© 2010 Pearson Education, Inc. Lecture Outline Chapter 3 College Physics, 7 th Edition Wilson /...

© 2010 Pearson Education, Inc.

Lecture Outline

Chapter 3

College Physics, 7th Edition

Wilson / Buffa / Lou

Chapter 3Motion in Two Dimensions


Units of Chapter 3• Components of Motion

• Vector Addition and Subtraction

• Projectile Motion

• Relative Velocity


3.1 Components of MotionAn object in motion on a plane can be located using two numbers—the x and y coordinates of its position. Similarly, its velocity can be described using components along the x- and y-axes.


3.1 Components of Motion

The velocity components are:

The magnitude of the velocity vector is:



The components of the displacement are then given by:

Note that the x- and y-components are calculated separately.



The equations of motion are:

When solving two-dimensional kinematics problems, each component is treated separately. The time is common to both.



If the acceleration is not parallel to the velocity, the object will move in a curve:


3.2 Vector Addition and Subtraction

Geometric methods of vector addition

Triangle method:


3.2 Vector Addition and SubtractionThe negative of a vector has the same magnitude but is opposite in direction to the original vector. Adding a negative vector is the same as subtracting a vector.



Vector Components and the Analytical Component Method

If you know A and B, here is how to find C:



The components of C are given by:

Equivalently,



Vectors can also be written using unit vectors (i, j, k):


3.2 Vector Addition and SubtractionVectors can be resolved into components and the components added separately; then recombine to find the resultant.


Non-Collinear VectorsWhen 2 vectors are perpendicular, you must use

the Pythagorean theorem.

kmc

c

bacbac

8.10912050

5595Resultant 22

22222

95 km,E

55 km, N

Start

Finish

A man walks 95 km, East then 55 km, north. Calculate his RESULTANT DISPLACEMENT.The hypotenuse in Physics

is called the RESULTANT.

The LEGS of the triangle are called the COMPONENTS

Horizontal Component

Vertical Component

What about the direction?

In the previous example, DISPLACEMENT was asked for and since it is a VECTOR we should include a DIRECTION on our final answer.

NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE.

N

S

EW

N of E

E of N

S of W

W of S

N of W

W of N

S of E

E of S

N of E

What about the VALUE of the angle?

Just putting North of East on the answer is NOT specific enough for the direction. We MUST find the VALUE of the angle.

30)5789.0(

5789.095

55

1

Tan

sideadjacent

sideoppositeTan

N of E

55 km, N

95 km,E

To find the value of the angle we use a Trig function called TANGENT.

109.8 km

So the COMPLETE final answer is : 109.8 km, 30 degrees North of East

What if you are missing a component?Suppose a person walked 65 m, 25 degrees East of North. What

were his horizontal and vertical components?

EmCHopp

NmCVadj

hypopphypadj

hypotenuse

sideopposite

hypotenuse

sideadjacent

,47.2725sin65..

,91.5825cos65..

sincos

sinecosine

65 m25

H.C. = ?

V.C = ?

The goal: ALWAYS MAKE A RIGHT TRIANGLE!

To solve for components, we often use the trig functions since and cosine.

ExampleA bear, searching for food wanders 35 meters east then 20 meters north.

Frustrated, he wanders another 12 meters west then 6 meters south. Calculate the bear's displacement.

3.31)6087.0(

6087.23

14

93.262314

1

22

Tan

Tan

mR

35 m, E

20 m, N

12 m, W

6 m, S

- =23 m, E

- =14 m, N

23 m, E

14 m, N

The Final Answer: 26.93 m, 31.3 degrees NORTH of EAST

R

ExampleA boat moves with a velocity of 15 m/s, N in a river which

flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north.

1.28)5333.0(

5333.015

8

/17158

1

22

Tan

Tan

smRv

15 m/s, N

8.0 m/s, W

Rv

The Final Answer : 17 m/s, @ 28.1 degrees West of North

ExampleA plane moves with a velocity of 63.5 m/s at 32 degrees South of East. Calculate

the plane's horizontal and vertical velocity components.

SsmCVopp

EsmCHadj

hypopphypadj

hypotenuse

sideopposite

hypotenuse

sideadjacent

,/64.3332sin5.63..

,/85.5332cos5.63..

sincos

sinecosine

63.5 m/s

32

H.C. =?

V.C. = ?

ExampleA storm system moves 5000 km due east, then shifts course at 40

degrees North of East for 1500 km. Calculate the storm's resultant displacement.

NkmCVopp

EkmCHadj

hypopphypadj

hypotenuse

sideopposite

hypotenuse

sideadjacent

,2.96440sin1500..

,1.114940cos1500..

sincos

sinecosine

91.8)364.0(

157.01.6149

2.964

14.62242.9641.6149

1

22

Tan

Tan

kmR

5000 km, E

40

1500 km

H.C.

V.C.

5000 km + 1149.1 km = 6149.1 km

6149.1 km

964.2 kmR

The Final Answer: 6224.14 km @ 8.91 degrees, North of East

3.3 Projectile MotionA projectile launched in an arbitrary direction may have initial velocity components in both the horizontal and vertical directions, but its acceleration is still downward.

