Physics 199BB The Physics of Baseball

Physics 199BB The Physics of Baseball Fall 2007 Freshman Discovery Course Alan M. Nathan 403 Loomis 333-0965 a-nathan@uiuc.edu Week 2 Week 2 1 The Flight of a Baseball • The goal: to develop an understanding of the trajectory of a baseball in flight – Pitched baseball – Batted baseball – Thrown baseball • First step: we need to go over some basic physics concepts Week 2 2 Position, Velocity, and Acceleration 1. Position: x,y,z • • • Units of length (m, ft, …) Trajectory completely known if we know the position of an object at every instant of time x(t),y(t),z(t) Position is a vector with three components r r = (x,y,z) Week 2 3 Position, Velocity, and Acceleration 2. Velocity vx,vy,vz • • Units of length/time (m/s, ft/s,mph,…) Velocity is the rate of change of position x dx vx   t dt Velocity is a vector with three components r r dr v = (v x ,v y ,v z )= dt r 2 speed = magnitude of v = v 2 +v 2 +v z x y Week 2 4 • Position, Velocity, and Acceleration 3. Acceleration ax,ay,az • • Units of length/time2 (m/s2, ft/s2,…) Acceleration is the rate of change of velocity v x dv x d 2 y ax    2 t dt dt Acceleration is a vector with three components • r 2r r dv d r a = (a x ,a y ,a z )=  2 dt dt Week 2 5 Motion with Constant Acceleration x = x0 + v0t + ½at2 v = v0 + at Special Case 1: a=0 x = x0 + v0t v = v0 Week 2 6 Special Case 2: Two-dimensional projectile motion with gravity ax = 0 (horizontal) ay = -g (vertical) g = 32.2 ft/s2 = 9.8 m/s2 • x = x0 + v0xt Week 2 vx = vx0 (constant) 7 • y = y0 + v0yt - ½gt2 vy = v0y - gt Detailed example: pitched baseball • Suppose a pitcher throws a baseball with an initial horizontal velocity of 90 mph at a height of 6 ft above home plate. How long does the pitch take to reach home plate? How much does the pitch drop vertically? x = x0 + v0xt x0=0 v0x = 90 mph y = y0 + v0yt - ½gt2 y0=6 ft v0y=0 Week 2 8 Detailed example: pitched baseball x = x0 + v0xt x0=0 v0x = 90 mph y = y0 + v0yt - ½gt2 y0=6 ft v0y=0 Want to find time T when x = 60.5 ft. Use that T to find y. But first…need a consistent set of units. Convert mph to f/s 90 mile/hour = 90 (mile/hour)(5280 ft/mile)(1/3600 hour/sec) =90*1.467=132.0 ft/s Week 2 9 Useful thing to remember • To convert mph to ft/s, multiply by 1.467 • To convert ft/s to mph, divide by 1.467 Week 2 10 Detailed example: pitched baseball x = x0 + v0xt x0=0 v0x = 90 mph y = y0 + v0yt - ½gt2 y0=6 ft v0y=0 Now solve to find T: 60.5 ft = 0 + 132T ft/s T=0.458 s Now solve to find y(T): Y = 6 ft + 0 -0.5*32.2 ft/s2*(0.458)2 s2 = 6 ft – 3.382 ft = 2.618 ft Ball drops ~3.4 ft! Week 2 11 Using Excel to Compute the Trajectory • divide up time into slices separate by dt • suppose x,y,vx,vy are known at time t • at time t+dt x(t+dt)=x(t)+vx(t)*dt y(t+dt)=y(t)+vy(t)*dt vx(t+dt)=vx(t)+ax(t)*dt vy(t+dt)=vy(t)+ay(t)*dt • for case at hand values known at t=0: – x0,y0,v0x,v0y – ax=0 ay=-g Week 2 12 Detailed example: batted baseball • Suppose the baseball is hit at an initial height of 3 ft off the ground at a speed of 100 mph and an angle of 35o to the horizontal. – How far does it travel? – How