REGAN PHY34210 PHYS 34210 PHYSICS I Notre Dame, London Programme, Fall 2013 Prof. Paddy Regan Dept. of Physics, University of Surrey, Guildford, GU2 7XH, UK E-Mail: [email protected] 1 • • • Course & General Information REGAN PHY34210 Lectures, usually, Tuesdays 2.15-5.00 – first lecture Tues 27th August 2010 – One ‘makeup’ lecture Mon. 30th Sept. 5.15 – 8pm (no class on Tues. 29th Oct) Grading – 3 x 2 hour class examinations • Exam 1 : Tues. 24th September (30%); • Exam 2: Tues 5th November (35%), • Exam 3: Tues. 26th November (35%) Some information about Prof. Paddy Regan FInstP CPhys: – National Physical Lab. & University of Surrey Chair Professor in Radionuclide Metrology, ( staff since 1994). – BSc University of Liverpool (1988); DPhil University of York (1991). – Adjunct Assoc. Prof. at ND London 2002-7; Full Professor from 2007 - present – Held post-doctoral research positions at: • University of Pennsylvania, Philadelphia, USA (1991-2) • Australian National University, Canberra, Australia (1992-4); • Yale University (sabbatical researcher 2002 ; Flint Visiting Research Fellow 2004 – 2013) – Co-author of >200 papers in nuclear physics; supervised 25 PhD students so far + 100 Masters. – Led RISING and PreSPEC projects (major nuclear physics research project at GSI, Germany). – Married (to a nurse), 4 kids. – Understands gridiron, baseball, (ice) hockey etc., regular visitor to US (and other countries) – Still plays squash and golf (poor, 27); formerly football (soccer), cricket & a bit of rugby (union). – Occasional half marathons for the mental health charity, MIND (see – http://uk.virginmoneygiving.com/Paddy-James-Clare-Regan – Have also done some (physics related) media work in the UK and USA, see e.g., – http://www.bbc.co.uk/news/world-asia-pacific-12744973 2 REGAN PHY34210 Course textbook, Fundamentals of Physics, Halliday, Resnick & Walker, published by Wiley & Sons. Now in 9th Edition. http://www.wiley.com/WileyCDA/WileyTitle/productCd-EHEP001575.html 3 Course Timetable (2013) PART 1 Lect 1: 27 Aug (Cp 1,2) Lect 2: 03 Sept (Cp 3,4) Lect 3: 10 Sept (Cp 5,6) Lect 4: 17 Sept (revision) Lect 5: 24 Sept Exam 1 PART 2 * Lect 6: Mon. 30th Sept. (Ch. 7,8) 5.15 - 8pm * Lect 7: 01 Oct (Ch 9,10) * Lect 8: 08 Oct. (Ch 11,12) * Lect 9: 15 Oct. (revision) REGAN PHY34210 4 PART 3 Lect 11: 12 Nov. (13,14) Lect 12: 19 Nov. (15,16) Lect 13: Weds. 20 Nov (17,18) 5.15-8pm. Lect 14: 26 Nov Exam 3 Break, no lect. 22nd Oct. No lect. 29th Oct (resched). * Lect 10: 5th Nov Exam 2 Course notes and past papers/solns can be found at the following link: http://personal.ph.surrey.ac.uk/~phs1pr/lecture_notes/notre_dame/ • 1: Measurement 1st Section: REGAN PHY34210 – Units, length, time, mass • 2: Motion in 1 Dimension – displacement, velocity, acceleration • 3: Vectors – adding vectors & scalars, components, dot and cross products • 4: Motion in 2 & 3 Dimensions – position, displacement, velocity, acceleration, projectiles, motion in a circle, relative motion • 5: Force and Motion: Part 1 – Newton’s laws, gravity, tension • 6: Force and Motion: Part 2 – Friction, drag and terminal speed, motion in a circle 5 2nd Section: REGAN PHY34210 • 7: Kinetic Energy and Work – Work & kinetic energy, gravitational work, Hooke’s law, power. • 8: Potential Energy and Conservation of Energy – Potential energy, paths, conservation of mechanical energy. • 9: Systems of Particles – Centre of mass, Newton’s 2nd law, rockets, impulse, • 10: Collisions. – Collisions in 1 and 2-D • 11 : Rotation – angular displacement, velocity & acceleration, linear and angular relations, moment of inertia, torque. • 12: Rolling, Torque and Angular Momentum – KE, Torque, ang. mom., Newton’s 2nd law, rigid body rotation 6 3rd section: REGAN PHY34210 • 13: Equilibrium and Elasticity – equilibrium, centre of gravity, elasticity, stress and strain. • 14: Gravitation – Newton’s law, gravitational potential energy, Kepler’s laws. • 15: Fluids – density and pressure, Pascal’s principle, Bernoulli’s equation. • 16 : Oscillations – Simp. Harm. Mo. force and energy, pendulums, damped motion. • 17 & 18 : Waves I and II – Types of Waves, wavelength and frequency, interference, standing waves, sound waves, beats, Doppler effect. 7 Recommended Problems and Lecture Notes. REGAN PHY34210 Problems are provided at the end of each book chapter. Previous years examinations papers will also be provided with solutions (later) for students to work through at their leisure. No marks will be give for these extra homework problems Final grade will come from the three class exams. Full lecture notes can be found on the web at http://www.ph.surrey.ac.uk/~phs1pr/lecture_notes/phy34210_13.ppt and http://www.ph.surrey.ac.uk/~phs1pr/lecture_notes/phy34210_13.pdf 8 1: Measurement REGAN PHY34210 Physical quantities are measured in specific UNITS, i.e., by comparison to a reference STANDARD. The definition of these standards should be practical for the measurements they are to describe (i.e., you can’t use a ruler to measure the radius of an atom!) Most physical quantities are not independent of each other (e.g. speed = distance / time). Thus, it often possible to define all other quantities in terms of BASE STANDARDS including length (metre), mass (kg) and time (second). 9 SI Units REGAN PHY34210 10 The 14th General Conference of Weights and Measures (1971) chose 7 base quantities, to form the International System of Units (Systeme Internationale = SI). There are also DERIVED UNITS, defined in terms of BASE UNITS, e.g. 1 Watt (W) = unit of Power = 1 Kg.m2/sec2 per sec = 1 Kg.m2/s3 Scientific Notation In many areas of physics, the measurements correspond to very large or small values of the base units (e.g. atomic radius ~0.0000000001 m). This can be reduced in scientific notation to the ‘power of 10’ ( i.e., number of zeros before (+) or after (-) the decimal place). e.g. 3,560,000,000m = 3.56 x 109 m = 3.9 E+9m & 0.000 000 492 s = 4.92x10-7 s = 4.92 E-7s Prefixes For convenience, sometimes, when dealing with large or small units, it is common to use a prefix to describe a specific power of 10 with which to multiply the unit. e.g. 1000 m = 103 m = 1E+3 m = 1 km 0.000 000 000 1 m = 10-10 m = 0.1 nm • • • • • • • • • REGAN PHY34210 1012 = Tera = T 109 = Giga = G 106 = Mega = M 103 = Kilo = k 10-3 = milli = m 10-6 = micro = m 10-9 = nano = n 10-12 = pico = p 10-15 = femto = f 11 Converting Units REGAN PHY34210 It is common to have to convert between different systems of units (e.g., Miles per hour and metres per second). This can be done most easily using the CHAIN LINK METHOD, where the original value is multiplied by a CONVERSION FACTOR. NB. When multiplying through using this method, make sure you keep the ORIGINAL UNITS in the expression e.g., 1 minute = 60 seconds, therefore (1 min / 60 secs) = 1 and (60 secs / 1 min) = 1 Note that 60 does not equal 1 though! Therefore, to convert 180 seconds into minutes, 180 secs = (180 secs) x (1 min/ 60 secs) = 3 x 1 min = 3 min. 12 Length (Metres) REGAN PHY34210 13 Original (1792) definition of a metre (meter in USA!) was 1/10,000,000 of the distance between the north pole and the equator. Later the standards was changed to the distance between two lines on a particular standard Platinum-Iridium bar kept in Paris. (1960) 1 m redefined as 1,650,763.73 wavelengths of the (orange/red) light emitted from atoms of the isotope 86Kr. (1983) 1 m finally defined as the length travelled by light in vacuum during a time interval of 1/299,792,458 of a second. • To Andromeda Galaxy • Radius of earth • Adult human height • Radius of proton ~ 1022 m ~ 107 m ~2m ~ 10-15 m Time (Seconds) REGAN PHY34210 Standard definitions of the second ? Original definition 1/(3600 x 24) of a day, 24 hours = 1day, 3600 sec per hours, thus 86,400 sec / day, 3651/4 days per year and 31,557,600 sec per year. But, a day does not have a constant duration! (1967) Use atomic clocks, to define 1 second as the time for 9,192,631,770 oscillations of the light of a specific wavelength (colour) emitted from an atom of caesium (133Cs) From HRW, p6 14 Mass (Kg, AMU) REGAN PHY34210 1 kg defined by mass of Platinum-Iridium cylinder near to Paris. Masses of atoms compared to each other for other standard. Define 1 atomic mass unit = 1 u (also sometimes called 1 AMU) as 1/ 12 the mass of a neutral carbon-12 atom. 1 u = 1.66054 x 10-27 kg Orders of Magnitude It is common for physicists to ESTIMATE the magnitude of particular property, which is often expressed by rounding up (or down) to the nearest power of 10, or ORDER OF MAGNITUDE, e.g.. 140,000,000 m ~ 108m, 15 Estimate Example 1: REGAN PHY34210 A ball of string is 10 cm in diameter, make an order of magnitude estimate of the length, L , of the string in the ball. 4 3 Volume of string, V d L r 3 r radius of ball 10cm/2 5cm 0.05m 2 r assume cross- sectionof string~ 3mmsquare d d 4 3 2 2 V 0.05m 3m m L 0.003m L 3 4 0.05m 3 4 0.05 0.05 0.05m 3 L 3 2 0.003 0.003m 2 0.003m L 55.5m 60m 16 REGAN PHY34210 E.g., 2: Estimate Radius of Earth (from the beach.) From P ythagoras , d 2 r 2 r h r 2 2rh h 2 2 d d 2 2rh h 2 , but h r d 2 2rh h r r is the angle through which the sun moves around the earth during the time between the ‘two’ sunsets (t ~ 10 sec). t t ( 10sec) 360(deg) 3600 o (deg) (deg) 0 . 04 360o 24 hours 24 60 60sec 86,400 2h Now, from trigonometry,d r tan thusr 2 tan2 2rh r tan2 2 2m 4m 6 if h human height ~ 2m, thensubstituting, r 8 10 m 2 o 7 tan (0.04 ) 4.9 10 (acceptedvalue for earthradius 6.4x108 m!) 17 2: Motion in a Straight Line REGAN PHY34210 Position and Displacement. To locate the position of an object we need to define this RELATIVE to some fixed REFERENCE POINT, which is often called the ORIGIN (x=0). In the one dimensional case (i.e. a straight line), the origin lies in the middle of an AXIS (usually denoted as the ‘x’-axis) which is marked in units of length. x = -3 -2 -1 0 1 2 3 Note that we can also define NEGATIVE co-ordinates too. The DISPLACEMENT, Dx is the change from one position to another, i.e., Dx= x2-x1 . Positive values of Dx represent motion in the positive direction (increasing values of x, i.e. left to right looking into the page), while negative values correspond to decreasing x. Displacement is a VECTOR quantity. Both its size (or ‘magnitude’) AND direction (i.e. whether positive or negative) are important. 18 REGAN PHY34210 19 Average Speed and Average Velocity We can describe the position of an object as it moves (i.e. as a function of time) by plotting the x-position of the object (Armadillo!) at different time intervals on an (x , t) plot. The average SPEED is simply the total distance travelled (independent of the direction or travel) divided by the time taken. Note speed is a SCALAR quantity, i.e., only its magnitude is important (not its direction). From HRW REGAN PHY34210 20 The average VELOCITY is defined by the displacement (Dx) divided by the time taken for this displacement to occur (Dt). x x Dx vav 2 1 t2 t1 Dt The SLOPE of the (x,t) plot gives average VELOCITY. Like displacement, velocity is a VECTOR with the same sign as the displacement. The INSTANTANEOUS VELOCITY is the velocity at a specific moment in time, calculated by making Dt infinitely small (i.e., calculus!) Dx dx v lim Dt 0 Dt dt Acceleration Acceleration is a change in velocity (Dv) in a given time (Dt). The average acceleration, aav, is given by dx2 dx1 v2 v1 Dv dt dt aav t 2 t1 t 2 t1 Dt The instantaneous ACCELERATION is given by a, where, dv d dx d 2 x a 2 dt dt dt dt SI unit of acceleration is metres per second squared (m/s2) REGAN PHY34210 HRW 21 REGAN PHY34210 22 Constant Acceleration and the Equations of Motion For some types of motion (e.g., free fall under gravity) the acceleration is approximately constant, i.e., if v0 is the velocity at time t=0, then By making the assumption that the acceleration is a constant, we can derive a set of equations in terms of the following quantities v v0 a aav t 0 x x0 thedisplacement v0 initialvelocity(at timet 0) v a t velocityat timet acceleration (constant) timetakenfromt 0 Usually in a given problem, three of these quantities are given and from these, one can calculate the other two from the following equations of motion. REGAN PHY34210 Equations of Motion (for constant a). x x0 v v0 at (1) recallingthatvav , then t 0 v0 v since by definition, vav , 2vav v v0 2 1 substituting into (1) for v0 gives vav v0 at , 2 d x x0 1 2 x x0 v0t at (2) , note v v0 at 2 dt combining(1) and (2) to eliminatet , a and v0 gives 1 v v 2a ( x x0 ) (3) ; x x0 v0 v t (4) ; and 2 1 2 x x0 vt at (5) 2 2 2 0 23 Alternative Derivations (by Calculus) REGAN PHY34210 24 dv d 2 x a 2 by definit ion dv a dt dt dt dv a dt and thusfor a constant v at C , evaluatedby knowing thatat t 0, v v0 v0 a 0 C thus, v v0 at (1) dx v dx v dt , v not constant,but v0 is, therefore dt by substitution, dx v0 atdt dx v0 dt a t dt 1 2 integrating gives, x v0t at C , calculateC by 2 knowing x x0 at t t0 C x0 1 2 and x x0 v0t at (5) 2 Free-Fall Acceleration REGAN PHY34210 At the surface of the earth, neglecting any effect due to air resistance on the velocity, all objects accelerate towards the centre of earth with the same constant value of acceleration. This is called FREE-FALL ACCELERATION, or ACCELERATION DUE TO GRAVITY, g. At the surface of the earth, the magnitude of g = 9.8 ms-2 Note that for free-fall, the equations of motion are in the y-direction (i.e., up and down), rather than in the x direction (left to right). Note that the acceleration due to gravity is always towards the centre of the earth, i.e. in the negative direction, a= -g = -9.8 ms-2 25 Example REGAN PHY34210 26 A man throws a ball upwards with an initial velocity of 12ms-1. (a) how long does it take the ball to reach its maximum height ? (a) since a= -g = -9.8ms-2, Therefore, time to max height from initial position is y0=0 and v v0 0 12m s1 at the max. height vm a x=0 v v0 at t a 9.8m s2 1.2s (b) what’s the ball’s maximum height ? 2 2 2 1 v v 2 a ( y y ) ( 12 ) 2 a y 0 & v 0 , v 12 m s , (b) 0 0 0 2 2 1 2 v v 0 ( 12 m s ) 2 0 a g 9.8m s y 7.3m 2 2a 2 9.8m s REGAN PHY34210 ( c) How long does the ball take to reach a point 5m above its initial release point ? v0 12m s1 , a 9.8m s2 , y y0 5m 1 2 1 1 from y y0 v0t at 5m 12m s t 9.8m s 2t 2 2 2 assuming SI units, we havea quadratic equation,4.9t 2 12t 5 0 recallingat 2 bt c 0 solutionsare given by b b 2 4ac t t 0.5s AND 1.9s 2a Note that there are TWO SOLUTIONS here (two different ‘roots’ to the quadratic equation). This reflects that the ball passes the same point on both the way up and again on the way back down. 27 3: Vectors REGAN PHY34210 • Quantities which can be fully described just by their size are called SCALARS. – Examples of scalars include temperature, speed, distance, time, mass, charge etc. – Scalar quantities can be combined using the standard rules of algebra. • A VECTOR quantity is one which need both a magnitude (size) and direction to be complete. – Examples of vectors displacement, velocity, acceleration, linear and angular momentum. – Vectors quantities can be combined using special rules for combining vectors. 