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REGAN PHY1033 2015 Fundamentals of Physics: On-line lecture notes Prof. Paddy Regan Dept. of Physics, University of Surrey, Guildford, GU2 7XH, UK E-Mail: [email protected] 1 REGAN PHY1033 2015 Course textbook, Fundamentals of Physics, Halliday, Resnick & Walker, published by Wiley & Sons. . http://www.wiley.com/WileyCDA/WileyTitle/productCd-EHEP001575.html 2 • 1: Measurement REGAN PHY1033 2015 – 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 3 • 7: Kinetic Energy and Work REGAN PHY1033 2015 – 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 4 REGAN PHY1033 PHY34210 2015 • 13: Equilibrium and Elasticity – equilibrium, centre of gravity, elasticity, stress and strain. • 14: Gravitation (and Electromagnetism) – Newton’s law, gravitational potential energy, Kepler’s laws 5 1: Measurement REGAN PHY1033 2015 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). 6 SI Units REGAN PHY1033 2015 7 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 PHY1033 2015 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 8 Converting Units REGAN PHY1033 2015 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. 9 Length (Metres) REGAN PHY1033 2015 10 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 PHY1033 2015 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 11 Mass (Kg, AMU) REGAN PHY1033 2015 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, 12 Estimate Example 1: REGAN PHY1033 2015 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 assume cross - section of string ~ 3mm square 2 r d d 4 3 2 2 V 0.05m 3mm 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 13 REGAN PHY1033 2015 E.g., 2: Estimate Radius of Earth (from the beach!) From Pythagoras , 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 ( 10 sec) 360(deg) 3600 o (deg) (deg) 0 . 04 360o 24 hours 24 60 60 sec 86,400 2h Now, from trigonome try, d r tan thus r 2 tan 2 2rh r tan 2 2 2m 4m 6 if h human height ~ 2m, then substituti ng, r 8 10 m 2 o 7 tan (0.04 ) 4.9 10 (accepted value for earth radius 6.4x10 8 m! ) 14 2: Motion in a Straight Line REGAN PHY1033 2015 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. 15 REGAN PHY1033 2015 16 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 PHY1033 2015 17 The average VELOCITY is defined by the displacement (Dx) divided by the time taken for this displacement to occur (Dt). x2 x1 Dx vav t 2 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 t2 t1 t2 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 PHY1033 2015 HRW 18 REGAN PHY1033 2015 19 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 the displaceme nt v0 initial velocity (at time t 0) v a t velocity at time t accelerati on (constant) time taken from t 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 PHY1033 2015 Equations of Motion (for constant a). x x0 v v0 at (1) recalling that vav , then t 0 v0 v since by definition , vav , 2vav v v0 2 1 substituti ng 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 eliminate t , 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 20 Alternative Derivations (by Calculus) REGAN PHY1033 2015 21 dv d 2 x a 2 by definition dv a dt dt dt dv a dt and thus for a constant v at C , evaluated by knowing that at 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 substituti on, dx v 0 at dt dx v0 dt a t dt 1 2 integratin g gives, x v0t at C , calculate C by 2 1 knowing x x0 at t t0 C x0 and x x0 v0t at 2 (5) 2 Free-Fall Acceleration REGAN PHY1033 2015 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 22 Example REGAN PHY1033 2015 23 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 12ms 1 at the max. height vm a x=0 v v0 at t a 9.8ms 2 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 ms , (b) 0 0 0 2 2 1 2 v v 0 ( 12 ms ) a g 9.8ms 2 y 0 7.3m 2 2a 2 9.8ms REGAN PHY1033 2015 ( c) How long does the ball take to reach a point 5m above its initial release point ? v0 12ms 1 , a 9.8ms 2 , y y0 5m 1 2 1 1 from y y0 v0t at 5m 12ms t 9.8ms 2t 2 2 2 assuming SI units, we have a quadratic equation, 4.9t 2 12t 5 0 recalling at 2 bt c 0 solutions are 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. 24 3: Vectors REGAN PHY1033 2015 • 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. 25 b REGAN PHY1033 2015 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. 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. r a b c a b c 26 s a b a s b b a a s b a a b c s r c b = a ' s r REGAN PHY1033 2015 Subtracting Vectors, Negative Vectors b is the same magnitude as b but in the opposite direction. b d a b s d a b a b s a b b a s as with usual algebra, we can re - arrange vector equations, 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!) 27 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 this can also be written in MAGNITUDE - ANGLE NOTATION as a a a , tan 2 x 2 y ay ax REGAN PHY1033 2015 28 a ax a cos x REGAN PHY1033 2015 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 respective ly. 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 ĵ y 1 k̂ iˆ 1 ˆ a a xiˆ a y ˆj a z k , b bx iˆ by ˆj bz kˆ then if s a b using vector addition by components s (a x bx )iˆ (a y by ) ˆj (az bz )kˆ s xiˆ s y ˆj s z kˆ x 29 Vector Multiplication REGAN PHY1033 2015 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 vecto r notation, a.b a x iˆ a y ˆj a z kˆ . bx iˆ by ˆj bz kˆ but since cos0o 1 and cos 90o 0 expanding this reduces to a.b a x bx a y by a z bz since iˆ.iˆ iˆ. ˆj iˆ.kˆ ˆj. ˆj kˆ.kˆ 1 (11cos0 o 1) and ˆj.kˆ 0 (11cos 90o 0) 30 Vector (‘Cross’) Product The VECTOR PRODUCT of two vectors a and b REGAN PHY1033 2015 31 produces a third vector whose magnitude is given by c ab sin 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 x iˆ a y ˆj a z kˆ bx iˆ by ˆj bz kˆ but a xiˆ bxiˆ a xbxiˆ 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 PHY1033 2015 Example 1: Add the following three vectors 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 32 x 2 Example 2: REGAN PHY1033 2015 What are the (a) scalar and (b) vector products of the two vectors aˆ 2iˆ 3 ˆj 4kˆ and bˆ 4iˆ 20 ˆj 12kˆ (a) Scalar Product : a . b ab cos 2iˆ 3 ˆj 4kˆ . 