2349 uct-physics-dga 8 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 Units Physics is based on measurement and experiment. Units allow us to express these measurements in a standard way. Base units for basic quantities like distance, time and mass. Derived units convenient combinations of base units. 20150220.2349 uct-physics-dga 9 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 Base units Length Sizes, lengths, distances are measured in metres (abbreviated m). (L) Time Time intervals and durations are measured in seconds (abbreviated s). (T) Mass Masses are measured in kilograms (abbreviated kg). (M) 20150220.2349 uct-physics-dga 10 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 Base units: Length The metre is defined as the distance light travels in vacuum in 1/299792458 seconds. Note that the speed of light is now, by definition, c = 299792458 ms−1 . 20150220.2349 uct-physics-dga 11 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 Base units: Time The second is defined using the characteristic frequency of a particular kind of caesium atom in an atomic clock, as 9192631770 periods. 20150220.2349 uct-physics-dga 12 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 Base units: Mass The kilogram is defined as the mass of a certain platinum-iridium alloy cylinder kept at the International Bureau of Weights and Measures in France. 20150220.2349 uct-physics-dga 13 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 Derived units Derived units are convenient combinations of the base units. Often they are given their own names. The combinations of units are termed the dimensions of the unit: the base units have dimensions of L, T and M. The phrase [energy] can be read as “the dimensions of energy”. 20150220.2349 uct-physics-dga 14 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 Derived units: example For example, we have [energy] = [mass][length]2 ML2 = 2 [time]2 T The unit of energy is the joule, abbreviated J. Note that we can treat the combination of units algebraically. 20150220.2349 uct-physics-dga 15 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 Use of units All quantities in physics have units associated with them. In some, the combinations of base units all cancel out; such quantities are termed dimensionless. All quantities must thus be given with their units; they are otherwise meaningless numbers. For example, a distance might be expressed as 10.42 m. 20150220.2349 uct-physics-dga 16 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 Use of units (2) Two quantities can only be equated if they have the same units. We can use this as a rough check on the consistency of our equations: F = ma [F] = [m][a] ML L = M T2 T2 20150220.2349 uct-physics-dga 17 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 Units: important points A number referring to a physical quantity is meaningless without units. Both sides of an equation must have the same units. Use powers of 10 and prefixes to scale units. 20150220.2349 uct-physics-dga 18 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 Units: SI prefixes Factor 24 10 1021 1018 1015 1012 109 106 103 102 101 20150220.2349 Prefix yotta zetta exa peta tera giga mega kilo hecto deka uct-physics-dga Symbol Y Z E P T G M k h da Factor −1 10 10−2 10−3 10−6 10−9 10−12 10−15 10−18 10−21 10−24 Prefix deci centi milli micro nano pico femto atto zepto yocto Symbol d c m µ n p f a z y 19 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 1 End 01 20150220.2349 uct-physics-dga 20 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2 Ch 2: Motion along a straight line Linear kinematics. 20150220.2349 uct-physics-dga 21 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-2 1-d Coordinate system 20150220.2349 uct-physics-dga 22 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-2 Position Vector Position of a particle is given by position vector. A vector is a quantity with length and direction. We write a vector either as boldface ~x or with arrow over it 20150220.2349 uct-physics-dga x 23 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-2 Position Vector in terms of unit vector î is a unit vector in the +x direction. x = x î In general, position vector is called r 20150220.2349 uct-physics-dga 24 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-2 Unit vectors The unit vectors have unit length. In Cartesian coordinates, the unit vectors in the x, y and z directions are called î, ĵ and k̂. Sometimes they are called x̂, ŷ and ẑ. Occasionally they are called ê x , êy and êz 20150220.2349 uct-physics-dga 25 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-2 Displacement as a Vector Change is difference between final state and initial state. Displacement is the change in position of an object. ∆r = x2 − x1 = ∆x î 20150220.2349 uct-physics-dga 26 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-2 Displacement vector for three situations 20150220.2349 uct-physics-dga 27 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-2 Touchstone Example 2-1: Displacements Three position pairs are shown. Which represent negative displacements? 20150220.2349 uct-physics-dga 28 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-3 Graph of x(t) for stationary object at x = −2 m 20150220.2349 uct-physics-dga 29 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-3 Position-time graph for walking student 20150220.2349 uct-physics-dga 30 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-3 Graph of x(t) for walking student 20150220.2349 uct-physics-dga 31 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-3 Velocity as a vector In 3-d a vector has three components, eg v = v x î + vy ĵ + vz k̂ In 1-d a vector only can have one component v = v x î Sometimes in 1-d we drop the unnecessary subscript v = v î 20150220.2349 uct-physics-dga 32 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-3 Average Velocity hvi = 20150220.2349 uct-physics-dga ∆x ∆x x2 − x1 = î = î ∆t ∆t t2 − t1 33 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-3 Getting average velocity from slope of graph For t1 = 1.0 s, t2 = 1.5 s, average velocity hvi = 2.04 î ms−1 20150220.2349 uct-physics-dga 34 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-3 Average speed hsi = 20150220.2349 uct-physics-dga total distance ∆t 35 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-3 Instantaneous velocity Instantaneous velocity is the rate at which the position vector changes with time, at a given instant. ∆x dx = ∆t→0 ∆t dt v = lim The derivative with of position vector with respect to time. (Calculus). Derivative of a vector? Easy. v = v x î = dx dx î = dt dt So v x = dx/dt 20150220.2349 uct-physics-dga 36 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-3 Touchstone example 2-2: Out of gas You drive at 70 km/h for 8.4 km along a road, then run out of gas. For 30 min you walk another 2.0 km to gas station. What is . . . your average velocity from beginning of drive to arrival at gas station? 20150220.2349 uct-physics-dga 37 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-3 Touchstone example 2-2: Out of gas You drive at 70 km/h for 8.4 km along a road, then run out of gas. For 30 min you walk another 2.0 km to gas station. What is your average velocity from beginning of drive to arrival at gas station? hv x i = ∆x/∆t = 10.4 km/0.62 h = 16.8 km h−1 20150220.2349 uct-physics-dga 38 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-4 Average acceleration Acceleration is a change in velocity. Average acceleration over time interval (t1 , t2 ) hai = v2 − v1 ∆v = t2 − t1 ∆t For 1-d motion hai = 20150220.2349 uct-physics-dga v2x − v1x î t2 − t1 39 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-4 Instantaneous acceleration Instantaneous acceleration is the rate of change of the velocity at that instant. ∆v dv = ∆t→0 ∆t dt a = lim 20150220.2349 uct-physics-dga 40 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-4 An upward elevator journey 20150220.2349 uct-physics-dga 41 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-5 Constant acceleration 20150220.2349 uct-physics-dga 42 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-5 Kinematic equation (1) for constant acc. a = a x î = hai v2x − v1x î t2 − t1 v2x − v1x ax = t2 − t1 hai = Solve for v2x v2x = v1x + a x (t2 − t1 ) = v1x + a x ∆t or ∆v x = a x ∆t 20150220.2349 uct-physics-dga 43 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-5 Kinematic equation (1): v = v0 + a x t For constant acceleration, recall that ∆v x = a x ∆t We can write velocity as a function of time. Initially, at t = 0 let v = v0 , and at ANY later time t let v = v(t) v(t) = v0 + a x t 20150220.2349 uct-physics-dga 44 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-5 Constant acceleration: x as function of t hvi = hv x i î = ∆x x2 − x1 î = î ∆t t2 − t1 The x-component of average velocity is hv x i = x2 − x1 t2 − t1 and solving for x2 gives x2 = x1 + hv x i(t2 − t1 ) 20150220.2349 uct-physics-dga 45 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-5 Constant acc.: average velocity Velocity increases linearly with time. Reasonable to assume that average velocity is average of initial and final instantaneous velocity 20150220.2349 uct-physics-dga 46 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-5 Kinematic equation (2) for constant acc. Recall x2 = x1 + hv x i(t2 − t1 ) and hv x i = (v1x + v2x )/2 substitute v2x = v1x + a x (t2 − t1 ) to get hv x i = (1/2)([v1x + v1x + a x (t2 − t1 )] = v1x + (1/2)a x (t2 − t1 ) x2 − x1 = v1x (t2 − t1 ) + (1/2)a x (t2 − t1 )2 or ∆x = v1,x ∆t + (1/2)a x (∆t)2 20150220.2349 uct-physics-dga 47 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-5 Kinematic equation (2): x = x0 +v0 t+(1/2)a x t2 Initially at t = 0 let x = x0 and v = v0 Consider interval (0, t) of duration ∆t = t − 0 = t. Express position as a function of time x = x(t) x = x0 + v0 t + (1/2)a x t2 x has quadratic dependence on t. Parabolic graph. 