PHY 5200 Mechanical Phenomena Projectile MotionPHY 5200 Mechanical Phenomena Newton’s Laws of Motion Click to editClaude Master style A title Pruneau Physics and Astronomy Department Wayne State University Dec 2005. Click Claude to edit A Pruneau Master subtitle style Physics and Astronomy Wayne State University 1 Content • Projectile Motion – – – – Air Resistance Linear Air Resistance Trajectory and Range in a Linear Medium Quadratic Air Resistance • Charge Particle Motion – Motion of a Charged Particle in a Uniform Field – Complex Exponentials – Motion in a Magnetic Field Description of Motion with F=ma • F=ma, as a law of Nature applies to a very wide range of problems whose solution vary greatly depending on the type of force involved. • Forces can be categorized as being “fundamental” or “effective” forces. • Forces can also be categorized according to the degree of difficulty inherent in solving the 2nd order differential equation F = m a. – Function of position only – Function of speed, or velocity – Separable and non-separable forces • In this Chapter – Separable forces which depend on position and velocity. – Non separable forces. . motion of a golf ball.g. • Air Resistance is however often non-negligible and must be accounted for to properly describe the trajectories of projectiles.Air Resistance • Air Resistance is neglected in introductory treatment of projectile motion. • One also need a way/technique to determine whether air resistance is important in any given situation. – While the effect of air resistance may be very small in some cases. it can be rather important and complicated e. • Not a good approximation for motion of a wing (airplane) additional force involved called “lift”. .no sideways force. • True for spherical objects. we will only consider cases where the force is antiparallel to the velocity .Basic Facts • Air resistance is known under different names – Drag – Retardation Force – Resistive Force • Basic Facts and Characteristics – – – – Not a fundamental force… Friction force resulting from different atomic phenomena Depends on the velocity relative to the embedding fluid. a good and sufficient approximation for many other objects. Direction of the force opposite to the velocity (typically). – Here.Air Resistance . • • Measurements reveal f(v) is complicated . one can write as a good approximation: f (v) = bv + cv 2 = flin + fquad .especially near the speed of sound… At low speed. ˆ v ! ˆ f = ! f (v)v ! ! w = mg ! ˆ f = ! f (v)v • Where ! v ! v ˆ= ! v ! f(v) is the magnitude of the force.Air Resistance .Drag Force • Consider retardation force strictly antiparallel to the velocity. Definitions f (v ) = bv + cv 2 = flin + fquad flin ! bv Viscous drag • Proportional to viscosity of the medium and linear size of object.Air Resistance .g. canon ball. Inertial • Must accelerate mass of air which is in constant collision. fquad ! cv 2 For a spherical projectile (e.6 " 10 #4 N i s / m 2 $ = 0. • Proportional to density of the medium and cross section of object.25 N i s / m 4 . baseball. drop of rain): b = !D c = ! D2 Where D is the diameter of the sphere !and " depend on the nature of the medium At STP in air: ! = 1. consider fquad flin = cv !D = v = 1.Air Resistance .6 # 10 3 ms2 bv " 2 ( ) $ ! 1: linear case & Dv % &" 1: quadratic case ' • Example: Baseball and Liquid Drops • A baseball has a diameter of D = 7 cm. • Millikan Oil Drop Experiments. and travel at speed of order v=5 m/s. D=1. either of the linear or quadratic terms can be neglected.6 m/s !1 Neither term can be neglected.5 mm and v=5x10-5 m/s. fquad flin fquad flin fquad flin ! 