4iˆ 20 ˆj 12kˆ recalling only iˆ.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 Product, a b 2iˆ 3 ˆj 4kˆ 4iˆ 20 ˆj 12kˆ recalling a 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 . 11 iˆ 10 ˆj 13kˆ 33 REGAN PHY1033 2015 34 4: Motion in 2 and 3 Dimensions 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 the position of a particle can be descibed in Cartesian co - ordinates by r xiˆ yˆj zkˆ , then the DISPLACEME NT 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 then Dr r2 r1 (9 (3))iˆ (2 2) ˆj (8 5)kˆ 12iˆ 3kˆ Velocity and Acceleration REGAN PHY1033 2015 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 vxiˆ v y ˆj vz 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 dv y ˆ dvz ˆ a i j k dt dt dt dt dt 35 Projectile Motion REGAN PHY1033 2015 36 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. The initial projectile velocity (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 PHY1033 2015 Horizontal Motion 37 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 (equation of 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 v y v0 sin 0 gt and v y2 v0 sin 0 2 g y y0 v0 Max. height occurs when v0 sin v y 0, i.e., v0 sin 0 gt v0 cos vy REGAN PHY1033 2015 38 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 substituti ng for the time between th e 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 PHY1033 2015 The Horizontal Range The range, R x x0 is defined when the projectile hits the ground 1 2 i.e., when y y0 , then R 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 gR 2 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 (note assumes y y0 ) v0 sin (y0,x0) vy=0 Max height v0 cos Range 39 Example REGAN PHY1033 2015 40 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 the range 150m is given v02 sin 2 0 by R , v0 50ms 1 , g 9.8ms 2 g gR 9.8ms- 2 150m 1470 sin2 0 2 1 1 v0 50ms 50ms 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 maximum when sin2 0 max 1, i.e., when 2 0 900 , g thus 0max 45 Rmax o 50ms 1 50ms 1 255m 840 feet! 2 9.8ms Uniform Circular Motion REGAN PHY1033 2015 v 41 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 The PERIOD OF REVOLUTION time taken for the particle to go around the circle. If the speed (i.e., the magnitude of the velocity for UCM) of the particle v, the time taken is, by definition Circumference 2r T velocity v Proof for Uniform Circular Motion REGAN PHY1033 2015 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 vxp ˆ dv v dy p ˆ v dx p ˆ i j . Accelerati on is a i j v dt r dt r dt r r but dy p v yp v cos and dx p v xp v sin yp r 42 v dt dt v dy p ˆ v dx p 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 magnitude, a a x a y r r v 2 sin ay r sin ACCELERATI ON DIRECTION from tan tan , 2 a x v cos cos r i.e., , accelerati on is along the radius.... towards centre of 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 PHY1033 2015 p rAp 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 43 5: Force and Motion (Part 1) REGAN PHY1033 2015 44 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 PHY1033 2015 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. 45 Force REGAN PHY1033 2015 46 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 PHY1033 2015 47 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 PHY1033 2015 48 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 170 N x 0 A FC 28 0 170 and FA sin 47 FC sin 28 FB o o FB 220 0.731 170 0.47 241 N FB ?? N HRW p79 The Gravitational Force REGAN PHY1033 2015 49 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 taking components in the vertical direction ( y positive correspond s to upward ) then Fgy ma y m g In vector form, we have Fg Fg ˆj mg ˆj 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 PHY1033 2015 50 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 Normal Force, N y component, ma y N Fg N mg N m a y g i.e., if block is at rest then N m 0 g mg i.e. same magnitude as gravitatio nal force but in opposite direction. Note the NORMAL FORCE is ‘normal’ (i.e. perpendicular) to the surface. Gravitatio nal Force, Fg mg REGAN PHY1033 2015 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 may Fg N mg 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. 51 REGAN PHY1033 2015 Tension 52 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 Ma Mx T y components ( a downward thus amy ) M T Fnet , y mamy T mg FgM Mg T magnitudes must be equal, aMx amy a m mamy Ma Mx mg a Fnet , x T Ma Mx Fgm mg mg M m mg Mmg M T M m M m Newton’s Third Law REGAN PHY1033 2015 53 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 Normal Force, 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 Gravitatio nal Force, Fg mg REGAN PHY1033 2015 Example N N T cos50o T y 54 T x 40 o 40 o mg = g x 15kg Question ? What is the tension in the string ? T N Fg ma 0 x component, T 0 mg sin Fx ,net 0 T mg sin 15kg 9.8ms 2 sin 400 94.5 N y component, 0 N mg cos 0 N 112.6 N 50o 40 o mg mg sin 40o 6: Force and Motion (Part II) REGAN PHY1033 2015 55 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 PHY1033 2015 56 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 PHY1033 2015 57 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 , the drag force will be cancel the downward accelerati on. In general, Fnet , y may D Fg For ' terminal speed' , a y 0, thus 2 Fg 1 2mg 2 CrAvt Fg 0 vt 2 CrA CrA Forces in Uniform Circular Motion REGAN PHY1033 2015 58 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 PHY1033 2015 (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. 