03 lecture outline part i
DESCRIPTION
university physicsTRANSCRIPT
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
Copyright © 2012 Pearson Education Inc.
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)
Copyright © 2012 Pearson Education Inc.
• 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
Copyright © 2012 Pearson Education Inc.
Position vector
• The position vector from the origin to point P has components x, y, and z.
Copyright © 2012 Pearson Education Inc.
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.
Copyright © 2012 Pearson Education Inc.
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.
Copyright © 2012 Pearson Education Inc.
Average acceleration
• The average acceleration during a time interval ∆t is defined as the velocity change during ∆t divided by ∆t.
Copyright © 2012 Pearson Education Inc.
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.
Copyright © 2012 Pearson Education Inc.
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.
Copyright © 2012 Pearson Education Inc.
Projectile Motion
Examples of projectile motion. Notice the effects
of air resistance
Copyright © 2012 Pearson Education Inc.
Projectile Motion
A projectile is an object moving in two
dimensions under the influence of Earth's gravity; its path is a
parabola.
Copyright © 2012 Pearson Education Inc.
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.
Copyright © 2012 Pearson Education Inc.
Understood by analyzing the horizontal and vertical motions
separately.
Projectile Motion
Copyright © 2012 Pearson Education Inc.
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.
Copyright © 2012 Pearson Education Inc.
Projectile Motion
The speed in the x-direction is constant; in the y-direction the
object moves with constant acceleration g.
Copyright © 2012 Pearson Education Inc.
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.
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
Projectile motion is motion with constant acceleration in two dimensions,
Usually where vertical acceleration is g and is down.
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
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.
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
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
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
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.
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
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)
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
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.
vy0 = 0 m/s!
We KNOW:
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
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.
∆y = -50.0 m!We KNOW:
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
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.
∆x = +90.0 m!We KNOW:
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
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.
ax = 0!
ay = -9.8!
We KNOW:
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
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.
For Horizontal:
∆x = +90.0 m = vxt
vx = 90.0/t
How to get t ??
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
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.
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
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.
Copyright © 2012 Pearson Education Inc.
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.
Copyright © 2012 Pearson Education Inc.
Solving Problems Involving Projectile Motion
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.
Copyright © 2012 Pearson Education Inc.
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.