Physics 218 Alexei Safonov - PowerPoint Presentation

Slide1

Physics 218

Alexei Safonov

Lecture 2:

Units

and

Vectors

KinematicsSlide2

Checklist

Yesterday:Homework for Chapter 1 submitted via MasteringPre-lectures and check-points completed before 8:00AM todayWednesday:

Pre-lectures and checkpoints for 1-Dim motion

Sunday:

Homework for Chapter 2 dueSlide3

Today

Talk more about vectorsOperations with vectors: scalar and vector productsWill also revisit some of the checkpoints from the pre-lectures

Second half of the lecture:

Kinematics in 1 dimensional caseSlide4

Clickers Setup

Turn on your clicker (press the power button)Set the frequency:Press and hold the power button

Two letters will be flashing

If it’s not “BD”, press “B” and then “D”

If everything works, you should see

“Welcome” and “Ready”Slide5

Clicker Question 1

Do you have your i>clicker with you today?

Yes

No

Maybe

I like pudding

We use the “BD” frequency in this classSlide6

Math Prelecture

Math Preview:Most people got through it fine

Most concerns were about the “sine theorem”

View this as a test of your math skills:

Take action

to catch up in the areas which this review found to be problematic for youSlide7

Pre-Lecture Question

You look it up and find that there are 2.54 centimetres in one inchThe

motorcycle engine on a Kawasaki Ninja 1000 has a displacement of 1043 cubic-centimeters (cm3). In order to calculate its engine displacement in cubic-inches (in3) what unit conversion factor would you use to multiply the given displacement?

A. 1

in

3

/ 2.54 cm

3

B. 2.54

cm

3 / 1 in3 C. 1 in3 / 16.4 cm3 D. 16.4 cm3 / 1 in3 Slide8

Pre-Lecture Question

You look it up and find that there are 2.54 centimeters in one inch

The

motorcycle engine on a Kawasaki Ninja 1000 has a displacement of 1043 cubic-centimeters (cm

3

). In order to calculate its engine displacement in cubic-inches (in

3

) what unit conversion factor would you use to multiply the given displacement?

A. 1

in

3

/ 2.54 cm3 B. 2.54 cm3 / 1 in3 C. 1 in3 / 16.4 cm3 D. 16.4 cm3 / 1 in3 Slide9

Specifying a Vector

Two equivalent ways:

Components

V

x

and

V

y

Magnitude

V

and angle

qSwitch back and forthMagnitude of V |V| = (vx2 + vy2)½ Pythagorean Theoremtanq = vy /vxEither method is fine, pick one that is easiest for you, but be able to use bothSlide10

Unit Vectors

Another notation for vectors:

Unit Vectors denoted

i

,

j

,

k

x

z

y

^

j

^

i

^

kSlide11

Unit Vectors

Similar notations, but with

x, y, z

x

z

y

^

j

^

i

^

kSlide12

Vector in Unit Vector NotationSlide13

General Addition Example

Add two vectors using the

i

-hats,

j

-hats and

k

-hatsSlide14

Simple Multiplication

Multiplication of a vector by a scalarLet’s say I travel 1 km east. What if I had gone 4 times as far in the same direction?

→

Just stretch it out, multiply the magnitudes

Negatives:

Multiplying by a negative number turns the vector aroundSlide15

Subtraction

Subtraction is easy: It’s the same as addition but turning around one of the vectors. I.e., making a negative vector is the equivalent of making the head the tail and vice versa. Then add:Slide16

Vector Question

Vector A has a magnitude of 3.00 and is directed parallel to the negative y-axis and vector B has a magnitude of 3.00 and is directed parallel to the positive y-axis. Determine the magnitude and direction angle (as measured counterclockwise from the positive x-axis) of vector C, if C=A−B.

A. C=0.00

(its direction is undefined)

B. C

= 3.00;

θ = 270

o

C. C

= 3.00;

θ = 90o D. C = 6.00; θ = 270o F. C = 6.00; θ = 90oSlide17

Vector Question

Vector A has a magnitude of 3.00 and is directed parallel to the negative y-axis and vector B has a magnitude of 3.00 and is directed parallel to the positive y-axis. Determine the magnitude and direction angle (as measured counterclockwise from the positive x-axis) of vector C, if C=A−B.