28 b REGAN PHY34210 Adding Vectors Geometrically Any two vectors can be added using the VECTOR EQUATION, where the sum of s a b vectors can be worked out using a triangle. r a b c a b c s a b Note that two vectors can be added together in either order to get the same result. This is called the COMMUTATIVE LAW. Generally, if we have more than 2 vectors, the order of combination does not affect the result. This is called the ASSOCIATIVE LAW. 29 a a s s b a a b b b a c s r c b = a ' s r REGAN PHY34210 Subtracting Vectors, Negative Vectors b is thesame magnitudeas b but in theoppositedirection. b d a b s s a b b a s as with usual algebra, we can d a b a b re - arrangevectorequations, e.g., d a b d b a Note that as with all quantities, we can only add / subtract vectors of the same kind (e.g., two velocities or two displacements). We can not add differing quantities e.g., apples and oranges!) 30 REGAN PHY34210 Components of Vectors A simple way of adding vectors can y be done using their COMPONENTS. The component of a vector is the ay projection of the vector onto the x, y (and z in the 3-D case) axes in the a sin Cartesian co-ordinate system. Obtaining the components is known as RESOLVING the vector. The components can be found using the rules for a right-angle triangle. i.e. a x a cos and a y a sin thiscan also be writtenin MAGNIT UDE- ANGLE NOT AT IONas a a a , tan 2 x 2 y ay ax 31 a ax a cos x REGAN PHY34210 Unit Vectors A UNIT VECTOR is one whose magnitude is exactly equal to 1. It specifies a DIRECTION. The unit vectors for the Cartesian co-ordinates x,y and z are given by,iˆ, ˆj and kˆ respectively. The use of unit vectors can make the addition/subtraction of vectors simple. One can simply add/subtract together the x,y and z components to obtain the size of the resultant component in that specific direction. E.g, z1 ˆj y 1 kˆ iˆ 1 ˆ a a x iˆ a y ˆj a z k , b bx iˆ by ˆj bz kˆ thenif s a b using vectoraddition by components s (a x bx )iˆ (a y by ) ˆj (a z bz )kˆ s x iˆ s y ˆj s z kˆ x 32 Vector Multiplication REGAN PHY34210 There are TWO TYPES of vector multiplication. One results in a SCALAR QUANTITY (the scalar or ‘dot’ product). The other results in a VECTOR called the vector or ‘cross’ product. For the SCALAR or DOT PRODUCT, a.b a cos b a b cos also, a . b b . a In unit vector notation, a.b a x iˆ a y ˆj a z kˆ . bx iˆ by ˆj bz kˆ but since cos0o 1 and cos90o 0 expandingthisreduces to a.b a x bx a y by a z bz since iˆ.iˆ iˆ. ˆj iˆ.kˆ ˆj. ˆj kˆ.kˆ 1 (11cos0o 1) and ˆj.kˆ 0 (1 1 cos90o 0) 33 Vector (‘Cross’) Product The VECTOR PRODUCT of two vectors a and b REGAN PHY34210 34 produces a third vector whose magnitude is given by c absin The direction of the resultant is perpendicular to the plane created by the initial two vectors, such that is the angle between the two initial vectors also a b b a and a b a xiˆ a y ˆj a z kˆ bxiˆ by ˆj bz kˆ but a x iˆ bxiˆ a xbx iˆ iˆ 0 and a xiˆ by ˆj a xby iˆ ˆj a xby kˆ, thus a b a y bz by a z iˆ a z bx bz a x ˆj a xby bx a y kˆ b c a REGAN PHY34210 Example 1: Add thefollowing threevectors ˆ a i 4 ˆj , b 3iˆ 2 j , c iˆ 2 ˆj ˆ r a b c i 4 ˆj 3iˆ 2 j iˆ 2 ˆj r 3iˆ 4 ˆj rx 3 , ry 4 y r b 4 r 3 4 5 , tan 53.1o 3 2 a c 35 x 2 Example 2: REGAN PHY34210 Whatare the(a) scalar and (b) vect orproductsof the two vect ors aˆ 2iˆ 3 ˆj 4kˆ and bˆ 4iˆ 20 ˆj 12kˆ (a) Scalar P roduct: a . b ab cos 2iˆ 3 ˆj 4kˆ . 4iˆ 20 ˆj 12kˆ recallingonlyiˆ.iˆ ˆj.ˆj kˆ.kˆ 1, iˆ.ˆj iˆ.kˆ ˆj.kˆ 0 then a . b a x bx a y by a z bz (2.4) (3. 20) (4.12) a . b 8 - 60 - 48 -100 (b) Vector P roduct,a b 2iˆ 3 ˆj 4kˆ 4iˆ 20 ˆj 12kˆ recallinga b a y bz by a z iˆ a z bx bz a x ˆj a x by bx a y kˆ a b (3).(12) (20).(4) iˆ (4).(4) (12).(2) ˆj (2).(20) (4)(3) kˆ 36 80 iˆ (16 24) ˆj (40 12) kˆ a b 44 iˆ 40 ˆj 52 kˆ 4 . 11iˆ 10 ˆj 13kˆ 36 4: Motion in 2 and 3 Dimensions REGAN PHY34210 37 The use of vectors and their components is very useful for describing motion of objects in both 2 and 3 dimensions. Position and Displacement If in general theposit ionof a particlecan be descibed in Cartesianco - ordinatesby r xiˆ yˆj zkˆ , then theDISP LACEMENT is ˆ Dr r2 r1 x2 i y2 ˆj z 2 k x1i y1 ˆj z1kˆ x x i y y ˆj z z kˆ 2 1 2 1 2 1 Dxiˆ Dyˆj Dzkˆ ˆ ˆ ˆ e.g., if r1 3i 2 j 5k and r2 9iˆ 2 ˆj 8kˆ thenDr r2 r1 (9 (3))iˆ (2 2) ˆj (8 5)kˆ 12iˆ 3kˆ Velocity and Acceleration REGAN PHY34210 The average velocity is given by Dr Dxiˆ Dyˆj Dzkˆ Dx ˆ Dy ˆ Dz ˆ vav i j k Dt Dt Dt Dt Dt While the instantaneous velocity is given by making Dt tend to 0, i.e. dr d ( xiˆ yˆj zkˆ) dx ˆ dy ˆ dz ˆ v i j k dt dt dt dt dt v v x iˆ v y ˆj v z kˆ Similarly, the average acceleration is given by, v2 v1 Dv Dvxiˆ Dv y ˆj Dvz kˆ Dvx ˆ Dv y ˆ Dvz ˆ aav i j k Dt Dt Dt Dt Dt Dt While the instantaneous acceleration is given by dv d (vxiˆ v y ˆj vz kˆ) dvx ˆ dvy ˆ dvz ˆ a i j k dt dt dt dt dt 38 Projectile Motion REGAN PHY34210 39 The specialist case where a projectile is ‘launched’ with an initial velocity, v0 and a constant free-fall acceleration, g . Examples of projectile motion are golf balls, baseballs, cannon balls. (Note, aeroplanes, birds have extra acceleration see later). We can use the equations of motion for constant acceleration and what we have recently learned about vectors and their components to analyse this type of motion in detail. T heinitialproject ilevelocity (at t 0) is v0 v0 x iˆ v0 y ˆj where v0 x v0 cos and v0 y v0 sin More generally, v vxiˆ v y ˆj v0 v0 sin v0 cos where vx v cos and v y v sin REGAN PHY34210 Horizontal Motion 40 In the projectile problem, there is NO ACCELERATION in the horizontal direction (neglecting any effect due to air resistance). Thus the velocity x x0 v0 xt (equationof motion 1) component in the x (horizontal) direction and v0 x v0 cos 0 , thus x x0 v0 cos 0 t remains constant throughout the flight, i.e., Vertical Motion 1 2 1 2 y y0 v0 y t gt v0 y sin 0 t gt 2 2 2 2 v y v0 sin 0 gt and v y v0 sin 0 2 g y y0 v0 Max.height occurs when v y 0, i.e., v0 sin 0 gt v0 sin v0 cos vy REGAN PHY34210 41 The Equation of Path for Projectile Motion Given that x x0 v0 cos 0 t and y y0 v0 sin 0 t 1 2 gt 2 substituting for thetimebetween the two upper equns. x x0 1 x x0 y y0 v0 sin 0 g v cos 2 v cos 0 0 0 0 2 1 x x0 y y0 tan 0 x x0 g 2 v0 cos 0 Note that this is an equation of the form y=ax+bx2 i.e., a parabola (also, often y0=x0=0.) 2 REGAN PHY34210 The Horizontal Range T herange, R x x0 is defined when theproject ilehits theground 1 2 i.e., when y y0 , thenR v0 cos 0 t and 0 v0 sin 0 t gt 2 R 1 R 0 v0 sin 0 g v0 cos 0 2 v0 cos 0 2 R sin 0 2v02 cos 0 sin 0 gR2 2 R 2 cos 0 2v0 cos 0 g v0 since in general 2 cos 0 sin 0 sin 2 0 , then v02 sin 2 0 R g (noteassumes y y0 ) v0 sin (y0,x0) vy=0 Max height v0 cos Range 42 Example REGAN PHY34210 43 At what angle must a baseball be hit to make a home run if the fence is 150 m away ? Assume that the fence is at ground level, air resistance is negligible and the initial velocity of the baseball is 50 m/s. Recalling that therange 150mis given v02 sin 2 0 by R , v0 50m s1 , g 9.8m s 2 g gR 9.8m s- 2 150m 1470 sin2 0 2 1 1 v0 50m s 50m s 2500 2 0 sin 1 0.588 36o OR 143o 0 18o or 71.5o R How far must the fence be moved back for no homers to be possible ? 2 R v0 sin 2 0 is a maximumwhen sin2 0 max 1, i.e., when 2 0 900 , g 1 1 50 m s 50 m s thus 0max 45o Rmax 255m 840 feet! 2 9.8m s Uniform Circular Motion REGAN PHY34210 v 44 A particle undergoes UNIFORM CIRCULAR v MOTION is it travels around in a circular arc at a a CONSTANT SPEED. Note that although the speed does not change, the particle is in fact a r ACCELERATING since the DIRECTION OF THE a VELOCITY IS CHANGING with time. The velocity vector is tangential to the instantaneous v direction of motion of the particle. The (centripetal) acceleration is directed towards the centre of the circle Radial vector (r) and the velocity vector (v) are always perpendicular T hePERIODOF REVOLUT ION timetakenfor theparticleto go around thecircle.If thespeed (i.e., themagnitudeof the velocit yfor UCM) of theparticle v, the timetakenis, by definition Circum ference 2r T velocity v Proof for Uniform Circular Motion REGAN PHY34210 yp xp ˆ ˆ ˆ ˆ v v x i v y j v sin i v cos j , but sin and cos , thus r r vy p ˆ vx p ˆ dv v dyp ˆ v dxp ˆ i j . Acceleration is a i j v dt r dt r dt r r but dyp v yp v cos and dxp v xp v sin yp r 45 v dt dt v dyp ˆ v dxp ˆ v 2 cos ˆ v 2 sin ˆ i j i j a xp r r r dt r dt v2 v2 2 2 2 2 cos sin , thus, a m agnitude, a a x a y r r v 2 sin a y r sin ACCELERAT ON I DIRECT ION from tan tan , 2 a x v cos cos r i.e., , acceleration is along theradius....towards centreof circle Relative Motion If we want to make measurements of velocity, position, acceleration etc. these must all be defined RELATIVE to a specific origin. Often in physical situations, the motion can be broken down into two frames of reference, depending on who is the OBSERVER. A ( someone who tosses a ball up in a moving car will see a different motion to someone from the pavement). REGAN PHY34210 rAp p vAB const rBp B rAB If we assume that different FRAMES OF REFERENCE always move at a constant velocity relative to each other, then using vector addition, i.e., acceleration d rpA rpB rBA rpA rpB rBA , v pA v pB vBA is the SAME for dt both frames of d v pA v pB vBA d v pB reference! a pA 0 a pB (if VAB=const)! dt dt 46 5: Force and Motion (Part 1) REGAN PHY34210 47 If either the magnitude or direction of a particle’s velocity changes (i.e. it ACCELERATES), there must have been some form of interaction between this body and it surroundings. Any interaction which causes an acceleration (or deceleration) is called a FORCE. The description of how such forces act on bodies can be described by Newtonian Mechanics first devised by Sir Isaac Newton (1642-1712).. Note that Newtonian mechanics breaks down for (1) very fast speeds, i.e. those greater than about 1/10 the speed of light c, c=3x108ms-1 where it is replaced by Einstein’s theory of RELATIVITY and (b) if the scale of the particles is very small (~size of atoms~10-10m), where QUANTUM MECHANICS is used instead. Newton’s Laws are limiting cases for both quantum mechanics and relativity, which are applicable for specific velocity and size regimes Newton’s First Law REGAN PHY34210 Newton’s 1st law states ‘ If no force acts on a body, then the body’s velocity can not change, i.e., the body can not accelerate’ This means that (a) if a body is at rest, it will remain at rest unless acted upon by an external force, it; and (b) if a body is moving, it will continue to move at that velocity and in the same direction unless acted upon by an external force. So for example, (1) A hockey puck pushed across a ‘frictionless’ rink will move in a straight line at a constant velocity until it hits the side of the rink. (2) A spaceship shot into space will continue to move in the direction and speed unless acted upon by some (gravitational) force. 48 Force REGAN PHY34210 49 The units of force are defined by the acceleration which that force will cause to a body of a given mass. The unit of force is the NEWTON (N) and this is defined by the force which will cause an acceleration of 1 m/s2 on a mass of 1 kg. If two or more forces act on a body we can find their resultant value by adding them as vectors. This is known as the principle of SUPERPOSITION. This means that the more correct version of Newton’s 1st law is ‘ If no NET force acts on a body, then the body’s velocity can not change, i.e., the body can not accelerate’ Mass: we can define the mass of a body as the characteristic which relates the applied force to the resulting acceleration. Newton’s 2nd Law REGAN PHY34210 50 Newton’s 2nd law states that ‘ The net force on a body is equal to the product of the body’s mass and the acceleration of the body’ We can write the 2nd law in the form of an equation: Fnet ma As with other vector equations, we can make three equivalent equations for the x,y and z components of the force. i.e., Fnet, x max , Fnet, y may and Fnet, z maz The acceleration component on each axis is caused ONLY by the force components along that axis. REGAN PHY34210 51 If the net force on a body equals zero and thus it has no acceleration, the forces balance out each other and the body is in EQUILIBRIUM. We can often describe multiple forces acting on the same body using a FREE-BODY DIAGRAM, which shows all the forces on the body. FA 220 N, FB ?, FC 170 N FA FB FC ma 0 FB FA FC FA 220 N components along the x and y axes cancel F F F 0 F cos 133 F cos 137 F cos133 220 0.682 cos 0.883 y FBy FAy FCy FA sin 47 FC sin FB o o Bx Ax Cx A C o 43o 47o Fc 170N x 0 A FC 170 28 and FA sin 47 FC sin 28 FB 0 o o FB 220 0.731 170 0.47 241 N FB ?? N HRW p79 The Gravitational Force REGAN PHY34210 52 The gravitational force on a body is the pulling force directed towards a second body. In most cases, this second body refers to the earth (or occasionally another planet). From Newton’s 2nd law, the force is related to the acceleration by Fg ma takingcomponentsin theverticaldirection ( y positivecorresponds to upward) then Fgy m ay m g ˆ ˆ In vectorform,we have Fg Fg j m g j mg A body’s WEIGHT equals the magnitude of the gravitational force on the body, i.e, W = mg. This is equal to the size of the net force to stop a body falling to freely as measured by someone at ground level. Note also that the WEIGHT MUST BE MEASURED WHEN THE BODY IS NOT ACCELERATING RELATIVE TO THE GROUND and that WEIGHT DOES NOT EQUAL MASS. Mass on moon and earth equal but weights not ge=9.8ms-2, gm=1.7ms-2 The Normal Force REGAN PHY34210 53 The normal force is the effective ‘push’ a body feels from a body to stop the downward acceleration due to gravity, for example the upward force which the floor apparently outs on a body to keep it stationary against gravity. General equation for block on a table is Fnet ma N Fg NormalForce, N y component,m ay N Fg N m g N m a y g i.e., if block is at rest then N m 0 g m g i.e. same magnitudeas gravitational forcebut in oppositedirection. Note the NORMAL FORCE is ‘normal’ (i.e. perpendicular) to the surface. Gravitational Force, Fg mg REGAN PHY34210 Example A person stands on a weighing scales in a lift (elevator!) What is the general solution for the persons measured weight on the scales ? a Fnet ma Fg N N Fy ,net m ay Fg N m g N N m a y g Fgy mg So, if lift accelerates upwards (or the downward speed decreases!) the persons weight INCREASES, if the lift accelerates downwards (or decelerates upwards) the persons weight DECREASES compared to the stationary (or constant velocity) situation. 54 REGAN PHY34210 Tension 55 Tension is the ‘pulling force’ associated with a rope/string pulling a body in a specific direction. This assumes that the string/rope is taught (and usually also massless). For a frictionless surface and a massless, frictionless pulley, what are the accelerations of the sliding and N hanging blocks and the tension in the cord ? x components, Fnet , x MaMx T y components( a downward thus amy ) M T Fnet , y m amy T m g FgM Mg T m magnitudesmust be equal, aMx amy a mg M m mg Mm g M T M m M m m amy MaMx m g a Fnet , x T MaMx Fgm mg Newton’s Third Law REGAN PHY34210 56 Two bodies interact when they push or pull on each other. This leads to Newton’s third law which states, ‘ When two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction ’ Sometimes this is differently stated as NormalForce, N ‘ for every action there is an equal but opposite reaction ’ The forces between two interacting bodies are called a ‘third-law pair forces’. e.g., Table pushes up block with force N, block pushes down table with force Fg, where Fg=N Gravitational Force, Fg mg REGAN PHY34210 Example N N T cos50o T y 57 T x 40 o 40 o mg = g x 15kg Question? Whatis the tensionin thestring? T N Fg ma 0 x component,T 0 m g sin Fx ,net 0 T m g sin 15kg 9.8m s 2 sin 400 94.5 N y component,0 N m g cos 0 N 112.6 N 50o 40 o mg mg sin 40o 6: Force and Motion (Part II) REGAN PHY34210 58 Friction: When two bodies are in contact, the resistance to movement between their surfaces is known as FRICTION. The properties of frictional forces are that if a force, F, pushes an object along a surface (e.g., a block along a surface), 1) If the body does not move, the STATIC FRICTIONAL FORCE, fs is equal in magnitude and opposite in direction to the component of the pushing force, F, along the surface. 