unit 2: motion in 2d

Unit 2: Motion in 2D

Textbook:

Chapter 3 & Chapter 4

Unit Objectives: Motion Models

1. Recognize that an object in free fall will accelerate at a constant rate of 9.8 m/s2 downward near the surface of the earth.

Use kinematic equations to determine velocity or position at any time

2. Determine which model (constant velocity or constant acceleration, or varying acceleration) is appropriate to describe the horizontal and vertical component of motion of an object

Unit Objectives: Projectiles

3. Use appropriate kinematic equations to determine the position or velocity of a projectile at a specific point. Sketch the graph of motion for projectiles

a) y-x, y-t, x-t, vx – t, vy- t, ax-t, ay-t

4. Given information about the initial velocity and height of a projectile, determine a) time of flight, b) the point where a projectile lands, and c) velocity at impact

Unit Objectives: Vectors

5. Graphical representation of Vectorsa) Given a vector, draw its components b) Recognize the magnitude and direction of a vector from

a vector diagram c) Determine the sum of 2 or more vectors graphically

6. Numerical Analysis of Vectors a) Given the magnitude and direction of a vector, determine

the components using trigb) Given the components of a vector, determine the

magnitude and direction using Pythagorean Theorem and trig

c) Determine the sum of 2 or more vectors using Pythagorean Theorem and trig

d) Represent by using unit vectors i, j, & k.

Unit Objectives: Relative Motion

7. Use vectors to perform relative velocity calculations

Textbook:

Chapter 2: Section 26

Chapter 3 & Chapter 4

Free Fall Acceleration

Acceleration due to the force of Earth’s gravity

Acceleration due to gravity at the surface of the Earth is -9.8 m/s2. Negative because it points down.

Graphs: x-t & v-t

Free Fall on the Moon

Acceleration of a falling object is constant regardless of mass or density

In 1971, the commander of Apollo 15 confirmed this concept by dropping a hammer and a feather. Both hit the ground at the same time.

Free Fall – Key Points

1) At max height, velocity is zero.

2) At a given height, velocity up is equal to velocity down.

3) Time up equals time down

Vectors: How much & which way?

When describing motion, often the questions asked are “How far?” or “How fast?”

However, for a person that is lost, “which way?” becomes more valuable.

Vectors answer both questions:

1 – How much (magnitude)?

2 – Which way (direction)?

Scalars vs Vectors

Scalars have magnitude only Quantity of something Distance, speed, time, mass, temperature

Vectors have both magnitude and direction displacement, velocity, acceleration

R

headtail

Direction of Vectors

The direction of a vector is represented by the direction in which the ray points.

This is typically given by an angle. Can also be given by using unit vectors

Ax

Magnitude of Vectors

The magnitude of a vector is the size of whatever the vector represents.

The magnitude is represented by the length of the vector. Symbolically, the magnitude is often represented as │A │

AIf vector A represents a displacement of three kilometers to the north… B

Then vector B, which is twice as long, would represent a displacement of six kilometers to the north!

Polar Notation

Magnitude and direction of the vector are stated separately. Magnitude is a positive number and the angle

is made with the positive x-axis

v = 5 m/s at 135˚

Rectangular Notation

Defining a vector by its components y-component: vector projection parallel to y-axis x-component: vector projection parallel to x-axis

Vx

Vy

Converting Polar & Rectangular

A

B

R A + B = R

Graphical Addition of Vectors

Vectors are added graphically together head-to-tail. The sum is called the resultant. The inverse, is called the equilibrant .

Component Addition of Vectors

1) Resolve each vector into its x- and y-components.Ax = Acos Ay = AsinBx = Bcos By = Bsin

2) Add the x-components together to get Rx and the y-components to get Ry.

3) Use the Pythagorean Theorem to get the magnitude of the resultant.

4) Use the inverse tangent function to get the angle.

