Transcript Chapter 4 Two-Dimensional Kinematics
Chapter 4 Two-Dimensional Kinematics
Learning Objectives • Motion in Two Dimensions • Projectile Motion: Basic Equations • Zero Launch Angle • General Launch Angle • Projectile Motion: Key Characteristics PowerPoint presentations are compiled from
Walker 3 rd Edition Instructor CD-ROM
and Dr. Daniel Bullock’s own resources
4-1 Motion in Two Dimensions
If velocity is constant, motion is along a straight line:
4-1 Motion in Two Dimensions
Motion in the x- and y-directions should be solved separately:
4-2 Projectile Motion: Basic Equations
Assumptions:
•
ignore air resistance
•
g = 9.81 m/s 2 , downward
•
ignore Earth’s rotation If y-axis points upward, acceleration in x-direction is zero and acceleration in y-direction is -9.81 m/s 2
4-2 Projectile Motion: Basic Equations
The acceleration is independent of the direction of the velocity:
4-2 Projectile Motion: Basic Equations
These, then, are the basic equations of projectile motion:
4-3 Zero Launch Angle
Launch angle: direction of initial velocity with respect to horizontal
4-3 Zero Launch Angle
In this case, the initial velocity in the y-direction is zero. Here are the equations of motion, with
x
0 = 0 and y 0 = h:
4-3 Zero Launch Angle
Eliminating t and solving for y as a function of x: This has the form y = a + bx 2 , which is the equation of a parabola.
The landing point can be found by setting y = 0 and solving for x:
4-3 Zero Launch Angle Example
x
v
0
x t h v h
0
y
v
0
y
0
t
1 2
g t
2 1 2
g t
2
h
v
0
y t
1 2
g t
2
solve for time
,
t
2
h
g t
2
t
2 2
h g
t
1 2
g t
2 2
g h
now that we have an expression for time, t we can find the range x
h = 1 m v 0X = 2 m/s = 0 x = ?
t x
v
0
x t
2
h g
v f
x
v
0
x
We could also find the final velocity, v f
v
f
v fx
x
v fy
y v fy
v f
2
v fx v fy
2
v
0
y
gt
0 9 .
8
m s
2 2
h g
2
m s
0 .
45
s
v f
v fx
v
0
x
2
m s v fy
??
4 .
5
m s
2 1
m
9 .
8
m s
2 0 .
9
m
4 .
8
m s
fx
2
m s
2
fy
2 4 .
5
m
2
s
Motion in more than one dimension.
http://www.physicsclassroom.com/mmedia/vectors/mzi.html
straight line (line of site) the monkey begins to fall ad the precise moment when the ball leaves the barrel of the gun The ball and the monkey arrive at the point marked by the red dot at the same time path of bullet path of monkey Two types of motion occurring !
Monkey: Object under constant acceleration in 1 – dimension y x Bullet: The motion of the bullet is a combination of motion with a constant velocity (along the x-axis) and motion with constant acceleration (along the y-axis)
To start analyzing this problem we must find the x, and y components of the velocity: y v i
v i
v ix i
v iy j
v ix adj v iy opp x sin cos
opp hyp adj hyp
v iy v i v ix v i
v iy
v ix
v i v i
sin cos Let’ start by looking at an example where the launch angle is zero
4-4 General Launch Angle
In general, v 0x = v 0 cos θ and v 0y = v 0 sin
θ
This gives the equations of motion:
4-4 General Launch Angle
Snapshots of a trajectory; red dots are at t = 1 s, t = 2 s, and t = 3 s
Let’s work an example that is a launch angle different from 0, but it returns to the same height that it is launched.
A pirate ship is 560 m from a fort defending the harbor entrance of an island. A defense cannon, located at sea level, fires balls at initial speed v 0 = 82 m/s At what angle, the ship?
x
v
0 from the horizontal must a ball be fired to hit
x t
v
0 cos
t
We need an expression for the time, t
x
v
0
v
0
x t
cos
t
We need an expression for the time, t
h
y
h = y –y 0 y
0
v
0
y t
1 2
gt
2 since the projectile will = 0 so our equation becomes: 0
v
0
y t
1 2
gt
2
v
0 sin
t
1 2
gt
2 0
gt
v
0 sin
t
1 2
gt
2 from here we can solve for time, t factor out a t, and it cancels: 2
v
0 sin
t
2
v
0 sin
g
0
v
0 sin 1 2
gt x
v
0 1 2
gt
cos
t v
0 sin now substitute this back into:
x
v
0 cos
t
v
0 cos 2
v
0 sin
g
2
v
2 0
g
cos sin from trigonometry: sin 2 2 sin cos
x
v
2 0
g
sin 2 this is called the Range equation!