2) The magnitude of the frictional force, fs, has a maximum value, f s,max, which is given by f s,max=msN where ms is the coefficient of static friction. 3) If the body begins to move along the surface, the magnitude of the frictional force reduces to fk=mkN, where mk is the coefficient of kinetic friction. Drag Force and Terminal Speed REGAN PHY34210 59 When a body passes through a fluid (i.e., gas or a liquid) such as a ball falling through air, if there is a relative velocity between the body and the fluid, the body experiences a DRAG FORCE which opposes this relative motion and is in the opposite direction to the motion of the body (i.e., in the direction which the fluid flows relative to the body). The magnitude of this drag force is related 1 2 to the relative speed of the body in the fluid D C r Av by a DRAG COEFFICIENT, C, which is 2 experimentally determined. The magnitude of the drag force is given by the expression for D, which depends on the fluid density (i.e. mass per unit volume, r), the effective cross-sectional area, A (i.e. the cross-sectional area perpendicular to the direction of the velocity vector), and the relative speed, v. REGAN PHY34210 60 Note that the drag coefficient. C, is not really a constant, but rather a quantity associated with a body which can varies with the speed, v. (for the purposes of this course, however, assume C = constant). The direction of the drag force is opposed to the motion of the object through the fluid. If a body falls through air, the drag force due to the air resistance will start at zero (due to zero velocity) at the start of the fall, increasing as the downward velocity of the falling body increases. Ultimately, thedrag force will be cancelthedownward acceleration. In general, Fnet , y m ay D Fg For ' terminalspeed', a y 0, thus 2 Fg 1 2m g 2 CrAvt Fg 0 vt 2 CrA CrA Forces in Uniform Circular Motion REGAN PHY34210 61 Recalling that for a body moving in a circular arc or radius, r, with constant speed, v, the MAGNITUDE of the ACCELERATION, a, is given by a = v2/r, where a is called the centripetal acceleration. We can say that a centripetal force accelerates a body by changing the direction of that body’s velocity without changing its speed. Note that this centripetal force is not a ‘new’ force, but rather a consequence of another external force, such as friction, gravity or tension in a string. Examples of circular motion are (1) Sliding across your seat when your car rounds a bend: The centripetal force (which here is the frictional force between the car wheels and the road) is enough to cause the car to accelerate inwards in the arc. However, often the frictional force between you and your seat is not strong enough to make the passenger go in this arc too. Thus, the passenger slides to the edge of the car, when its push (or normal force) is strong enough to make you go around the arc. REGAN PHY34210 (2) the (apparent) weightlessness of astronauts on the space shuttle. Here the centripetal force (which causes the space shuttle to orbit the earth in a circular orbits) is caused by the gravitational force of the earth on all parts of the space shuttle (including the astronauts).The centripetal force is equal on all areas of the astronauts body so he/she feels no relative extra pull etc. on any specific area, giving rise to a sensation of weightlessness. Note that the magnitude of the centripetal FORCE is given, (from Newton’s second law) by : F = ma = m v2/r Note that since the speed, radius and mass are all CONSTANTS so is the MAGNTIUDE OF THE CENTRIPETAL FORCE. However, DIRECTION IS NOT CONSTANT, varying continuously so as to point towards the centre of a circle. 62 REGAN PHY34210 Example: r At what constant speed does the roller coasters have to go to ‘loop the loop’ of radius r ? At the top of the loop, the free body forces on the roller coaster are gravity (downwards) and the normal force (also inwards). The total acceleration is also inwards (i.e., in the downwards direction). Fg N Fy ,net N Fg m a y , limit at N 0 (no contact!) v2 v2 thus, Fg m a y m. m g g v gr r r i.e., independent of mass! 63 7: Kinetic Energy and Work REGAN PHY34210 One way to describe the motion of objects is by the use of Newton’s Laws and Forces. However, an alternative way is describe the motion in terms of the ENERGY of the object. The KINETIC ENERGY (K) is the energy associated with the MOTION of an object. It is related to the mass and velocity of a body by K= 1/2 mv2 , where m and v are the mass and velocity of the body. The SI unit of energy is the Joule (J) where 1 Joule = 1kg.m2s-2. Work: `Work is the energy transferred to or from an object by means of a force acting on it. Energy transferred to the object is positive work, while energy transferred from the object is negative work.’ For example, if an object is accelerated such that it increases its velocity, the force has ‘done work’ on the object. 64 Work and Kinetic Energy REGAN PHY34210 65 The work done (W) on an object by a force, F, causing a displacement, d, is given by the SCALAR PRODUCT, W = F.d =dFcos where Fcos is the component of the force along the object’s displacement. This expression assumes a CONSTANT FORCE (one that does not change in magnitude or direction) and that the object is RIGID (all parts of the object move together). Example: If an object moves in a straight line with initial velocity, v0 and is acted on by a force along a distance d during which the velocity increases to v due to an acceleration, a, from Newton’s 2nd Law the magnitude of the force is given by F = max . From the equations of motion v2=vo2+2axd . By substituting for the acceleration, ax, we have, Fx 1 2 2 1 2 1 2 d v v0 , ax Fx d m v m v0 DK work done 2ax m 2 2 1 2 1 2 mv mv 0 DK work done is the Work-Kinetic Energy Theorem 2 2 Work Done by a Gravitational Force. REGAN PHY34210 66 If an object is moved upwards against gravity, work must be done. Since the gravitational force acts DOWNWARDS, and equals Fgr=mg , the work done in moving the object upwards in the presence of this force is W=F.d = mg . d where d is the (vector) displacement in the upward direction, (which we assume is the positive y-axis). Wgr mg.d m gdcos , mg and d are in oppositedirections, 180o , W m gd. sign shows gravity transfersKINET IC ENERGY to GRAVIT AT IONAL POT ENT IALENERGY. When theobject falls back down, 0 and W m gd sign impliesgravitational force transfersenergyT O the object from potentialenergy tokinetic. REGAN PHY34210 Work Done Lifting and Lowering an Object. 67 If we lift an object by applying a vertical (pushing) force, F, during the upward displacement, work (Wa) is done on the object by this applied force. The APPLIED FORCE TRANSFERS ENERGY TO the object, while the GRAVITY TRANSFERS ENERGY FROM it. From the work - kineticenergy theorem, DK K f K i net workdone Wa Wg . If theobject is stationarybeforeand after thelift (v 0 at start and finish) then DK 0 Wa Wg Wa Wg and Wa m gdcos where is theangle between Fg (i.e.' downwards' ) and thedisplacement,d . If theobject is liftedup, 1800 and the work done by the appliedforce, Wa m gd. If theobject falls, 0o , Wa m gd Spring Forces and Hooke’s Law REGAN PHY34210 68 The spring force is an example of a VARIABLE FORCE. For a PERFECT SPRING, stretching or compressing gives rise to RESTORING FORCE which is proportional to the displacement of the spring from its relaxed state. This is written by Hooke’s Law (after 17th century British scientist) as Robert Hooke, Frestoring kd , where k springconstant,stiffnessof thespring. In the1 - d case, we can simply use the x - direction F kx The work done by a perfect spring can not be obtained from F.d, as the force is not constant with d. Instead, the work done over the course of the extension/compression must be summed incrementally. xf 1 x x Ws F j Dx , as Dx 0 then, Ws xif Fdx xif (kx)dx kx2 2 xi 1 2 1 2 1 1 2 2 2 Ws kx f kxi k xi x f if xi 0, Ws kx f 2 2 2 2 Work Done by an Applied Force REGAN PHY34210 69 During the displacement of the spring, the applied force, Fa, does work, Wa on the block and the spring restoring force, Fs does work Ws. T hechangein kineticenergy (of theblock attachedto thespring) due to thesetwo energy transfersis given by DK K f K i Wa Ws T hus, if DK 0 , Wa Ws If the block attached to a spring is stationary before and after its displacement, then the work done on the spring by the applied force is the negative of the work done on it by the spring restoring force. Work Done by a General Variable Force. REGAN PHY34210 T he work done by a forceaveragedover a distance,Dx, is DW j F j ,ave Dx. T otalwork done equals sum of all j th increments, W DW j F j ,ave Dx. As Dx 0 , W j j f F D x j ,ave xi F x dx x j , Dx 0 Workdone by 1 - D force AREA UNDERT HE CURVE of F ( x) against x. In 3 - D, F F iˆ F ˆj F kˆ. If F only depends on x, F on y and x y z x y Fz on z , theby SEP ARAT INGT HE VARIABLES if theparticle movesthroughan incremental displacement,dr dxiˆ dyˆj dzkˆ , theincrementof work in dr, dW, is given by dW F.dr Fx dx Fy dy Fz dz , then thetotalwork is rf xf yf zf ri xi yi zi W dW Fx dx Fy dy Fz dz 70 Work-Kinetic Energy Theorem with a General,Variable Force xf xf REGAN PHY34210 71 xf dv W F x dx m ax dx m dx dt xi xi xi dv dv dx dv using theCHAIN RULE, . v dt dx dt dx dv dv we can CHANGE T HE VARIABLE, m dx m v . dx m vdv dt dx xf vf vf dv W m dx m vdv m vdv dt xi vi vi 1 2 1 2 W m vf m vi K f K i DK , 2 2 which is the WORK- KINET ICENERGY T HEOREM Power REGAN PHY34210 72 POWER is the RATE AT WHICH WORK IS DONE. The AVERAGE POWER done due to a force responsible for doing work, W in a time period, Dt is given by Pave = W/D t . The INSTANTANEOUS POWER is given by dW P dt The SI unit of power = Watt (W), where 1 W= 1 J per sec=1 kg.m2/s3 Note that the imperial unit of horsepower (hp) is still used, for example for cars. 1hp = 746 W The amount of work done is sometimes expressed as the product of the power output multiplied by time taken for this. A common unit for this is the kilowatt-hour, where 1kWh = 1000x3600 J = 3.6 x106J = 3.6MJ. We can also describe the instantaneous power in terms of rate at which a force does work on a particle, dW F cos dx dx P F cos F cos v F .v dt dt dt REGAN PHY34210 Example 1: What is the total energy associated with a collision between two locomotives, at opposite ends of a 6.4km track accelerating towards each other with a constant acceleration of 0.26 m/s2 if the mass of each train was 122 tonnes (1 tonne =103kg) ? Using, v 2 v02 2ax x0 x x0 3.2 103 m, v0 0, a 0.26m s- 2 T he velocit yof the trainsat collisionis then v 2 0.26m s 2 3.2 103 m 40.8m s1 T hekineticenergyof each locomotiveis given by 1 2 1 5 1 2 K m v 1.2210 kg 40.8m s 108 J 2 2 T hus totalenergyof collisionis 2 K 200MJ 73 Example 2: REGAN PHY34210 74 If a block slide across a frictionless floor through a displacement of d 3iˆm -3m in the direction, while at the same time a steady (i.e. constant) force of F=(2i-6j) Newtons pushes against the crate, (a) How much work does the wind force do F 2iˆ 6 ˆj N on the crate during this displacement ? W F .d 2iˆ 6 ˆj N . 3iˆ m 6 J T hus, the' wind' forcedoes 6 J of NEGAT IVE WORK on thecrate i.e. it transfers 6 J of kineticenergy FROM T HECRAT E (b) If the crate had a kinetic energy of 10J at the start of the displacement, how much kinetic energy did it have at the end of the -3m ? Work-kineticenergy theorem,W ΔK 6 J K f 10J K f 6 10J 4 J i.e., block is slowed down by wind force. REGAN PHY34210 Example 3: If a block of mass, m, slides across a frictionless floor with a constant speed of v until it hits and compresses a perfect spring, with a spring constant, k. At the point where the spring is compressed such that the block is momentarily stopped, by what distance, x, is the spring compressed ? Using the work - energy theorem,the work done on theblock by thespring forceis 1 Ws kx2 . T he work is also relatedto thechange 2 in kineticenergy of theblock,i.e.,W ΔK K f K i 1 2 1 2 m kx 0 m v x v 2 2 k v k x v=0 m 75 REGAN PHY34210 8: Potential Energy & Conservation of Energy Potential energy (U) is the energy which can be associated with configuration of a systems of objects. One example is GRAVITATIONAL POTENTIAL ENERGY, associated with the separation between two objects attracted to each other by the gravitational force. By increasing the distance between two objects (e.g. by lifting an object higher) the work done on the gravitational force increases the gravitational potential energy of the system. Another example is ELASTIC POTENTIAL ENERGY which is associated with compression or extension of an elastic object (such as a perfect spring). By compressing or extending such a spring, work is done against the restoring force which in turn increases the elastic potential energy in the spring. 76 Work and Potential Energy REGAN PHY34210 77 In general, the change in potential energy, DU is equal to the negative of the work done (W) by the force on the object (e.g., gravitational force on a falling object or the restoring force on a block pushed by a perfect spring), i.e., DU=-W Conservative and Non-Conservative Forces If work, W1, is done, if the configuration by which the work is done is reversed, the force reverses the energy transfer, doing work, W2. If W1=-W2, whereby kinetic energy is always transferred to potential energy, the force is said to be a CONSERVATIVE FORCE. The net work done by a conservative force in a closed path is zero. The work done by a conservative force on a particle moving between 2 points does not depend on the path taken by the particle. NON-CONSERVATIVE FORCES include friction, which causes transfer from kinetic to thermal energy. This can not be transferred back (100%) to the original mechanical energy of the system. REGAN PHY34210 Determining Potential Energy Values xf xf xi xi W F ( x)dx , DU F ( x)dx . For GRAVIT .P OT .ENERGY, yf yf yi yi DU F ( y )dy yf m gdy m g dy m gy f yi yi DU grav m g y f yi m gDy Only CHANGESin gravit at ional P ot .energy are meaningful, i.e., it is usual to defineU i 0 at yi , then U y m gy For theELAST ICP OT ENT IALENERGY, xf xf xf DU elas F ( x)dx kxdx k xdx xi xi xi 1 k x2 2 xf xi 1 k x 2f xi2 . P ot energyis relative,thus we chose 2 1 2 U 0 at xi 0 T hen, U x kx , x is extension/compression. 