600 ! ˆ f = ! cv 2 v • A drop of rain has D = 1 mm and v=0.Linear or Quadratic • Often. ! 10 "7 ! ! f = !bv . • To determine whether this happens in a specific problem. • One finds fquad flin !R! Dv" # Reynolds Number .Air Resistance .Reynolds Number • The linear term drag is proportional to the viscosity. # • The quadratic term is related to the density of the fluid. $. one gets coupled y vs x motion ! 2 2! ˆ = ! c vx f = ! cv 2 v + vy v . 2 2 !x = ! c vx mv + vy vx 2 2 !y = mg ! c vx mv + vy vy • By contrast. thus: ! ! ! " mv = mg ! bv !x = !bvx mv !y = mg ! bvy mv A 1st order differential equation • Furthermore.Case 1: Linear Air Resistance • • Consider the motion of projectile for which one can neglect the quadratic drag term. Two separate differential equations Uncoupled.y. it is separable in coordinates (x.z). for f(v)~v2. From the 2nd law of Newton: y x ˆ v ! ! w = mg ! ! f = ! bv ! ! ! ! "" mr = F = mg ! bv • Independent of position. ! ! Assume the only relevant force is the drag force. f = !bv Obviously. one must solve: Clearly: b m dv !x = x = ! kvx v dt dvx ! vx = " k ! dt with dvx = ! kdt vx ln vx = ! kt + C Which can be re-written: vx (t ) = vx 0 e! t /" ! =1/ k = m /b Velocity exhibits exponential decay .Case 1: Linear Air Resistance . the object will slow down !x = ! v b vx m • • • • Define (for convenience): k = Thus.Horizontal Motion • • • • Consider an object moving horizontally in a resistive linear medium. Assume vx = vx0. x = 0 at t = 0. Case 1: Linear Air Resistance . integrate dx dt ! = x(t ) # x(0) " dt ! 0 • One gets x (t ) = x (0) + $ vx 0 e! t " /# dt " 0 t t =0+% & ! vx 0# e ! t " /# ' (o t x(t ) = x! 1 " e" t /# x! $ vx 0# ( ) x(t ) = x! 1 " e" t /# x! $ vx 0# ( ) vx (t ) = vx 0 e! t /" .Horizontal Motion (cont’d) • Position vs Time. One then has terminal speed. . y x ! ! f = !bv ˆ v ! ! w = mg !y = mg ! bvy mv • Gravity accelerates the object down. Implies terminal speed is different for different objects. 0 = mg ! bvy mg vter = vy (a = 0) = b Note dependence on mass and linear drag coefficient b.Vertical Motion with Linear Drag • Consider motion of an object thrown vertically downward and subject to gravity and linear air resistance. the speed increases until the point when the retardation force becomes equal in magnitude to gravity. Equation of vertical motion for linear drag • The equation of vertical motion is determined by !y = mg ! bvy mv • Given the definition of the terminal speed. • One can write instead mg vter = b !y = !b vy ! vterm mv • Or in terms of differentials ( ) ) mdvy = !b vy ! vterm dt • Separate variables ( dvy vy ! vterm • Change variable: bdt =! m u = vy ! vterm du = dvy du bdt =! = ! kdt u m where k= b m . we get du bdt =! = ! kdt u m du ! u = " k ! dt ln u = ! kt + C u = Ae! kt u = vy ! vterm vy ! vter = Ae! t /" with ! =1/ k = m /b Now apply initial conditions: when t = 0. vy = vy0 This implies v ! v = Ae!0 /" = A y0 ter The velocity as a function of time is thus given by vy = vter + vy 0 ! vter e! t /" ( vy = vy 0 e! t /" + vter 1 ! e! t /" ( ) ) .Equation of vertical motion for linear drag (cont’d) • • • • • • • • So we have … Integrate Or… Remember So. 5 3 95.3 .2 2 86. consider vy0=0. one has Whereas for t!" vy = vy 0 As the simplest case. I.Equation of vertical motion for linear drag (cont’d) • • • • We found vy = vy 0 e! t /" + vter 1 ! e! t /" vy = vy 0 ( ) At t=0. vy = vter 1 ! e! t /" ( ) time percent of t/tau vter 0 0. dropping an object from rest.e.0 4 98.2 5 99.0 1 63. and initial velocity vy = vy0. One