20150220.2349 uct-physics-dga 48 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-5 Kinematic equation (3): v2 −v20 = 2a x (x− x0 ) v − v0 = at (v + v0 )/2 = (x − x0 )/(t − 0) Eliminate t by multiplying v2 − v20 = 2a x (x − x0 ) 20150220.2349 uct-physics-dga 49 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 2-5 Fan on low-friction cart Cart at rest. Fan turns on at t = 1.2 s. Fan up to full thrust at t = 1.4 s. Parabolic graph as cart slows, turns around, speeds up. 20150220.2349 uct-physics-dga 50 UCT EEE4106Z Introductory Nuclear Physics 2015 ConcepTest: Truck on track Graph shows position of truck as function of time. PINK: Truck speeds up all the time GREEN: Truck slows down all the time YELLOW: Truck speeds up, then slows down BLUE: Truck travels at constant velocity Answer: 20150220.2349 uct-physics-dga 51 UCT EEE4106Z Introductory Nuclear Physics 2015 ConcepTest: Trains A and B Graph shows position of two trains as function of time. PINK: At tB both have same velocity GREEN: Both trains speed up all the time YELLOW: Both trains have same velocity some time before tB BLUE: Somewhere on graph, both have same acceleration WHITE: A is always ahead of B Answer: 20150220.2349 uct-physics-dga 52 UCT EEE4106Z Introductory Nuclear Physics 2015 ConcepTest: Throwing a ball upwards A ball is at rest in your hand(1). You swing your arm up, and let go (2). The ball flies up to its highest point (3), then comes back to the ground (4). Which is the correct velocity-time graph? Answer: 20150220.2349 uct-physics-dga 53 UCT EEE4106Z Introductory Nuclear Physics 2015 Chapter 02: Problems 20150220.2349 uct-physics-dga 54 UCT EEE4106Z Introductory Nuclear Physics 2015 Ex: Acceleration of head of rattlesnake The head of a rattlesnake has acceleration 50 ms−2 . If a car could have this acceleration, how long would it take from rest to reach velocity of 100 km/hr? Find car final speed in SI units km v = 100 km/hr = 100 hr ! 1000 m 1 km ! ! 1 hr = 27.8 m/s 3600 s v = v0 + a x t gives t = (v − v0 )/a x . Use v0 = 0 t= 20150220.2349 uct-physics-dga (27.8 − 0) m s−1 = 0.55 s 50 m s−2 55 UCT EEE4106Z Introductory Nuclear Physics 2015 Ex: constant acceleration A body has uniform acceleration. Its velocity is 12.0 cm/s when x coordinate is 3.00 cm. If 2.00 s later x = −5.00 cm, what is acceleration? 1-d motion: x = x î v = v î a = a î At t = 0.00 s, x0 = +3.00 cm, v0 = +12.0 cm/s At t = 2.00 s, x = −5.00 cm x = x0 + v0 t + (1/2)at2 So 20150220.2349 → (1/2)at2 = x − x0 − v0 t a = 2(x − x0 )/t2 − 2v0 /t uct-physics-dga 56 UCT EEE4106Z Introductory Nuclear Physics 2015 Example: (continued) a = 2(x − x0 )/t2 − 2v0 /t a=2 (−5.00 − 3.00) cm 12.0 cm s−1 − 2 = −16.00 cm s−2 (2.00)2 s2 2.00 s So a = −16.00 î cm s−2 20150220.2349 uct-physics-dga 57 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC p44 Niagra Falls plunge CLRC p 44 Height of Niagra is 48 m. How long to fall? Velocity at bottom? Assume 1-d motion g = 9.8 m s−2 DOWN 20150220.2349 uct-physics-dga 58 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC p44 Niagra Falls (2) Choice of origin? Choose origin x = 0 at bottom. Positive direction? Choose positive upwards. THEN x1 = 48 m and x2 = 0 m AND acceleration a x = −9.8 î m s−2 r ∆t = t2 − t1 = 20150220.2349 uct-physics-dga 2(x2 − x1 ) = ax r 2(−48) m = 3.1 s −9.8 m s−2 59 UCT EEE4106Z Introductory Nuclear Physics 2015 Ratio of distances like odd integers Prove the famous result due to Galileo, that for an object released from rest, undergoing constant acceleration, the ratio of distances traveled in successive time intervals are in the ratio of the odd integers, eg 1, 3, 5, . . . , 2n − 1 20150220.2349 uct-physics-dga 60 UCT EEE4106Z Introductory Nuclear Physics 2015 Proof Time interval ∆t. Time after nth interval is tn = n∆t Distance xn = 21 atn2 Distance traveled in nth interval is (∆x)n = xn − xn−1 (∆x)n = 12 a(∆t)2 [(n2 − (n − 1)2 ] = 12 a(∆t)2 (2n − 1) 20150220.2349 uct-physics-dga 61 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 3 Ch 3: Forces, motion in straight line Linear dynamics. A force causes an acceleration. 20150220.2349 uct-physics-dga 62 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 3-2 Newton’s First Law If no force acts on a body at rest, it remains at rest, or if it is moving, it continues moving with constant velocity. Question: In which cases is there no force acting: a book on a table a book sliding across a table a puck sliding on the ice a dropped ball a ball thrown by an astronaut in a satellite the moon an electron moving in an atom? 20150220.2349 uct-physics-dga 63 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 3-4 Measuring forces Subject a given mass to force, measure acceleration. SI unit of force is the Newton. A force of 1 N will give a mass of 1 kg an acceleration of 1 m s−2 . Use extension of a spring (Hooke’s law) Use a transducer (eg. piezoelectric crystal) 20150220.2349 uct-physics-dga 64 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 3-5 Defining and measuring mass Gravitational mass: use a balance to compare against a known mass. Inertial mass: the constant of proportionality between force and acceleration. Gravitational and inertial masses the same to within 1 part in 1012 (Eötvos experiment) 20150220.2349 uct-physics-dga 65 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 3-6 Newton’s Second Law for a single force F = ma When a single force acts on an object, it causes the object to accelerate in the direction of the force. The amount of acceleration is given by the force divided by the mass of the object. If force of gravity on an object is proportional to its gravitational mass, and its inertial mass is equal to its gravitational mass, then all objects, whatever their (nonzero mass), have the same acceleration due to gravity. (Galileo’s dropped balls (at Pisa?)) 20150220.2349 uct-physics-dga 66 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 3-7 Combining forces along a line Forces are vectors. Total force is the vector sum of the individual forces. Add them like vectors. In 1-d, this is algebraic (easy). 20150220.2349 uct-physics-dga 67 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 3-9 Gravitational forces; free-fall motion F = mg where g = 9.81 m s−2 DOWN YOU choose the positive direction. That sets the sign of g. g = g î. If you choose positive x UP, then g = −9.81 m s−2 If you choose positive x DOWN, then g = +9.81 m s−2 20150220.2349 uct-physics-dga 68 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 3-10 Newton’s third law If object A exerts a force on object B, then object B exerts a force back on object A. The forces have equal magnitude but act in opposite directions. FB→A = −FA→B (Action-reaction pair). Horse and cart. 20150220.2349 uct-physics-dga 69 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 3-10 End 03 20150220.2349 uct-physics-dga 70 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 4 Ch 4: Vectors Vectors needed to describe motion in 2-d and 3-d. 20150220.2349 uct-physics-dga 71 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3 Motion in 2-d and 3-d We can extend the previous coordinate-based treatment. However, it is useful to introduce other objects to represent the various physical quantities. SCALAR: Quantity that is represented at a point by one number. Ordinary rules of algebra apply. e.g. temperature, energy, mass. VECTOR: Quantity with magnitude and direction. e.g. velocity, acceleration, momentum, electric field. 20150220.2349 uct-physics-dga 72 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.1 Vectors: notation An arrow is used to depict the vector in drawings, with the arrow giving the direction, and the length the magnitude. Note that the magnitude is always a = |a| ≥ 0. The information is in the direction and length of the vector — we can shift it around if we don’t change these. 