2 DU elas 78 Conservation of Mechanical Energy REGAN PHY34210 79 T hemechanicalenergyis thesum of kineticand pot entialenergies, Emech K U . If thesystemis isolatedfromits environment and no externalforcecauses any internalenergy changes, DK W & DU W , DK DU K f K i U f U i K f U f K i U i i.e, T hesum of thekineticand pot entialenergies ( themechanicalenergy)is thesame for all statesof an isolatedsystem, i.e. theMECHANICALENERGY of an ISOLAT ED SYST EM where thereare only conservative forcesis CONST ANT . T hisis theP RINCIP LEOF CONSERVAT ION OF MECHANICAL ENERGY (note,conservation is due to CONSERVAT IVE FORCES ). T hiscan also be writtenas DEmech DK DU 0 The Potential Energy Curve REGAN PHY34210 For the1 - D case, the work done,W , by a force,F , movingan object through a displacement,Dx equals, FDx , therefore, the potentialenergy can be writtenas DU x dU x DU x W FDx F Dx dx e.g ., Hooke's Law, if theelastic potentialis given by, 1 2 U x kx thendifferentiatinggives, F kx 2 also, in thegravitational case, U x m gh F m g In the general, the force at position x, can be calculated by differentiating the potential curve with respect to x (remembering the -ve sign). F(x) is minus the SLOPE of U(x) as a function of x 80 REGAN PHY34210 Turning Points For conservative forces, the mechanical energy of the system is conserved and given by, U(x) + K(x) = Emec where U(x) is the potential energy and K(x) is the kinetic energy. Therefore, K(x) = Emec-U(x). Since K(x) must be positive ( K=1/2mv2), the max. value of x which the particle has is at Emec=U(x) (i.e., when K(x)=0). Note since F(x) = - ( dU(x)/dx ) , the force is negative. Thus the particle is ‘pushed back. i.e., it turns around at a boundary. 81 K 0 at ymax , Emec mgymax F ( y) dU ( y) m g dy Emec K ( y ) U ( y ) 1 2 m v m gy 2 Equilibrium Points REGAN PHY34210 82 Equilibrium Points: refer to points where, dU/dx=-F(x)=0. Neutral Equilibrium: is when a particle’s total mechanical energy is equal to its potential energy (i.e., kinetic energy equals zero). If no force acts on the particle, then dU/dx=0 (i.e. U(x) is constant) and the particle does not move. (For example, a marble on a flat table top.) Unstable Equilibrium: is a point where the kinetic energy is zero at precisely that point, but even a small displacement from this point will result in the particle being pushed further away (e.g., a ball at the very top of a hill or a marble on an upturned dish). Stable Equilibrium: is when the kinetic energy is zero, but any displacement results in a restoring force which pushes the particle back towards the stable equilibrium point. An example would be a marble at the bottom of a bowl, or a car at the bottom of a valley. REGAN PHY34210 U(x) x D B C A Particles at A,B, C and D are in at equilibrium points where dU/dx = 0 A,C are both in stable equilibrium ( d 2U/dx2 = +ve ) B is an unstable equilibrium ( d 2U/dx2 = -ve ) D is a neutral equilibrium ( d 2U/dx2 = 0 ) 83 Work Done by an External Force REGAN PHY34210 Previously we have looked at the work done to/from an object. We can extend this to a system of more than one object. Work is the energy transferred to or from a system by means of an external force acting on that system. No friction (conservative forces) W DK DU DEmec Including friction From Newt ons2 nd law, F f k m a , t heforce(t husacceleraton) i is const ant , t hereforewe can use v 2 v02 2ad v 2 v02 f k and By substit ution, F m 2d 1 2 1 2 Fd m v m v0 f k d DK f k d 2 2 84 Conservation of Energy REGAN PHY34210 85 This states that ‘ The total energy of a system, E, can only change by amounts of energy that are transferred to or from the system. ’ Work done can be considered as energy transfer, so we can write, W DE DEmec DEth DEin DEmec is thechangein mechanicalenergy,DEth is thechangein thermal energy(i.e.,heat)and DEin is thechangein internalenergyof thesystem. If a system is ISOLATED from it surroundings, no energy can be transferred to or from it. Thus for an isolated system, the total energy of the system can not change, i.e., DE DE DE DE 0 mec th in Emec , 2 Emec ,1 DEth DEin Another way of writing this is, which means that for an isolated system, the total energies can be related at different instants, WITHOUT CONSIDERING THE ENERGIES AT INTERMEDIATE TIMES. REGAN PHY34210 Example 1: A child of mass m slides down a helter skelter of height, h. Assuming the slide is frictionless, what is the speed of the child at the bottom of the slide ? h=10m From theCONSERVAT ION OF MECHANICALENERGY, Emec ,i Emec , f U i K i U f K f 1 2 U i m gh , U f 0, K i 0, K f m v 2 1 2 m gh 0 0 m v v 2 gh 2 Note that thisis thesame speed that thechild would have if it fell directlyfroma height h. 86 Example 2: REGAN PHY34210 A man of mass, m, jumps from a ledge of height, h above the ground, attached by a bungee cord of length h L. Assuming that the cord obeys Hooke’s law and has a spring constant, k, what is the general solution for the maximum extension, x, of the cord ? By CONSERVAT ION OF MECHANICALENERGY, DK DU 0 , if v 0 at topand bottom,K i K f 0 DK 0 87 L x m 1 2 also, DU DU grav DU elas m g Dy 0 kx 2 1 2 1 2 kx m gL x m gL m gx kx m gx m gL 0 , 2 2 solving thisquadratic equation, x mg m g2 2km gL k , x ve root 9: Systems of Particles REGAN PHY34210 88 Centre of Mass (COM): The COM is the point that moves as though all the mass of a body were concentrated there. For 2 particlesof mass, m1 and m2 separatedby d , if theorgin of x - axis coincides with theparticleof mass m1 , thecent reof mass of thesystemis m2 xcom d . Moregenerally,if m1 is at x1 and m2 is at x2 , theCOM m1 m2 m1 x1 m2 x2 m1 x1 m2 x2 is defined by xcom where M is the totalmass m1 m2 M T hegeneralformfor a n-particlesystem is given by m1 x1 m2 x2 m3 x3 m4 x4 1 n xcom mi xi , similarly,for 3 - D M M i 1 1 n 1 n ycom mi yi and zcom mi zi . In vectorform,if r xiˆ yˆj zkˆ M i 1 M i 1 n 1 then rcom xcomiˆ ycom ˆj zcom kˆ and rcom m r ii M i 1 Centre of Mass for Solid Bodies REGAN PHY34210 89 Solid objectshaveso manyparticles(atoms)that theycan be considered to be made up of manyinfinitessimallysmall MASS ELEMENT S,dm. 1 1 1 T hen, xcom xdm , ycom ydm , zcom zdm , M M M Often, theintegralsare simplifiedassuming a UNIFORMDENSIT Y( r ) M dm where r , where dV is the volumeoccupiedby mass, dm. V dV 1 1 1 substituting, xcom xdm xrdV xdV and similarly, M rV V 1 1 ycom ydV , zcom zdV , V V Note that thecentreof mass need not necessarily lie in thevolumeof the object (for examplea doughnut or an igloo). Newton’s 2nd Law for a System of Particles. Mrcom m1r1 m2 r2 m3 r3 mn rn differentiating with respect totime, dmn rn mvn we get since, dt drcom Mvcom m1v1 m2 v2 m3v3 mn vn M dt dvn an differentiatingonceagain, and recalling dt and Newton's 2 nd law, dvcom Macom m1a1 m2 a2 m3 a3 mn an M dt Fcom F1 F2 F3 Fn REGAN PHY34210 90 Linear Momentum REGAN PHY34210 91 T heLINEARMOMENT UMis defined by p mv dp d mv m dv F ma (for m is constant ). dt dt dt Thus we can re-write Newton’s 2nd law as ‘ The rate of change of the linear momentum with respect to time is equal to the net force acting on the particle and is in the direction of the force.’ For a systemof particles,thesystemhas a totallinear momentum,P which is the vectorsum of theindividual particle linear momenta,i.e., P p1 p2 p3 pn m1v1 m2 v2 m3v3 mn vn P M vcom The linear momentum of a system of particles is equal to the product of the total mass of the system, M, and the velocity of the centre of mass, Conservation of Linear Momentum REGAN PHY34210 92 dP dvcom Since, F m macom , in a closed system,if the dt dt net externalforceis zero,and no particlesenteror leave the dP system,then, Fnet 0 P constant i.e., Pi Pf dt This is the law of CONSERVATION OF LINEAR MOMENTUM which we can write in words as ‘In no net external force acts on a system of particles, the total linear momentum, P , of the system can not change.’ also, leading on from this, ‘ If the component of the net external force on a system is zero along a specific axis, the components of the linear momentum along that axis can not change.’ Varying Mass: The Rocket Equation REGAN PHY34210 93 For rockets, the mass of the rocket is is not constant, (the rocket fuel is burnt as the rocket flies in space). For no gravitational/drag forces, By conservaton i of momentum,Pi Pf T heinitialP of therocketplus theexhaust fuel equals the P of theexhaust product splus the P of therocketafter timeinterval,dt. Mv dMU M dM v dv a) time = t M v if vrel is therelativespeed between the rocketand theexhaust product s(and dM - ve) , then,v dv vrel U Mv dMU M dM vrel U Mv dM v dv vrel M dM v dv Mv dMv dMdv dMvrel Mv Mdv dMv dMdv 0 dMvrel Mdv dM dv vrel M dt dt b) time = t+dt -dmM+dm U dv if R is therateof mass loss, then,we obt ain Rvrel M Ma dt v+dv 1st rocket equation Rvrel is called theT HRUST (T ) of therocketengine. REGAN PHY34210 94 M is themass at timet and a is theacceleration, T Ma, which is Newt on's 2 nd law. T o find the velocityas themass changes, dM Mdv vrel dM dv vrel M vf Mf vi Mi Integrat ing, we obt ain, dv -vrel dM where vi and v f M are theinitialand final rocket vel ocit ies,corresponding to rocket masses of M i and M f respectively. 1 Since, in general, dx ln x , then x M i 2nd rocket Mf v rel ln v f vi vrel ln M f ln M i vrel ln M equation Mi f thus increasein velocity greatestfor small M f (use of mult i- st age rockets!) Internal Energy Changes and External Forces REGAN PHY34210 Energy can be transferred ‘inside a system’ between internal and mechanical energy via a force, F. (Note that up to now each part of an object has been rigid). In this case, the energy is transferred internally, from one part of the body to another by an external force. T hechangein internalenergyof thesystemis given by, ΔEint Fd cos where d is thedisplacement of theCENT REOF MASS and is the angle between the directionsof theforce F and displacement d . T heassociatedchangein theMECHANICALENERGY is then ΔE mec ΔK ΔU Fd cos 95 10: Collisions REGAN PHY34210 96 ‘A collision is an isolated event in which two or more colliding bodies exert forces on each other for a short time.’ Impulse For a head on collision between two bodies, the3rd forcepair, -F(t) F(t) F(t) and -F(t)acts between the two at time,t. F(t)is a T IME- VARYING FORCE. From Newton's 2 nd law, theseforceswill change thelinear momenta of both bodies. T heamountby which p changesdepends on thetime interval,Δt , during which these forcesact. From Newton's 2 law dp F t dt nd pf tf pi ti dp F t dt IMP ULSE,J T heIMP ULSEis theCHANGE IN LINEARMOMENT UMof thebody acted on by F(t) (right handside). T hisis also equal to theproduct of thestrengthand durationof theappliedforce,and theAREA UNDERT HECURVE of F t versust. REGAN T heIMP ULSE-LINEARMOMENT UMT HEOREMstates PHY34210 that thechangein thelinear momentumof each body in a collisionis equal to theIMP ULSE thatactson thatbody, i.e., p f pi Dp J Since, impulse is a VECT OR, we can also write thisin componentform, p f x pix Dp x J x , p f y pi y Dp y J y , p f z piz Dp z J z If Fave is the timeaveragedforceover a period, Δt , the magnitudeof theimpulse is given by J Fave Dt E.g A 140g is pitched with a horizontal speed of vi=39m/s. If it is hit back in the opposite direction with the same magnitude of speed what is the impulse, J, which acts on the ball ? J p f pi mv f vi takingtheinitialvelocitydirectionas the NEGAT IVEdirection,J 0.1439 39kgm s1 10.9kgm s1 97 REGAN PHY34210 98 Momentum and Kinetic Energy in Collisions In any collision, at least one of the bodies must be moving prior to the collision, meaning that there must be some amount of kinetic energy in the system prior to the collision. During the collision, the kinetic energy and linear momentum are changed by the impulse from the other colliding body. If the total kinetic energy of the system is equal before and after collision, it is said to be an ELASTIC COLLISION. However, in most everyday cases, some of this kinetic energy is transferred into another form of energy such as heat or sound. Collisions where the kinetic energies are NOT CONSERVED are known as INELASTIC COLLISIONS. In a closed system, the total linear momentum, P of the system can not change, even though the linear momentum of each of the colliding bodies may change. REGAN PHY34210 By CONSERVAT ION OF LINEARMOMENT UM,Pi Pf 99 totalmomentumbeforecollision totalmomentumaftercollision For a 2 BODY COLLISION, p1,i p2,i p1, f p2, f m1v1,i m2 v2,i m1v1, f m2 v2, f For a COMP LET ELYINELAST ICCOLLISION,the two particles stick aftercollision(e.g., a rugby t ackle!) , then m1v1,i m1 m2 V For an isolat edsystem,the velocit yof thecentreof mass can not change in a collisionas thesystemis isolat edand thereis no net ext ernalforce. Recalling P Mvcom m1 m2 vcom m1v1 m2 v2 P p1 p2 vcom m1 m2 m1 m2 Elastic Collisions in 1-D REGAN PHY34210 100 In an elastic collision, the total energy before the collision is equal to the total kinetic energy after the collision. Note that the kinetic energy of each body may change, but the total kinetic energy remains constant. after elastic collision before elastic collision m1, v1,f m2, v2,f m1, v1,i m2, v2,i=0 For a head - on collision between two billiard balls, with mass, m2 at rest. By conservation of linear momentum,m1v1,i m1v1, f m2 v2, f m1 v1,i v1, f m2 v2, f (1- D case, magnitudesalongsame axis). In an elasticcollision the totalkineticenergyis conserved 1 1 1 m1v12,i m1v12, f m2 v22, f m2 v22, f m1 v1,i v1, f v1,i v1, f 2 2 2 m1 m2 2m1 which leads to, v1, f v1,i and v2, f v1,i m1 m2 m1 m2 Note thatv2 ,f is always positive(i.e.m2 is always pushed forward). REGAN PHY34210 For 1 - D elasticcollisions, v1, f 101 m1 m2 2m1 v1,i & v2, f v1,i m1 m2 m1 m2 T heselead to thefollowinglimitingcases. 1) Equal masses, m1 m2 (e.g. poolballs) : v1, f 0 , v2, f v1,i i.e., for a head - on collision between equal masses, theprojectilestops followingcollisionand the targetmovesoff with theprojectile' s velocity. 2) Massive target,m2 m1 (e.g., golf ball on a cannonball) : 2m1 v1,i i.e., light projectilebounces back with v1, f v1,i , v2 ,f m2 similar velocity(but oppositedirection)to incomingprojectile. Heavy target movesforwards with small velocity. 3) Massive projectile m1 m2 (e.g., cannonball on golf ball) : v1,f v1,i , v2 ,f 2v1,i i.e. heavyprojectilecontinuesforwards at approx. unchangedvelocity,light target movesoff with twice theprojectilevelocity Example 1: REGAN PHY34210 102 Nuclear reactors require that the energies of neutrons be reduced by nuclear collisions with a MODERATOR MATERIAL to low energies (where they are much more likely to take part in chain reactions). If the mass of a neutron is 1u~1.66x10-27kg, what is the more efficient moderator material, hydrogen (mass = 1u) or lead (mass~208u)? Assume the neutron-moderator collision is head-on and elastic. We want theMAXIMUMtransferof kineticenergy FROM T HE NEUT RONfor a single collisionas a functionof moderatormass. T heinitialand final kinetic 1 1 mn vn2,i and K f mn vn2, f 2 2 K i K f vn2,i vn2, f T hefractionalenergyloss per collisionis F Ki vn2,i energiesof theorginaland scatteredneutronare K i For a closed neutron- nucleus collision& themoderatingnucleus initiallyat rest, fromcons.of lin. mom. vn , f vn , i mn mMOD 4mn mMOD , thusF therefore, 2 mn mMOD mn mMOD F 4 / 4 1 for hydrogen proton(NB. water H 2 O) and ~ 4/208~ 1/50 for Pb! Example 2: The Ballistic Pendulum A ballastic pendulum uses the transfer of energy to measure the speed of bullets fired into a wooden block suspended by string. vbul REGAN PHY34210 Mblock By conservation of linear momentum,mbul vbul mbul M block vblock Also know thatis theblock systemis closed, we can assume a conservation of mechanicalenergy,then 1 2 mbul M block vblock mbul M block gh 2 where h is theincreasein height of theblock as it swings upwards. mbul vbul 1 mbul M block mbul M block gh 2 mbul M block 2 2 vbul mbul M block 2 gh 2 m M v bul block bul 2 mbul mbul 2 gh h 103 1-D Collisions with a Moving Target REGAN PHY34210 104 By conservation of linear momentum, m1v1,i m2 v2,i m1v1, f m2 v2, f m1 v1,i v1, f m2 v2,i v2, f For an elasticcollision,kineticenergyis conserved,thus 1 1 1 1 m1v12,i m2 v22,i m1v12, f m2 v22, f 2 2 2 2 m1 v12,i v12, f m2 v22, f v22,i m1 v1,i v1, f v1,i v1, f m2 v2, f v2,i v2, f v2,i before elastic collision m1, v1,i m2, v2,i solving thesesimultaneous equations,we obtain thegeneralrelations, v1, f m1 m2 2m2 2m1 m2 m1 v1,i v2 ,i & v2 , f v1,i v2 ,i m1 m2 m1 m2 m1 m2 m1 m2 T hesubscripts1 and 2 are arbitrary.Note,if we set v2,i 0 (stationary target)we obtain thepreviousresults of v1, f m1 m2 v1,i m1 m2 & v2 , f 2m1 v1,i m1 m2 Collisions in Two Dimensions REGAN PHY34210 105 When two bodies collide, the m2, v2,f y impulses of each body on the other m2, v2,i determine the final directions following 2 x the collision. If the collision is not 1 head-on (i.e. not the simplest 1-D case) in a closed system, momentum remains m1, v1,i conserved, thus, for an elastic collision m1, v1,f where Ktot,I=Ktot,f , we can write, 1 1 1 1 2 2 2 P1,i P2,i P1, f P2, f and m1v1,i m2 v2,i m1v1, f m2 v22, f 2 2 2 2 For a 2-D glancing collision, the collision can be described in terms of momentum components. For the limiting case where the body of m2 is initially at rest, if the initial direction of mass, m1 is the x-axis, then, x axis, m1v1,i m1v1, f cos1 m2 v2, f cos 2 y axis, 0 m1v1, f sin 1 m2v2, f sin 2 For an elastic collision,m1v12,i m1v12, f m2 v22, f 11: Rotation REGAN PHY34210 Most motion we have discussed thus far refers to translation. Now we discuss the mechanics of ROTATION, describing motion in a circle. First, we must define the standard rotational properties. A RIGID BODY refers to one where all the parts rotate about a given axis without changing its shape. (Note that in pure translation, each point moves the same linear distance during a particular time interval). A fixed axis, known as the AXIS OF ROTATION is defined by one that does not change position under rotation. Each point on the body moves in a circular path described by an angular displacement D. The origin of this circular path is centred at the axis of rotation. 