Wednesday, August 27, 2008

Vectors Review

Announcements

  Lab fee? Chris, Liam, Greg?

Lab books due today with Kinematics Graphing lab.

Scalars vs Vectors

  Scalars have magnitude only  Distance, speed, time, mass Vectors have both magnitude and direction  displacement, velocity, acceleration tail 

R

head

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.

A

 x

Magnitude of Vectors

A    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 │ If vector A represents a displacement of three miles to the north… Then vector B, which is twice as long, would represent a displacement of six miles to the north!

B

Equal Vectors

 Equal vectors have the same length and direction, and represent the same quantity (such as force or velocity).

Inverse Vectors

 Inverse vectors have the same length, but opposite direction.

A -A

Graphical Addition of Vectors

   Vectors are added graphically together head-to tail.

The sum is called the resultant.

The inverse of the sum is called the equilibrant

B A

A + B = R

R

Component Addition of Vectors

1) 2) 3) 4) Resolve each vector into its x- and y components.

A x = Acos  A y = Asin  B x = Bcos  B y = Bsin  etc.

Add the x-components together to get R x and the y-components to get R y .

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

Use the inverse tangent function to get the angle.

•

Sample problem:

Add together the following graphically and by component, giving the magnitude and direction of the resultant and the equilibrant.

– Vector A: 300 m @ 60 o – Vector B: 450 m @ 100 o – Vector C: 120 m @ -120 o

Thursday, August 28, 2008

Unit Vectors

Announcements

  Lab fee? Chris, Liam, Greg?

HW Quiz: Chapter 2  Problems 30, 36, 39, 42, 49, 52  Roll the die!

Consider Three Dimensions

Polar Angle z Azimuthal Angle x a x a z f 

a

a y y xy Projection

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.

i

ˆ points in the x direction

j

ˆ points in the y direction

k

ˆ points in the z direction

Unit Vectors

z x

i k j

y

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:

i

ˆ

j

ˆ 

k

30,-60,40 m

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

.

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).

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

 Given the vector

r



r i x

ˆ 

r j y

ˆ 

r k z

ˆ

r



r x

2

r y r z

2

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?

Friday, August 29, 2008

Position, Velocity, and Acceleration Vectors in Multiple Dimensions

1 Dimension 2 or 3 Dimensions     x: position  x: displacement v: velocity a: acceleration In Unit Vector Notation     r: position  r: displacement v: velocity a: acceleration     r = x i + y j + z k  r =  x i +  y j +  z k v = v x a = a x i + v i + a y y j + v j + a z z

k k

Sample problem:

The position of a particle is given by

r

= (80 + 2t)

i

– 40

j

- 5t 2

k.

Derive the velocity and acceleration vectors for this particle. What does motion “look like”?

j Sample problem:

with x = t 3 A position function has the form

r

– 6 and y = 5t - 3. = x

i

+ y a) Determine the velocity and acceleration functions.

b) Determine the velocity and speed at 2 seconds.

Miscellaneous

   Let’s look at some video analysis.

Let’s look at a documentary.

Homework questions?

Tuesday, September 2, 2008

Multi-Dimensional Motion with Constant (or Uniform) Acceleration

Sample Problem:

A baseball outfielder throws a long ball. The components of the position are x = (30 t) m and y = (10 t – 4.9t

2 ) m a) Write vector expressions for the ball’s position, velocity, and acceleration as functions of time.

Use unit vector notation!

b) Write vector expressions for the ball’s position, velocity, and acceleration at 2.0 seconds.

Sample problem:

a velocity of 4

i

A particle undergoing constant acceleration changes from – 3

j

to a velocity of 5

i

displacement during this time period?

+

j

in 4.0 seconds. What is the acceleration of the particle during this time period? What is its

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/s trajectory.

2 for the entire

Trajectory of Projectile

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

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

y x x t t

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

V y V x t t

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

a y a x t t

Remember… To work projectile problems…

 …resolve the initial velocity into components.

V

o

V

o,y

= V

o

sin

  V o,x = V o cos 

Sample problem:

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

Sample problem:

A cannon is fired at a 15 o 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?