59 REGAN PHY1033 2015 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., independen t of mass! 60 7: Kinetic Energy and Work REGAN PHY1033 2015 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. 61 Work and Kinetic Energy REGAN PHY1033 2015 62 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 , a x Fx d mv mv0 DK work done 2a x m 2 2 1 2 1 2 mv mv0 DK work done is the Work-Kinetic Energy Theorem 2 2 Work Done by a Gravitational Force. REGAN PHY1033 2015 63 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 mgd cos , mg and d are in opposite directions , 180o , W mgd . sign shows gravity tr ansfers KINETIC ENERGY to GRAVITATIO NAL POTENTIAL ENERGY. When the object falls back down, 0 and W mgd sign implies gravitatio nal force transfers energy TO the object from potential energy to kinetic. REGAN PHY1033 2015 Work Done Lifting and Lowering an Object. 64 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 - kinetic energy the orem, DK K f K i net work done Wa Wg . If the object is stationary before and after the lift (v 0 at start and finish) then DK 0 Wa Wg Wa Wg and Wa mgd cos where is the angle between Fg (i.e. ' downwards' ) and the displaceme nt, d . If the object is lifted up, 1800 and the work done by the applied force, Wa mgd . If the object falls, 0 o , Wa mgd Spring Forces and Hooke’s Law REGAN PHY1033 2015 65 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 spring constant, stiffness of the spring. In the 1 - 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 PHY1033 2015 66 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. The change in kinetic energy (of the block attached to the spring) due to these two energy tra nsfers is given by DK K f K i Wa Ws Thus, 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. The work done by a force averaged over a distance, Dx, is REGAN PHY1033 2015 DW j F j ,ave Dx. Total work 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 Work done by 1 - D force AREA UNDER THE CURVE of F ( x) against x. In 3 - D, F Fx iˆ Fy ˆj Fz kˆ. If Fx only depends on x, Fy on y and Fz on z , the by SEPARATING THE VARIABLES if the particle moves through an incrementa l displaceme nt, dr dxiˆ dyˆj dzkˆ , the increment of work in dr, dW , is given by dW F.dr Fx dx Fy dy Fz dz , then the total work is rf xf yf zf ri xi yi zi W dW Fx dx Fy dy Fz dz 67 Work-Kinetic Energy Theorem with a General,Variable Force xf xf REGAN PHY1033 2015 68 xf dv W F x dx max dx m dx dt xi xi xi dv dv dx dv using the CHAIN RULE, . v dt dx dt dx dv dv we can CHANGE THE VARIABLE, m dx mv . dx mvdv dt dx xf vf vf dv W m dx mvdv m vdv dt xi vi vi 1 2 1 2 mv f mvi K f K i DK , 2 2 which is the WORK - KINETIC ENERGY THEOREM W Power REGAN PHY1033 2015 69 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 PHY1033 2015 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 0.26ms- 2 The velocity of the trains at collision is then v 2 0.26ms 2 3.2 103 m 40.8ms 1 The kinetic energy of each locomotive is given by 1 2 1 5 1 2 K mv 1.22 10 kg 40.8ms 108 J 2 2 Thus total energy of collision is 2 K 200MJ 70 Example 2: REGAN PHY1033 2015 71 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 ˆ 6 ˆj N F 2 i on the crate during this displacement ? W F .d 2iˆ 6 ˆj N . 3iˆ m 6 J Thus, the ' wind' force does 6 J of NEGATIVE WORK on the crate i.e. it transfe rs 6 J of kinetic energy FROM THE CRATE (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-kinetic energy the orem, W ΔK 6 J K f 10 J K f 6 10 J 4 J i.e., block is slowed down by wind force. REGAN PHY1033 2015 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 the orem, the work done on the block by the spring force is 1 2 Ws kx . The work is also related to the change 2 in kinetic energy of the block, i.e., W ΔK K f K i 1 1 m kx2 0 mv 2 x v 2 2 k v k x v=0 m 72 REGAN PHY1033 2015 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. 73 Work and Potential Energy REGAN PHY1033 2015 74 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 PHY1033 2015 Determining Potential Energy Values xf xf xi xi W F ( x)dx , DU F ( x)dx . For GRAVIT. POT. ENERGY, yf yf yi yi DU F ( y )dy yf mg dy mg dy mgy f yi yi DU grav mg y f yi mgDy Only CHANGES in gravitatio nal Pot. energy are meaningful , i.e., it is usual to define U i 0 at yi , then U y mgy For the ELASTIC POTENTIAL ENERGY, xf xf xf DU elas F ( x)dx kxdx k xdx xi xi xi 1 k x2 2 xf xi 1 DU elas k x 2f xi2 . Pot energy is relative, thus we chose 2 1 2 U 0 at xi 0 Then, U x kx , x is extension/ compressio n. 2 75 Conservation of Mechanical Energy REGAN PHY1033 2015 76 The mechanical energy is the sum of kinetic and potential energies, Emech K U . If the system is isolated from its environmen t and no external force causes any internal energy 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, The sum of the kinetic and potential energies ( the mechanical energy) is the same for all states of an isolated system, i.e. the MECHANICAL ENERGY of an ISOLATED SYSTEM where there are only conservati ve forces is CONSTANT. This is the PRINCIPLE OF CONSERVATI ON OF MECHANICAL ENERGY (note, conservati on is due to CONSERVATI VE FORCES ). This can also be written as DEmech DK DU 0 The Potential Energy Curve REGAN PHY1033 2015 For