20150220.2349 uct-physics-dga 73 UCT EEE4106Z Introductory Nuclear Physics 2015 Vectors: notation Usually, in print, vectors are denoted by bold Roman type, e.g. a is a vector. The magnitude is denoted by the same symbol, in non-bold italic type, e.g. a is the magnitude of a. HRW6 uses an arrow over an italic symbol to denote a vector: ~a. In script, we may denote a vector as: 20150220.2349 uct-physics-dga 74 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.2 Vectors: addition Adding vectors geometrically, eg by scale drawing. s=a+b 20150220.2349 uct-physics-dga 75 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.2 Vectors: addition is commutative Use geometry to show some properties of vectors. For instance, vector addition is commutative: s=a+b=b+a 20150220.2349 uct-physics-dga 76 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.2 Vector addition — properties a+b=b+a (commutative) (a + b) + c = a + (b + c) (associative) For each b there is −b: b + (−b) = 0 a − b = a + (−b) (subtraction) 20150220.2349 uct-physics-dga 77 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.2 Vectors: multiplication by a scalar q = pa magnitude: q = pa. direction of q is direction of a. units: [q] = [p] [a]. p(a + b) = pa + pb (distributive) 20150220.2349 uct-physics-dga 78 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.3 Components of vectors Manipulating vectors geometrically is tedious. Using a (rectangular) coordinate system, we can use components to manipulate vectors algebraically. A component is the projection of a vector on an axis of the coordinate system. 20150220.2349 uct-physics-dga 79 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.3 Components of vectors (2) We project a onto the axes. a x and ay are the components of a. (We say we have resolved the vector). 20150220.2349 uct-physics-dga 80 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.3 Components — notes The components depend on the coordinate system — this can be chosen to suite the problem. The components are signed (“head-tail”). Components have units ! 3 components in 3-d (e.g. a x , ay , az ). The components are unchanged by a translation of the axes (but are changed by a rotation). 20150220.2349 uct-physics-dga 81 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.3 Polar coordinates The direction of a can be measured by the angle from the x-axis: Obviously, a x = a cos θ 20150220.2349 uct-physics-dga and ay = a sin θ 82 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.3 Polar coordinates (2) Also, a= tan θ = q a2x + a2y ay ax i.e. given components in one coordinate system, we can transform them to components in the other. 20150220.2349 uct-physics-dga 83 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.3 Projecting vectors We can use similar methods to project vectors onto any direction. If we have a coordinate axis parallel to this direction, the components of the original vector and the projected vector are the same. 20150220.2349 uct-physics-dga 84 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.4 Unit vectors Components are most useful when used with unit vector notation. A unit vector is a vector with magnitude 1: Only the direction is important. The unit vector has no units.(!) Notation: the unit vector in the direction a is â. It follows that â = a/a or a = a â. 20150220.2349 uct-physics-dga 85 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.4 Unit vectors and components We denote the unit vectors along the x, y, z axes by î, ĵ, k̂ Suppose a has components a x , ay , az . Then a x î is a vector along the x-axis, etc. Then a = a x î + ay ĵ + az k̂ 20150220.2349 uct-physics-dga 86 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.5 Adding vectors (once more . . . ) Suppose s = a + b. Then: [ a + b = a x î + ay ĵ + az k̂ + b x î + by ĵ + bz k̂ = (a x + b x )î + (ay + by )ĵ + (az + bz )k̂ =s So, s x = a x + b x , etc: we can add vectors by adding their components. 20150220.2349 uct-physics-dga 87 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.5 Adding vectors by components 20150220.2349 uct-physics-dga 88 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.7 Multiplication and Vectors We have seen that we can multiply a vector by a scalar. In unit vector notation this becomes: pa = pa x î + pay ĵ + paz k̂ There are two ways of “multiplying” two vectors: the scalar product or dot product the vector product or cross product 20150220.2349 uct-physics-dga 89 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.7 The scalar product We define p = a · b = ab cos θ. Pronounced “a dot b”. The product is a scalar. For a parallel to b, a · b = ab For a perpendicular to b, a · b = 0 a·b=b·a 20150220.2349 uct-physics-dga 90 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.7 Scalar product in unit vector form The scalar product can easily be calculated using the component form of vectors. a · b = (a x î + ay ĵ + az k̂) · (b x î + by ĵ + bz k̂) = a x b x î · î + a x by î · ĵ + a x bz î · k̂ + . . . = a x b x 1 + a x by 0 + a x bz 0 + . . . Thus, a · b = a x b x + ay by + az bz 20150220.2349 uct-physics-dga 91 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.7 Scalar product in unit vector form (2) a · a = a x a x + ay ay + az az = a2x + a2y + a2z = a2 i.e., a= √ a·a p = a x a x + ay ay + az az (Pythagorus!) 20150220.2349 uct-physics-dga 92 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.7 Scalar product in unit vector form (3) a · b = ab cos θ a·b cos θ = ab a·b = √ √ a·a b·b Note: this is the way to determine the angle between two vectors. 20150220.2349 uct-physics-dga 93 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.7 Vector product This is also known as the cross product. This is a vector (pronounced “a cross b”): c=a×b with magnitude: c = ab sin θ direction: perpendicular to plane containing a and b, and given by a right hand rule. 20150220.2349 uct-physics-dga 94 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.7 Vector product — right hand rule The direction of a × b is given by a right-hand rule. 20150220.2349 uct-physics-dga 95 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.7 Properties of vector product |a × b| = ab if a ⊥ b |a × b| = 0 if a k b c = a × b given by right hand rule c0 = b × a has opposite direction. i.e. a × b = −b × a The vector product is NOT commutative! 20150220.2349 uct-physics-dga 96 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW 3.7 Evaluation of vector product The definition of the cross product yields: î × ĵ = k̂ ĵ × î = −k̂ ĵ × k̂ = î k̂ × ĵ = −î k̂ × î = ĵ î × k̂ = −ĵ î × î = ĵ × ĵ = k̂ × k̂ = 0 Then: â × b̂ = (ay bz − by az )î + (az b x − bz a x )ĵ + (a x by − b x ay )k̂ 20150220.2349 uct-physics-dga 97 UCT EEE4106Z Introductory Nuclear Physics 2015 End 04 20150220.2349 uct-physics-dga 98 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 4 Ch 5: Net force and 2-d motion We apply vectors to the study of motion in two and three dimensions. How does the position of a body change with time? What will happen to a body is determined by the acceleration, and depends on the initial velocity and initial position. We use vectors to represent these directional quantities. 20150220.2349 uct-physics-dga 99 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.2 Position as a vector A position vector gives the position of a point with respect to the origin at time t. Suppose the point has coordinates (x, y, z). These coordinates are functions of the time; as time varies the point moves along a path or trajectory. Then the position vector of the point is r(t) = x(t) î + y(t) ĵ + z(t) k̂ 20150220.2349 uct-physics-dga 100 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.2 Position and displacement vectors Position vector tells you where a particle is. Displacement describes a change in position. The position vector is given by a displacement, from the origin, to the position of the particle. 20150220.2349 uct-physics-dga 101 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.2 Displacement At some time t1 the point is at r1 , and at t2 is at r2 . The displacement ∆r of the point during the time interval ∆t = t2 − t1 is the change in position: ∆r = r2 − r1 = (x2 − x1 ) î + (y2 − y1 ) ĵ + (z2 − z1 ) k̂ = ∆x î + ∆y ĵ + ∆z k̂ 20150220.2349 uct-physics-dga 102 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.2 Displacement vector as a function of time 20150220.2349 uct-physics-dga 103 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.2 Change in displacement 20150220.2349 uct-physics-dga 104 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.3 Average velocity The average velocity v̄ over the time interval ∆t gives us a measure of how fast displacement is changing: v̄ = = ∆r r2 − r1 = ∆t ∆t ∆x ∆y ∆z î + ĵ + k̂ ∆t ∆t ∆t Note that this average will depend on ∆t, so we must specify the time interval. 20150220.2349 uct-physics-dga 105 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.3 Velocity As ∆t gets smaller and approaches zero (∆t → 0), the average velocity approaches a definite value, the velocity (also called the instantaneous velocity) : ∆r dr ≡ ∆t→0 ∆t dt v = lim We call this the derivative of the position vector with respect to time. Note that it is independent of ∆t. The velocity is tangent to the path. Python: avgvel.py 20150220.2349 uct-physics-dga 106 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.3 Velocity — notation v= dr dx dy dz = î + ĵ + k̂ dt dt dt dt = v x î + vy ĵ + vz k̂ Velocity is a vector. The components of v are given by the derivatives of the coordinates x, y and z with respect to time. The magnitude of v is called the speed. 