Transcript Physics 131: Lecture 4 Notes

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Physics 151: Lecture 4, Pg 1
Physics 151: Lecture 4
Today’s Agenda
3-D Kinematics :
Review motion vs time graphs
Kinematics in 2 or 3 dimensions
Independence of x and y components
Projectile motion, baseball example
Physics 151: Lecture 4, Pg 2
Review of 1-D Motion :

For constant acceleration we found:
1
x  x 0  v 0 t  at 2
2
v  v 0  at
a  const
x
v

vav
A few other useful formulas :
1
v av  (v 0  v)
2
v 2  v 02  2a(x  x 0 )
t
a
t
t
Physics 151: Lecture 4, Pg 3
Lecture 4, ACT 1
2-D Motion

Alice and Bill are playing air hockey on a table with no
bumpers at the ends. Alice scores a goal and the puck
goes flying off the end of the table. Which diagram best
describes the path of the puck ?
Alice
A)
Bill
B)
C)
Physics 151: Lecture 4, Pg 4
See text: 4-1
3-D Kinematics (Chapter #4)

The position, velocity, and acceleration of a particle in 3
dimensions can be expressed as:
r= xi+yj+zk
v = v x i + vy j + vz k
a = a x i + ay j + az k
•
(i , j , k unit vectors )
We have already seen the 1-D kinematics equations.
x  x(t )
dx
v
dt
dv d 2 x
a
 2
dt dt
Physics 151: Lecture 4, Pg 5
See text: 4-2 and 4-3
3-D Kinematics
• For 3-D, we simply apply the 1-D equations to each of the
component equations.
x  x(t )
vx 
ax 

dx
dt
vy 
d2x
dt
y  y( t )
2
ay 
dy
dt
vz 
d2y
dt
z  z( t )
2
az 
dz
dt
d2z
dt 2
Which can be combined into the vector equations:
r = r(t)
v = dr / dt
a = d2r / dt2
Physics 151: Lecture 4, Pg 6
See text: 4-4
3-D Kinematics

So for constant acceleration we can integrate to get:
a = const
v = v0 + a t
r = r0 + v0 t + 1/2 a t2
(where a, v, v0, r, r0, are all vectors)
Physics 151: Lecture 4, Pg 7
See text: 4-5
2-D Kinematics

Most 3-D problems can be reduced to 2-D problems when
acceleration is constant;
Choose y axis to be along direction of acceleration.
Choose x axis to be along the “other” direction of
motion.

Example: Throwing a baseball (neglecting air resistance).
Acceleration is constant (gravity).
Choose y axis up: ay = -g.
Choose x axis along the ground in the direction of the
throw.
Physics 151: Lecture 4, Pg 8
“x” and “y” components of motion are
independent !

A man on a train tosses a ball straight up in the air.
View this from two reference frames:
Reference frame
on the moving train.
y motion: a = -g y
x motion: x = v0t
Reference frame
on the ground.
Physics 151: Lecture 4, Pg 9
Projectile Motion.

If I set something moving near the earth, it reduces
to a 2d problem we call projectile motion.

Use a coordinate system with x along the ground, y
vertical with respect to the ground. (Notice no
change in third direction.)

Equations of motion reduce to:
X: Dx = voxt
ax = 0
Y: y = yo + voyt – g t2
y positive upwards
Physics 151: Lecture 4, Pg 10
Problem 1:

Sammy Sosa clobbers a fastball toward center-field.
You are checking out your new fancy radar gun which
can detect ball velocity, i.e. speed and direction. You
measure that the ball comes off the bat with initial
velocity is 36.5 m/s at an angle of 30o above
horizontal. Since Sammy was hitting a high fastball,
you estimate that he contacted the ball about one
meter off of the ground. You know the dimensions of
Wrigley field and the center-field wall is 371 feet
(113m) from the plate and is 10 feet (3m) high. You
decide to demonstrate your superfast math and
physics skills by predicting whether Sammy get a
home run before the play is decided.
Physics 151: Lecture 4, Pg 11
Problem 1:
1)
We need to find how high the ball is at a
distance of 113m away from where it starts.
Animation
v
yo
h

D
Physics 151: Lecture 4, Pg 12
2)
Problem 1:
This is a problem in projectile motion.
Choose y axis up.
Choose x axis along the ground in the direction of the
hit.
Choose the origin (0,0) to be at the plate.
Say that the ball is hit at t = 0, x = xo = 0, y = yo = 1m
y
v
h

