Week 4 Assignments: - HW #3 Q13,Q15,Q18, P26, P28, P37, P41, P42, P44, P58

- MasteringPhysics: - Assignment #3

Week 4 Reading: Chapter 4 - Giancoli

This Week’s Announcements:

Lecture 7: 2/3-D Motion

* HW #2 due today by the end of class

Due: 9/15

* HW #1 will be returned in recitation

Vector Algebra - Vector Addition (2D):

B A+B

Ax Bx

By

Ay

A = Ax i + Ay j

B = Bx i + By j A+B = (Ax +Bx )i + (Ay+By )j

θ

A

- components add properly when the tail of vector B is attached to the head of vector A and the resultant vector (A+B) is drawn from the tail of A to the head of B

^ i ^ j k ^ , , - unit vectors: vector of length 1 along the x, y and z axes respectively

i j

Clicker Question:

1) Vectors A and B are labeled below. Which answer correctly represents B + A?

a)

b) e)

d)

c)

A B

Clicker Question:

1) Vectors A and B are labeled below. Which answer correctly represents B + A?

a)

b) e)

d)

c)

A B

A B

Vector Algebra

- Vector Subtraction:

A

B A+B

Ax -Bx -By

Ay - components subtract properly when the tail of vector B is attached to the tail of vector A and the resultant vector (A-B) is drawn from the head of B to the head of A

A = Ax i + Ay j

B = Bx i + By j

A-B = (Ax -Bx )i + (Ay-By )j

θ

- subtraction is just addition of the negative

- the negative of a vector is just the vector pointed in the opposite direction

-B

A-B

A B

A-B

Vector Algebra

- Vector Multiplication (by a number, n):

- n x vector = same vector just lengthen by n

A: 3A:

Clicker Question:

2) Vectors A and B are labeled below. Which answer correctly represents A - 3B?

a)

b)

e)

d)

c)

A

B A

3B

A - 3B

Clicker Question:

2) Vectors A and B are labeled below. Which answer correctly represents A - 3B?

a)

b)

e)

d)

c)

A

B A

3B

A - 3B

Example: Vector Subtraction

- Uniform circular motion:

v4 v3

r3 r4

Δv21 Δv21 = v2 - v1

Δv43

- even with |v| not changing there still is an acceleration - direction of acceleration is always towards the center (centripedal)

- |a| = v2/r

Δr43

V:

R: r

r Δr

Δv v

Δv43 = v4 - v3

sinθ = Δv/v

sinθ = Δr/r

Δv21 Δt

= a21

θ

θ

θ

θ

θ

- divide by Δt:

θ

Δr21 r1

r2

v2

v1

Projectiles: Velocity (Vector)

v0

v0

v

v

v

gt

gt

gt

#1:

#1

#2

#3

#2:

#3: v0

v

gt

v0 v gt

v0

v v0 gt

v = v0 + (- gt)

{ {

vector 1 + vector 2

- Vectors provide a shorthand to manipulate all dimensions simultaneously

Projectiles: Velocity

v0x

v0y

vx

vx

vx

vy

vy vy

v0

v

v v

#1

#2

#3 v0x v0x v0x v0x

#1 #2 #3

#1

#2

#3

vx = v0x vx = v0x vx = v0x vy = v0y-gt vy = v0y-gt = 0 vy = v0y-gt

- Couple the two together by the time

#2: #3: #1:

Neglecting air resistance

+

- Do two separate 1-D kinematics problems

vy

vy

vy

Projectiles: Position

X:

Y:

v0x

v0y v0

(No acceleration in this direction)

x = x0 + v0x t x = v0 cos(θ) t

X:

Y:

θ

(constant acceleration due to gravity in this direction)

⇒

y = y0 + v0y t + 1/2 ay t2 y-y0 = v0 sin(θ) t - 1/2 g t2 ⇒

y - y0 = tan(θ) x - x2 g

2 v02 cos2(θ)

Parabolic in SPACE

X

Y The time is the same for both dimensions

(x0 = 0)

(x0 = 0)

Clicker Question:

3) Ice skaters, ballet dancers and basketball players executing vertical leaps often give the illusion of “hanging” almost motionless near the top of the leap. To see why this is, consider a leap that takes an athlete up a vertical distance h. Of the total time spent in the air, what fraction of it is spent in the upper half (x > 1/2 h)?

a) 25 %

b) 33.3 %

e) 75 %

d) 70.7 %

c) 50 %

Clicker Question:

3) Ice skaters, ballet dancers and basketball players executing vertical leaps often give the illusion of “hanging” almost motionless near the top of the leap. To see why this is, consider a leap that takes an athlete up a vertical distance h. Of the total time spent in the air, what fraction of it is spent in the upper half (x > 1/2 h)?

a) 25 %

b) 33.3 %

e) 75 %

d) 70.7 %

c) 50 %

!

Range = x for a given launch angle θ and initial velocity, v0

θ

y-y0 = tan(θ) x - x2 g

2 v02 cos2(θ)

v0

- to find distance traveled solve for y-y0 = 0 (assumes object launches and lands at the same level)

R = (v02/g) x 2 cos(θ)sin(θ)

R = (v02/g) x sin(2θ)

- optimal launch angle = 45o

Projectiles: Position

- again only true when object launches and lands at the same level)