106 Summary of Rotational Variables REGAN PHY34210 107 All rotational variables are defined relative to motion about a fixed axis of rotation. The ANGULAR POSITION, , of a body is then the angle between a REFERENCE LINE, which is fixed in the body and perpendicular to the rotation axis relative to a fixed direction (e.g., the x-axis). If is in radians, we know that =s/r where s is the length of arc swept out by a radius r moving through an angle . (Note counterclockwise represent increase in positive . axis of Radians are defined by s/r and are thus rotation pure, dimensionless numbers without units. The circumference of a circle (i.e., a full arc) s=2r, thus in radians, the angle swept out by a single, full reference revolution is 360o = 2r/r=2. Thus, line r 1 radian = 360 / 2 = 57.3o s x = 0.159 of a complete revolution. The angular displacement, D represents the change in PHY34210 the angular position due to rotational motion. In analogy with the translational motion variables, other angular motion variables can be defined in terms of the change (D), rate of change ( ) and rate of rate of change ( ) of the angular position. REGAN Angular position(radians), Angular displacement (radians), 108 s r D 2 1 D 2 1 AverageAngular Velocity(radian per second), av Dt t 2 t1 Instantane ous Angular Velocity(rad/s), AverageAngular Acceleration, (radiansper s 2 ), Instantane ous Angular Acceleration, (rad/s2 ), d dt D av Dt d av dt Relating Linear and Angular Variables REGAN PHY34210 109 For the rotation of a rigid body, all of the particles in the body take the same time to complete one revolution, which means that they all have the same angular velocity,, i.e., they sweep out the same measure of arc, d in a given time. However, the distance travelled by each of the particles, s, differs dramatically depending on the distance, r, from the axis of rotation, with the particles with the furthest from the axis of rotation having the greatest speed, v. at and ar are the tangential and radial accelerations respectively. We can relate the rotational and linear variables using the following (NB.: RADIANS MUST BE USED FOR ANGULAR VARIABLES!) ds d dv d r d s r ; v r r ; at r r dt dt dt dt dt v 2 r Radial componentof theacceleration is ar r 2 r r 2r 2 P eriodof revolution, T v 2 Rotation with Constant Acceleration REGAN PHY34210 For translational motion we have seen that for the case of a constant acceleration, we can derive a series of equations of motion. By analogy, for CONSTANT ANGULAR ACCELERATION, there is a corresponding set of equations which can be derived by substituting the translational variable with its rotational analogue. TRANSLATIONAL v v0 at 1 2 x x 0 v0t at 2 v 2 v02 2ax x0 v v0 x x0 t 2 1 2 x x0 vt at 2 ROTATIONAL 0 t 1 2 0 t t 2 2 02 2 0 0 0 t 2 1 2 0 t t 2 110 Example 1: REGAN PHY34210 A grindstone rotates at a constant angular acceleration of =0.35rad/s2. At time t=0 it has an angular velocity of 0=-4.6rad/s and a reference line on its horizontal at the angular position, 0=0. 111 ref. line for 0=0 axis of rotation (a) at what time after t=0 is the reference line at =5 revs ? 1 2 0 0t t 2 : 5rev 10 rad; 0 0 ; 0 4.6rad / s ; 0.35rad / s 2 4.6 1 10π-0 -4.6t 0.35 t 2 t 2 4.62 4 0.175 10 2 0.175 4.6 6.56 32s 0.35 Note that while 0 is negative, is positive. Thus the grindstone starts rotating in one direction, then slows with constant deceleration before changing direction and accelerating in the positive direction. At what time does the grindstone momentarily stop to reverse direction? t 0 a 0 4.6rad / s 13s 2 0.35rad / s Kinetic Energy of Rotation REGAN PHY34210 112 For a compositebody which we can treatas a collectionof masses, mn , movingat differentspeeds, vn , thekineticenergyis 1 1 1 1 2 2 2 K m1v1 m2 v2 m3v3 mn vn2 2 2 2 n 2 1 1 2 2 2 K mn rnn . BUT is constant,thus K mn rn . 2 n n 2 Now we can define I mn rn2 where n I is theMOMENTOF INERT IAor ROT AT IONALINERT IA 1 2 Kineticenergy of rotationis given by K I 2 T husin general,a smaller momentof inertiameansless work is needed to be done (i.e.less K ) for rotation ot takeplace. Calculating to the Rotational Moment of Inertia For a rigid body, I mn rn2 REGAN PHY34210 113 where r is theperpendicular distanceof then th n part iclefrom therotationaxis. For a continuousbody, I mn rn2 r 2 dm. n The Parallel-Axis Theorem To calculate I if the moment of inertia about a parallel axis passing through the body’s centre of mass is known, we can use I=Icom+Mh2, where, M= the total mass of the body, h is the perpendicular distance between the parallel centre of mass axis and the axis of rotation and Icom is the moment of inertia about the centre of mass axis. If h 2 a 2 b 2 , I r 2 dm x a y b dm 2 2 I x 2 y 2 dm 2a xdm 2b ydm a 2 b 2 dm 1 Now, since x y R and since xcom x dm, assuming we M take thecentreof mass as theorgin, thenby definition, 2 2 2 2b xdm 2b ydm 0 I R 2 dm a 2 b 2 dm I com h 2 M Example 2: REGAN PHY34210 114 The HCl molecule consists of a hydrogen atom (mass 1u) and a chlorine atom (mass 35u). The centres of the two atoms are separated by 127pm (=1.27x10-10m). What is the moment of inertia, I, about an axis perpendicular to the line joining the two atoms which passes through the centre of mass of the HCl molecule ? We can locatethecent reof mass of the2 - particle systemusing xcom m x m2 x2 1 1 M m1 m2 a Cl d-a com H If the x co - ordinatefor thecent reof mass x 0 then, mCl a mH d a mH 0 a d . Now mCl mH mCl mH I com mi ri 2 mH d a mCl a 2 2 d rotation axis i 2 2 mH mH mCl 2 mH 1 35 2 mH d d mCl d d I u 127pm mCl mH mCl mH 1 35 mCl mH I 15,250u. pm2 (not eunits for rotationalmomentsof inertiafor molecules). Torque and Newton’s 2nd Law REGAN PHY34210 The ability of a force, F, to rotate an object depends not just on the magnitude of its tangential component, Ft but also on how far the applied force is from the axis of rotation, r. The product of Ft r =Frsin is called the TORQUE (latin for twist!) . 115 F Ft Frad r O r T ORQUE, r F sin rFt AND r sin F r F . r is the perpendicular distancebetween O and a line running throughF . r is theMOMENTARM OF T HE FORCE F. SI unit of T orqueis Nm, which are equivalent to theunit of work W F.d . W in Joules, in Nm. Relatingthe tangential force to the tangential acceleration, Ft m at . T orqueactingon theparticleis Ft r m at r , since, at r , m rr m r2 I τ net Iα Newton's 2nd law for rotation. Work and Rotational Kinetic Energy From the WORK- KINET ICENERGY T HEOREM, 1 2 1 2 m vf m vi W 2 2 1 1 2 2 since v rω , then mr f mr i W 2 2 Recallingfor a single - particlebody, I m r2 , DK K f K i 1 2 1 2 then W DK I f Ii 2 2 Workdone, W Fs Ft rD D Workdone in an angular displacement 1 to 2 2 is given by W d 1 dW d P OWERis given by P dt dt REGAN PHY34210 116 REGAN PHY34210 117 12: Rolling, Torque and Angular Momentum Rolling: Rolling motion (such as a bicycle wheel on the ground) is a combination of translational and rotational motion. O COM motion. R O P P S A wheelrollingat a CONST ANTSP EED, means that t hespeed of the centreof mass, vcom is constant.In a timeintervaldt, thecentreof mass travelsthesame distanceas thedistance theoutside of the wheel moves throughan arc of length,s R , where R is the wheel radius and is its angular displacement. d T heangular speed of the wheel about its centreis , while thespeed dt ds d R d of thecentreof mass is given by vcom R R dt dt dt REGAN PHY34210 118 The kinetic energy of rolling. R A rolling object has two types of COM kinetic energy, a rotational O O motion. kinetic energy due to the rotation about the centre of P P S mass of the body and translational kinetic energy due to the translation of its centre of mass. We can viewthesituationas pure rotationabout an axis through he t point,P. 1 T hekineticenergyof thisrotationis given by K I P 2 where I p is the 2 momentof inertia through he t pointof contact with theground, P. From thePARALLELAXIST HEOREM,we can write 1 I P I com MR 2 and thus, K I com MR 2 2 2 1 1 1 1 2 2 2 2 K I com M R I com Mvcom 2 2 2 2 N REGAN 119 Rolling Down a Ramp PHY34210 If a wheel rolls at a constant speed, it R has no tendency to slide. However, if this Fg sin P Fg cos wheel is acted upon by a net force (such as gravity) this has the effect of speeding Fg up (or slowing down) the rotation, causing an acceleration of the centre of mass of the system, acom along the direction of travel. It also causes the wheel to rotate faster. These accelerations tend to make the wheel SLIDE at the point, P, that it touches the ground. If the wheel does not slide, it is because the FRICTIONAL FORCE between the wheel and the slide opposes the motion. Note that if the wheel does not slide, the force is the STATIC FRICTIONAL FORCE ( fs ). Since therotationalfrequencyis given by R vcom , then d R d vcom by differentiatingboth sides, acom R dt dt N Rolling down a ramp (cont.) For a uniform body of mass, M and radius, R, rolling smoothly (i.e. not sliding) down a ramp Fg sin tilted at angle, (which we define as the x-axis in this problem), the translational acceleration down the ramp can be calculated, from REGAN PHY34210 120 R P Fg Fg cos theforcecomponentsalong theslope, Fx ,net Macom, x f s Mg sin where f s m s N m s Mg cos . Rot.formof Newton's 2 nd law is Fr I . T heonly forcecausing a rollingmotionin thefigure is theFRICT IONat point P . T hegravitational and Normalforcesall act through theCOM and thus haveR 0. acom net I com Fr f s R. For smoothrolling, (notesign) R I com acom, x I com I com . acom, x f s Macom, x Mg sin R R R R2 I com acom, x g sin Macom, x Mg sin acom 2 R I com 1 MR 2 The Yo-Yo: If a yo - yo rolls down a distanceh it loses gravitational potentialenergy,m gh. T hisis transferred into kinetic REGAN PHY34210 T 1 2 energyin both translational K trans m v and rotational R0 R 2 1 2 K rot I forms.As the yo - yo climbs back up thestring, 2 it loses thiskineticenergy and transfersit back topotentialenergy. Mg T heexpressionfor thevalue of theacceleration of the yo - yo rolling down thestringcan be calculatedassuming Newton's 2 nd law (as for a body rollingdown a hill) with thefollowingassumptions. (1) the yo - yo rolls directlydown thestring(i.e. 900 ). (2) the yo - yo rolls around theaxle with radius R0 , not theouter radius, R. (3) the yo - yo is slowed by thetensionin thestringrather tha n friction. g T hisanalysisleads to theexpression, acom I 1 com2 MRO 121 Example 1: A uniform ball of mass M=6 kg and radius R rolls smoothly from rest down a ramp inclined at 30o to the horizontal. (a) If the ball descends a vertical height of 1.2m to reach the bottom of the ramp, what is the speed of the ball at the bottom ? REGAN PHY34210 122 1.2m By conservation of mechanicalenergy, K i U i K f U f Mgh 0 0 K rot K trans 1 1 2 Mvcom I com 2 . 2 2 2 MR 2 for a sphere, 5 2 1 1 1 12 1 1 2 2 2 2 vcom 2 2 Mgh Mvcom I com Mvcom MR 2 Mvcom Mvcom 2 2 2 25 5 R 2 For smoothrolling,vcom R and subsit uting, I com v 2 com gh 7 10 vcom 10gh 4.1m s1 (note,Mass independent, marble 7 and bowling ball reach bottomat same time!) REGAN PHY34210 Example 1 (cont): 123 (b) A uniform ball, hoop and disk, all of mass M=6 kg and radius R roll smoothly from rest down a ramp inclined at 30o to the horizontal. Which of the three objects reaches the bottom of the slope first ? 1.2m 2 1 2 T hemomentsof inertiafor a sphere MR ; disk MR 2 ; and hoop MR 2 . 5 2 T hefractionof kineticenergy which goes intoT RANSLAT IONAL MOT ION, 2 Mvcom vcom 2 f 1 . In general,I com MR , wit h a constantand 2 1 2 I com R 2 Mv 1 2 2 com f 1 2 2 Mvcom 2 v 2 2 com 1 1 2 2 Mvcom 2 MR R 1 1 1 2 For hoop, 1, f 0.5 ; For disk, , f 0.66* ; For sphere, , f 0.71. 