motion in two or three dimensions - education.fcps.org 3.pdfintroduction • when the race car goes...

Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley

PowerPoint® Lectures forUniversity Physics, Twelfth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by James Pazun

Chapter 3

Motion in Two or Three Dimensions


Goals for Chapter 3

• To study position, velocity, and acceleration vectors

• To apply position, velocity, and acceleration insights to projectile motion

• To extend our linear investigations to uniform and non-uniform circular motion

• To investigate relative velocity


Homework

3.12, 3.21, 3.33, 3.41, 3.48, 3.56, 3.63, 3.69, 3.75, 3.85, 3.89 (Bonus)


Introduction

• When the race car goes around a curve, is it accelerating? If it is, in what directions?

• If I drop a bullet, does it hit the ground at the same time as one I fired downrange toward a target?

• This deeper look at kinematics will include curved motion as well.


Position relative to the origin—Figure 3.1

• An overall position relative to the origin can have components in x, y, and z dimensions.

• The path for a particle is generally a curve.


Position, Velocity and Acceleration Vectors

Coordinates in the x,y,z plane:

• (x,y,z)

• Unit Vectors

• Position Vector

x

y

z

ˆ ˆ ˆ

ˆ

x y z= + +

=

r i j krrr


Average velocity—Figure 3.2• The average velocity between two points will have the same direction as

the displacement.

tttav ΔΔ

=−−

=rrrv

12

12


Instantaneous velocity—Figure 3.3If we bring the two points closer together until Δt goes to zero we get the instantaneous velocity.

dtd

tt

rrv =ΔΔ

=→Δ

lim0


Velocity Components—Figure 3.4Any velocity vector can be described by its components. To calculate the instantaneous velocity with components we can take the time derivatives of the x, y and z components of the displacement vector.

x

y

z

dxvdtdyvdtdzvdt

=

=

=


Component CalculationsWe can also get the instantaneous velocity by taking the derivative of the displacement vector itself.

2 2 2

ˆˆ ˆ

tan

x y z

y

x

dr dx dy dzv i j kdt dt dt dt

v v v v v

vv

α

= = + +

= = + +

=


A robotic vehicle, or rover, is being used to explore the surface of Mars. The landing craft is the origin of coordinates and the surrounding Martian surface lies in the xy-plane. The rover, which we represent as a point, has x- and y-coordinates that vary with time according tox = 2.0 m – (0.25 m/s2)t2

y = (1.0 m/s)t + (0.025 m/s3)t3

Find the rover’s coordinates and its distance from the lander at time t = 2.0 s. Find the rover’s displacement and average velocity vectors during the interval from t = 0.0 s to t = 2.0 s.Derive a general expression for the rover’s instantaneous velocity vector, and find the instantaneous velocity at t = 2.0 s. Express the instantaneous velocity in component form and also in terms of magnitude and direction.

Components of velocity


Answer:


The acceleration vector—Figure 3.6The acceleration vector can result in a change in either the magnitude OR the direction of the velocity.

ˆ ˆˆ ˆ ˆ ˆ

yx zx y z

yx z

dvdv dva a adt dt dt

dvdv dv dx dy dza i j k i j kdt dt dt dt dt dt

= = =

= + + = + +

2 1

2 1

0lim

av

t

v v vat t t

v dvat dtΔ →

− Δ= =

− ΔΔ

= =Δ


Calculating average and instantaneous accelerationLet’s look again at the motions of the robotic rover. We found that the components of instantaneous velocity at any time t arevx = dx/dt = (-0.25 m/s2)(2t)vy = dy/dt = 1.0 m/s + (0.025 m/s3)(3t2)And that the velocity vector is

( ) ( )2 32ˆ ˆ ˆ ˆ0.50 1.0 0.075m m m

sx y s sv v i v j ti t j⎡ ⎤= + = − + +⎣ ⎦

Find the components of the average acceleration in the interval from t =0.0s to t = 2.0s.Find the instantaneous acceleration at t = 2.0s.


a║ and a┴ Components of AccelerationWhen an object moves along a curved path, its often useful to describe the acceleration in terms of its parallel (a║) and perpendicular (a┴) components. These components are parallel and perpendicular to the velocity of the object.