the 1 - D case, the work done, W , by a force, F , moving an object thr ough a displaceme nt, Dx equals, FDx , therefore , the potential energy can be written as DU x dU x DU x W FDx F Dx dx e.g., Hooke' s Law, if the elastic potential is given by, 1 2 U x kx then differenti ating gives, F kx 2 also , in the gravitatio nal case, U x mgh F mg 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 77 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. REGAN PHY1033 2015 78 K 0 at ymax , Emec mgymax dU ( y ) F ( y) mg dy Emec K ( y) U ( y) 1 2 mv mgy 2 Equilibrium Points REGAN PHY1033 2015 79 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 PHY1033 2015 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 ) 80 Work Done by an External Force REGAN PHY1033 2015 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 Newtons 2 nd law, F f k ma , the force (thus accelerati on) is constant, therefore we can use v 2 v02 2ad v 2 v02 f k and By substituti on, F m 2d 1 2 1 2 Fd mv mv0 f k d DK f k d 2 2 81 Conservation of Energy REGAN PHY1033 2015 82 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 the change in mechanical energy, DEth is the change in thermal energy (i.e., heat) and DEin is the change in internal energy of the system. 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 PHY1033 2015 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 the CONSERVATI ON OF MECHANICAL ENERGY, Emec,i Emec, f U i K i U f K f 1 2 U i mgh , U f 0, K i 0, K f mv 2 1 2 mgh 0 0 mv v 2 gh 2 Note that this is the same speed that the child would have if it fell directly from a height h. 83 Example 2: REGAN PHY1033 2015 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 CONSERVATI ON OF MECHANICAL ENERGY, DK DU 0 , if v 0 at top and bottom, K i K f 0 84 L x m 1 2 DK 0 also, DU DU grav DU elas mg Dy 0 kx 2 1 2 1 2 kx mg L x mgL mgx kx mgx mgL 0 , 2 2 solving this quadratic equation, x mg mg 2 2kmgL k , x ve root 9: Systems of Particles REGAN PHY1033 2015 85 Centre of Mass (COM): The COM is the point that moves as though all the mass of a body were concentrated there. For 2 particles of mass, m1 and m2 separated by d , if the orgin of x - axis coincides with the particle of mass m1 , the centre of mass of the system is m2 xcom d . More generally, if m1 is at x1 and m2 is at x2 , the COM m1 m2 m1 x1 m2 x2 m1 x1 m2 x2 is defined by xcom where M is the total mass m1 m2 M The general form for a n-particle system is given by xcom m1 x1 m2 x2 m3 x3 m4 x4 1 M M ycom 1 M n m y i 1 i i and zcom 1 M n m x , similarly, i 1 i i for 3 - D mi zi . In vector form, if r xiˆ yˆj zkˆ n i 1 1 ˆ ˆ ˆ then rcom xcomi ycom j zcom k and rcom M mi ri n i 1 Centre of Mass for Solid Bodies REGAN PHY1033 2015 86 Solid objects have so many particles (atoms) that they can be considered to be made up of many infinitess imally small MASS ELEMENTS, dm. 1 1 1 Then, xcom xdm , ycom ydm , zcom zdm , M M M Often, the integrals are simplified assuming a UNIFORM DENSITY ( r ) M dm where r , where dV is the volume occupied by mass, dm. V dV 1 1 1 substituti ng, xcom xdm xrdV xdV and similarly, M rV V 1 1 ycom ydV , zcom zdV , V V Note that the centre of mass need not necessaril y lie in the volume of the object (for example a doughnut or an igloo). Newton’s 2nd Law for a System of Particles. Mrcom m1r1 m2 r2 m3r3 mn rn differenti ating with respect to time, dmn rn mvn we get since, dt drcom Mvcom m1v1 m2 v2 m3v3 mn vn M dt dvn an differenti ating once again, 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 PHY1033 2015 87 Linear Momentum REGAN PHY1033 2015 88 The LINEAR MOMENTUM is defined by p mv dp d mv mdv 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 system of particles, the system has a total linear momentum, P which is the vector sum of the individual 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 PHY1033 2015 89 dP dvcom Since, F m macom , in a closed system, if the dt dt net external force is zero, and no particles enter or 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 PHY1033 2015 90 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 conservati on of momentum, Pi Pf The initial P of the rocket plus the exhaust fuel equals the P of the exhaust products plus the P of the rocket after time interval, dt. Mv dMU M dM v dv if vrel is the relative speed between th e rocket and the exhaust products (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 dM dv 0 dMvrel Mdv vrel M dt dt if R is the rate of mass loss, then, we obtain Rv rel M a) time = t M v b) time = t+dt -dmM+dm U dv Ma dt v+dv 1st rocket equation Rv rel is called the THRUST (T ) of the rocket engine. REGAN PHY1033 2015 91 M is the mass at time t and a is the accelerati on, T Ma , which is Newton' s 2 nd law. To find the velocity as the mass changes, dM Mdv vrel dM dv vrel M vf Integratin g, we obtain, Mf dv -v rel vi Mi dM where vi and v f M are the initial and final rocket vel ocities, correspond ing to rocket masses of M i and M f respective ly. Since, in general, 1 x dx ln x , then Mi Mf v rel ln v f vi vrel ln M f ln M i vrel ln M Mi f thus increase in velocit y greatest for small M f (use of multi 2nd rocket equation - stage rockets! ) Internal Energy Changes and External Forces REGAN PHY1033 2015 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. The change in internal energy of the system is given by, ΔEint Fd cos where d is the displaceme nt of the CENTRE OF MASS and is the angle between th e directions of the force F and displaceme nt d . The associated change in the MECHANICAL ENERGY is then ΔE mec ΔK ΔU Fd cos 92 10: Collisions REGAN PHY1033 2015 93 ‘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 tw o bodies, the 3rd force pair, F(t) and -F(t) acts between th e two at time, t. F(t) is a TIME - VARYING FORCE. -F(t) F(t) From Newton' s 2 nd law, these forces will change the linear momenta of both bodies. The amount by which p changes depends on the time interval, Δt , during which the se forces act. nd From Newton' s 2 law dp F t dt pf tf dp F t dt IMPULSE, J pi ti The IMPULSE is the CHANGE IN LINEAR MOMENTUM of the body acted on by F(t) (right hand side). This is also equal to the product of the strength and duration of the applied force, and the AREA UNDER THE CURVE of F t versus t. REGAN The IMPULSE -LINEAR MOMENTUM THEOREM states PHY1033 2015 that the change in the linear momentum of each body in a collision is equal to the IMPULSE that acts on that body, i.e., p f pi Dp J Since, impulse is a VECTOR, we can also write this in component form, 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 time averaged force over a period, Δt , the magnitude of the impulse 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 taking the initial velocity direction as the NEGATIVE direction, J 0.1439 39kgms1 10.9kgms1 94 REGAN PHY1033 2015 95 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 PHY1033 2015 By CONSERVATI ON OF LINEAR MOMENTUM, Pi P f 96 total momentum before collision total momentum after collision 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 COMPLETELY INELASTIC COLLISION, the two particles stick after collision (e.g., a rugby tack le! ) , then m1v1,i m1 m2 V For an isolated system, the velocity of the centre of mass can not change in a collision as the system is isolated and there is no net external force. Recalling P Mvcom m1 m2 vcom m1v1 m2 v2 P p p2 vcom 1 m1 m2 m1 m2 Elastic Collisions in 1-D REGAN PHY1033 2015 97 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 tw o billiard balls, with mass, m2 at rest. By conservati on of linear momentum, m1v1,i m1v1, f m2 v2, f m1 v1,i v1, f m2 v2, f (1 - D case, magnitudes along same axis). In an elastic collision the total kinetic energy is 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 that v2 ,f is always positive (i.e. m2 is always pushed forward). REGAN PHY1033 2015 For 1 - D elastic collisions , v1, f 98 m1 m2 2m1 v1,i & v2, f v1,i m1 m2 m1 m2 These lead to the following limiting cases. 1) Equal masses, m1 m2 (e.g. pool balls) : v1, f 0 , v2, f v1,i i.e., for a head - on collision between equal masses, the projectile stops following collision and the target moves off with the projectile ' s velocity. 2) Massive target, m2 m1 (e.g., golf ball on a cannon ball) : 2m1 v1,i i.e., light projectile bounces back with v1, f v1,i , v2 ,f m2 similar ve locity (but opposite direction) to incoming projectile . Heavy targ et moves forwards with small velocity. 3) Massive projectile m1 m2 (e.g., cannon ball on golf ball) : v1,f v1,i , v2 ,f 2v1,i i.e. heavy projectile continues forwards at approx. unchanged velocity, light targ et moves off with twic e the projectile velocity. Example 1: REGAN PHY1033 2015 99 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 the MAXIMUM transfer of kinetic energy FROM THE NEUTRON for a single collision as a function of moderator mass. The initial and final kinetic 1 1 mn vn2,i and K f mn vn2, f 2 2 K i K f vn2,i vn2, f The fractional energy loss per collision is F Ki vn2,i energies of the orginal and scattered neutron are K i For a closed neutron - nucleus collision & the moderating nucleus initially at rest, from cons. of lin. mom. vn , f vn ,i mn mMOD 4mn mMOD , thus F therefor e, 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 100 PHY1033 2015 Mblock By conservati on of linear momentum, mbul vbul mbul M block vblock Also know that is the block system is closed, we can assume a conservati on of mechanical energy, then 1 2 mbul M block vblock mbul M block gh 2 where h is the increase in height of the block 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 2 gh bul block bul 2 mbul mbul h 1-D Collisions with a Moving Target REGAN 101 PHY1033 2015 By conservati on 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 elastic collision, kinetic energy is conserved, thus 1 1 1 1 before elastic collision m1v12,i m2 v22,i m1v12, f m2 v22, f 2 2 2 2 m1, v1,i m1 v12,i v12, f m2 v22, f v22,i m2, v2,i m1 v1,i v1, f v1,i v1, f m2 v2, f v2,i v2, f v2,i solving these simultaneo us equations, we obtain the general relations, m1 m2 2m2 2m1 m2 m1 v1, f v1,i v 2 , i & v2 , f v1,i v2 ,i m1 m2 m1 m2 m1 m2 m1 m2 The subscripts 1 and 2 are arbitrary. Note, if we set v2,i 0 (stationar y target) we obtain the previous results of m m2 2m1 v1, f 1 v1,i & v2, f v1,i m1 m2 m1 m2 Collisions in Two Dimensions REGAN 102 PHY1033 2015 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 cos 1 m2v2, f cos 2 y axis, 0 m1v1, f sin 1 m2v2, f sin 2 For an elastic collision, m1v12,i m1v12, f m2v22, f 11: Rotation REGAN 103 PHY1033 2015 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. Summary of Rotational Variables REGAN 104 PHY1033 2015 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 PHY1033 2015 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 displaceme nt (radians), Average Angular Velocity (radian per second), Instantane ous Angular Velocity (rad/s), Average Angular Accelerati on, (radians per s 2 ), Instantane ous Angular Accelerati on, (rad/s 2 ), 105 s r D 2 1 D 2 1 av Dt t 2 t1 d dt D av Dt d av dt Relating Linear and Angular Variables REGAN 106 PHY1033 2015 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 component of the accelerati on is ar r 2 r r 2r 2 Period of revolution , T v 2 Rotation with