gets ( ) y0 = !" vy 0 ! vter + C C = y0 + ! vy 0 " vter • The position is thus given by ( ) ( ) x y ! ! f = !bv ˆ v ! ! w = mg y = y0 + vter t + ! vy 0 " vter 1 " e" t /! ( )( ) .Equation of vertical motion for linear drag (cont’d) • • • Vertical position vs time obtained by integration! Given vy = vter + vy 0 ! vter e! t /" ( ) The integration yields y = vter t ! " vy 0 ! vter e! t /" + C • Assuming an initial position y=y0. Equation of vertical motion for linear drag (cont’d) • Note that it may be convenient to reverse the direction of the y-axis. the position may thus be written y x ! ! f = !bv • ˆ v ! ! w = mg y = y0 ! vter t + " vy 0 + vter 1 ! e ( )( ! t /" ) . Assuming the object is initially thrown upward. Equation of motion for linear drag (cont’d) • Combine horizontal and vertical equations to get the trajectory of a projectile. y(t ) = y0 + vy 0 + vter vx 0 # x & x + vter! ln % 1 " vx 0! ( $ ' . and substitute in the second equation. x(t ) = vx 0! 1 " e" t /! y(t ) = y0 ! vter t + " vy 0 + vter 1 ! e! t /" ( ( ) )( ) • To obtain an equation of the form y=y(x). solve the 1st equation for t. 1 100 34500 24500 y (m) Linear friction No friction x (m) .Example: Projectile Motion m tau vx0 vy0 vter 5 50 2 200 490 b vx0*tau (vy0+vter)*tau vter*tau 0. one has x(t ) = vxot y(t ) = vyot ! 0. one has 0= vy 0 + vter vx 0 # R & R + vter! ln % 1 " vx 0! ( $ ' A transcendental equation .cannot be solved analytically .Horizontal Range • In the absence of friction (vacuum).98 2 2 t • The range in vacuum is therefore Rvac = 2 vxo vyo g • For a system with linear drag. R ! vter" + + % + % . ( ) Neglect orders beyond We now get This leads to 0= !3 2 3 ) R 1# R & 1# R & . consider a Taylor expansion of the logarithm in 0 = Let != R vxo" 2 1 3 We get ln(1 ! " ) = ! " + 1 2 " + 3 " + .. x0 x0 * x0 - vy 0 + vter vx 0 R=0 R= 2 vxo vyo g ! 2 R2 3vxo" R ! Rvac " $ 2 4 vyo ' 2 Rvac = Rvac & 1 " 3vxo# 3 vter ) % ( . ( ( v " 2 v " 3 v " $ ' $ ' + ..Horizontal Range (cont’d) • • • • • • • If the the retardation force is very weak… R ! vxo! vy 0 + vter vx 0 # R & R + vter! ln % 1 " vx 0! ( $ ' So. it is NOT separable in x. it is usually a better approximation to consider the drag force is quadratic ! 2! f = ! cv v • Newton’s Law is thus ! ! 2! " mv = mg ! cv v • Although this is a first order equation.Quadratic Air Resistance • For macroscopic projectiles.z components of the velocity. .y. ! ct * v0 v $ v" ' v0 v(t ) = v(! ) = v0 v0 = 1 + cv0t / m 1 + t / ! v0 = v0 / 2 1+! /! with • • Solving for v Note: for t=%.Horizontal Motion with Quadratic Drag • We have to solve m dv = ! cv 2 dt Separation of v and t variables permits independent integration on both sides of the equality… t • Rearrange m dv = ! cdt v2 v • Integration dv! m " 2 = # c " dt ! v! vo 0 v where v = vo at t = 0. • Yields ) 1 1. # 1& m % ! ( = m + ! . != m cvo . Horizontal Motion with Quadratic Drag (cont’d) • Horizontal position vs time obtained by integration … x(t ) = x + " v(t ! )dt ! 0 t = v0# ln(1 + t / # ) • • Never stops increasing By contrast to the “linear” case. one must include both the linear and quadratic terms. . x (t ) = v0! ln(1 + t / ! ) • In realistic treatment. v(t ) = v0 1+ t /! x (t ) = vx 0! 1 " e" t /! • • • Which saturates… Why? ! ? ( ) The retardation force becomes quite weak as soon as v<1. m Terminal velocity achieved for vter = mg c dv = mg ! cv 2 dt For the baseball of our earlier example. y. # " vter & %+ . this yields ~ 35 m/s or 80 miles/hour 2 Rewrite in terms of the terminal velocity dv = g " 1 ! v % $ dt v2 ' # & ter • Solve by separation of variables dv = gdt v2 1! 