2 5 Sphere rolls fast est,followed by thedisk. Any size marble will beat disk. REGAN PHY34210 124 Torque was defined previously for a rotating rigid body as =rFsin. More generally, torque can be defined for a particle moving along ANY PATH relative to a fixed point. i.e. the path need not be circular. z z rxF = F redrawn at origin O O x x F r r F r y F y T he torqueis defined by r F . T hedirectionof the torqueis found using the vectorcross product right - hand rule, (i.e.perpendicular toboth r and F ). T heMAGNIT UDEOF T HET ORQUE is given by rF sin r F rF where r r sin and F F sin . REGAN PHY34210 125 Angular Momentum A particleof mass m, with velocity v (i.e. with linear momentum,p mv ), has an ANGULAR MOMENT UM given by l r p mr v . z z rxp = l l O y r p redrawn at origin p p O x y r p r x p T heangular moment umdirect ionis given by t hevect orcross product (t heright - hand rule shows t hatl is t o bot h r and p, v ). T hemagnit udeof t heangular moment um(in unit s of kg.m2 /s Js ) is given by l rp sin r p p r where r r sin and p p sin . Newton’s 2nd Law in Angular Form. REGAN PHY34210 126 dp Newton's 2 nd law in translationalform can be writtenas Fnet dt If theangular momentumof a particleis given by l r p mr v Differentiatingboth sides with respect totimegives, dv dr dl d mr v m r v mr a v v dt dt dt dt v v 0 since these vectorsare parallel(sin 0) dl mr a r ma r Fnet ri Fi net dt i i.e. therateof changeof angular momentumwith respect totime dl is equal to the vectorsum of torquesactingon theparticle net . dt REGAN 127 For a SYST EM OF P ART ICLES,the totalangular momentum, PHY34210 L is theVECT OR SUM of theangular momenta,l of theindividual particles, n n n dli dL i.e., L l1 l2 l3 ln li net ,i dt i 1 dt i 1 i 1 Only EXT ERNALtorqueschange theT OT ALANGULAR MOMENT UM( L ) i.e., thosedue to forceson theparticlesfromexternalbodies. If net is the NET EXT ERNALT ORQUE, i.e. the vectorsum of all dL externaltorques, τ net , we obtain a formfor Newton's 2 nd law : dt The net external torque, net acting on a system is equal to the rate of change of the total angular momentum of the system ( L ) with time. REGAN 128 For a given part iclein a rigid body rot at ingabout a fixed PHY34210 axis, t he magnitudeof t heangular moment umof a mass element Δmi , is l ri pi sin 900 ri Dmi vi . T heangular moment umcomponentparallelt o t herot at ionDm (z) axis is liz li sin ri sin Dmi vi ri Dmi vi T hecomponentfor t heENT IREBODY is t hesum of t heseelement alcont ribut ons i z r r pi y x n i.e., L z liz Dmi vi ri Dmi r i ri Dmi r2i i 1 i 1 i 1 i 1 n n n n is a CONST ANTfor all point son t herot at ingbody and Dmi r2i I , i 1 t he momentof inert iaof t hebody about a fixed axis, we can writ e, Lz I Usually t he ' z ' is dropped,assuming t hat L is about therot at ionaxis. REGAN dPHY34210 L Since thenet torqueis relatedto thechangein angular momentumby, τ net , dt dL if NO NET T ORQUEactson thesystem,then 0 and thus dt T HE ANGULAR MOMENT UMOF T HESYST EM IS CONSERVED. Conservation of Angular Momentum 129 T hismeans that thenet angular momentumat timeti , is equal to thenet angular momentumof thesystemat some other time ,tf . We can thussay thatif thenet externaltorqueactingon a systemis zero, the angular momentumof thesystem,L remainsconstant,no matter what changes takeplace WIT HINthesystem. Similarly,if theCOMP ONENT of thenet externaltorqueon a systemalonga fixed axis is zero, then thecomponentof angular momentumalong thataxis can not change,no matter what takesplace within the system. T heconservation law can be writtenin algebraic formas I ii I f f . T hismeans thatif themomentof inertiaof a systemdecreases,its rotational speed increasesto compensate, (e.g., pirouetting skaters,neutronstarsand nuclei!) Example1: Pulsars (Rotating Neutron Stars) Crab nebula, SN remnant observed by chinese in 11th century before REGAN PHY34210 after! SN1987A Pulsars have similar periodicities ~0.1-1s. Vela supernova remnant, pulsar period ~0.7 secs 130 REGAN PHY34210 131 Rotational period of crab nebula (supernova remnant) =1.337secs Lighthouse effect Star quakes optical I ii k .MRi2 i I f f k .MR 2f f , k constant Tf i R f Ri Ri , T periodof rotation f Ti Ri ( sun) ~ 7 108 m, Ti ( sun) ~ 2.5 106 s, T f ( pulsar) ~ 1s 1s R f ~ 7 10 m ~ 400Km. 6 2.5 10 s 8 x-ray PULSAR = PULSAting Radio Star (neutron-star) REGAN PHY34210 TRANSLATIONAL ROTATIONAL dp dl Force, F ma Torque, I r F dt dt Mom entum Linear , p mv Ang. Mom. l I r p dp dL nd Newtons 2 Law, F net dt dt Conservation, LinearMom. Angular mom. dp dL Fnet 0 net 0 dt dt 132 13: Equilibrium and Elasticity REGAN PHY34210 133 An object is in ‘equilibrium’ if p=Mvcom and L about an any axis are constants (i.e. no net forces or torques acts on the body). If both equal to zero, the object is in STATIC EQUILIBRIUM. If a body returns to static equilibrium after being moved (by a restoring force, e.g., a marble in a bowl) it is in STABLE EQUILIBRIUM. If by contrast a small external force causes a loss of equilibrium, it has UNSTABLE EQUILIBRIUM (e.g., balancing pennies edge on). dp F net 0 (i.e.balanceof forces)for dt T RANSLAT IONAL EQUILIBRIUM and dL net 0 (i.e.balanceof torques)for dt ROT AT IONALEQUILIBRIUM The Centre of Gravity REGAN PHY34210 The gravitational force acts on all the individual atoms in an object. In principle these should all be added together vectorially. However, the situation is usually simplified by the concept of the CENTRE OF GRAVITY (cog), which is the point in the body which acts as though all of the gravitational force acts through that point. If the acceleration due to gravity, g, is equal at all points of the body, the centre of gravity and the centre of mass are at the same place. 134 Elasticity REGAN PHY34210 135 A solid is formed when the atoms which make up the solid take up regular spacings known as a LATTICE. In a lattice, the atoms take up a repetitive arrangement whereby they are separated by a fixed, well defined EQUILIBRIUM DISTANCE (of ~10-9->10-10m) from their NEAREST NEIGHBOUR ATOMS. The lattice is held together by INTERATOMIC FORCES which can be modelled as ‘inter-atomic springs’. This lattice is usually extremely rigid (i.e., the springs are stiff). Note that all rigid bodies are however, to some extent ELASTIC. This means that their dimensions can be changes by pulling, pushing, twisting and/or compressing them. STRESS is defined as the DEFORMING FORCE PER UNIT AREA= F/A, which produced a STRAIN, which refers to a unit deformation. The 3 STANDARD type of STRESS are (1) tensile stress ->DL/L (stretching) ; (2) shearing stress -> Dx/L (shearing) ; and (3) hydraulic stress -> DV/V (3-D compression). REGAN PHY34210 136 STRESS and STRAIN are PROPORTIONAL TO EACH OTHER. The constant of proportionality which links these two quantities is know as the MODULUS OF ELASTICITY, where STRESS = MODULUS x STRAIN F T heSTRESSon an object for simple tensionor compression is given by , A where F is themagnitudeof theforceappliedperpendicularly tothearea A (T hisalso defines thepressure at thatpoint). T heSTRAIN is theunit deformation. For tensile stress, this is a dimensionless ΔL corresponding to the fractionalchange inthe length quantitydefined by L of theobject ( L is theorginallength,ΔL is theextension). T heYOUNG'S MODULUS ( E ) for tensile or F compressive stressis defined by E A F For SHEARING, thestressis still , but A F ΔL L+ DL L F is parallel Δx t o theplaneof thearea. T hestrain is now , leading l F Dx L t o theSHEAR MODULUS,(G) where G A l HYDRAULICSTRESSis defined as thefluid pressure P, REGAN PHY34210 L F Dx F V ΔV (i.e.forceper unit area).T hestrain is defined as , V V-DV whereV is theinitial volumeand ΔV is the volumechange. DV T heBULK MODULUS( B ) is defined by P B V 137 F DV REGAN PHY34210 If we plots stress as a function of strain, for an object, over a wide range, there is a linear relationship. This means that the sample would regain its original dimensions once the stress was removed (i.e., it is ‘elastic’). However, if the stress is increases BEYOND THE YIELD STRENGTH, Sy,of the specimen, it will become PERMANENTLY DEFORMED. If the stress is increased further, it will ultimately reach its ULTIMATE STRENGTH, Su, where the specimen breaks/ruptures. 138 Su (rupture) Sy (perm. deformed) Strain (Dl/l) Example 1: REGAN 139 PHY34210 F=62kN A cylindrical stainless steel rod has a radius r = 9.5mm and length, L = 81cm. A force of 62 kN stretches along its length. (a) what is the stress on the rod ? F F 6.2 104 N 8 2 stress 2 2 . 2 10 Nm 2 3 A r 9.5 10 m A l= 81cm F=62kN (b) If the Young’s modulus for steel is 2.2 x are the elongation and strain on the cylinder ? Dl F l From thedefinitionof Young's modulus, E Δl stress l A E 0.81m 2.2 108 Nm 2 4 Δl 8 . 9 10 m 11 2 2.2 10 Nm Dl 8.9 10 4 m strain 1.110 4 0.11% l 0.81m 1011 Nm-2, what 14: Gravitation REGAN PHY34210 140 Isaac Newton (1665) proposed a FORCE LAW which described the mutual attraction of all bodies with mass to each other. He proposed m1m2 that each particle attracts any other particle via the F G 2 GRAVITATIONAL FORCE with magnitude given by r G=6.67x10-11N.m2/kg2=6.67x10-11m3kg-1s-2 is the gravitational constant ‘Big G’ (as opposed to ‘little g’ the acceleration due to gravity). m1 The two particles m1 and m2 mutually attract with a force of magnitude, F. m1 attracts m2 with equal magnitude F but opposite sign to the attraction of m2 to m1. Thus, r F and -F form a third force pair, which only depends on F the separation of the particles, r, not their specific positions. m2 F is NOT AFFECTED by other bodies between m1 and m2. THE SHELL THEOREM: While the law described PARTICLES, if the distances between the masses are large, the objects can be estimated to be point particles. Also, ‘a uniform, spherical shell of matter attracts a particle outside the shell as if all the shell’s mass were concentrated at its centre’. Gravitation Near the Earth’s Surface REGAN PHY34210 141 The earth can be thought of a nest of shells, and thus all its mass can be thought of as being positioned at it centre as far as bodies which lie outside the earth’s surface are concerned. Assuming theearthis a UNIFORM SP HERE of mass M , themagnitudeof thegravit ational forcefrom theearthon a particleof mass m, at a distancer , Mm from theearth's centreis given by : Fgrav G 2 r If theparticleis released,it will accelerateto theearth's centreunder gravit y with a GRAVIT AT IONAL ACCELERAT ON, I a g , whose magnitudeis given Mm GM by Fgrav m ag G 2 a g 2 . T hus theacceleration due to gravit y r r depends on the' height'at whichan object is droppedfrom. average ag at earth’s surface = 9.83 ms-2 altitude = 0 km ag at top of Mt. Everest = 9.80ms-2 altitude = 8.8 km ag for space shuttle orbit = 8.70 ms-2 altitude = 400km REGAN PHY34210 142 We have assumed the free fall acceleration g equal the gravitational acceleration, ag, and that g=9.8ms-2 at the earth’s surface, In fact, the measured values for g differ. This is because • The earth is not uniform. The density of the earth’s crust varies. Thus g varies with position at the earth’s surface. • The earth is not a sphere. The earth is an ellipsoid, flattened at the poles and extended at the equator. (rpolar is ~21km smaller than requator). Thus g is larger at poles since the distance to the core is less. • The earth is rotating. The rotation axis passes through a line joining the north and south poles. Objects on the earth surface anywhere apart these poles must therefore also rotate in a circle about this axis of rotation (joining the poles), and thus have a centripetal acceleration directed towards the centre of the circle mapped out by this rotation. Centripetal Acceleration at Earth’s Surface T henormalforceon a surface object is from Fnet m ar N m ag m 2 R N T henormalforce,N is equal to the weight, m g m g m ag m ω 2 R g a g 2 R REGAN PHY34210 R m mag R is theradius which theobject rot at esaround and is therot at ionalvelocity. R is max.at theequator, R Rearth 6.37106 m. Δθ 2 rads can be estimatedfrom Δt 24 3600s acentr 2 R 0.034m s 2 (cf.m ag 9.8m s- 2 ) i.e. verysmall comparedto m ag . Assuming the weight equals thegravitational acceleration is usually (on earthat least!) well justified. S ‘above’ view, looking from pole, RN m 143 Gravitation Inside the Earth REGAN PHY34210 144 ‘A uniform shell of matter exerts no NET force on a particle located inside it.’ Therefore, a particle inside a sphere only feels a net gravitational attraction from the portion of the sphere inside the radius at which it is at. No net Force m r Net force In the example on the left, for r = M/V = constant R a planet of radius, R and total mass M. An object of mass m, which burrows downwards such that it is now at a distance r from the centre of the planet (with r < R ). The object will experience a gravitational attraction from the mass of the planet inside the ‘shell’ of radius r and none from the portion of the planet between radii r and the outer radius R. M ins rV r 43 r 3 and since theforceexperienced by theparticledue to the GM ins m Gr 43 r 3 4Grr m mass inside theshell is F i.e. Fnet kr 2 2 r r 3 Gravitational Potential Energy REGAN PHY34210 145 T hegravitational potentialenergyis defined by theexpression, Mm U G and defined to be zero at infiniteseparation(r ). r PROOF In general,work done is W F r .dr , F r .dr F r dr cos , R Mm cos cos1800 -1 F r .dr G 2 dr W F r .dr r R W G R Mm 1 GMm GMm GMm dr GMm dr 0 R r 2 r r2 R R R W WORK REQUIRED to movea mass, m froma distance R out to. Since potentialenergyand work done are relatedby thegeneralexpression, GMm DU W U U R , U R W R Potential Energy and Force REGAN PHY34210 146 Gravityis a conservative forceand changesin thegrav.potentialenergy onlydependon theinitialand final positions,NOT T HE PAT HT AKEN. Since we can derive thegravitational potentialenergyfrom theexpressionfor theforce,theconverseis also true.F dU d Mm Mm G G dr dr r r2 Escape Speed (Velocity) A mass m projectileleavinga mass M planetof radius R has an ESCAP E SP EED,v. T hiscauses theobject t omoveup with const antspeed, v against gravity,unt ilit slows down tov 0 at infinitedistance. 1 2 GMm m v ; T hegravitational potentialenergy, U . 2 R At infinitedistance,K U 0 (i.e.zero velocit yand at thezero potentialenergy T hekineticenergy, K configuration for r ). T husfrom theprincipleof conservaton i of energy K U 1 2 GMm 2GM earth 1 m vesc 0 v ; v 11 . 2 km s esc esc 2 R R Johannes Kepler’s (1571-1630) Laws • THE LAW OF ORBITS: All planets move in elliptical orbits with the sun at one focus. • THE LAW OF AREAS: A line that connects a planet to the sun sweeps out equal areas in the plane of the planet’s orbits in equal times. i.e., dA/dt=constant. • THE LAW OF PERIODS: The square of the period of any planet around the sun is proportional to the cube of the semi-major axis of the orbits. REGAN PHY34210 a b a b 147 The Law of Orbits REGAN PHY34210 148 If M >> m,the centre of Rp Ra mass of the planet-sun system is approximately m at the centre of the sun. r The orbit is described by the length of the M semi-major axis, a f f’ and the eccentricity ea ea parameter, e. The eccentricity is defined by the fact that the each a focus f and f’ are distance ea from the centre of the ellipse. A value of e=0 corresponds to a perfectly circular orbit. Note that in general, the eccentricities of the planetary orbits are small (for the earth, e=0.0167). Rp is called the PERIHELION (closest distance to the sun); Ra is the APHELION (further distance). Ra r Rp y ea 149 r f REGAN PHY34210 ea f’ x f a In generalfor an ellipse, theeccentricity is defined by , rmax rmin rmax rmin f’ Ra R p Ra R p If we taketheorigin of theco - ordinatesas thefocus, f , r0 r r r rmax 0 , rmin 0 1 cos 1 1 In Cart esianco - ordinates r x 2 y 2 , x r cosθ, y r sin θ Ellipseis defined by theequation, 1 2 x 2 2r0 x y 2 r02 1 1 A Lengt hof majoraxis 2 a rmax rmin RP Ra r0 1 1 r A T helength of theSEMI - MAJOR AXIS 0 2 2 1 2r0 2 1 REGAN PHY34210 The Law of Areas 150 If theDA is thearea swept out in timeDt , ΔA can be EST IMAT EDassuming the wedge of area swept out is a T RIANGLEof height,r , and base s rD . p T hearea swept out is approximately p 1 1 DA base height .rDθ.r. 2 2 T hisexpressionbecomesmoreexact for smaller values of ΔA . As D 0 and ΔA 0, r D m DA M ΔA dA 1 2 d 1 2 r r Δt dt 2 dt 2 where is theangular speed of therotatingline connectingthesun and planet (i.e., rotationalvelocityof planetaround thesun). T heang. mom.of theplanetaround thestar is, L rp r m v r m r m r2 dA 1 2 L r dt 2 2m dA . T hus,if L is conserved, constant. dt REGAN PHY34210 The Law of Periods For a circular orbit,using Newton's 2 nd law, Mm v2 rω G 2 mg m m m rω 2 r r r m 2 F m agrav 151 r M Mm G 2 m rω 2 GM r 3ω 2 r 2π T heperiodof revolution, T , substit uting in we get ω 2 2 3 4 π 4 3 2 r . For ellipse, r a semi - majoraxis. GM r T 2 T GM T 2 4π 2 Exact version of law predicts 3 a G M m T2 2 i.e., 3 constantfor M m, ( Ta 3 3.010-34 yr2 m -3for solar system). a REGAN PHY34210 Satellites, Orbits and Energies 152 T hepotentialenergy of systemis given by GMm U , U 0 for infiniteseparation r T heKINET ICENERGY OF A CICRULARLY ORBIT INGSAT ELLIT E,via Newt on's 2 nd law is K(r) r Etot(r) GMm v2 1 GMm F 2 m K m v2 r r 2 2r =-K(r) U T herefore,K for a sat ellit ein a circular orbit. 2 T he totalmechanicalenergyis given by E U K GMm GMm GMm E K r 2r 2r i.e. thetotalenergyis equal to the NEGAT IVEOF T HE KINET ICENERGY GMm For an ellipt icalorbit substitutea (semi- majoraxis length)for r , ie. E 2a Example 1: REGAN PHY34210 153 A satellite in a circular orbit at an altitude of 230km above the earth’s surface as a period of 89 minutes. From this information, calculate the mass of the earth ? 