The magnitude of the velocity is changed by a║.

The direction of v is changed by a┴.


Parallel or perpendicular component?

What acceleration component is responsible for:

• Superman slowing down a speeding train?

• Mark Martin accelerating while passing Ricky Martin along a straight stretch of I-70?

• Superman spinning Ricky Martin in a circular path at a constant speed.


Consider the different vectors—Figure 3.12

• Notice the acceleration vector change as velocity decreases, remains the same, or increases.


Calculating parallel and perpendicular components of acceleration

The acceleration of the rover in the previous examples is 0.58 m/s2. Use the figure below to calculate the parallel and perpendicular components of the acceleration.


Acceleration of a skierA skier moves along a ski-jump ramp as shown. The ramp is straight from point A to point C and curved from point C onward. The skier picks up speed as she moves downhill from point A to point E, where her speed is maximum. She slows down after passing point E. Draw the direction of the acceleration vector at points B, D, E, and F.


Homework

3.12, 3.21, 3.33, 3.41, 3.48, 3.56, 3.63, 3.69, 3.75, 3.85, 3.89 (Bonus)


Projectile Motion

Projectile: any body that is given an initial velocity and then follows a trajectory determined entirely by gravity and air resistance.

Air Resistance and the curvature of the earth will be ignored unless otherwise stated.

x

y

aa

=

=


Kinematic Equations for Free Fall, x and y “flavors”.X and Y motion are separable

x y

20 0

12

y oy

y

v v gt

y y v t gt

= −

= + −0 0

x ox

x

v vx x v t== +


Projectile Motion Equations

cos

sinxo o o

oy o o

v v

v v

α

α=

=

Insert the component

equations into the x and y flavors of

the kinematic equations:

( )

( ) 2

cos1sin2

cossin

o o

o o

x o o

y o o

x v t

y v t gt

v vv v gt

α

α

αα

=

= −

=

= −


Projectile motion—Figure 3.15

The horizontal range is R and the max height is h.


We can get a lot of information from these equations:The distance r of the projectile from the origin at any time is:

The projectile’s speed is:

The direction of the velocity from the positive x-axis is:

The trajectories shape in terms of x and y by eliminating t where the initial positions are zero is:

( ) 20 2 2

0 0

tan2 cos

gy x xv

αα

= −

2 2x yv v v= +

tan y

x

vv

α =

2 2r x y= +


The effects of wind resistance—Figure 3.20

• Cumulative effects can be large.

• Peak heights and distance fall.

• Trajectories cease to be parabolic.


Acceleration of a skier, continued

Let’s consider the skier from the previous example. What is her acceleration at points G, H, and I in the figure after she flies off the ramp?


A body projected horizontallyA motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0 m/s. Find the motorcycle’s position, distance from the edge of the cliff, and velocity after 0.50 s.


Height and range of a baseball A batter hits a baseball so that it leaves the bat with an initial speed v0 = 37.0 m/s at an initial angle α0 = 53.1o, at a location where g = 9.80 m/s2.Find the position of the ball, and the magnitude and direction of its velocity, when t = 2.00 s.Find the time when the ball reaches the highest point of its flight and find its height h at this point.Find the horizontal range R.


Tranquilizing the falling monkey

How should the zookeeper aim?


Height and range of a projectileFor a projectile launched with speed v0 at initial angle α0 (between 0 and 90o), derive general expressions for the maximum height h and horizontal range R. For a given v0, what value of α0 gives maximum height? What value gives maximum horizontal range?


Different initial and final heights.You toss a ball from your window 8.0 m above the ground. When the ball leaves your hand, it is moving at 10.0 m/s at an angle of 20o

below the horizontal.How far horizontally from your window will the ball hit the ground?