20150220.2349 uct-physics-dga 107 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.3 Velocity vector 20150220.2349 uct-physics-dga 108 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.4 Acceleration The velocity can, of course, change with time. What is its rate of change? We can define the average acceleration in much the same way as we defined the average velocity. 20150220.2349 uct-physics-dga 109 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.4 Average acceleration We define the average acceleration ~ā over the time interval ∆t as ~ ~ ~ ~ā = ∆v = v2 − v1 ∆t ∆t = ∆vy ∆v x ∆vz î + ĵ + k̂ ∆t ∆t ∆t Note that this is defined using the instantaneous velocity, not the average velocity. 20150220.2349 uct-physics-dga 110 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.4 Acceleration As ∆t gets smaller and approaches 0 (∆t → 0), the average acceleration approaches a finite value, the acceleration (also called the instantaneous acceleration) : ~ ∆v ∆t→0 ∆t ~a = lim ≡ d~v dt = dvy dv x dvz î + ĵ + k̂ dt dt dt = a x î + ay ĵ + az k̂ 20150220.2349 uct-physics-dga 111 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.4 Acceleration — notes Acceleration is a vector. ~v2 and ~v1 can have the same direction even though ~a is nonzero. (how?) ~v can still change if ~a = 0 instantaneously. (how?) Units of m/s2 . 20150220.2349 uct-physics-dga 112 UCT EEE4106Z Introductory Nuclear Physics 2015 Special cases — constant ~v Then ~a = 0 so ~v = ~v̄ for all ∆t. Then, ~ ~r1 − ~r0 ∆r = = const ≡ ~v0 ∆t ∆t i.e., the displacement ~r1 − ~r0 = ~v0 ∆t or ~r1 = ~r0 + ~v0 ∆t 20150220.2349 uct-physics-dga 113 UCT EEE4106Z Introductory Nuclear Physics 2015 Motion at constant ~v Take t0 = 0, rename ~r1 → ~r, t1 → t Then: ~r = ~r0 + ~v0 t This gives the path (or trajectory) for constant ~v: a straight line. 20150220.2349 uct-physics-dga 114 UCT EEE4106Z Introductory Nuclear Physics 2015 Special cases — constant ~a If ~a is constant, then ~a = ~ā for all ∆t. i.e, in some time interval, ~ ~v1 − ~v0 ∆v = = const ≡ ~a0 ∆t ∆t so ~v1 = ~v0 + ~a0 ∆t 20150220.2349 uct-physics-dga 115 UCT EEE4106Z Introductory Nuclear Physics 2015 Motion at constant ~a (2) We can then show ~r1 = ~r0 + ~v0 ∆t + 12 ~a0 (∆t)2 Take t0 = 0, t1 = t, rename ~r1 → ~r Then: ~r(t) = ~r0 + ~v0 t + 12 ~a0 t2 ~v(t) = ~v0 + ~a0 t The position varies quadratically in time. 20150220.2349 uct-physics-dga 116 UCT EEE4106Z Introductory Nuclear Physics 2015 Note — constant ~a We can always write the vector equation 1 ~r(t) = ~r0 + ~v0 t + ~a0 t2 2 in component form, once we have introduced coordinates: x = x0 + v x0 t + 12 a x0 t2 y = y0 + vy0 t + 12 ay0 t2 z = z0 + vz0 t + 12 az0 t2 20150220.2349 uct-physics-dga 117 UCT EEE4106Z Introductory Nuclear Physics 2015 Pop and drop Strobe photo of two projectiles. Horizontal and vertical components of velocity are independent. 20150220.2349 uct-physics-dga 118 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.5,4.6 Special case — projectile motion Now specialise to constant acceleration due to gravity — also known as free fall. Take the y-axis as “up”. Then, in the absence of other forces such as air resistance, ~a = −g ĵ where g = 9.81 ms−2 (near the surface of the Earth). 20150220.2349 uct-physics-dga 119 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.5,4.6 Projectile motion — initial condition At time t = 0, the velocity is ~v0 at position ~r0 . y v 0 j 20150220.2349 uct-physics-dga θ x 120 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.5,4.6 Projectile motion — initial condition ĵ and ~v0 define a plane. The acceleration lies only in this plane. Thus the motion is only in the plane. Thus we can reduce it to 2-d by choosing x-axis in the plane. 20150220.2349 uct-physics-dga 121 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.5,4.6 What is the position at time t? Strategy: resolve ~v into x and y (horizontal and vertical) components; then solve! y x = x0 + (v0 cos θ) t v x = v0 cos θ v0 sin θ y = y0 + (v0 sin θ) t − 12 g t2 vy = v0 sin θ − gt 20150220.2349 uct-physics-dga v 0 θ x v0 cos θ 122 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.5,4.6 What is the trajectory? Eliminating t from the previous component equation: y − y0 = (x − x0 ) tan θ − g (x − x0 )2 2v20 cos2 θ — the trajectory is a parabola. 20150220.2349 uct-physics-dga 123 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.5,4.6 The range The position at which the projectile returns to its initial height is the range. i.e. R = x − x0 for y − y0 = 0: Thus, R = 0 (trivially), or: 2v20 2v20 2 R= cos θ tan θ = cos θ sin θ g g = v20 sin 2θ g This is maximum for θ = 45◦ (sin 2θ = 1). 20150220.2349 uct-physics-dga 124 UCT EEE4106Z Introductory Nuclear Physics 2015 Example: projectile motion A projectile has launch speed of 20 m s−1 . What is maximum range? Calculate the launch angle needed to hit target 39 m downrange and clear a 3 m high wall that is 35 m downrange. 20150220.2349 uct-physics-dga 125 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.9 Relative motion Observer floating in river sees boat sailing south. Observer on shore sees boat sailing south-east. Motion depends on coordinate frame. 20150220.2349 uct-physics-dga 126 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 4.9 Relative motion (2) boat rBR rBS river R RS shore The velocity of the river observed from the shore vRS is constant. (From shore): rBS (t) = RRS (t) + rBR (t) (Velocity): vBS (t) = VRS + vBR (t) (Acceleration): aBS = aBR 20150220.2349 uct-physics-dga 127 UCT EEE4106Z Introductory Nuclear Physics 2015 Single or double cricket throw? A cricket fielder can throw a ball a maximum range of 50 m. Neglect air resistance. If he needs to throw a range R < 50 m, for which R is it quicker to make a single throw of range R; or a double throw of range R/2, using a second fielder who has a time lag of 0.4 s to catch and throw. What is speed of throw v0 ? R = (v20 /g) sin 2θ. For θ = π/4 max range Rmax = v20 /g v0 = (gRmax )1/2 20150220.2349 uct-physics-dga 128 UCT EEE4106Z Introductory Nuclear Physics 2015 Single throw: Choose v0 , and θ s given by R = (v20 /g) sin 2θ s sin 2θ s = R/Rmax Get . . . θ s R = v0 cos θ s t s so ts = 20150220.2349 uct-physics-dga R v0 cos θ s 129 UCT EEE4106Z Introductory Nuclear Physics 2015 Double throw: Choose v0 , and θd given by R/2 = (v20 /g) sin 2θd sin 2θd = R/2Rmax Get . . . θd td = R/2 R/2 + tlag + v0 cos θd v0 cos θd Investigate for what (if any) R is td < t s . Use computer. 20150220.2349 uct-physics-dga 130 UCT EEE4106Z Introductory Nuclear Physics 2015 End 05 20150220.2349 uct-physics-dga 131 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5, 6 Dynamics: force and motion Why does a body accelerate? Newton (1642-1727) showed that this arose from an interaction between the body and its surroundings. 20150220.2349 uct-physics-dga 132 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.2 Newton’s Laws Newton put his ideas in the form of three laws, paraphrased below: 1. A body will move with constant velocity unless a (net) force is impressed on it. 2. The acceleration is related to the force, a = F/m. 3. Force is an interaction. 20150220.2349 uct-physics-dga 133 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.2 Inertia The first law (N1) implies that bodies possess the property of inertia, i.e. a resistance to acceleration. The motion of the body will persist unless a force acts on it. This persistence implies there is no “natural state of rest” of a body (i.e.no absolute rest). 20150220.2349 uct-physics-dga 134 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.2 Inertial reference frames Suppose a body is moving at a constant velocity, with no external force acting on it. Then we can change the velocity by viewing it from another reference frame, moving at some other constant velocity. This is called an inertial reference frame. There is no natural state of rest, so both reference frames are “correct”: all inertial reference frames are equivalent. (This is the Galilean principle of relativity). 20150220.2349 uct-physics-dga 135 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.2 Inertial reference frames If a frame is not inertial: Bodies will accelerate. Experiments can show if frame is inertial. Is the surface of the Earth an inertial frame? Not for vertical motion (acceleration due to gravity). Approximately for horizontal motion (over small areas). 20150220.2349 uct-physics-dga 136 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.3 Force Force is the agency which produces an acceleration and changes the state of motion of a body. We can compare forces by comparing the acceleration of different bodies. Acceleration is a vector — is force? This has to be established experimentally — it is. The acceleration must then depend on the net or resultant force — the vector sum of the forces acting on the body. (Principle of superposition of forces). 