2 vter • Integration yields ! v $ vter arctanh # =t g " vter & % • Solve for v ! gt $ v = vter tanh # & " vter % 2 vter ) ' ! gt $ * ( y= ln cosh • Integrate to find g ) ( . down.Vertical Motion with Quadratic Drag • • • • Measuring the vertical position. Quadratic Draw with V/H motion • Equation of motion ! ! ˆ m"" r = mg ! cv 2 v ! ! = mg ! cvv • With y vertically upward 2 2 !x = ! c vx mv + vy vx 2 2 !y = ! mg ! c vx mv + vy vy . Motion of a Charge in Uniform Magnetic Field • • • Another “simple” application of Newton’s 2nd law… Motion of a charged particle. vz ! B = ( 0. B ) ( ) ! v = vy B. Z The force is ! ! ! F = qv ! B ! B • The equation of motion ! ! ! " = qv ! B mv • • The 2nd reduces to a first order Eq. ! vx B. x ! v y Components of velocity and field ! v = v x . v y . in a uniform magnetic field. 0. pointing in the z-direction. q. B. 0 ( ) . Motion of a Charge in Uniform Magnetic Field (cont’d) • Three components of the Eq of motion !x = qBvy mv !y = ! qBvx mv !z = 0 mv • Define x y vz = constant ( v . v ) ! transverse velocity != qB m Cyclotron frequency • Rewrite !x = ! vy v !y = "! vx v Coupled Equations Solution in the complex plane … . Complex Plane y (imaginary part) Representation of the velocity vector ! = vx + ivy vy i = !1 x (real part) O vx . Why and How using complex numbers for this? • Velocity ! = vx + ivy !=v !x + iv !y ! • Acceleration • Remember Eqs of motion !x = ! vy v !y = "! vx v • We can write • Or !=v !x + iv !y = " vy # i" vx = #i" vx + ivy ! ! = "i#! ! ( ) . Why and How using … (cont’d) • Equation of motion ! = "i#! ! • Solution ! = Ae" i# t • Verify by substitution d! = "i# Ae" i# t = "i#! dt . large or small). It satisfies • • And is indeed a general solution for df ( z ) = kf ( z ) d Aekz = k Aekz dz ( ) ( ) dz So we were justified in assuming # is a solution of the Eqs of motion. .Complex Exponentials • Taylor Expansion of Exponential 2 3 z z ez = 1 + z + + + ! 2! 3! • • The series converges for any value of z (real or complex. Complex Exponentials (cont’d) The exponential of a purely imaginary number is 2 3 4 i ! i ! i ! ( ) ( ) ( ) e! = 1 + i! + + + +! 2! 3! 4! where & is a real number Separation of the real and imaginary parts .since i2=-1. i3=-I # !2 !4 & # & !3 e = %1 " + " !( + i %! " + !( 2! 4! 3! $ ' $ ' ! cos! We get Euler’s Formula sin ! e = cos! + i sin ! i! . Complex Exponentials (cont’d) • Euler’s Formula implies ei& lies on a unit circle. e = cos! + i sin ! y i! cos! ei! 1 O sin ! ! x cos 2 ! + sin 2 ! = 1 . Complex Exponentials (cont’d) • A complex number expressed in the polar form A = ae = a cos! + ia sin ! where a and & are real numbers y i! a cos! A = ae a O i! a 2 cos 2 ! + a 2 sin 2 ! = a 2 ! a sin ! Amplitude Phase x ! = Ae " i# t !i" t ! = Ae " i# t = ae i ($ " # t ) Angular Frequency . Solution for a charge in uniform B field • • vz constant implies z(t ) = zo + vzot The motion in the x-y plane best represented by introduction of complex number. ! = x + iy Greek letter “xi” • • The derivative of ' Integration of # !=x ! + iy ! = vx + ivy = " ! ! = # "dt = # Ae$ i% t dt iA # i" t ! = e + constant " x + iy = Ce! i" t + ( X+iY) . Solution for a charge in uniform B field (cont’d) x + iy = Ce! i" t + ( X+iY) Redefine the z-axis so it passes through (X.Y) x + iy = Ce! i" t y which for t = 0. implies C = xo + iyo xo + iyo 2 2 xo + yo Motion frequency != qB m O x x + iy !t . Solution for a charge in uniform B field (cont’d) x (t ) + iy(t ) = Ce! i" t z(t ) = zo + vzot != qB m y xo + iyo 2 2 xo + yo O !t x x + iy Helix Motion .