Sample problem:

y

 derive the trajectory equation.

x

 2

v o

2

g

cos 2    

x

2

Sample problem:

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

R

 2

v o

2

g



Sample problem:

Show that maximum range is obtained for a firing angle of 45 o .

R

 2

v o

2

g



Wednesday, September 3, 2008 Monkey Gun

Announcements

 Homework policy change.

Will the projectile always hit the target presuming it has enough range? The target will begin to fall as soon as the projectile leaves the gun.

Friday, September 5, 2008 Review of Uniform Circular Motion

Uniform Circular Motion

   Occurs when an object moves in a circle without changing speed. Despite the constant speed, the object’s velocity vector is continually changing; therefore, the object must be accelerating.

The acceleration vector is pointed toward the center of the circle in which the object is moving, and is referred to as centripetal acceleration.

Vectors in Uniform Circular Motion

v v a a

a = v

2

/ r

a a v v

Sample Problem

The Moon revolves around the Earth every 27.3 days. The radius of the orbit is 382,000,000 m. What is the magnitude and direction of the acceleration of the Moon relative to Earth?

 Sample problem: Space Shuttle astronauts typically experience accelerations of 1.4 g during takeoff. What is the rotation rate, in rps, required to give an astronaut a centripetal acceleration equal to this in a simulator moving in a 10.0 m circle.

Wednesday, September 10, 2008 Radial and Tangential Acceleration

Tangential acceleration

   Sometimes the speed of an object in circular motion is not constant (in other words, it’s not uniform circular motion).

An acceleration component is tangent to the path, aligned with the velocity. This is called tangential acceleration.

The centripetal acceleration component causes the object to continue to turn as the tangential component causes the radius or speed to change.

Tangential Acceleration

tangential component (a T ) v If tangential acceleration exists, the orbit is not stable.

a radial or centripetal component (a r or a c )

Sample Problem: Given the figure at right and tangential acceleration components if o behaving?

2  =  5.00 m

a

Sample problem: Suppose you attach a ball to a 60 cm long string and swing it in a vertical circle. The speed of the ball is 4.30 m/s at the highest point and 6.50 m/s at the lowest point. Find the acceleration of the ball at the highest and lowest points.

Sample problem:

A car is rounding a curve on the interstate, slowing from 30 m/s to 22 m/s in 7.0 seconds. The radius of the curve is 30 meters. What is the acceleration of the car?

Thursday, September 11, 2008 Relative Motion

Relative Motion

  When observers are moving at constant velocity relative to each other, we have a case of relative motion.

The moving observers can agree about some things, but not about everything, regarding an object they are both observing.

Consider two observers and a particle. Suppose observer B is moving relative to observer A.

P

particle

A

observer

B

observer

v

rel

Also suppose particle P is also moving relative to observer A.

P

v

A particle In this case, it looks to A like P is moving to the right at twice the speed that B is moving in the same direction.

A

observer

B

observer

v

rel

However, from the perspective of observer B… P

v

B

v

A particle it looks like P is moving to the right at the same speed that A is moving in the opposite direction, and this speed is half of what A reports for P.

-v

rel

A

observer

B

observer

v

rel

The velocity measured by two observers depends upon the observers’ velocity relative to each other.

P

v

B

v

A

-v

rel

A

v

B

v

observer A particle

= v

A

= v

B

– v

rel

+ v

rel

B

observer

v

rel

Sample problem:

Now show that although velocity of the observers is

different

, the acceleration they measure for a third particle is

the same

provided v

v

B =

v

A rel

v

rel is constant. Begin with

Galileo’s Law of Transformation of Velocities

 If observers are moving but not accelerating relative to each other, they agree on a third object’s acceleration, but not its velocity!

Inertial Reference Frames

  Frames of reference which may move observers find the same value for the acceleration of a third moving particle.

Inertial reference frames are moving at constant velocity relative to each other. It  at rest.

Newton’s Laws hold only in inertial

Sample problem:

How long does it take an automobile traveling in the left lane at 60.0km/h to pull alongside a car traveling in the right lane at 40.0 km/h if the cars’ front bumpers are initially 100 m apart?

Sample problem:

A pilot of an airplane notes that the compass indicates a heading due west. The airplane’s speed relative to the air is 150 km/h. If there is a wind of 30.0 km/h toward the north, find the velocity of the airplane relative to the ground.