20150220.2349 uct-physics-dga 137 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.4 Inertial mass Inertia is a property of an object: we use mass to measure inertia. We can use a given force to calibrate the mass of a body in terms of its acceleration, i.e. for the same force m1 a2 = m2 a1 20150220.2349 uct-physics-dga 138 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.5 Newton’s second law The net force on a body is equal to the product of mass and acceleration. X Fi = Fnet = ma Note that Fnet is the (vector) sum of the forces acting on the body. The unit newton is introduced for force: 1 N = 1 kg m s−2 . 20150220.2349 uct-physics-dga 139 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.5 Newton’s second law The law F = ma can be decomposed into components in a coordinate system: F x = ma x Fy = may Fz = maz 20150220.2349 uct-physics-dga 140 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.5 Newton’s second law If the net force acting on a body is 0, the acceleration of the body is zero. The body is then said to be in equilibrium. The net force involves the external force acting on a body. Internal forces do not affect the body as a whole. 20150220.2349 uct-physics-dga 141 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.7 Newton’s third law Force is an interaction between two bodies. If A exerts a force on B, then B exerts a force on A, and FA on B = −FB on A Note that these forces act on different bodies (and so don’t “cancel”). 20150220.2349 uct-physics-dga 142 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.6 Some common forces — force fields There are only a few fundamental forces, which we can regard as fields, i.e. they are defined at all points in space. e.g. the gravitational field, F = −G m1 m2 r̂ r2 where G = 6.67 × 10−11 N m2 kg−2 . 20150220.2349 uct-physics-dga 143 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.6 Some common forces — weight We call the magnitude of the force due to gravity acting on a body its weight. The force acting on a body on (or near) the surface of the Earth is F = −mgĵ where the y-axis has been chosen as “up”, and g = 9.81 m s−2 . Thus its weight is W = mg. 20150220.2349 uct-physics-dga 144 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.6 The normal force If we stand on the floor we do not accelerate towards the centre of the Earth. This is because the floor pushes back. This force is called a normal force (because it is normal or perpendicular to the surface). When a body presses against any solid surface, the surface exerts a normal force N on the body. 20150220.2349 uct-physics-dga 145 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.6 Some common forces — normal force (2) For a body on a horizontal surface we have: N − mg = may If the system is in equilibrium N − mg = 0 N Fg 20150220.2349 so N = mg y N x Fg uct-physics-dga 146 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.5 Free body diagram For problem solving, it is often useful to draw a free body diagram. We can then introduce a coordinate system and solve the equations using components. y N F1 N F1 θ F2 x F2 θ W W 20150220.2349 uct-physics-dga 147 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.5 Some common forces — tension If a cord/rope/string attached to a body is pulled, the cord exerts a force called a tension on the body. The tension is directed along the cord. T T’ F T2 T1 W2 W1 20150220.2349 uct-physics-dga 148 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.5 Some common forces — the spring A spring exerts a force with component F = −kx. This is known as Hooke’s law. 20150220.2349 uct-physics-dga 149 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 5.5 Some common forces — friction If a body slides (or attempts to slide) along a surface, there may be a force of friction acting along the surface, in the direction opposite to the direction of (intended) motion. 20150220.2349 uct-physics-dga 150 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 6.1 - 6.2 Friction There are two types of frictional force: For a stationary body, a static frictional force fS acts in equilibrium with an applied force F parallel to the surface. If F is large enough, the body moves, and experiences a kinetic frictional force fK . a=0 N N F fS uct-physics-dga F’ fK W 20150220.2349 a W 151 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 6.1 - 6.2 Friction — notes The magnitude of the static frictional force has a maximum value fS ,max = µS N . This defines the coefficient of static friction µS . If the magnitude of the external force parallel to fS is greater than fS ,max , the body will slide. If the body slides, the kinetic frictional force has the magnitude fK = µK N . This defines the coefficient of kinetic friction µK 20150220.2349 uct-physics-dga 152 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 6.1 - 6.2 Coefficients of friction µS and µK : depend on both body and surface; are dimensionless; must be determined by experiment. 20150220.2349 uct-physics-dga 153 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 6.1 - 6.2 Coefficients of friction — examples Rubber on concrete Steel on steel Metal on metal (lubricated) Teflon on Teflon 20150220.2349 uct-physics-dga µS µK 1.0 0.74 0.15 0.04 0.8 0.57 0.06 0.04 154 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 6.3 Drag forces A body moving in a fluid (air, liquid) experiences a drag force D which opposes the relative motion. The force is in the direction of fluid flow relative to the body. Over some range of velocity, the force acting on a body in air is D = 21 C ρ A v2 (C is the drag coefficient, ρ the air density, A the effective crosssectional area). 20150220.2349 uct-physics-dga 155 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 6.3 Falling body with drag D increases with v X W F = D + W = ma independent ofv Body accelerates until D = W = mg ie mg = 12 CρA v2t where the magnitude of the terminal velocity is s vt = 20150220.2349 uct-physics-dga 2mg CρA 156 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 6.2 Coin sliding on book - free body diagram 20150220.2349 uct-physics-dga 157 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 6 Initial velocity of braking car Car brakes from v0 to a stop in distance 290 m. Coefficient of friction of tyres on road µk = 0.60. Calculate v0 . 20150220.2349 uct-physics-dga 158 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5, HRW6 6 Initial velocity of braking car (2) V: N ĵ − mgĵ = 0 or N = mg H: − fk î = ma x î or a x = − fk /m but fk = µk N = µk mg So a x = −µk g = −(0.60)(10) = −6 m s−2 1-d motion with constant acceleration: v2 − v20 = 2a x (x − x0 ) with v = 0 and (x − x0 ) = 290 m. v20 = −2a x (x − x0 ) = −(2)(−6)(290) = 3480 v0 = 59.0 m s−1 (note: 3 sig. fig.) 20150220.2349 uct-physics-dga 159 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5-7, HRW6 6.4 Circular motion: centripetal acceleration Circular motion at uniform speed, but velocity changes. Acceleration a = (v2 /R)(−r̂) ie towards centre (centripetal) 20150220.2349 uct-physics-dga 160 UCT EEE4106Z Introductory Nuclear Physics 2015 Derivation of a = −(v2 /R)r̂ Circle of radius r = R. Arc length is s. Angle (in radians) θ = s/R Speed v = ds/dt = d(Rθ)/dt = Rω. Speed constant ⇒ ds/dt constant ⇒ dθ/dt = ω constant. Angular velocity ω is a vector. If motion is in (x, y)-plane, anticlockwise looking down onto plane, the Right Hand rule gives ω = ωk̂. Remember î × ĵ = k̂. 20150220.2349 uct-physics-dga 161 UCT EEE4106Z Introductory Nuclear Physics 2015 Derivation of a = −(v2 /R)r̂ (2) Particle going in circle (of radius R in (x, y)-plane) has position described by (x, y) or (r, φ). Note r = R and x2 + y2 = r2 . Variables x, y and φ change with time. φ = ωt and x = r cos φ = r cos ωt, y = r sin φ = r sin ωt Useful unit vectors: r̂ = cos φî + sin φĵ. φ̂ = − sin φî + cos φĵ 20150220.2349 uct-physics-dga (note cos2 φ + sin2 φ = 1 ) (Check r̂ · φ̂ = 0) 162 UCT EEE4106Z Introductory Nuclear Physics 2015 Derivation of a = −(v2 /R)r̂ (3) Velocity v changes direction, but has constant magnitude. v = = a = = = vφ̂ = Rωφ̂ Rω(− sin φî + cos φĵ) dv/dt Rω(− cos φωî − sin φωĵ) −Rω2 r̂ v2 = − r̂ [v = ωR] R [dφ/dt = ω] Centripetal acceleration towards centre of circle. 20150220.2349 uct-physics-dga 163 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5-7, HRW6 6-4 Centripetal force 20150220.2349 uct-physics-dga 164 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5-7, HRW6 6-4 Centripetal force: friction from tyres Skidding starts at f = µN V: N − mg = 0 H: −µN = −mv2 /R N = mg µ = v2 /gR 20150220.2349 uct-physics-dga 165 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 5-7, HRW6 6-4 Centripetal force: banking No friction µ = 0. Bank road at angle θ. v = Max speed to go round. V: N cos θ − mg = 0 H: −N sin θ = −mv2 /R N = mg/(cos θ) v2 = gR tan θ 20150220.2349 uct-physics-dga 166 UCT EEE4106Z Introductory Nuclear Physics 2015 End 06 20150220.2349 uct-physics-dga 167 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7 Energy Often, we don’t have enough information to solve Newton’s laws of motion, or the solution is very complicated. The concept of energy leads to useful techniques to solve problems. One of the most profound observations of physics is that the total energy of an isolated system remains constant. 