2 3 4 rd 2 r assuming M m From Kepler's 3 law : T GM r R h whereR 6.37106 m theearth's radius and G 6.6710-11 m3kg 1s 2 4π R h 6.6 10 m 4 M earth 2 2 11 3 1 2 G T 6 . 67 10 m kg s 89 60 s M earth 6 1024 kg 2 3 2 6 3 15: Fluids REGAN PHY34210 154 Fluids (liquids and gases), by contrast with solids, have the ability to FLOW. Fluids push to the boundary of the object which holds them. DENSIT Y( r ) : is theratioof themass Δm and thesize of thevolume elementΔV , i.e., r DDMV . For a uniformdensity ρ MV . Density has SI units of kg/m3 . In general, the density of liquids does not vary (they are incompressible); gases are readily compressible. Pressure: The pressure at any point in a fluid is defined by the limit of the expression, p = DF /DA as DA is made as small as possible. If the force is UNIFORM over a FLAT AREA, A, we can write p=F/A The pressure in a fluid has the same value no matter what direction the pressure WITHIN the fluid is measured. Pressure is a SCALAR quantity (i.e.,independent of direction). The SI unit of pressure is the PASCAL (Pa) where 1 Pa=1Nm-2. Other units of pressure include ‘atmospheres’ (atm), torr (mmHg) and lbs/in2 where 1 atm = 1.01x105Pa=760 torr = 760 mm Hg =14.7 lb/in2 REGAN PHY34210 Fluids at Rest For a tank of water open to air. The water pressure increases with depth below the air-water interface, while air pressure decreases with height above the water. If the water and air are at rest, their pressures are called HYDROSTATIC PRESSURES. AIR WATER 155 y=0 A F1 y F2 mg 1 y2 For wat erin ST AT ICEQUILIBRIUM, (stationary and theforceson it balance) F1 and F2 are theforcesat thetopand bottomof a cylinderof water with t op and bottomat depths y1 and y2 respectively and cross- sectionalarea, A. T hemass of water in cylinder m rV ρA y1 y2 , V volumeof cylinder. Balancingtheforcesgives F2 F1 m g where F1 p1 A and F2 p2 A. F2 F1 m g p2 A p1 A rA y1 y2 g p2 p1 r y1 y2 g If y1 surface level, y2 h depth, p1 p0 atmospheric pressure ph p0 rgh (P rinciplebehind theMERCURY BAROMET ER) The pressure at a point in a fluid in static equilibrium depends on the depth of that point but NOT on any horizontal dimension of the fluid. Pascal’s Principle REGAN PHY34210 156 ‘ a change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container ’ i.e. squeezing a tube of toothpaste at one end pushes it out the other. We can write Pascal’s principle as Dp=Dpext, i.e. the change in pressure in the liquid equals the change in the applied external pressure. This is the basis behind the concept of the HYDRAULIC LEVER. Fi Fo A downward force on one platform (the load Ao ‘input piston’) causes a change in Ai pressure of the INCOMPRESSIBLE do LIQUID, resulting in the movement of a second platform (the ‘output piston’). di Fo For equilibrium, there must be a downward force due to a load on the output piston which balances the upward force, Fo. The Hydraulic Lever T heappliedforce Fi , and thedownward Fi REGAN PHY34210 157 Ao do load Fo force Fo , from theload on theright hand piston Ai producesa changein thepressure of theliquid Fi Fo Ao Dp Fo Fi Ai Ao Ai di Fo If theinput pistonmovesdown a distanced i , theoutput pistonmovesupwards a distanced o such thatthesame volumeV , of theincompressible fluid is displaced at both pistons,then, Ai V Ai d i Ao d o d o d i if Ao Ai , d i d o Ao i.e., theoutput pistonmovesa smaller distance than theinput piston. Ao Ai d i Fi d i T heOUT P UT WORK is given by W Fo d o Fi Ai Ao With a hydraulic lever, a given force applied over a given distance can be transformed to a greater force over a smaller distance. Archimedes’ Principle REGAN PHY34210 158 ‘when a body is fully or partially submerged in a fluid, a BUOYANT FORCE, Fb from the surrounding fluid acts on the body. The force is directed upwards and has a magnitude equal to the weight, mfg of the fluid that has been displaced by the body. This net upward bouyant force exists because the water pressure around the submerged body increases with depth below the surface (Dp=rgh). Thus the pressure at the bottom of the object is larger than at the top. If a body submerged in a fluid has a greater density that then fluid, there is a net force downwards (Fg>Fb), while if the density is less than the fluid, there will be net force upwards (since Fb<Fg). For a body to float in a fluid, the magnitude of the bouyant force Fb, equals to the magnitude of the gravitational force Fg or, the magnitude of the gravitational force on the body is equal to the weight, mfg of the fluid which has been displaced by the body. The weight of a body in fluid is the APPARENT WEIGHT (Wapp) where Wapp= actual weight - magnitude of bouyant force. Since floating bodies have Fb=mg, their apparent weight is ZERO! Flow of Ideal Fluids In Motion REGAN PHY34210 159 The flow of real fluids is very complicated mathematically. Often matters are simplified by assuming an IDEAL FLUID. This requires 4 basic assumptions: A) STEADY FLOW: in steady flow, at a fixed point, the velocity of the moving fluid does not change in magnitude or direction. B) INCOMPRESSIBLE FLOW: This assumes the fluid has constant and fixed density (i.e. it is incompressible). C) NONVISCOUS FLOW: Viscosity is a measure of how resistive a fluid is to flow and is analogous to friction in solids. For example, honey has a higher viscosity than water). An object moving through an ideal, non-viscous fluid experiences NO VISCOUS DRAG force (i.e. no resistive force due to the viscosity of the fluid). D) IRROTATIONAL FLOW: In irrotational flow, a body can not rotate about its own centre of mass as it flows in the fluid. (Note that this does not mean that it can not move in a circular path). The Equation of Continuity REGAN PHY34210 160 Everyday experience tells us that the velocity of a fluid emerging from a tube depends on the cross-sectional areas of the tube. (For example, you can speed up the water exiting a hose by squeezing the end). If we have a tube of cross-sectional area, A1, v1 which narrows to area A2. In a time interval, v2 A2 Dt, a volume DV of fluid enters the tube, with A1 velocity v1. Since the fluid is ideal, and thus incompressible, the same volume of fluid must exit the smaller end of the tube with velocity, v2, some time interval later. The volume of fluid element at both ends (DV) is given by the product of the cross-sectional area (A) and the length it flows (Dx). Also, by definition, v=Dx/Dt thus, Applyingt o bot h ends of t he t ube segment ,DV A1v1Dt A2 v2 Dt A1v1 A2 v2 EQUAT IONOF CONT INUIT Y . T hiscan be rewrit t enas RV Av const ant where RV is t heVOLUME FLOW RAT E. If t hedensit y of t hefluid is const ant , t he MASS FLOW RAT E,Rm rRV rAv const ant Bernoulli’s Equation REGAN PHY34210 161 If an ideal (incompressible) fluid flows through a p2 ,v2 tube at a steady rate. If in time Dt , a volume of fluid, DV enters the tube and an identical p1 ,v1 y2 volume emerges from the other end. If y1, v1 and p1 are elevation, speed and pressure of the y1 fluid entering the tube and y2, v2 and p2 are the same quantities for the fluid emerging from the other end of the tube. These quantities are related by the BERNOULLI’S EQUATION, which states, p1 12 rv12 rgy1 p2 12 rv22 rgy2 which can be re - writtenas p 12 rv 2 rgy constant.If thefluid does not changeelevationin its flow, then y1 y2 y 0, Bernoulli's equn. reduces to p1 12 rv12 p2 12 rv22 . For fluids at rest v1 v2 0, Bernoulli's equn becomes p2 p1 rg y1 y2 If the speed of a fluid element increases as it travels along a horizontal streamline, the pressure of the fluid must decrease and vice versa. Proof of Bernoulli’s Equation REGAN PHY34210 T he work - kineticenergy theoremstatesW DK . i.e., thenet workdone on thesystemequals thechange in kineticenergy ΔK , resultingfrom thechangein fluid 162 p2 ,v2 p1 ,v1 y2 y1 speed between the ends of the tube. ΔK 12 Δm v22 12 Δm v12 12 ρΔV v22 v12 Δm ρΔV is thefluid mass enteringand leaving theoutput in timeinterval,Δt . T he work done on thesystemis due to the work done by thegrav.force on thefluid elementof mass Dm, during its verticallift from theinput to theoutput level. i.e. Wg Dm g y2 y1 rDVg y2 y1 . T he- sign arises since themotionand gravitational forceare in oppositedirections. Workis also done ON thesystemby pushing thefluid through he t tube at theentranceand BY thesystem to push forwardfluid at theemergingend. Since, in general,the work done is given by W FDx pAΔx pDV T he NET work done is W p p2 ΔV p1 ΔV p2 p1 ΔV . T he work - kinetic theoremW Wg W p ΔK rgDV y2 y1 ΔV p2 p1 12 rDV v22 v12 which gives, rg y2 y1 p2 p1 12 r v22 v12 Bernoulli' s equation 16: Oscillations REGAN PHY34210 163 Oscillations describe motions which are repetitive. An important property of oscillatory motion is its FREQUENCY, f, which describes the number of oscillations per second. The SI unit for frequency is the Hertz (Hz), where 1 Hz = 1 oscillation per second =1 s-1. Motion which regularly repeats is called periodic or harmonic motion. T heperiod,T , is the timeto completeone oscillation, where T 1f . For SIMPLEHARMONICMOT ION,the timedependenceof the displacement of a particlex, is given by xt xm cost where xm is theAMP LIT UDE(maximumvalue of x). ANGULAR FREQUENCYdefined by xm cost xm cos t T . Since cos cos 2 , thisleads to therelation hat t 2Tπ 2f . is theP HASE ANGLE of themotion,which is determinedby the displacement and velocityof theparticleat timet 0. t is called theP HASE of themotion. REGAN 164 The Velocity of Simple Harmonic Motion PHY34210 The velocity of a particle undergoing simple harmonic motion can be found by differentiating the displacement, x(t) with respect to time. dxt d xm cost vt xm sin t vm sin t dt dt where xm vm is called theVELOCIT YAMPLIT UDE. Note, in SHM, the magnitude of the velocity is greatest when the displacement is smallest and vice versa, since cos( )=sin(+/2) The Acceleration of Simple Harmonic Motion The acceleration can be found by differentiating the velocity in SHM, dvt d xm sin t at 2 xm cost dt dt 2 xm am is known as theacceleration amplitude. at 2 xt , which is thesignatureequation for SHM ‘ In SIMPLE HARMONIC MOTION, the acceleration a(t), is proportional to the displacement x(t), but opposite in sign, and the two quantities are related by the square of the angular frequency 2 ’ The Force Law for Simple Harmonic Motion REGAN PHY34210 From Newton's 2 nd law, F m a F m a m -ω2 x kx which is HOOKE'S LAW! T hisis CONSIST ENT with theidea of a REST ORINGFORCE.Substituting for theSPRING CONST ANT , thenfor a Simple HarmonicMotion, k m 2 Re - arrangingthisgives that theangular frequencyfor a simple harmonic k oscillatoris relatedto thestrengthof thespringconstantby m 2π m 2 ω k ‘ Simple Harmonic Motion is the motion which is described by a particle of mass m subject to a force which is proportional to the displacement of the particle but opposite in sign’ T hus thePERIODof oscillation for a linear oscillatoris T A LINEARHARMONICOSCILLAT ORdescribes a system where the forceF x , is proportion al to x (rather th an someotherpower of x). 165 Energy in Simple Harmonic Motion REGAN PHY34210 166 T heP OT ENT IALENERGY of a linear oscillatoris given by 1 2 1 U t kx k xm2 cos2 t 2 2 T heKINET ICENERGY of thesystemis given by 1 2 1 K t m v m 2 xm2 sin 2 t 2 2 T heMECHANICALENERGY, E, is defined as 1 1 k xm2 cos2 t m 2 xm2 sin 2 t 2 2 recallingthatfor SHM, k mω 2 , thensince cos2 sin 2 1, E U K 1 2 1 2 2 2 E U K kxm cos t sin t kxm 2 2 Therefore, the mechanical energy of a linear oscillator is constant and time independent. Angular Simple Harmonic Motion REGAN PHY34210 167 An ANGULAR SIMPLE HARMONIC fixed end PENDULUM (also known as a torsion wire TORSION PENDULUM) is an angular version of the linear simple harmonic +m oscillator. The disk oscillates in the reference horizontal plane, with the reference line line, =0 - m oscillating with an oscillation amplitude m. The torsion wire twists, thereby storing potential energy in the same way that a compressed spring does in the linear SHO case. The torsion wire also gives rise to the RESTORING TORQUE, . For angular simple harmonicmotion,rot at ingthedisk through an angle fromits rest posit ion(at 0) causes a REST ORINGT ORQUE given by τ -θ , where is called theT ORSIONCONST ANT . By analogy with t hesimple harmonicoscillatorcase, theP ERIOD of an angular simple harmonicoscillatoris given by T 2π I κ Simple Pendulums REGAN PHY34210 A SIMPLE PENDULUM has a bob of mass m hanging from a massless string of constant length l, fixed at the other end to that which the bob is attached. τ r F rF l Fg sin lm g sin From Newton's 2 nd law, I lm g sin I mom.of inertia, angular acceleration of thependulum at angular displacement, . lm g I T 2 I lm g , I m l2 l T s=r Fgsin Fgcos Fg For small angular displacement,in radians, 2 l m T he sign indicatesthat thetorque , acts to REDUCE theangular displacement, . lm g sin I lm g I 168 simple pendulums have SHM ONLY for small values of . T 2 l (no m dependence) g Real (‘Physical’) Pendulums REGAN PHY34210 O Real pendulums exhibit similar behaviour to simple pendulums, but the restoring component of the gravitational force, Fgsin, has a moment arm of distance h from the pivot point. h is the distance from the pivot point to the centre of mass of the object. h s=r 169 cog Fgsin Fgcos Fg For a physicalpendulum,for small amplitudes, I the periodis given by T 2 . (For thesimple pendulum, I m l2 ). m gh For real pendulums I differs for each case. For a uniformrod of lengt hl , I through the cent re(of mass) is 121 m l2 . T heP ARALLELAXIST HEOREM gives thatfor a pivotat one end of therod, I 0 1 12 m l m 2 l 2 2 1 2 ml 3 I m l2 2l 8π 2l i.e. can measure since T 2 2 2 g l m gh 3m g 2 3g 3T 2 g directly using Focault’s pendulum Damped Simple Harmonic Motion REGAN PHY34210 170 If the motion of an oscillator is reduced as a result of an external force, the oscillator and its motion are described as DAMPED. If thedampingforceis proportion al to the velocity of theof theoscillating system,then Fd bv , where,b is a DAMPINGCONST ANT .T heminus sign indicatesthat thisforceopposesthemotion. From Newton's 2 nd law, Fnet m a kx bv spring, constant, k Fk=-kx m Fb=-bv water tank d 2x dx m 2 kx b 0 . T hesolutionfor this2 nd order differential equation dt dt k b2 where ω' is theangular m 2m is given by xt xm e bt 2 m cos ' t k frequencyfor thedampedoscillator. For b 0 (no damping)ω' as in SHM. m If thedampingconstantis small,i.e., b km , then ω' . T heamplitudefor a dampedoscillator xm e bt 2 m bt 2 m 2 1 Emec t 2 k xme const. Forced Oscillations and Resonances REGAN PHY34210 171 If a body oscillates without an external force on the body, the body is said to undergo FREE OSCILLATION. However, if there is an external force periodically pushing the system (such as someone pushing a swing), this is known as FORCED or DRIVEN OSCILLATION. 2 angular frequencies are associated with a system undergoing forced oscillations, namely the (i) NATURAL ANGULAR FREQUENCY ( ) of the system, which is the frequency at which the system would oscillate if it was disturbed and left to oscillate freely; and the (ii) ANGULAR FREQUENCY OF THE EXTERNAL DRIVING FORCE (d ) which is the angular frequency of the force causing the driven oscillations. If =d, the system is said to be ‘in resonance’. If this condition is achieved, the velocity amplitude, vm is maximised (and so approximately is the displacement, xm). 