03 lecture outline part i

Copyright © 2012 Pearson Education Inc.

PowerPoint® Lectures forUniversity Physics, Thirteenth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by Wayne Anderson

Chapter 3

Motion in Two or Three Dimensions

Part I


Goals for Chapter 3

• Vectors to represent the position of a body

– Velocity vector using the path of a body

– Acceleration vector of a body

• Describe the curved path of projectile

• Investigate circular motion

• (Describe the velocity of a body as seen from different frames of reference)


• What determines where a batted baseball lands?• If a cyclist is going around a curve at constant speed, is he

accelerating?• How is the motion of a particle described by different moving

observers?• We need to extend our description of motion to two and three

dimensions.

Introduction


Position vector

• The position vector from the origin to point P has components x, y, and z.


Average velocity—Figure 3.2

• The average velocity between two points is the displacement divided by the time interval between the two points, and it has the same direction as the displacement.


Instantaneous velocity• The instantaneous velocity

is the instantaneous rate of change of position vector with respect to time.

• The components of the instantaneous velocity are vx = dx/dt, vy = dy/dt, and vz = dz/dt.

• The instantaneous velocity of a particle is always tangent to its path.


Average acceleration

• The average acceleration during a time interval ∆t is defined as the velocity change during ∆t divided by ∆t.


Instantaneous acceleration

• The instantaneous acceleration is the instantaneous rate of change of the velocity with respect to time.

• Any particle following a curved path is accelerating, even if it has constant speed.

• The components of the instantaneous acceleration are ax = dvx/dt, ay = dvy/dt, and az = dvz/dt.


Direction of the acceleration vector

• The direction of the acceleration vector depends on whether the speed is constant, increasing, or decreasing, as shown in Figure 3.12.


Projectile Motion

Examples of projectile motion. Notice the effects

of air resistance


Projectile Motion

A projectile is an object moving in two

dimensions under the influence of Earth's gravity; its path is a

parabola.


Projectile motion—Figure 3.15• Projectile is any body with initial velocity that follows path

determined by gravity (& air resistance).

• Begin by assuming “g” is constant (near ground), & neglectingair resistance, Earth’s curvature, rotation, & motion.


Understood by analyzing the horizontal and vertical motions

separately.

Projectile Motion


Projectile Motion

•Photo shows 2 balls starting to fall at the same time.

•Yellow ball on right has an initial speed in the x-direction.

•Red ball on left has vx0 = 0

•Vertical positions of the balls are identical at identical times

•Horizontal position of yellow ball increases linearly in time.


Projectile Motion

The speed in the x-direction is constant; in the y-direction the

object moves with constant acceleration g.


The x and y motion are separable—Figure 3.16

We can analyze projectile motion as horizontalmotion with constantvelocity and vertical motion with constant acceleration:

ax = 0

and

ay = −g.


Solving Problems Involving Projectile Motion

Projectile motion is motion with constant acceleration in two dimensions,

Usually where vertical acceleration is g and is down.



1. Read the problem carefully, and choose the object(s) you are going to analyze.

2. Draw a diagram.

3. Choose an origin and a coordinate system.

4. Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g.

5. Examine the x and y motions separately.



6. List known and unknown quantities.

Remember that vx never changes, and that vy = 0 at the highest point.

•7. Plan how you will proceed. •Use the appropriate equations; you may have to combine some of them, or find time from one set and use in the other



Example Driving off a cliff.

A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.





We WANT:

vx (m/s)





vy0 = 0 m/s!

We KNOW:





∆y = -50.0 m!We KNOW:





∆x = +90.0 m!We KNOW:





ax = 0!

ay = -9.8!

We KNOW:





For Horizontal:

∆x = +90.0 m = vxt

vx = 90.0/t

How to get t ??


Projectile Motion

If object is launched at initial angle θ0 with the horizontal, analysis is similar except that the initial velocity has a vertical component.



Example: A kicked football.

A football is kicked at an angle θ0 = 37.0° with a velocity of 20.0 m/s, as shown.

It travels through the air (assume no resistance) and lands at the ground.


Example: A kicked football.

Calculate

a) the maximum height,

b) the time of travel before the football hits the ground,

c) how far away it hits the ground.

d) the velocity vector at the maximum height,

e) the acceleration vector at maximum height.

03 lecture outline part i

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