Constant Acceleration REGAN 107 PHY1033 2015 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 Example 1: 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. REGAN 108 PHY1033 2015 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 32 s 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 109 PHY1033 2015 For a composite body which we can treat as a collection of masses, mn , moving at different speeds, vn , the kinetic energy is 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 the MOMENT OF INERTIA or ROTATIONAL INERTIA 1 2 Kinetic energy of rotation is given by K I 2 Thus in general, a smaller moment of inertia means less work is needed to be done (i.e. less K ) for rotation t o take place. Calculating to the Rotational Moment of Inertia For a rigid body, I mn rn2 REGAN 110 PHY1033 2015 where r is the perpendicu lar distance of the nth n particle from the rotation axis. For a continuous body, 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 the centre of mass as the orgin, then by 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 111 PHY1033 2015 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 locate the centre of mass of the 2 - particle system using xcom m x m2 x2 1 1 M m1 m2 a Cl d-a com H If the x co - ordinate for the centre of 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 127 pm mCl mH mCl mH 1 35 mCl mH I 15,250u. pm 2 (note units for rotational moments of inertia for molecules) . Torque and Newton’s 2nd Law REGAN 112 PHY1033 2015 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!) . F Ft Frad r O r TORQUE, r F sin rFt AND r sin F r F . r is the perpendicu lar distance between O and a line running through F . r is the MOMENT ARM OF THE FORCE F. SI unit of Torque is Nm, which are equivalent to the unit of work W F.d . W in Joules, in Nm. Relating the tangentia l force to the tangentia l accelerati on, Ft mat . Torque acting on the particle is Ft r mat r , since, at r , mrr mr 2 I τ net Iα Newton' s 2 nd law for rotation. Work and Rotational Kinetic Energy From the WORK - KINETIC ENERGY THEOREM, 1 2 1 2 DK K f K i mv f mvi W 2 2 1 1 2 2 since v rω , then mr f mr i W 2 2 Recalling for a single - particle body, I mr 2 , 1 2 1 2 then W DK I f Ii 2 2 Work done, W Fs Ft rD D Work done in an angular displaceme nt 1 to 2 2 is given by W d 1 dW d POWER is given by P dt dt REGAN 113 PHY1033 2015 REGAN 114 PHY1033 2015 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 wheel rolling at a CONSTANT SPEED, means that the speed of the centre of mass, vcom is constant. In a time interval dt , the centre of mass travels the same distance as the distance the outside of the wheel moves through an arc of length, s R , where R is the wheel radius and is its angular displaceme nt. d The angular speed of the wheel about its centre is , while the speed dt ds d R d of the centre of mass is given by vcom R R dt dt dt REGAN 115 PHY1033 2015 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 view the situation as pure rotation about an axis through t he point, P. 1 I P 2 where I p is the 2 moment of inertia through t he point of contact wi th the ground, P. The kinetic energy of this rotation is given by K From the PARALLEL AXIS THEOREM, we can write 1 I P I com MR and thus, K I com MR 2 2 2 1 1 1 1 2 2 K I com 2 M R I com 2 Mvcom 2 2 2 2 2 N REGAN 116 Rolling Down a Ramp PHY1033 2015 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 the rotational frequency is given by R vcom , then d R d vcom by differenti ating both 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 117 PHY1033 2015 R P Fg Fg cos the force components along the slope, Fx ,net Macom, x f s Mg sin where f s m s N m s Mg cos . Rot. form of Newton' s 2 nd law is Fr I . The only force causing a rolling motion in the figure is the FRICTION at point P. The gravitatio nal and Normal forces all act throug h the COM and thus have R 0. acom net I com Fr f s R. For smooth rolling, (note sign) 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 distance h it loses gravitatio REGAN 118 nalPHY1033 2015 T potential energy, mgh. This is transferr ed into kinetic 1 2 energy in both trans lational K trans mv and rotational 2 1 2 K rot I forms. As the yo - yo climbs back up the string, 2 it loses this kinetic energy and transfers it back to potential energy. The expression for the value of the accelerati on of the yo - yo rolling R0 R down the string can be calculated assuming Newton' s 2 nd law (as for a body rolling down a hill) with the following assumption s. (1) the yo - yo rolls directly down the string (i.e. 900 ). (2) the yo - yo rolls around the axle with radius R0 , not the outer radius, R. (3) the yo - yo is slowed by the tension in the string rather tha n friction. g This analysis leads to the expression , acom I 1 com2 MRO Mg 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 119 PHY1033 2015 1.2m By conservati on of mechanical energy, 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 smooth rolling, vcom R and subsitutin g, I com v 2 com gh 7 10 vcom 10 gh 4.1ms 1 (note, Mass independen t, marble 7 and bowling ball reach bottom at same time! ) REGAN 120 PHY1033 2015 Example 1 (cont): (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 The moments of inertia for a sphere MR ; disk MR 2 ; and hoop MR 2 . 5 2 The fraction of kinetic energy whi ch goes into TRANSLATIO NAL MOTION, 2 Mvcom vcom 2 f 1 . In general, I com MR , with a constant and 2 1 2 I com R 2 Mv 1 2 2 com 1 2 f 1 2 Mv 2 com 2 Mvcom 2 vcom MR 2 R 1 2 2 1 1 1 2 For hoop, 1, f 0.5 ; For disk, , f 0.66* ; For sphere, , f 0.71. 