Projectile - Any object which projected by some means and continues to move due to its own inertia (mass). Its parabolic in path of motion.

3.3 Projectile Motion

An object projected horizontally has an initial velocity in the horizontal direction, and acceleration (due to gravity) in the vertical direction. The time it takes to reach the ground is the same as if it were simply dropped.




Horizontally launched vs dropped.

See how the x component is different from the y?

This is ALWAYS true for projectile motion.

Solve the x separate from the y.

3.3 Projectile MotionThe vertical motion is the same as if the object were thrown straight up or down with the same initial y velocity, and the horizontal velocity is constant.



The range of a projectile is maximummaximum (if there is no air resistance) for a launch angle of 45°.

What else do you notice?


3.3 Projectile MotionWith air resistance, the range is shortened, and the maximum range occurs at an angle a little less than 45°. Its still projectile motion but now the relationship is not parabolic. Complex algebra or calculus.


Horizontally Launched ProjectilesProjectiles which have NO upward trajectory and NO initial

VERTICAL velocity.

0 /oyv m s

constantox xv v

Horizontally Launched ProjectilesTo analyze a projectile in 2 dimensions we need 2

equations. One for the “x” direction and one for the “y” direction. And for this we use kinematic #2.

212oxx v t at

oxx v t

Remember, the velocity is CONSTANT horizontally, so that means the acceleration is ZERO. Range is only solved for in the x direction. Constant velocity -> Essentially DIRT! d=rt.

212y gt

Remember that since the projectile is launched horizontally, the INITIAL VERTICAL VELOCITY is equal to ZERO. Time is always solved for in the y direction ONLY.

Horizontally Launched Projectiles

Example: A plane traveling with a horizontal velocity of 100 m/s is 500 m above the ground. At some point the pilot decides to drop some supplies to designated target below. (a) How long is the drop in the air? (b) How far away from point where it was launched will it land?

What do I know?

What I want to know?

vox=100 m/s t = ?y = 500 m x = ?voy= 0 m/s

g = -9.8 m/s/s

2 2

2

1 1500 ( 9.8)2 2

102.04

y gt t

t t

10.1 seconds(100)(10.1)oxx v t 1010 m

Vertically Launched Projectiles

Component Magnitude Direction

Horizontal Constant Constant

Vertical Decreases up, 0 @ top, Increases down

Changes

Horizontal Velocity is constant

Vertical Velocity decreases on the way upward

Vertical Velocity increases on the way down

NO Vertical Velocity at the top of the trajectory.

Vertically Launched ProjectilesSince the projectile was launched at a angle, the

velocity MUST be broken into components!!!

cos

sinox o

oy o

v v

v v

vo

vox

voy

Vertically Launched ProjectilesThere are several

things you must consider when doing these types of projectiles besides using components. If it begins and ends at ground level, the “y” displacement is ZERO: y = 0

Vertically Launched ProjectilesYou will still use kinematic #2, but YOU MUST use

COMPONENTS in the equation.

cos

sinox o

oy o

v v

v v

vo

vox

voy

oxx v t 212oyy v t gt

ExampleA place kicker kicks a football with a velocity of 20.0 m/s

and at an angle of 53 degrees.

(a) How long is the ball in the air?

(b) How far away does it land?

(c) How high does it travel?

cos

20cos53 12.04 /

sin

20sin 53 15.97 /

ox o

ox

oy o

oy

v v

v m s

v v

v m s

v o=20

.0 m

/s

ExampleA place kicker kicks a

football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(a) How long is the ball in the air?

What I know What I want to know

vox=12.04 m/s t = ?voy=15.97 m/s x = ?

y = 0 ymax=?

g = - 9.8 m/s/s

2 2

2

1 0 (15.97) 4.92

15.97 4.9 15.97 4.9

oyy v t gt t t

t t t

t

3.26 s

Example

A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(b) How far away does it land?


vox=12.04 m/s t = 3.26 svoy=15.97 m/s x = ?

y = 0 ymax=?

g = - 9.8 m/s/s

(12.04)(3.26)oxx v t 39.24 m

Example

A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(c) How high does it travel?

CUT YOUR TIME IN HALF!


vox=12.04 m/s t = 3.26 svoy=15.97 m/s x = 39.24 m

y = 0 ymax=?

g = - 9.8 m/s/s

2

2

12

(15.97)(1.63) 4.9(1.63)

oyy v t gt

y

y

13.01 m

3.4 Relative Velocity

Velocity may be measured in any inertial reference frame. At top, the velocities are measured relative to the ground; at bottom they are measured relative to the white car.


3.4 Relative VelocityIn two dimensions, the components of the velocity, and therefore the angle it makes with a coordinate axis, will change depending on the point of view.


What if the runner’s velocity is vbr?

What if the walker’s velocity is vbs?

Review of Chapter 3• Two-dimensional motion is analyzed by considering each component separately. Time is the only common factor and only solved for in y.


Review of Chapter 3

• Vector components:


Review of Chapter 3

• Range is the maximum horizontal distance traveled.

• Relative velocity is expressed relative to a particular reference frame.


© 2010 Pearson Education, Inc. Lecture Outline Chapter 3 College Physics, 7 th Edition Wilson /...

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