Physics 199BB The Physics of Baseball - PowerPoint

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					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 speed = magnitude of v = v 2 +v 2 +v 2 x y z
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

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