2 5 Sphere rolls fastest, followed by the disk. Any size marble will beat disk. REGAN 121 PHY1033 2015 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 The torque is defined by r F . The direction of the torque is found using the vector cross product right - hand rule, (i.e. perpendicu lar to both r and F ). The MAGNITUDE OF THE TORQUE is given by rF sin r F rF where r r sin and F F sin . REGAN 122 PHY1033 2015 Angular Momentum A particle of mass m, with velo city v (i.e. with linear momentum, p mv ), has an ANGULAR MOMENTUM 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 The angular momentum direction is given by the vector cross product (the right - hand rule shows that l is to both r and p, v ). The magnitude of the angular momentum (in units 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 123 PHY1033 2015 dp Newton' s 2 nd law in transla tional form can be written as Fnet dt If the angular momentum of a particle is given by l r p mr v Differenti ating both sides with respect to time gives, dv dr dl d mr v m r v mr a v v dt dt dt dt v v 0 since these vectors are parallel (sin 0) dl mr a r ma r Fnet ri Fi net dt i i.e. the rate of change of angular momentum with respect to time dl is equal to the vector sum of torques acting on the particle net . dt REGAN 124 For a SYSTEM OF PARTICLES, the total angular momentum, PHY1033 2015 L is the VECTOR SUM of the angular momenta, l of the individual 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 EXTERNAL torques change the TOTAL ANGULAR MOMENTUM ( L ) i.e., those due to forces on the particles from external bodies. If net is the NET EXTERNAL TORQUE, i.e. the vector sum of all dL external torques, τ net , we obtain a form for 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 125 For a given particle in a rigid body rotating about a fixed PHY1033 2015 axis, the magnitude of the angular momentum of a mass element Δmi , is l ri pi sin 900 ri Dmi vi . z r The angular momentum component parallel to the rotation Dm r (z) axis is liz li sin ri sin Dmi vi ri Dmi vi The component for the ENTIRE BODY is the sum of these elemental contributi ons pi y x n 2 i.e., L z liz Dmi vi ri Dmi r i ri Dmi ri i 1 i 1 i 1 i 1 n n n n is a CONSTANT for all points on the rotating body and Dmi r2i I , i 1 the moment of inertia of the body about a fixed axis, we can write, Lz I Usually th e ' z ' is dropped, assuming that L is about the rotation axis. REGAN PHY1033 2015 dL Since the net torque is related to the change in angular momentum by, τ net , dt dL if NO NET TORQUE acts on the system, then 0 and thus dt THE ANGULAR MOMENTUM OF THE SYSTEM IS CONSERVED. This means that the net angular momentum at time ti , is equal to the net angular Conservation of Angular Momentum 126 momentum of the system at some other time , t f . We can thus say that if the net external torque acting on a system is zero, the angular momentum of the system, L remains constant, no matter wha t changes take place WITHIN the system. Similarly, if the COMPONENT of the net external torque on a system along a fixed axis is zero, then the component of angular momentum along that axis can not change, no matter wha t takes place within th e system. The conservati on law can be written in algebraic form as I ii I f f . This means that if the moment of inertia of a system decreases, its rotational speed increases to compensate , (e.g., pirouettin g skaters, neutron stars and nuclei! ) REGAN 127 PHY1033 2015 Example1: Pulsars (Rotating Neutron Stars) Crab nebula, SN remnant observed by chinese in 11th century before after! SN1987A Pulsars have similar periodicities ~0.1-1s. Vela supernova remnant, pulsar period ~0.7 secs REGAN 128 PHY1033 2015 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 period of rotation f Ti Ri ( sun ) ~ 7 108 m, Ti ( sun) ~ 2.5 106 s, T f ( pulsar ) ~ 1s 1s R f ~ 7 10 m ~ 400 Km. 6 2.5 10 s 8 x-ray PULSAR = PULSAting Radio Star (neutron-star) REGAN 129 PHY1033 2015 TRANSLATIONAL ROTATIONAL dp dl Force , F ma Torque, I r F dt dt Momentum, Linear p mv Ang. Mom. l I r p dp dL nd Newtons 2 Law, F net dt dt Conservati on, Linear Mom. Angular mom. dp dL Fnet 0 net 0 dt dt 13: Equilibrium and Elasticity REGAN 130 PHY1033 2015 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. balance of forces) for dt TRANSLATIO NAL EQUILIBRIU M and dL net 0 (i.e. balance of torques) for dt ROTATIONAL EQUILIBRIU M The Centre of Gravity REGAN 131 PHY1033 2015 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. Elasticity REGAN 132 PHY1033 2015 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 133 PHY1033 2015 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 The STRESS on an object for simple tension or compressio n is given by , A where F is the magnitude of the force applied perpendicu larly to the area A (This also defines the pressure at that point). The STRAIN is the unit deformatio n. For tensil e stress, this is a dimensionl ess ΔL correspond ing to the fractional change in the length quantity defined by L of the object ( L is the orginal length, ΔL is the extension) . F REGAN 134 PHY1033 2015 The YOUNG'S MO DULUS ( E ) for tensil e or L F ΔL L+ compressiv e stress is defined by E DL A L F F For SHEARING , the stress is still , but F is parallel A Dx F Δx to the plane of the area. The strain is now , leading l F Dx L to the SHEAR MODULUS, (G) where G F A l HYDRAULIC STRESS is defined as the fluid pressure P, V ΔV (i.e. force per unit area). The strain is defined as , V V-DV DV where V is the initial volume and ΔV is the volume change. DV The BULK MODULUS ( B) is defined by P B V REGAN 135 PHY1033 2015 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. Su (rupture) Sy (perm. deformed) Strain (Dl/l) Example 1: REGAN 136 PHY1033F=62kN 2015 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 the definition of 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 137 PHY1033 2015 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 138 PHY1033 2015 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 the earth is a UNIFORM SPHERE of mass M , the magnitude of the gravitatio nal force from the earth on a particle of mass m, at a distance r , Mm from the earth' s centre is given by : Fgrav G 2 r If the particle is released, it will accelerate to the earth' s centre under gravity with a GRAVITATIO NAL ACCELERATI ON, a g , whose magnitude is given Mm GM by Fgrav mag G 2 a g 2 . Thus the accelerati on due to gravity r r depends on the ' height' at which an object is dropped from. 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 139 PHY1033 2015 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 The normal force on a surface object is from Fnet mar N mag m 2 R N The normal force, N is equal to the weight, mg mg mag m ω 2 R g a g 2 R REGAN 140 PHY1033 2015 R m mag R is the radius which the object rotates around and is the rotational velocity. R is max. at the equator, R Rearth 6.37 106 m. Δθ 2 rads can be estimated from Δt 24 3600s acentr 2 R 0.034ms 2 (cf. mag 9.8ms - 2 ) i.e. very small compared to mag . Assuming the weight equals the gravitatio nal accelerati on is usually (on earth at least! ) well justified. S ‘above’ view, looking from pole, RN m Gravitation Inside the Earth REGAN 141 PHY1033 2015 ‘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 the force experience d by the particle due to the GM insm Gr 43 r 3 4Grr m mass inside the shell is F i.e. Fnet kr 2 2 r r 3 Gravitational Potential Energy REGAN 142 PHY1033 2015 The gravitatio nal potential energy is defined by the expression , Mm U G and defined to be zero at infinite separation (r ). r PROOF In general, work done is W F r .dr , F r .dr F r dr cos , R Mm cos cos 180 0 -1 F r .dr G 2 dr W F r .dr r R Mm 1 GMm GMm GMm W G 2 dr GMm 2 dr 0 r r R R r R R R W WORK REQUIRED to move a mass, m from a distance R out to . Since potential energy and work done are related by the general expression , GMm DU W U U R , U R W R Potential Energy and Force REGAN 143 PHY1033 2015 Gravity is a conservati ve force and changes in the grav. potential energy only depend on the initial and final positions, NOT THE PATH TAKEN. Since we can derive the gravitatio nal potential energy from the expression for the force, the converse is also true. F dU d Mm Mm G G dr dr r r2 Escape Speed (Velocity) A mass m projectile leaving a mass M planet of radius R has an ESCAPE SPEED, v. This causes the object to move up with constant speed, v against gravity, until it slows down to v 0 at infinite distance. 1 2 GMm mv ; The gravitatio nal potential energy, U . 2 R At infinite distance, K U 0 (i.e. zero velocity and at the zero potential energy configurat ion for r ). Thus from the principle of conservati on of energy The kinetic energy , K K U 1 2 GMm 2GM earth 1 mvesc 0 v ; v 11 . 2 kms 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 144 PHY1033 2015 a b a b The Law of Orbits REGAN 145 PHY1033 2015 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 r f ea REGAN 146 PHY1033 2015 ea f’ x f a In general for an ellipse, the eccentrici ty is defined by , rmax rmin rmax rmin f’ Ra R p Ra R p If we take the origin of the co - ordinates as the focus, f , r0 r0 r0 r rmax , rmin 1 cos 1 1 In Cartesian co - ordinates r x 2 y 2 , x r cos θ, y r sin θ Ellipse is defined by the equation, 1 2 x 2 2r0 x y 2 r02 2r0 1 1 A Length of major axis 2 a rmax rmin RP Ra r0 2 1 1 1 r A The length of the SEMI - MAJOR AXIS 0 2 2 1 The Law of Areas REGAN 147 PHY1033 2015 If the DA is the area swept out in time Dt , ΔA can be ESTIMATED assuming the wedge of area swept out is a TRIANGLE of height, r , and base s rD . p The area swept out is approximat ely p 1 1 DA base height .rDθ.r. 2 2 m r This expression becomes more exact D DA for smaller va lues of ΔA . M As D 0 and ΔA 0, ΔA dA 1 2 d 1 2 r r Δt dt 2 dt 2 where is the angular speed of the rotating line connecting the sun and planet (i.e., rotational velocity of planet around the sun). The ang. mom. of the planet around the star is, L rp r mv r mr mr 2 dA 1 2 L r dt 2 2m dA . Thus, if L is conserved, constant. dt The Law of Periods For a circular orbit, using Newton' s 2 nd law, REGAN 148 PHY1033 2015 m Mm v2 rω r F magrav G 2 mg m m mrω 2 r r r M Mm G 2 mrω 2 GM r 3ω 2 r 2π The period of revolution , T , substituti ng in we get ω 2 2 3 4 π 4 3 2 r . For ellipse, r a semi - major axis. GM r T 2 T GM 2 T 2 4π 2 Exact vers ion of law predicts 3 a G M m T2 2 i.e., 3 constant for M m, ( Ta 3 3.0 10-34 yr 2 m -3for solar system). a Satellites, Orbits and Energies The potential energy of system is given by GMm U , U 0 for infinite separation r The KINETIC ENERGY OF A CICRULARLY ORBITING SATELLITE, via Newton' s 2 nd law is REGAN 149 PHY1033 2015 K(r) r Etot(r) GMm v2 1 GMm F 2 m K mv2 r r 2 2r =-K(r) U Therefore, K for a satellite in a circular orbit. 2 The total mechanical energy is given by E U K GMm GMm GMm E K r 2r 2r i.e. the total energy is equal to the NEGATIVE OF THE KINETIC ENERGY GMm For an elliptical orbit substitute a (semi - major axis length) for r , ie. E 2a Example 1: REGAN 150 PHY1033 2015 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 where R 6.37 106 m the earth' s radius and G 6.67 10-11 m 3 kg 1s 2 4π 2 R h 3 6.6 106 m 4 2 M earth 2 2 11 3 1 2 G T 6 . 67 10 m kg s 89 60 s M earth 6 10 24 kg 3