17: Waves - Part 1 REGAN PHY34210 172 Waves describe situations where the energy of the system is spread out over the space through which it passes. This is in contrast to particles which imply a tiny concentration of matter which is capable of transmitting energy by moving from one place to another. There are THREE main types of waves: a) Mechanical Waves: These are governed by Newtons’s Law and can only exist within a medium (such as a taut string, water, air, etc.) b) Electromagnetic Waves: These are massless objects which require no medium to travel in. All EM-waves travel through vacuum at the same, constant speed (‘the speed of light, c=3x108ms-1). Examples of EM waves are visible light, UV and IR radiation, radio-waves, x-rays and gamma-rays. (the only difference between these waves is their wavelength and their mode of origin, (atomic, nuclear etc.) c) Matter Waves: These are quantum descriptions of subatomic particles such as electrons, protons etc. They are described by the ‘de Broglie’ wavelength, dependent on the particle’s (lin.) momentum. Transverse and Longitudinal Waves REGAN PHY34210 173 One way to investigate wave motion is to look at the WAVEFORM, which describes the shape of the wave (i.e. y=f(x)). Alternatively, one can monitor the motion of a particular element of the wave medium (e.g., a string) as function of time (i.e., y=f(t)). In cases where them displacement of (for example) each element in an oscillating string is perpendicular to the direction of travel of the wave, the wave said to be TRANSVERSE (i.e. a transverse wave, such as waves on a string.) By contrast, if the displacement is parallel to the direction of motion of the wave (as in sound waves), the motion is described at LONGITUDINAL (i.e., transmitted via a longitudinal wave such as sound). Wavelength and Frequency REGAN PHY34210 174 To completely describe a wave on a string (and the motion of any element along its length) a function which describes the shape of the wave as a function of time t, is required. This means we need a function of the form, y = f (x,t) , where y is the displacement in the ‘up-down’ direction and x is the position along the string. For a sinusoidal wave, thedisplacement y, as a functionof timet , for an elementat positionx along thestringis y( x, t ) ym sin kx t HRW p374 The amplitude (ym) is the magnitude of the maximum displacement. The phase is the argument of ( kx-t ). As the wave passes through a string element at a position, x, its phase changes linearly with time. Wavelength and Angular Wave Number REGAN PHY34210 175 The wavelength l , of a wave is the distance (parallel to the waves direction of travel) between repetitions of the shape of the wave. T heshape of he t wavecan be seen for t 0 as y x, t 0 ym sin kx t ym sin kx By definition, thedisplacement, y is thesame at both ends of a single wavelength, thus y x,0 y x l ,0 ym sin kx ym sin k x l Since thefunctionsin repeatsitself every2 radians, thusif kx kx kl kl 2 T heANGULAR WAVENUMBER, k HRW p374 2 l Period, Angular Frequency and Frequency REGAN PHY34210 176 We can monitor the time dependence of the displacement of a fixed position on a vibrating string. This can be done by taking x=0. HRW p375 T hedisplacement of thestringat x 0 as a functionof timeis then y 0, t ym sin 0 t ym sin t sin sin y 0, t ym sin t The PERIOD OF OSCILLATION T, is defined as the time for any string element to move through one oscillation. The displacement at both end of the period of oscillation are, by definition, equal. Thus, ym sin t1 ym sin t1 T ym sin t1 T 2 T hiscan only be trueif ωT 2π Angular Frequency, T 1 ω T hefrequencyis defined as f T 2π Speed of a Travelling Wave REGAN PHY34210 If the wave travelsin thex - direction,we can define the Δx dx WAVE SP EED v as theratioof . If we takea Δt dt fixed pointon themovingwaveformwhich is defined by havingthesame displacement y, then thephase of the waveformmust remainconstantsince y ym sin kx t . T hephase,kx ωt const . differentiating with respect dx dx to timegives k -ω0 v dt dt k 2 2 2π l T Re calling k and ω v 2 l f l T T l 177 Wave Speed on a Stretched String REGAN PHY34210 178 For a wave to pass through a medium, the particles in the medium must oscillate as the waves passes through. The medium must have mass (so the particles have kinetic energy = 1/2mv2) and elasticity (for potential energy = 1/2 kx2). The mass and elasticity of the medium determine how fast the wave can travel through the medium. A small stringelementof length Δl , formsa circular arc of Dl F radius R subtending an angle 2θ. If a force with a R v magnitudeequal to the tensionin thestringpulls O tangentially tothe two ends, thehorizontalcomponents cancel.T he verticalcomponentssum, providingtherestoringforceF of magnitudeof thisforceis, F 2τ sin θ 2τ θ τ Δl R . T hemass of theelement Δm μDl , where m is thelinear densityof thestring. The speed of a wave T hestringelementmovesin a circle, thus a v2 R . v2 F m a leads to m a μDl v R m R Dl on an ideal stretched string only depends on the string’s tension and linear density. Energy and Power of a Travelling Wave REGAN PHY34210 2 179 T hekineticenergy of a stringelementof mass dm is dK 12 dm u whereu is theT RANSVERSE SP EED of theoscillating stringelement.u dy dt u ym coskx t . energyis given by dK 1 2 Using thelinear density, dm μ.dx, thekinetic μ.dx ym 2 cos2 kx t . T herateat which he t kineticenergy of a stringelementis thus dK 1 dx 2 2 2 μ. ym cos2 kx t 12 μv ym cos2 kx t dt dt theAVERAGE rateat whichkineticenergy transportedis given by dK 1 2 2 μv ym cos2 kx t ave 14 mv 2 ym2 averageof cos2 X 12 dt Recallingfor an oscillating system,theaveragekineticand potentialenergies are equal, theAVERAGE P OWERis theaveragerate which thetotal dK 2 2 mechanicalenergy transmittedby thewave i.e., P 2 12 mv ym dt ave The Principle of Superposition REGAN PHY34210 180 The principle of superposition states that when several effects occur simultaneously, their net effect is the sum of the individual effects. y' x, t y1 x, t y1 x, t Mathematically, this means, Overlapping waves add algebraically to produce a RESULTANT or NET WAVE. Note however, that overlapping HRW p383 waves do not in any wave affect each others travel. Interference of Waves If 2 sinusoidal waves of the same wavelength and amplitude overlap, the resultant wave depends on the relative PHASES of the waves. If they are perfectly ‘in phase’ they will add coherently, doubling the displacement observed for individual waves. By contrast, if they are completely out of phase (peaks of one wave matched by troughs of the other), they will completely cancel out resulting in a ‘flat’ string. REGAN 181 If y1 x, t ym sinkx t and y2 x, t ym sinkx t PHY34210 These waves have the same frequency determined by , wavelength from l and amplitude ym . They differ only by the phase constant . HRW p384 From the principle of superposition, y ' x, t ym sin kx t ym sin kx t Since, sin sin 2 sin cos 2 2 ym sin kx t ym sin kx t y ' x, t 2 ym cos 2 sin kx t 2 amplitudeof resultantwave 2 ym cos 2 and phaseangle 2 . If 0, the two initial waves are ' in phase'the waves interferefully ' constructively'.If radians, the waves are completelyout of phaseand interferecompletely, DEST RUCT IVELY. If 2 sinusoidal waves of the same amplitude and wavelength travel in the same direction along a stretched string, they interfere to produce a resultant sinusoidal wave travelling in that direction. Phasors REGAN PHY34210 182 y’=y1+y2 Waves can be represented in vector form using the idea of PHASORS. This is a vector whose amplitude is represented by the length which is equal to the magnitude of the wave and which rotates around the origin of a set of Cartesian co-ordinates. The angular speed of the phasor about the origin is equal to the angular frequency, of the wave. As the phasor rotates about the origin, y’m its projection, y1 onto the vertical axis y2 y2 y m,2 varies sinusoidally between +ym and -ym. 2 waves which travel along the same y1 ym,1 y1 string in the same direction can be added using a PHASOR DIAGRAM. If y1 x, t ym,1 sin kx t and y2 x, t ym, 2 sin kx t , the resultantis of theform y ' x, t ym' sin kx t where ym' and can be found using thePHASOR DIAGRAM.Adding vectorially thephasors y1 x, t and y2 x, t at any instant,themagnitudeof theresultantequals y ' x, t and is theangle between the resultantand thephasorfor y1 x, t . Standing Waves REGAN PHY34210 183 If two sinusoidal waves travel in opposite directions along a string, their sum can be found using the principle of superposition. There are specific places along the resultant wave which DO NOT MOVE, known as NODES. Halfway between neighbouring nodes (the ‘anti-nodes’) the amplitude of the resultant wave is maximised. Such wave patterns are called STANDING WAVES since the wave patterns do not move in the x-direction (i.e. they are stationary left to right). HRW If two sinusoidal waves of the same amplitude and wavelength travel in opposite directions along a stretched string, their intereference with each other produces a standing wave. Analysis of Standing Waves REGAN PHY34210 184 T he t wo combiningwaves which makeup t hest anding wave are y1 x, t ym sin kx t and y2 x, t ym sin kx t . From t heprincipleof superposition, y ' x, t y1 x, t y2 x, t y ' x, t ym sin kx t ym sin kx t . Recallingsin sin 2 sin cos 2 2 y ' x, t 2 ym sin kxcost Note, this does not describe a travelling wave, but rather a standing wave. • The absolute value of [2ymsin(kx)] is the amplitude of oscillation at x. • The amplitude varies with position for a standing wave. T heamplitudeis zero if sin kx 0. i.e., for integern where kx nπ 2π λ x amplitudeis zero (i.e., nodes) occur at x n 2λ . For a standing waves, adjacent nodes are separatedby a distanceof half the wavelength l2 . Similarly,themax.amp.is 2 y for sin kx 1 kx n 12 x n 12 l2 Standing Waves and Resonance REGAN PHY34210 185 A standing wave can be set up by allowing a wave to be reflected at a boundary of a string. The interference of the original (incident) and reflected wave can interfere to give rise to a standing wave. (Note that for ‘hard’ reflection, the reflection point must be a fixed node.) HRW If a taught string is fixed at both ends (such as in a guitar) and a continual sinusoidal wave is sent down from one end, it will be subsequently reflected at the other end. The reflected wave and the next transmitted wave will interfere. If more waves are continually sent from the generator, many such waves can add coherently. REGAN PHY34210 186 At certain frequencies, such behaviour results in STANDING WAVE PATTERNS on the string. Such standing waves ‘RESONATE’ at fixed ‘RESONANT FREQUENCIES’. (Note that if the string is oscillated at a non-resonant frequency, a standing wave is NOT set up.) 2L A standing wave can be set up on a stringof length L, by a wave if λ n v v where T hiscorresponds to resonancefrequencies given by f n 2L l v is the wavespeedalong thestring. n 1 is called thefundamental mode or ' first harmonic'. n 2 is called thesecond harmonic,n 3 the thirdharmonicand so on. T hefrequencies associatedwith these modes are oftengiven the symbols, f1,f2 , f 3, , f n Example 1: REGAN PHY34210 187 Two identical sinusoidal waves moving in the same direction along a stretched string interfere with each other. The amplitude of each wave is 9.8 mm and the phase difference between them is 100o. (a) What is the amplitude of the resultant wave due to the interefence between these two waves ? ym' 2 ym cos 2 2 9.8 mm. cos 500 13 mm (b) What phase difference (in both radians and in fractions of wavelength) will give a resultant wave amplitude of 4.9 mm ? Since ym' 2 ym cos 2 4.9mm 2 9.8mm cos 2 4.9 mm cos 2 0.25 2.6 radians 2 9.8 mm single wavelength corresponds to 2 , thus,in wavelengths, the phase differenceis given by rads 2 rads / wavelength 2.6 2 0.42 wavelengths Example 2: REGAN PHY34210 188 Two sinusoidal waves y1(x,t) and y2(x,t) have the same wavelength and travel together in the same direction along a string. Their amplitudes are y1,m=4.0 mm and y2,m=3.0mm and their phase constants are 0 and /3 respectively. What are the amplitude, y’ and phase constant of the resulting wave ? ym2 / Adding thehorizontalcomponents y'mh ym1 cos0 ym 2 cos3 4 3 cos3 5.5m m Adding the verticalcomponents 3 ym1 y’m ym1 y'mv ym1 sin 0 ym 2 sin 3 0 3 sin 3 2.6m m From P ythagorastheorem,theresultantwave has an amplitudeof y'm 5.52 2.62 m m 6.1m m ym2 -1 2.6 T hephaseconstantis tan 0.44 rads 5.5 y ' x, t 6.1m msin kx t 0.44rads 18: Waves - Part 2 REGAN PHY34210 189 ray Sound Waves: can be generally defined as longitudinal waves whose oscillations are parallel to the direction of travel through some medium (such as air). P wavefronts planes ray If a point source, P, emits sound waves, wavefronts and rays describe the direction of travel of the waves. Wavefronts correspond to surfaces over which the wave has the same displacement value. Rays are lines drawn perpendicular to wavefronts which indicate the direction of travel of the waves. Note that in real bodies, wavefronts spread out in 3 dimensions in a spherical pattern. Far from the point source the wavefronts can appear as planes or straight lines to an observer. Speed of Sound REGAN PHY34210 190 The speed of any mechanical wave depends on the physical properties of the medium through which it travels. As a sound wave passes through air, we can associate a potential energy with periodic compressions and expansions of small volume elements. The BULK MODULUS is the property which determines the volume change in a material when exposed to an external pressure (p=F/A). Dp Recalling that t heBULK MODULUSis defined by B DV V where ΔV/V is thefract ionalchangein volumeproducedby a changein pressure Δp. Since thesigns of Δp and ΔV are always oppositea minus sign is included to make B a positivequanity. T hespeed of sound for a longitudinal wave in a medium is give by v B r where r is thedensity of Vair(20oC)=343 ms-1 themedium. Vwater(20oC)=1482 ms-1 Vsteel = 5941 ms-1 The Doppler Effect REGAN PHY34210 191 The Doppler effect describes how sound waves from a point source (such as a car or train or star or galaxy!)) are apparently shifted in frequency for an observer which is moving relative to that source. T hegeneralexpressionfor theDOPP LERFORMULAfor sound waves is v vD where f is theemittedfrequencyof thesource and f ' f v vS f ' is thedetectedfrequencyby theobserver.v is thespeed of sound through theair, vD is therelativespeed of thedetector(or ' observer') relativeto an air - fixed frameand vS is thesource speed relativeto thesame air - fixed referenceframe.(Notein most cases, either thesource or thedetectoris stationary, i.e., vD or vS 0). When the motion of the detector or source are towards each other, the sign on its speed gives an UPWARD SHIFT IN FREQUENCY. When the motion of the detector or source are away from each other the sign on its speed gives a DOWNWARD SHIFT IN FREQUENCY. Example 1 REGAN PHY34210 192 A rocket moves at a speed of 242 m/s through stationary air directly towards a stationary pole while emitting sound waves at a source frequency of f =1250Hz. (a) What frequency is measured by a detector attached to the pole ? v vD 343m s1 0 4245Hz. T he- sign on bottom f ' f 1250 1 1 343m s 242m s v vS gives an INCREASE in observedfrequencyfor relativemotion tow ards source. (b) If the some of the sound waves reflect from the pole back to the rocket, what frequency f ’’does the rocket detect for the echo ? v vD 343m s1 242m s1 4245 7240Hz. T he sign on topgives f ' f 1 343m s 0 v vS an INCREASE in observedfrequencyfor relativemotion tow ards source (pole).