20150220.2349 uct-physics-dga 168 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7 Energy is a familiar concept Vague idea: kinetic energy plus potential energy is constant 20150220.2349 uct-physics-dga 169 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7.1 What is energy Energy is a number (a scalar) which can be associated with a system of objects — it labels the state of the system. There are many “kinds” of energy. We start with the state of motion of an object. The form of energy in this case is kinetic energy, 1 K = mv2 2 20150220.2349 uct-physics-dga 170 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9-2, HRW6 7.3 Kinetic energy The faster an object moves, the more kinetic energy it has; the more massive an object is, the more kinetic energy it has. The unit of kinetic energy is the joule. Symbol J. 1 J = 1 kg m2 s−2 . 20150220.2349 uct-physics-dga 171 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9-3, HRW6 7.3 Work We can make an object accelerate, or decelerate, by exerting an external force on it. This increases or decreases the kinetic energy of the object. We can say that kinetic energy is transferred to, or from, the object by the agency of the force. This transfer of energy is referred to as work. 20150220.2349 uct-physics-dga 172 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9-2, HRW6 7.3 Work and kinetic energy v0 v F d Consider a body sliding on a horizontal surface, under action of a constant force, no friction. After displacement d v2x = v2x0 + 2a x d Fx = v2x0 + 2 d m 1 mv2x − 12 mv2x0 = F x d 2 20150220.2349 uct-physics-dga 173 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9-2, HRW6 7.3 Work and kinetic energy We can thus identify the work as W = F x d = F cos θ d. i.e. the work is given by the scalar product, ~ · ~d W=F Thus forces which are perpendicular to the direction of motion do no work. 20150220.2349 uct-physics-dga 174 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 7.3 HRW6 Sample Problem 7-2: Sliding the safe Safe m = 225 kg. Floor µ = 0. Displacement d = 8.50 m. Spy one pushes F1 = 12.0 N, 30◦ below horizontal. Spy two pulls F2 = 10.0 N, 40◦ above horizontal. 20150220.2349 uct-physics-dga 175 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 7.3 a) Calculate net work on safe by F1 and F2 . W1 = = = W2 = = = W = 20150220.2349 uct-physics-dga F1 · d = F1 d cos φ1 (12.0 N)(8.5 m) cos 30◦ 88.33 J F2 · d = F2 d cos 400 (10.0 N)(8.5 m) cos 40◦ 65.11 J W1 + W2 = 153 J 176 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 7.3 b) What is work done on safe by weight, and by normal force? Wg = Fg · d = mg d cos π/2 = 0 WN = N · d = N d cos π/2 = 0 If force is orthogonal to the displacement, it does no work. 20150220.2349 uct-physics-dga 177 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9-2, HRW6 7.3 Work-kinetic energy theorem For a body moving horizontally, even with a varying acceleration (or force): W = 21 mv2x − 12 mv2x0 = K f − Ki = ∆K 20150220.2349 uct-physics-dga 178 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7.3 Work and kinetic energy with friction Same system, now with friction and a constant applied force. v0 v F fk d After displacement d F x d = 12 mv2x − 21 mv2x0 + fk d = ∆K + fk d 20150220.2349 uct-physics-dga 179 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7.3 Work and kinetic energy with friction Part of the net work done on the system by the external work is dissipated into thermal energy by the action of the frictional force: ∆Eth = fk d and so W = ∆ K + ∆Eth in this case. 20150220.2349 uct-physics-dga 180 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7.4 Work done by the gravitational force Body projected upwards. The weight is a constant, external, force acting on the body. f yf ∆r yi φ i mg 20150220.2349 uct-physics-dga 181 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7.4 Work done by gravity is: Wg = mg ∆y cos φ = −mg ∆y = K f − Ki = 21 m(v2f − v2i ) <0 20150220.2349 uct-physics-dga 182 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7.4 Work done by the gravitational force (2) Body in free fall. yi φ i mg ∆r yf 20150220.2349 f uct-physics-dga 183 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7.4 Work done by gravity is: Wg = mg ∆y cos φ = mg ∆y = K f − Ki = 21 m(v2f − v2i ) >0 For a “round trip” (up and down to same spot) Wg = 0. 20150220.2349 uct-physics-dga 184 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7.4 Lifting an object ~ to lift an object. Apply a force F Take the object at rest before and after. 20150220.2349 uct-physics-dga 185 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7.4 F mg ~v f = ~vi = 0 ∆K = K f − Ki = 0 =W = Wa + Wg i.e. Wa = −Wg : (work done by the applied force) = - (work done by the weight). 20150220.2349 uct-physics-dga 186 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7.4 Work done by a variable force In one dimension, W= xf Z F x (x)dx xi In 3-d In one dimension, W= f Z ~ r) · d~r F(~ i 20150220.2349 uct-physics-dga 187 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9-5, HRW6 7.5 Force of a spring For an ideal spring, the force is proportional to the displacement of the free end of the spring from a relaxed state ~ = −k ~d F or, with an x axis along the spring axis, F = −k ∆x This is known as Hooke’s law. 20150220.2349 uct-physics-dga 188 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9-5, HRW6 7.5 Work done by a spring What is the work done by a spring force on some system, as the spring changes its extension from xi to x f ? Since the force varies with x, the work has to be evaluated by integration, Z xf Ws = F dx xi =− xf Z kx dx xi = 12 kxi2 − 12 kx2f 20150220.2349 uct-physics-dga 189 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9-5, HRW6 7.5 Work and springs Special case (spring initially relaxed):take xi = 0, x f = x. work done by the spring is W s = − 21 kx2 . work done by an applied force on the spring is Wa = −W s = 12 kx2 . the work done in compression ( x < 0) and extension ( x > 0) for the same magnitude of displacement is the same. 20150220.2349 uct-physics-dga 190 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7.7 Work and power It is often of interest to know “how fast” work is done. The rate at which work is done is measured by power. The average power developed by a force in a time interval ∆t is defined as P̄ = W ∆t Power is measured in watts: 1 W = 1 J/s. Alternately, energy, especially electrical energy, is sometimes measured in Ws (or, e.g. kWh). 20150220.2349 uct-physics-dga 191 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 9, HRW6 7.7 Power The instantaneous power is defined as P= dW dt For a constant force over an infinitesimal time interval dt, P= 20150220.2349 uct-physics-dga ~ · d~r F ~ · d~r = F ~ · ~v =F dt dt 192 UCT EEE4106Z Introductory Nuclear Physics 2015 End 07 20150220.2349 uct-physics-dga 193 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10-1, HRW6 8.1 Potential energy Potential energy is associated with the configuration of a system of objects exerting forces on one another. e.g. gravitational potential energy: state of separation of gravitating bodies. elastic potential energy: state of compression or extension of a spring. 20150220.2349 uct-physics-dga 194 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10-2, HRW6 8 Potential energy A force does work and transfers, e.g., energy to part of a system and increases its kinetic energy, W = ∆K = K f − Ki We now argue that energy is transferred from part of the system which decreases its potential energy: W = −∆U = −(U f − Ui ) 20150220.2349 uct-physics-dga 195 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10, HRW6 8 Kinetic and potential energy The forces within a system can transfer energy between kinetic and potential forms. We will see that we can define a mechanical energy Emec . For a system with no external forces acting (an isolated system) ∆Emec = ∆U + ∆K 20150220.2349 uct-physics-dga 196 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10-4, HRW6 8.3 Work done by gravitational force 20150220.2349 uct-physics-dga 197 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10-4, HRW6 8.3 Gravitational potential energy Gravitational force: change in gravitational potential energy. Take ĵ vertically up. Particle m, initially thrown up, is slowed down and stops, then falls. It rises distance ∆y ∆U == −Wg = −Fg · ∆y ĵ = −(−mgĵ) · ∆y ĵ = +mg ∆y 20150220.2349 uct-physics-dga 198 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10-4, HRW6 8.3 Work done by spring force 20150220.2349 uct-physics-dga 199 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10-4, HRW6 8.3 Elastic potential energy of compressed spring Spring force: change in elastic potential energy. A moving mass hits a spring, and gets slowed down ∆U = −W s = − xf Z (−kx) dx xi = 12 k x2f − 12 k xi2 Take origin at point where spring is uncompressed: At x = 0, Ui = 0 ∆U = U f − 0 = 12 k x2f U(x) = 12 k x2 20150220.2349 uct-physics-dga 200 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10-3, HRW6 8 Defining a potential energy Only changes in potential energy are meaningful. Still, we can define a potential energy relative to some arbitrary reference point. Gravitational force (y up). Take U = 0 at yi = 0 Then U(y) = mgy Spring force. Take U = 0 for the relaxed state, xi = 0 Then 20150220.2349 uct-physics-dga U(x) = 21 k x2 201 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10, HRW6 8 Recognising a potential energy Take a system of two or more objects, interacting via forces. Change the configuration: forces do work W1 ; energy is transferred between kinetic energy and another form. If the change is reversed: forces do work W2 to reverse the energy change. If the system is reversible, and W1 = −W2 , the force is conservative and the other form of energy is potential energy. 