A. C=0.00

(its direction is undefined)

B. C

= 3.00;

θ = 270

o

C. C

= 3.00;

θ = 90

o D. C = 6.00; θ = 270o F. C = 6.00; θ = 90oSlide18

First way:

Scalar Product

or

Dot Product

Why Scalar Product?

Because the result is a scalar (just a number)

Why a Dot Product?

Because we use the notation

A

.

BA.B = |A||B|CosQHow do we Multiply Vectors?Slide19

A

.

B = |A||B|Cos

Q

First Question:

x

z

y

^

j

^

i

^

kSlide20

A

.

B = |A||B|Cos

Q

First Question:

x

z

y

^

j

^

i

^

k

1

0

-1

unit vector

k

Slide21

Harder ExampleSlide22

Vector Cross Product

This is the last way of multiplying vectors we will see

Direction from the “right-hand rule”

Swing from A into B!Slide23

Vector Cross Product Cont…

Multiply out, but use the Sinq to give the magnitude, and RHR to give the direction

+

_

x

z

y

^

j

^

i

^

kSlide24

Cross Product ExampleSlide25

Vector Product

Calculate the vector

product

C

=

A

x

B

:

Bold font means vector (same as having an arrow on the top)

Vector A points in positive y direction and has magnitude of 3

Vector B points in negative x direction and has magnitude of 3Which is the correct way to calculate C?A. C = 3j x (-3i) = - 9k B. C = 3j x (-3i) = +9kC. C = 3 x (-3) x sin (90o) = - 9D. C = 3 x (-3) x sin (270o

) = + 9Slide26

Scalar Product

Calculate the scalar product A⋅B:

11.6

12.0

14.9

15.4

19.5Slide27

Vector Product

Calculate the scalar product A⋅B:

11.6

into the page

11.6 out of the page

12.0 into the page

12.0 out of the page

14.9 into the page

14.9 out of the page

15.4 into the page

15.4 out of the pageSlide28

Kinematics in 1 dimensionSlide29

Kinematics: Describing Motion

Interested in two key ideas:How

objects move as a function of time

Kinematics

Chapters 2 and 3

Why

objects move the way they do

Dynamics

Do this in Chapter 4 and laterSlide30

Chapter 2: Motion in 1-Dimension

Velocity & AccelerationEquations of Motion DefinitionsSome calculus (derivatives)

Wednesday:

More calculus (integrals)

ProblemsSlide31

Notes before we begin

This chapter is a good example of a set of material that is best learned by doing examples

We’ll do some examples today

Lots more next time…Slide32

Lecture Thoughts from FIP

I'm used to using math-based approaches to find velocity, acceleration, and position, so understanding how to use physics, and why we should use physics instead of simply deriving and integrating is fuzzy to me

All physics problems are math problems with boundary conditions. You need to understand physics to correctly set boundary conditions, so you have a well defined math problem. Then it’s all math.

Questions of reading, understanding and interpreting graphs and their relationship with formulas:

Lots and lots of questions, so we will heavily focus on that todaySlide33

Equations of Motion

We want Equations that describe:

Where

am I as a function of time?

How fast

am I moving as a function of time?

What direction

am I moving as a function of time?

Is my velocity changing? Etc.Slide34

Motion in One Dimension

Where is the car? X=0

feet

at

t

0

=

0 sec

X=22

feet at t1=1 secX=44 feet at t2=2 secWe say this car has “velocity” or “Speed”Plot position vs. time. How do we get the velocity from the graph?Slide35

Motion in One Dimension Cont…

Velocity:

“Change in position during a certain amount of time”

Calculate from the

Slope

:

The

“Change in position as a function of time”

Change in Vertical

Change in HorizontalChange: DVelocity  DX/DtSlide36

Constant Velocity

Equation of Motion for this example:

X = bt

Slope is constant

Velocity is constant

Easy to calculate

Same everywhere