Physics 218

transcript

Physics 218

Alexei Safonov

Lecture 2: • Units and Vectors

• Kinematics

Checklist

• Yesterday:– Homework for Chapter 1 submitted via

Mastering– Pre-lectures and check-points completed

before 8:00AM today• Wednesday:

– Pre-lectures and checkpoints for 1-Dim motion

• Sunday:– Homework for Chapter 2 due

• Talk more about vectors– Operations with vectors: scalar and vector

products– Will also revisit some of the checkpoints from

the pre-lectures• Second half of the lecture:

– Kinematics in 1 dimensional case

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”

Clicker Question 1• Do you have your i>clicker with you today?

A) Yes

C) Maybe

D) I like pudding

We use the “BD” frequency in this class

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 you

Pre-Lecture Question• You look it up and find that there are 2.54 centimetres in one inch

• The 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 in3 / 2.54 cm3 B. 2.54 cm3 / 1 in3 C. 1 in3 / 16.4 cm3 D. 16.4 cm3 / 1 in3

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

Specifying a Vector• Two equivalent ways:

– Components Vx and Vy– Magnitude V and angle q

• Switch back and forth– Magnitude of V |V| = (vx

2 + vy2)½

• Pythagorean Theorem– tanq = vy /vx

• Either method is fine, pick one that is easiest for you, but be able to use both

Unit Vectors

Another notation for vectors:– Unit Vectors denoted i, j, k

kV jV iV V

direction z in the 1 means ˆdirectiony in the 1 means ˆdirection x in the 1 means ˆ

Unit VectorsSimilar notations, but with x, y, z

zV yV xV V

k as same theis ˆ

j as same theis ˆ

i as same theis ˆ

Vector in Unit Vector Notation

jΘ |V| iΘ |V| V

j V i V V

Θ|V|||VΘ|V|||V

ˆsinˆcos

sincos

General Addition Example

• Add two vectors using the i-hats, j-hats and k-hats

k km 0 j km 5 i km 10 D

Simple Multiplication

• Multiplication of a vector by a scalar–Let’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 around

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:

)V(- V V V 1212

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 = 270o C. C = 3.00; θ = 90o D. C = 6.00; θ = 270o F. C = 6.00; θ = 90o

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 = 270o C. C = 3.00; θ = 90o D. C = 6.00; θ = 270o F. C = 6.00; θ = 90o

• 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.B

• A.B = |A||B|CosQ

How do we Multiply Vectors?

A.B = |A||B|CosQ

First Question:

?j i isWhat

A.B = |A||B|CosQ

First Question:

?j i isWhat

D) unit vector k

Harder Example

notation?Vector Unit using BA isWhat

j B i B B

j A i A A

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!

SinB ACB A C

Vector Cross Product Cont…• Multiply out, but use

the Sinq to give the magnitude, and RHR to give the direction

)1(sin ˆˆˆ)1(sin ˆˆˆ)0(sin 0ˆˆ

SinB ACB A C

Cross Product Example

notation?Vector Unit using BA isWhat

j B i B B

j A i A A

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 3

• Which 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) = + 9

Scalar Product• Calculate the scalar product A B:⋅

A. 11.6B. 12.0C. 14.9D. 15.4E. 19.5

Vector Product

• Calculate the scalar product A B:⋅

A. 11.6 into the pageB. 11.6 out of the pageC. 12.0 into the pageD. 12.0 out of the pageE. 14.9 into the pageF. 14.9 out of the pageG. 15.4 into the pageH. 15.4 out of the page

KINEMATICS IN 1 DIMENSION

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 later

Chapter 2: Motion in 1-Dimension

• Velocity & Acceleration– Equations of Motion – Definitions– Some calculus (derivatives)

• Wednesday:– More calculus (integrals)– Problems

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…

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 today

Equations of MotionWe 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.

Motion in One Dimension• Where is the car?

– X=0 feet at t0=0 sec– X=22 feet at t1=1 sec– X=44 feet at t2=2 sec

• We say this car has “velocity” or “Speed”

• Plot position vs. time. How do we get the velocity from the graph?

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 Horizontal

• Change: D• Velocity DX/Dt

Constant Velocity

Equation of Motion for this example:

X = bt• Slope is constant• Velocity is constant

–Easy to calculate–Same everywhere

Physics 218

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