20150220.2349 uct-physics-dga 202 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10, HRW6 8 Potential energy and forces In order to define a potential energy in a system, changes must be reversible. Conservative force W1 = −W2 Non-conservative force e.g friction: kinetic energy transferred to thermal energy — many degrees of freedom. Dissipation of energy. 20150220.2349 uct-physics-dga 203 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10, HRW6 8 Conservative forces Net work done by a conservative force on a body moving on a closed path is zero. Work done by a conservative force between two points A and B does not depend on the path. 20150220.2349 uct-physics-dga 204 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10, HRW6 8.4 Mechanical energy We define a mechanical energy Emec = U + K For a system with no external forces acting (an isolated system) ∆Emec = ∆U + ∆K =0 i.e. ∆K = −∆U 20150220.2349 uct-physics-dga 205 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10-4, HRW6 8.4 Conservation of mechanical energy ∆K = −∆U Rearranging this: K1 + U1 = K2 + U2 i.e. the mechanical energy is the same for all states of the system — mechanical energy is conserved. 20150220.2349 uct-physics-dga 206 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10-4, HRW6 8.4 Kinetic and potential energy in oscillator 20150220.2349 uct-physics-dga 207 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10, HRW6 8.5 Force from potential energy For 1-d motion, and small ∆x: ∆U(x) = −W = F(x)∆x Thus force can be obtained from potential energy: F(x) = − 20150220.2349 uct-physics-dga dU(x) dx 208 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10, HRW6 8.5 Potential energy curve 20150220.2349 uct-physics-dga 209 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10, HRW6 8.6 External forces For external force acting on a mechanical system, work W done by the external force is W = ∆U + ∆K = ∆Emec If there are frictional forces acting, W = ∆Emec + ∆Ethermal = ∆Emec + fk d 20150220.2349 uct-physics-dga 210 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 10-7, HRW6 8.7 Conservation of energy The total energy of an isolated system cannot change. The total energy of a system can only be changed by transferring energy to or from the system, by e.g. work or heat. 20150220.2349 uct-physics-dga 211 UCT EEE4106Z Introductory Nuclear Physics 2015 Particle slides on hemispherical dome Starting at top, a particle slides down a frictionless hemispherical dome. Where does it leave the surface? FBD: in radial direction: N − mg cos θ = −mv2 /R As v increases, cos θ decreases, so N must decrease. When N = 0, particle leaves surface. At this point v2 = Rg cos θ 20150220.2349 uct-physics-dga 212 UCT EEE4106Z Introductory Nuclear Physics 2015 Conservation of energy 0 + mgR = (1/2)mv2 + mgR cos θ gR = gR cos θ( 21 + 1) cos θ = 2/3 Height is 2R/3 20150220.2349 uct-physics-dga 213 UCT EEE4106Z Introductory Nuclear Physics 2015 End 08 20150220.2349 uct-physics-dga 214 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 Ch 09 Systems of particles 9-1 A special point 9-2 The centre of mass 9-3 Newton’s 2nd law for a system of particles 9-4 Linear momentum 9-5 Linear momentum of a system of particles 9-6 Conservation of linear momentum 9-7 Systems with varying mass: rocket 9-8 External forces and internal energy changes 20150220.2349 uct-physics-dga 215 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW6 Ch 10 Collisions 10-1 What is a collision 10-2 Impulse and linear momentum 10-3 Momentum and kinetic energy in collisions 10-4 Inelastic collisions in one dimension 10-5 Elastic collisions in one dimension 10-6 Collisions in two dimensions 20150220.2349 uct-physics-dga 216 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 07 Translational Momentum 7-1 Collisions and explosions 7-2 Translational momentum of a particle 7-3 Isolated systems of particles 7-4 Impulse and momentum change 7-5 Newton’s laws and momentum conservation 7-6 Simple collisions and conservation of momentum 7-8 Conservation of momentum in two dimensions 7-8 System with mass exchange: rocket and ejected fuel 20150220.2349 uct-physics-dga 217 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC Ch 08 Extended systems 8-1 The motion of complex objects 8-2 Defining the position of a complex object 8-3 The effective position - centre of mass 8-4 Locating a system’s centre of mass 8-5 Newton’s laws for a system of particles 8-6 The momentum of a particle system 20150220.2349 uct-physics-dga 218 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7-1, HRW6 9-1 Systems of Particles 20150220.2349 uct-physics-dga 219 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.2 Centre of mass System consists of two masses which collide (and conserve momentum). How do we interpret Newton 1 in this context? initial p 1i 20150220.2349 p 2i=0 uct-physics-dga final p 1f p 2f 220 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.2 Centre of mass Define the centre of mass as the weighted mean position of the system’s mass: xcm = m1 x1 + m2 x2 m1 + m2 x x1 20150220.2349 uct-physics-dga x cm x 2 221 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.2 Centre of mass In general, for many particles:: xcm 20150220.2349 uct-physics-dga P mi xi = P mi 222 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.2 Centre of mass In 3-d: xcm 20150220.2349 P mi x i = P mi uct-physics-dga ~rcm P mi~ri = P mi ycm P mi yi = P mi zcm P mi zi = P mi 223 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 8-1, HRW6 9.2 Using symmetry to estimate centre of mass HRW6 §9-2 CP9-1 Where is CM? : a) originally b) after removal of 1 c) after removal of 1 d) after removal of 1 e) after removal of 1 f) after removal of all 20150220.2349 uct-physics-dga and 2 , 2 & 3 and 3 i 224 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 8-1, HRW6 9.2 Centre of mass of 3 point masses HRW6 §9-2 SP9-1 CLRC §8-1 TE8-1 m1 = 1.2 kg, m2 = 2.5 kg, m3 = 3.4 kg equilat. triangle, a = 140 cm xcm = (1/M)(m1 x1 + m2 x2 + m3 x3 ) xcm = (1/7.1)((1.2)(0) + (2.5)(140) + (3.4)(70)) ycm = . . . = 58 cm 20150220.2349 uct-physics-dga 225 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.2 Motion of the centre of mass Take derivative of ~rcm , use M = ~vcm = P mi : d~rcm 1 X d~ri 1 X = mi = mi~vi dt M dt M or, (looking ahead) use momentum p = mv M~vcm = 20150220.2349 uct-physics-dga X mi~vi = X ~pi = ~ptot 226 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.2 Force on centre of mass Take derivative of ~vcm : M~acm = Hence X 20150220.2349 uct-physics-dga X mi~ai = X ~i F ~ ext = M~acm = d~ptot F dt 227 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.2 Ballet grand jeté Centre-of-mass moves in parabola. Arm, leg movement lifts cm in body. Head travels nearly horizontally, appears to float. 20150220.2349 uct-physics-dga 228 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9 Momentum Momentum is another conserved quantity in isolated systems. It is especially useful for dealing with systems of several particles. 20150220.2349 uct-physics-dga 229 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9 Linear momentum The linear momentum of a body of mass m moving with a velocity ~v is defined as ~p = m~v Momentum is a vector. The units of momentum are kg m/s. (Despite its importance, momentum does not have a special name for its unit). 20150220.2349 uct-physics-dga 230 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.4 Linear momentum and force Newton’s second law can be restated using momentum: X ~ = d~p = d(m~v) F dt dt If the particle is isolated, time. (= m~a) P~ F = 0 and hence ~p is constant in This restates Newton’s first law; we can (trivially) say also that ~p is conserved in this case. 20150220.2349 uct-physics-dga 231 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.5 Linear momentum: two particles Two isolated particles, interacting via some force. X ~ 21 = d~p1 F dt and X ~ 12 = d~p2 F dt p2 F12 m2 20150220.2349 uct-physics-dga F21 p1 m1 232 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.5 Linear momentum: two particles By Newton’s third law, X and so ~ 21 + F X ~ 12 = 0 F d~p1 d~p2 d + = (~p1 + ~p2 ) = 0 dt dt dt so that ~p1 + ~p2 is conserved. 