long is it in the air? – How high does it go? Week 2 13 Detailed example: batted baseball • Suppose the baseball is hit at an initial height of 3 ft off the ground at a speed of 100 mph and an angle of 35o to the horizontal. y v0 v0 = 100 mph = 146.7 ft/s  = 35o v0x = v0 cos() v0y = v0 sin() x  Week 2 14 Batted Ball Example x = x0 + v0xt = x0 + v0tcos() y = y0 + v0tsin() - ½gt2 x0=0 y0=3 ft v0=146.7 ft/s =35o First step: How long T is ball in the air? Trick: when ball hits ground, vy=-v0y=-v0sin() Use vy=v0y-gT and solve for T, with vy=-voy. T=2v0y/g=2v0sin()/g=2*146.7*sin(350)/32.2=5.23 s Week 2 15 Batted Ball Example x = x0 + v0xt = x0 + v0tcos() y = y0 + v0tsin() - ½gt2 x0=0 y0=3 ft v0=146.7 ft/s =35o Second step: How far D did ball travel? Use D = v0Tcos() = 146.7*5.23*cos(35o) = 628.5 ft 16 Week 2 Batted Ball Example x = x0 + v0xt = x0 + v0tcos() y = y0 + v0tsin() - ½gt2 x0=0 y0=3 ft v0=146.7 ft/s =35o Third step: How high H did ball go? Maximum height occurs at time t=T/2 = 2.62 s. Plug into equation for y, using t=T/2 H = 6 + 146.7*2.62*sin(35o)-0.5*32.2*(2.62)2 = 115.9 ft Week 2 17 Now let’s use Excel to solve (just like before) • divide up time into slices separate by dt – dt needs to be “small” • suppose x,y,vx,vy are known at time t • at time t+dt – – – – x(t+dt)=x(t)+vx(t)*dt y(t+dt)=y(t)+vy(t)*dt vx(t+dt)=vx(t)+ax(t)*dt vy(t+dt)=vy(t)+ay(t)*dt • for case at hand values known at t=0: – x0,y0,v0x,v0y – ax=0 ay=-g Week 2 18 Some Useful Formulas (we won’t use these for anything) • Maximum distance D = v02sin(2)/g  v0  D = 665 ft   sin(2θ) 100 mph  2 • Maximum height H = v02sin2()/2g  v0  H = 333 ft  sin 2 (θ) 100 mph    • Time of flight T = 2v0sin()/g 2  v0  T = 9.1 s   sin(θ) 100 mph  o • D is largest when =45 • T and H are largest with =90o Week 2 19 2 Baseball Trajectories with Drag and Magnus Forces • Some additional physics concepts – Newton’s First Law • Objects at rest stay at rest and objects in motion continue to move at constant velocity if not acted upon by an external force • In other words, with no external force v is constant in both magnitude and direction – Newton’s Second Law   • Forces cause acceleration: a = F/m or F=ma Week 2 20 Forces on a Baseball in Flight • Gravity – Already discussed FMagnus  v • Drag (“air resistance”) Force – We will do this next • Magnus Force – We will do this later Fdrag mg Week 2 21 Baseball Trajectories with Drag • Fdrag= ½ CDAv2 –  is density of air • 1.23 kg/m3 at normal temp and pressure  v – A is cross sectional area of ball • A = R2 = 4.16 x 10-3 m2 – v = speed of ball – CD is drag coefficient Fdrag mg • A number between 0 and 1 • Approximately 0.5 for v<50 mph • See plot in Adair, p. 8, Fig. 2.1 – Direction of force is exactly opposite velocity Week 2 22 Drag Coefficient from Adair Week 2 23 Let’s estimate size of drag force • Let CD = ½, v=100 mph • FD = ½CDAv2 – Convert mph to m/s: 100 mph = 44.7 m/s – FD=1/2*1/2*1.23*4.16x10-3*(44.7)2 • FD = 2.56 N = 0.574 lb – By comparison, weight of ball is 5.1 oz • mg = 0.319 lb – We conclude that the drag is very important Week 2 last slide of Week 2 24