Sample: What is the value of “a” and “b”?

a = -3 & b = 10

Sample Problem

Add together the following graphically and by component, giving the magnitude and direction of the resultant and of the equilibrant. Vector A: 300 m @ 60o

Vector B: 450 m @ 100o

Vector C: 120 m @ -120o

Resultant: 599 m @ 1o

Equilibrant: 599 m @ 181o

Yet another sample!!!

Sprint (-6, -2) blocks

Unit Vectors

Unit vectors are quantities that specify direction only. They have a magnitude of exactly one, and typically point in the x, y, or z directions.

ˆ points in the x direction

ˆ points in the y direction

ˆ points in the z direction

i

j

k

Unit Vectors

z

y

x

i

jk

Unit Vectors

Instead of using magnitudes and directions, vectors can be represented by their components combined with their unit vectors.

Example: displacement of 30 meters in the +x direction added to a displacement of 60 meters in the –y direction added to a displacement of 40 meters in the +z direction yields a displacement of:

ˆˆ ˆ(30 -60 40 ) m

30,-60,40 m

i j k

Adding Vectors Using Unit Vectors

Simply add all the i components together, all the j components together, and all the k components together.

Sample Problem

Consider two vectors, A = 3.00 i + 7.50 j and B = -5.20 i + 2.40 j. Calculate C where C = A + B.

C = -2.20 i + 9.90 j

Sample Problem

You move 10 meters north and 6 meters east. You then climb a 3 meter platform, and move 1 meter west on the platform. What is your displacement vector? (Assume East is in the +x direction).

5 i + 10 j + 3 k

Suppose I need to convert unit vectors to a magnitude and direction?

Given the vector

2 2 2

ˆˆ ˆx y z

x y z

r r i r j r k

r r r r

Back to Sample Problem

You move 10 meters north and 6 meters east. You then climb a 3 meter platform, and move 1 meter west on the platform. How far are you from your starting point?

11.56 m

1 Dimension 2 or 3 Dimensions

x: position x: displacement v: velocity a: acceleration

r: position r: displacement v: velocity a: acceleration

r = x i + y j + z k r = x i + y j + z k v = vx i + vy j + vz k

a = ax i + ay j + az k

In Unit VectorNotation

Sample Problem

The position of a particle is given by r = (80 + 2t)i – 40j - 5t2k. Derive the velocity and acceleration vectors for this particle. What does motion “look like”?

v = 2 i - 10t k

a = -10 k

freefall

Another Sample

A position function has the form r = x i + y j with x = t3 – 6 and y = 5t - 3. What are the velocity and acceleration functions? What are the velocity and acceleration at t=2s?

v = 3t2 i + 5 j v(2) = 12 i + 5 j

a = 6t i a(2) = 12 i

Practice Problems

1- A baseball outfielder throws a long ball. The components of the position are x = (30 t) m and y = (10 t – 4.9t2) m Write vector expressions for the ball’s position, velocity, and

acceleration as functions of time. Use unit vector notation! Write vector expressions for the ball’s position, velocity, and

acceleration at 2.0 seconds.

2- A particle undergoing constant acceleration changes from a velocity of 4i – 3j to a velocity of 5i + j in 4.0 seconds. What is the acceleration of the particle during this time period? What is its displacement during this time period?

Projectiles

An object that moves in two dimensions under the influence of only gravity Accomplished by usually launching at an angle or

going off a flat surface with initial horizontal velocity.

Neglect air resistance Follow parabolic trajectory

Launch Angle

cosiix vv

siniiy vv

The components vix & viy are not necessarily positive. If an object is thrown downward, then viy is negative.

Projectiles & Acceleration

If you take an object and drop, it will fall straight down and not sideways ax = 0 &

ay=g = -9.8 m/s2

The vertical component of acceleration is just the familiar g of free fall while the horizontal is zero

Trajectory of Projectile

g

g

g

g

g

This shows the parabolic trajectory of a projectile fired over level ground.

Acceleration points down at 9.8 m/s2 for the entire trajectory.

Trajectory of Projectile

vx

vy

vy

vx

vx

vy

vx

vy

vx

The velocity can be resolved into components all along its path. Horizontal velocity remains constant; vertical velocity is accelerated.