D
x
Physics 151: Lecture 4, Pg 13
Problem 1

Variables
vo = 36.5 m/s
yo = 1 m
h = 3 m
 o = 30º
 D = 113 m
a = (0,ay)  ay = -g
t = unknown,
Yf – height of ball when x=113m, unknown,
our target
Physics 151: Lecture 4, Pg 14
Problem 1
3) For projectile motion,
 Equations of motion are:
y
vx = v0x
vy = v0y - g t
x = vx t
y = y0 + v0y t - 1/ 2 g t2
v
y0
And, use geometry to find vox and voy

v0y
v0x
Find
and
g
v0x = |v| cos .
v0y = |v| sin .
x
Physics 151: Lecture 4, Pg 15
4) Solve the problem,

Problem 1
Numbers:
 y(t) = (1.0 m) + (113 m)(tan 30) (0.5)(9.8 m/s2)(113 m)2/(36.5 m/s cos 30)2
= (1.0 + 65.2 - 62.6) m = 3.6 m
5) Think about the answer,

The units work out correctly for a height (m)

It seems reasonable for the ball to be a little over 3m high
when it gets to the fence.

Answer: since the wall is 3m high, and the ball is 3.26m high
when it gets there, Sammy gets a homer.
Physics 151: Lecture 4, Pg 16
Typical questions :
(projectile motion; for given v0 and )


What is the maximum height
the ball reaches (h) ?
y
P
v0
h

L
How long does it take to reach
maximum height ?
P

x
y
Would the answers above be any
h
different if the projectile was moving
only along y-axis (1-D motion) with the
initial velocity: v0 sin () ?
v0 sin() x
( A ) YES
( B ) NO
( C ) CAN’T TELL
Physics 151: Lecture 4, Pg 17
Typical questions :
(projectile motion; for given v0 and )
y
v0
h
P

L
x

What is the range of the ball (L) ?

How long does it take for ball to reach final point (P) ?
Physics 151: Lecture 4, Pg 18
Lecture 4, ACT 2
Motion in 2D

Two footballs are thrown from the same point on a
flat field. Both are thrown at an angle of 30o above
the horizontal. Ball 2 has twice the initial speed of
ball 1. If ball 1 is caught a distance D1 from the
thrower, how far away from the thrower D2 will the
receiver of ball 2 be when he catches it ?
(a) D2 = D1
(b) D2 = 2D1
(c) D2 = 4D1
Physics 151: Lecture 4, Pg 19
Problem 2 (correlated motion of 2 objects in 3-D)
 Suppose a projectile is aimed at a target at rest
placed at the same height. At the time that the
projectile leaves the cannon the target is released
from rest and starts falling toward ground.
Would the projectile miss or hit the target ?
( A ) MISS
( B ) HIT
t=0
PROJECTILE
t=0
( C ) CAN’T TELL
v0
TARGET
x
t = t1
y
Physics 151: Lecture 4, Pg 20
Problem 3 (correlated motion of 2 objects in 3-D)
 Suppose a projectile is aimed at a target at rest
somewhere above the ground as shown in Fig. below.
At the same time that the projectile leaves the cannon
the target falls toward ground.
Would the projectile now miss or hit the target ?
t=0
TARGET
y
v0
t=0
t = t1

PROJECTILE
x
Physics 151: Lecture 4, Pg 21
Solution (Problem 3)

Prove that the projectile will hit the target
TARGET: y-component
yT = y0 + v0y t - 1/ 2 g t2
yT = h - 1/ 2 g t2
PROJECTILE: y-component
yP = y0 + v0y t - 1/ 2 g t2
yP = v0 sin() t - 1/ 2 g t2
yP = yT
if P hits T :
when (t) xP = D !
v0 sin() t =h
t=0
TARGET
but: t = D / v0x = D / v0 cos()
y
t=t
v0
t=0
h
so:
v0 sin() / v0 cos() = h/D
tan() = h / D

PROJECTILE
x
D

But this nothing else than the
condition that the projectile is
aimed at the target !
Physics 151: Lecture 4, Pg 22
Problem from previous Exam-1
Two balls, projected at different
times so they don’t collide, have
trajectories A and B, as shown.
Which statement is true.
1. Initial speed of ball B must be greater than that of ball A.
2. Ball A is in the air for a longer time than ball B.
3. Ball B is in the air for a longer time than ball A.
4. Ball B has a greater acceleration than ball A.
5. Ball A has a greater acceleration than ball B.
Physics 151: Lecture 4, Pg 23
Recap of today’s lecture

Kinematics in 2 and 3 dimensions, Chapter 4.1-3

Reading :
» Chapter 4: Sections 4-5
Solutions of Homework #1:
» Will be available on the web:

www.phys.uconn.edu/~dutta/151_2006

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
Homework#2 (due next Fri. 9/15 by 5.00 pm
Physics 151: Lecture 4, Pg 24