Circular motionAn object in circular motion will always have a component of its acceleration perpendicular to its velocity. (a┴ or arad) This component will not change the velocity of the object, only the direction and is directed toward the center of the circle.When a particle moves in a circle with constant speed, the motion is called uniform circular motion. Because triangles in figure 3.28 (a) and (b) are similar, ratios of corresponding sides are equal, so:

The magnitude aav of the average acceleration during Δt is therefore:

1

1

v vs or v sv R RΔ Δ

= Δ = Δ

1av

v v sat R t

Δ Δ= =

Δ Δ


Circular motionThe magnitude of the instantaneous acceleration at point P1 is the limit of expression as we take point P2 closer and closer to point P1:

The limit of Δs/Δt is the speed v1 at point P1, but since the velocity is constant we can drop the subscript and we are left with:

In uniform circular motion, the magnitude a of the instantaneous acceleration is equal to the square of the speed v divided y the radius R of the circle. Its direction is perpendicular to v and inward along the radius.

1 1

0 0lim lim

t t

v vs saR t R tΔ → Δ →

Δ Δ= =

Δ Δ

2

radvaR

=

22rad

R vv so aT Rπ

= =


Determination of the centripetal acceleration

• Uniform circular motion and projectile motion compared and contrasted.

• For uniform circular motion, the acceleration is centripetal.

• Refer to Figure 3.29.


Centripetal acceleration on a curved road

A BMW Z4 roadster has a ‘lateral acceleration’ of 0.87g, which is (0.87)(9.8m/s2) = 8.5 m/s2. This represents the maximum centripetal acceleration that can be attained without skidding out of the circular path. If the car is traveling at a constant 40 m/s, what is the minimum radius of curve it can negotiate?


Non-uniform circular motion—Figure 3.30If the speed changes we call it non-uniform circular motion. If the velocity changes, the magnitude of arad and atanchanges. The parallel component is atan, and the perpendicular component is arad.

2

tanrad

d vva and aR dt

= =


Relative velocity—Figures 3.31 and 3.32

• Fighter jets do acrobatic maneuvers moving at hundreds of miles per hour relative to the crowd but are nearly stationary with respect to each other.

• A conductor and a passenger on a moving train may both be moving at a rapid clip toward their destination, yet, the conductor can still move toward your seat to check your ticket.


Relative Velocity in 1-DConsider a train traveling at 3 m/s to the right. A passenger traveling at 1 m/s to the right from the train’s frame of reference would appear to be moving at 4 m/s to the right from the perspective of the cyclist.Using the notation (this is were you explain P, B and A) from the figure, the position of the passenger (P) from the frame of reference of the cyclist (A) is found by adding the position of the passenger relative to the train (xP/B) to the position of the train relative to the cyclist (xB/A):xP/A = xP/B + xB/A


Relative velocity on a straight road—passing! Velocity is just the derivative of position: vP/A = vP/B + vB/ALets try an example.

You are driving north on a straight two-lane road at a constant 88 km/h. A truck traveling at a constant 104 km/h approaches you.What is the truck’s velocity relative to you?What is your velocity with respect to the truck?How do the relative velocities change after you and the truck have passed each other?


Relative velocity in two or three dimensions• Back to the passenger walking on a train:

We need to use vector addition to find the relative motion. We can use the same equation, but in this situation sketches are even more useful.


Flying in a crosswindThe compass of an airplane indicates that it is headed due north, and its airspeed indicator shows that it is moving through the air at 240 km/h. If there is a wind of 100 km/h from west to east, what is the velocity of the airplane relative to the earth?


Correcting for Crosswind

In what direction should the pilot head to travel due north?

What will be her velocity relative to the earth?


Homework

3.12, 3.21, 3.33, 3.41, 3.48, 3.56, 3.63, 3.69, 3.75, 3.85, 3.89 (Bonus)

motion in two or three dimensions - education.fcps.org 3.pdfintroduction • when the race car goes...

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