20150220.2349 uct-physics-dga 233 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.6 Linear momentum: conservation The total momentum of an isolated system of two (or more) particles is conserved. ~ptot = ~p1 + ~p2 = constant or equivalently, ~p1i + ~p2i = ~p1 f + ~p2 f 20150220.2349 uct-physics-dga 234 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7-8, HRW6 9-7 Systems with varying mass: rocket Rocket works by throwing away mass. F= 20150220.2349 uct-physics-dga d (mv) dt 235 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7-8, HRW6 9-7 Rocket equation Rocket of mass M travels at velocity v. Velocity of exhaust gas is U . In time dt, mass of rocket increases by dM . Notice dM < 0 pi = Mv No external forces so momentum conserved pi = p f p f = −dM U + (M + dM)(v + dv) 20150220.2349 uct-physics-dga 236 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7-8, HRW6 9-7 Rocket equation: thrust Mv = −dM U + Mv + dM v + M dv + dM dv vrel is velocity of rocket relative to exhaust gas. Relative velocities: (v + dv) = vrel + U or U = v + dv − vvel Mv = −dM v − dM dv + dM vrel +Mv + dM v + M dv + dM dv −dM vrel = M dv dM dv − vrel = M dt dt Rvrel = Ma where R = −dM/dt > 0 is positive burn rate, Rvrel is thrust. 20150220.2349 uct-physics-dga 237 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7-8, HRW6 9-7 Rocket equation: velocity −dM vrel = M dv dM dv = −vrel M Z vf Z Mf dM dv = −vrel M vi Mi Mi v f − vi = vrel ln Mf To get high v f requires high exhaust velocity vrel , or M f → 0, ie high fuel/payload ratio. 20150220.2349 uct-physics-dga 238 UCT EEE4106Z Introductory Nuclear Physics 2015 End 09 20150220.2349 uct-physics-dga 239 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9,10 Collisions Collision involves an interactions between two particles. Forces may be large, but time of interaction is short. Concept of impulse is needed. Impulse is related to change of momentum. Interaction causes an action-reaction pair of forces. Momentum is always conserved (no acceleration of centre of mass) Energy may be conserved; it is in elastic collisions 20150220.2349 uct-physics-dga 240 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9 Linear momentum The linear momentum of a body of mass m moving with a velocity ~v is defined as ~p = m~v Momentum is a vector. The units of momentum are kg m/s. (Despite its importance, momentum does not have a special name for its unit). 20150220.2349 uct-physics-dga 241 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.4 Linear momentum and force Newton’s second law can be restated using momentum: X ~ = d~p = d(m~v) F dt dt If the particle is isolated, time. (= m~a) P~ F = 0 and hence ~p is constant in This restates Newton’s first law; we can (trivially) say also that ~p is conserved in this case. 20150220.2349 uct-physics-dga 242 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 10.2 Impulse Over some small time interval ∆t ~ = ∆~p F ∆t Then ~ ∆t ∆~p = F ~ ∆t is known as the impulse. The term F 20150220.2349 uct-physics-dga 243 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.5 Linear momentum: two particles Two isolated particles, interacting via some force. X ~ 21 = d~p1 F dt X and ~ 12 = d~p2 F dt p2 F12 m2 20150220.2349 uct-physics-dga F21 p1 m1 244 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.5 Linear momentum: two particles By Newton’s third law, X and so ~ 21 + F X ~ 12 = 0 F d~p1 d~p2 d + = (~p1 + ~p2 ) = 0 dt dt dt so that ~p1 + ~p2 is conserved. 20150220.2349 uct-physics-dga 245 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.6 Linear momentum: conservation The total momentum of an isolated system of two (or more) particles is conserved. ~ptot = ~p1 + ~p2 = constant or equivalently, ~p1i + ~p2i = ~p1 f + ~p2 f 20150220.2349 uct-physics-dga 246 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 10.2 Impulse - momentum theorem ~ dt. Then Over some small time interval dt, d~p = F ∆~p = ~p f − ~pi = Z tf ~ dt F ti The impulse is defined by I= Z tf ~ dt = ∆~p F ti F ti 20150220.2349 tf uct-physics-dga t 247 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 10.2 Impulse - momentum theorem Treat force as an average over time: ~¯ = ∆~p I = F∆t F F ti 20150220.2349 uct-physics-dga tf t 248 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.6 Collisions Particle 2 exerts a force on particle 1: ∆~p1 = Z tf ~ 21 dt F ti At the same time particle 1 exerts a force on particle 2: ∆~p2 = Z tf ~ 12 dt F ti F F12 t F21 20150220.2349 uct-physics-dga 249 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.6 Momentum in collisions ~ 12 = −F ~ 21 By Newton’s third law, F Hence ∆~p1 = −∆~p2 ∆~p1 + ∆~p2 = 0 i.e. ~ptot = ~p1 + ~p2 = constant 20150220.2349 uct-physics-dga 250 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 10.3 Types of collisions Is energy conserved in collisions? Elastic collisions Kinetic energy is conserved during the collision Inelastic collisions Kinetic energy is not conserved during the collision Fully inelastic collisions No relative motion of particles after collision 20150220.2349 uct-physics-dga 251 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 10.3 Inelastic collisions After, m1 and m2 move together with ~v f . Then m1~v1i + m2~v2i = (m1 + m2 )~v f ~v f = m1~v1i + m2~v2i m1 + m2 1-d, m2 initially at rest: vf = 20150220.2349 uct-physics-dga m1 v1i m1 + m2 252 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 10.3 Elastic collisions Momentum and kinetic energy are constant in the collision: m1~v1i + m2~v2i = m1~v1 f + m2~v2 f 1 m v2 2 1 1i + 21 m2 v22i = 21 m1 v21 f + 12 m2 v22 f Best solved individually for each case. 20150220.2349 uct-physics-dga 253 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 10.3 Elastic collisions — 1d Special case: m2 initially at rest. 20150220.2349 uct-physics-dga v1 f = m1 − m2 v1i m1 + m2 v2 f = 2m1 v1i m1 + m2 254 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 10.3 Elastic collisions — 1d General case: 20150220.2349 v1 f = m1 − m2 2m2 v1i + v2i m1 + m2 m1 + m2 v2 f = m2 − m1 2m1 v1i + v2i m1 + m2 m1 + m2 uct-physics-dga 255 UCT EEE4106Z Introductory Nuclear Physics 2015 CLRC 7, HRW6 9.2 Centre of mass System consists of two masses which collide and conserve momentum. initial p 1i p 2i=0 final p 1f p 2f How do we interpret Newton 1 in this context? 20150220.2349 uct-physics-dga 256 UCT EEE4106Z Introductory Nuclear Physics 2015 . Review of introductory mechanics Velocity v = dx/dt. Acceleration a = dv/dt = d2 x/dt2 If non-relativistic v c, Newton’s law F = ma = dp/dt where momentum p = mv Momentum conserved: P = pi . For isolated system dP/dt = 0 R WorkR W = F · dx R In 1-d: W = F dx R W = m(dv/dt) dx = m(dv/dx)(dx/dt)dx = mv dv " #v Z vf 1 2 f 1 2 1 2 W= mv dv = mv = mv f − mvi 2 2 2 vi vi P R Kinetic energy Ek = mv2 /2 If conservative: Total energy E = Ek + U where R potential energy U is integral of force. U = − F·dx or F = −∇U 20150220.2349 uct-physics-dga 257 UCT EEE4106Z Introductory Nuclear Physics 2015 . Collisions A collision is a (rapid) interaction between two (or more) bodies. No external forces act during collision. Momentum always conserved in a collision. Energy always conserved in a collision. Kinetic energy may or may not be conserved In elastic collisions kinetic energy is conserved. (eg billiard ball collision). In inelastic collisions kinetic energy is not conserved. (eg two lumps of putty) 20150220.2349 uct-physics-dga 258 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW7 §9-10 Elastic collisions: 1-d Momentum and kinetic energy are conserved in the collision: m1~v1i + m2~v2i = m1~v1 f + m2~v2 f 1 m v2 2 1 1i + 21 m2 v22i = 21 m1 v21 f + 12 m2 v22 f Best solved individually for each case. 20150220.2349 uct-physics-dga 259 HRW7 §9-10 UCT EEE4106Z Introductory Nuclear Physics 2015 Elastic collisions: 1-d, v2 = 0 Special case: m2 initially at rest. v1 f = m1 − m2 v1i m1 + m2 v2 f = 2m1 v1i m1 + m2 Equal masses m1 = m2 : Maasive target m2 m1 : v1 f = 0 and v2 f = v1i v1 f ' −v1i and v2 f ' (2m1 /m2 )v1i Massive projectile m1 m2 : 20150220.2349 uct-physics-dga v1 f ' v1i and v2 f ' 2v1i 260 UCT EEE4106Z Introductory Nuclear Physics 2015 HRW7 §9-10 Elastic collisions: 1-d, general case General case: Algebra . . . gives . . . v1 f = m1 − m2 2m2 v1i + v2i m1 + m2 m1 + m2 v2 f = 2m1 m2 − m1 v1i + v2i m1 + m2 m1 + m2 OR view from centre of mass frame. 20150220.2349 uct-physics-dga 261 UCT EEE4106Z Introductory Nuclear Physics 2015 . Elastic collisions: 1-d, general case (1) 20150220.2349 uct-physics-dga 262 UCT EEE4106Z Introductory Nuclear Physics 2015 . Elastic collisions: 1-d, general case (2) 20150220.2349 uct-physics-dga 263 UCT EEE4106Z Introductory Nuclear Physics 2015 . Elastic collisions: 1-d, general case (3) 20150220.2349 uct-physics-dga 264 UCT EEE4106Z Introductory Nuclear Physics 2015 . Elastic collisions: 1-d, general case (4) 20150220.2349 uct-physics-dga 265 UCT EEE4106Z Introductory Nuclear Physics 2015 . Elastic collisions: 1-d, general case (5) 20150220.2349 uct-physics-dga 266 UCT EEE4106Z Introductory Nuclear Physics 2015 . Elastic collisions: 1-d, general case (6) 20150220.2349 uct-physics-dga 267 UCT EEE4106Z Introductory Nuclear Physics 2015 . Elastic collisions: 1-d, general case (7) 20150220.2349 uct-physics-dga 268 UCT EEE4106Z Introductory Nuclear Physics 2015 . Proton proton elastic collisions 20150220.2349 uct-physics-dga 269 UCT EEE4106Z Introductory Nuclear Physics 2015 * End EEE4106Z 01-02-00 Mechanics * 20150220.2349 uct-physics-dga 270