Trajectory Path of a Projectile

Position graphs for 2-D projectiles. Assume projectile fired over level ground.

x

y

t

y

t

x

Acceleration graphs for 2-D projectiles. Assume projectile fired over level ground.

t

ay

t

ax

Lets think about this!!!

A heavy ball is thrown exactly horizontally at height h above a horizontal field. At the exact instant that ball is thrown, a second ball is simply dropped from height h. Which ball hits firsts? (demo-x-y shooter)

Two Independent Motions

1) Uniform motion at constant velocity in the horizontal direction

2) Free-Fall motion in the vertical direction

tvx x

atvv iyfy

2

2

1attvy iy

yavv iyfy 222

Remember…To work projectile problems… …resolve the initial velocity into components.

VoVo,y = Vo sin

Vo,x = Vo cos

Practice Problems

1) A soccer player kicks a ball at 15 m/s at an angle of 35o above the horizontal over level ground. How far horizontally will the ball travel until it strikes the ground?

2) A cannon is fired at 100m/s at an 15o angle above the horizontal from the top of a 120 m high cliff. How long will it take the cannonball to strike the plane below the cliff? How far from the base of the cliff will it strike?

3. Students at an engineering contest use a compressed air cannon to shoot a softball at a box being hoisted straight up at 10 m/s by a crane. The cannon, tilted upward at 30 degree angle, is 100 m from the box and fires by remote control the instant the box leaves the ground. Students can control the launch speed of the softball by setting air pressure. What launch speed should the students use to hit the box?

mx 57.21

smv /1.45

mx

st

88.286

97.2

Range Equation

Derive the range equation for a projectile fired over level ground.

g

vR i 2sin2

tvx i )cos( cosiv

xt

cosiv

Rt

2cos

22

2

cossin0

iv

gR

iv

Riv

22

1 gttvy iy

2sincos220 gRRiv gRiv 2sin20

g

vR i 2sin2

Acceleration in 2-D

A runner is going around a track. She is initially moving with a velocity vector of (0.00, -8.00) m/s and her constant acceleration is (1.10, 1.10) m/s2. What is her velocity 7.23 seconds later. Round the final velocity components to the nearest 0.01 m/s.

smjiv f /)05.0,95.7(

Multidimensional Motion - Calculus

What is the velocity function of the plane?

What is the velocity at t = 2 seconds?

Just like in 1-D, take the derivative of the position function, to get the velocity function.

Take the double derivative to find acelleration…

Unit Vectors & Calculus

Treat the same way as you do with one dimensional motion

Take the derivative or integral for each unit vector

Reference Frames

Coordinate system used to make observations.

The woman is using the surface of the Earth as her reference frame. She considers herself and the train platform to be

stationary, while the train is moving to the right with positive velocity.

Reference Frames cont.

If now, the perception of motion is from Ted’s point of view (man in the train). He uses the inside of the train as his reference frame. He sees other people in the train as stationary and objects outside the train moving back with a negative velocity.

Reference Frames

There is no right or wrong reference frame. Must be clear about which reference frame is

being used to assess motion.

Reference Frame Conditions

1. The frames are oriented the same, with the x and y axes parallel to each other

2. The origins of frame A & B coincide at t=0.

3. All motion is in the xy-plane, so we don’t need to consider the z-axis

4. The relative velocity (of the frames) is constant. (a = 0)

Inertial Reference Frames

Inertial Reference Frames

Classical Mechanics are only valid in inertial reference frames. In other words, all observers would measure

the same acceleration for a moving body.

We will discuss this in more detail when we talk about Newton’s Laws of Motion

Relative Velocity

v = 15 m/s

Another Sample

Practice Problems

VBS = 3.35 m/s at 63.4 degrees

Police Car Chasing

A motorist traveling west at 77.5 km/h is being chased by a police car traveling at 96.5 km/h. What is the velocity of the motorist relative to the police car?

hkmvMG /5.77

hkmvPG /5.96

hkmvGP /5.96

vvv GPMGMP

hkmvMP /19