1. Physics 121, Sections 9, 10, 11, and 12 Lecture 5
Today’s Topics:
Homework 2: Due Friday Sept. 16 @ 6:00PM
Ch.3: # 2, 11, 18, 20, 25, 32, 36, 46, 50, and 56.
Chapter 3: Forces and motion along a line
Motion with constant acceleration
Falling objects
Apparent weight
Chapter 4: Motion in 2-D
Vectors
Kinematics
Projectile motion 1

2. Recap: constant acceleration in 1-D
For constant acceleration:
From which we know:

3. Lecture 5 ACT 1 1D Freefall
Alice and Bill are standing at the top of a cliff of height H. Both throw a ball with initial speed v0, Alice straight down and Bill straight up. The speed of the balls when they hit the ground are vA and vB respectively.
Which of the following is true:
vA < vB (b) vA = vB (c) vA > vB v0
Bill
Alice H vA vB

4. About Air resistance ? When a body moves through a fluid
A drag force due to friction take place
Like friction, its direction is opposite to the motion
This force increases dramatically with speed
It is proportional to v2 Fd mg v
For an object falling through air
As gravity accelerates it, v increases
When the magnitude of Fd equals the weight mg
Not net force  v becomes constant
The object has reached its terminal speed vt

5. Air resistance When a body reaches vt Fdmg vt
When a body reaches vt
Both forces are equal in magnitude
So the coefficient b is
And
The terminal speed vt varies mass, shape, size
Feather: m/s
Raindrop: m/s
Skydiver: m/s (spread-eagle)
Skydiver: m/s (diving)

6. Apparent weight The weight is read by a scale a
It is given by the magnitude of the normal force
The block pushes on the scale with a force equal but opposite to N (action-reaction) a N mg -N
The normal is found from the net force
If platform (and object+scale) is not accelerated
Net force is zero

7. Apparent weight N a If the “elevator” is accelerated Net force  0mg
If the “elevator” is accelerated
Net force  0
Choosing axis as illustrated a N mg -N
Apparent weight is
W’ >mg if a>0 (upward): “heavier”
W’ <mg if a>0 (downward): “lighter”

8. Chapter 4: Vectors
In 1 dimension, we can specify direction with a + or - sign.
In 2 or 3 dimensions, we need more than a sign to specify the direction of something:
To illustrate this, consider the position vector r in 2 dimensions.
Example: Where is Boston?
Choose origin at New York
Choose coordinate system
Boston is 212 miles northeast of New York or
Boston is 150 miles north and 150 miles east of New York
Boston
New York r

9. Vectors...
There are two common ways of indicating that something is a vector quantity:
Boldface notation: A
“Arrow” notation:
A = A A

10. Vectors: definition A A = C B C A = B, B = C
A vector is composed of a magnitude and a direction
examples: displacement, velocity, acceleration
magnitude of A is designated |A|
usually carries units
A vector has no particular position
Two vectors are equal if their directions and magnitudes match. A B C
A = C
A = B, B = C

11. Vectors and scalars: B A = -0.75 B A A scalar is an ordinary number.
a magnitude without a direction
may have units (kg) or be just a number
usually indicated by a regular letter, no bold face and no arrow on top.
Note: the lack of specific designation of a scalar can lead to confusion
The product of a vector and a scalar is another vector in the same direction but with modified magnitude. B
A = B A

12. Lecture 5, ACT 2 Vectors and Scalars
While I conduct my daily run, several quantities describe my condition
Which of the following is not a vector ?
(For bonus points, which answer has a reasonable magnitude listed ?)
A) my velocity (3 m/s)
B) my acceleration
downhill (30 m/s2)
C) my destination
(the pub - 100,000 m)
D) my mass (150 kg)

13. Converting Coordinate Systems
In circular coordinates the vector r = (r,q)
In Cartesian the vector r = (rx,ry) = (x,y)
We can convert between the two as follows: y x
(x,y)  r ry rx
arctan( y / x )
rx = x = r cos 
ry = y = r sin 
The magnitude (length) of r is found using Pythagoras’ theorem: r y x
r=|r| is the magnitude of the vector (does not depend on direction)

14. Vector addition: A = B + C B B A C C
The sum of two vectors is another vector.
A = B + C B B A C C

15. Vector subtraction: B - C = B + (-1)C B B - C -C C B
Vector subtraction can be defined in terms of addition.
B - C
= B + (-1)C B C B -C
B - C

16. Unit Vectors: A Unit Vector is a vector having length 1 and no units.
It is used to specify a direction.
Unit vector u points in the direction of U.
Often denoted with a “hat”: u = û U û x y z i j k
Useful examples are the cartesian unit vectors [ i, j, k ]
point in the direction of the x, y and z axes.
R = rxi + ryj + rzk

17. Vector addition using components:
Consider C = A + B.
(a) C = (Ax i + Ay j ) + (Bx i + By j ) = (Ax + Bx )i + (Ay + By )j
(b) C = (Cx i + Cy j )
Comparing components of (a) and (b):
Cx = Ax + Bx
Cy = Ay + By By C B Bx A Ay Ax

18. Lecture 5, ACT 3 Vector Addition
Vector B = {3,0}
Vector C = {1,-4}
What is the resultant vector, D, from adding A+B+C?
(a) {3,-4} (b) {4,-2} (c) {5,-2}

19. Review (1-D): vav A few other useful formulas :
For constant acceleration we found: x a v t
vav
A few other useful formulas :

20. 2-D Kinematics
For 2-D, we simply apply the 1-D equations to each of the component equations.
Which can be combined into the vector equations:

21. 2-D Kinematics So for constant acceleration we get: a = const
v = v0 + a t
r = r0 + v0 t + 1/2 a t2
(where a, v, v0, r, r0, are all vectors)

22. 3-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.

23. “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.

24. Projectile Motion.
If I set something moving near the earth, it reduces to a 2-D 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 /2 y positive upwards

25. Lecture 5, ACT 4 2-D Motion Alice Bill C) B) A)
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
Bill C) B) A)

26. Problem:
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.

27. Problem:
We need to find how high the ball is at a distance of 113m away from where it starts.  v h D yo

28. Problem: 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  v h D y x

29. Problem... Variables vo = 36.5 m/s yo = 1 m h = 3 m qo = 30º D = 113 m
a = (0,ay)  ay = -g
t = unknown,
Yf – height of ball when x=113m, unknown,
our target

30. Problem... For projectile motion, Equations of motion are:
vx = v0x vy = v0y - g t
x = vx t y = y0 + v0y t - 1/ 2 g t2 y x g  v
v0x
v0y y0
And, use geometry to find vox and voy
Find v0x = |v| cos .
and v0y = |v| sin .

31. Problem... Solve the problem,
The time to reach the wall is: t = D / vx (easy!)
Height at any time: y(t) = y0 + v0y t - g t2/ 2
Combining the two gives, y(t) = y0 + v0y D/vox - g D2/ (2vox2)
And substitute for vox and voy: y(t) = y0 + D tanq - g D2/ 2(vocosq)2
All are known quantities. Solved.
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
= ( ) m = 3.6 m

32. Problem...
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.
But, we haven’t yet answered the question
Since the wall is 3m high, and the ball is 3.26m high when it gets there, Sammy gets a homer.

33. Recap of today’s lecture
Homework 2: Due Friday Sept. 16 @ 6:00PM
Ch.3: # 2, 11, 18, 20, 25, 32, 36, 46, 50, and 56.
Chapter 3: Forces and motion along a line
Motion with constant acceleration
Falling objects
Apparent weight
Chapter 4: Motion in 2-D
Vectors
Kinematics
Projectile motion 27