DANGER WILL ROBINSON: This equation only works if the projectile returns to the same height that it was launched!!!
x
v
2 0 sin
g
0 .
816 2 sin 2
gx v
2 0 9 .
8
m
/
s
82
m
2 / 560
m s
2 2 54 .
sin 1 6 0 .
816
or
180 54 .
6 125 .
3 27
o or
63
o
Now let’s look at an example of projectile motion where the projectile lands at a different elevation from it’s launch height.
A stone is projected at a cliff of height h with an initial speed of 42 m/s directed at angle above the ground.
0 = 60 0 above the horizontal. The stone strikes A 5.5 s after launching. Find: h, the speed of the stone just before impacting A, and the maximum height H reached
total time, t = 5.5 s v 0 = 42 m/s = 60 0 Let’s start by finding the v and y components of the initial velocity, v 0 (Note: this is always a good place to start!!) sin
opp hyp
v
0
y v
0
v
0
y
v
0 sin 42
m s
sin 60 0 36 .
4
m s
cos
adj hyp
v
0
x v
0
v
0
x
v
0 cos 42
m s
cos 60 0 21
m s v
0
y
36 .
4
m s v
0
x
21
m s
Now we can proceed with finding the height of the cliff!
total time, t = 5.5 s v 0 = 42 m/s v 0y =36.4 m/s = 60 0 v 0x =21 m/s
v
0
y h
36 .
4
m v
0
x
21
m s s v
0
y t
1 2 200 .
2
m gt
2 148 .
36 2
m
.
4
m s
5 .
5
s
52
m
1 2 9 .
8
m s
2 5 .
5
s
2 Now let’s try and find the speed of the rock just before impact.
Remember that this will be the magnitude of the final velocity vector. And this vector has both an x and y component.
v 0 = 42 m/s total time, t = 5.5 s v f v 0y =36.4 m/s = 60 0 52m v 0x =21 m/s
v
Because there is no acceleration along the x axis: v the y component of the final velocity: 0x =v fx =21 m/s However in the y-direction there is acceleration so we must find
fy
v
0
y
gt
36 .
4
m s
9 .
8
s m
2 5 .
5
s
17 .
5
m s
This negative sign means the y-component is downward!
v f
fx fy
2 2 17 .
5
m
2
s
21
m s
2 27 .
3
m s
v 0 = 42 m/s total time, t = 5.5 s v f =27.3 m/s v 0y =36.4 m/s = 60 0 52m v 0x =21 m/s Now let’s find the maximum height, H! To do this you have to know that the instant the stone is at it’s maximum height the y component of the velocity equals zero, v yMAV H = 0. Using this information we can use:
v
2
fy
v
2 0
y
2
gH
2
gH
v
2 0
y
H
0
v
2 0
y
2
gH v
2 2 0
g y
36 .
4
m s
2 2 9 .
8
m s
2 67 .
5
m
= 35 0 = 3.3 km = 9.4 km During volcano eruptions, chunks of solid rock can be blasted out of the volcano; these projectiles are called volcanic bombs. At what initial speed would a bomb have to be ejected, at angle 0 =35 0 to the horizontal, from the vent at A in order to fall at the foot of the volcano at B, at vertical distance h=3.3 km and horizontal distance d=9.4 km ?
So we need to do is figure at what the initial speed of the bombs need to be in order to hit point B.
v 0 =??
= 35 0 = 3.3 km = 9.4 km We have an equation for range, or horizontal distance:
x
v
0
x t
v
0 cos 0
t
v
0
x
cos 0
t
We have, x = 9.4 km, we have 0 =35 0 what we need is the time, t, and this is where it get tricky! Let’s start by using:
y
y
0
h
v
0
y t
1 2
gt
2
h
v
0 sin 0
t
1 2
gt
2
v 0 =??
= 35 0 = 3.3 km = 9.4 km
h
v
0 sin 0
t
1 2 1 2
gt
2
gt
2
v
0 I’m going to bring the –h over and make this a quadratic equation sin 0
t
h
0 a b c
a x
2
b x
c
0 What is the solution to a quadratic equation?
v 0 =??
= 35 0 = 3.3 km
a
1 2
x
2
gt
2
b
v
0
x
c
sin 0 0
t t
v
0 sin 0
h
0
v
0 sin 0 2 2
x
b
a = ½ g b= v 0 sin 0 c = h 4 1 2
g
1 2
g
= 9.4 km
b
2 2
a
4
ac
this reduces to
v 0 =??
= 35 0 = 3.3 km = 9.4 km
t
v
0 sin 0
v
0 sin 0
v
0 sin 0 2 2 1 2 4 1 2
g g v
2 0 sin 2 0 2
gh g
now we can substitute this expression for time into our range equation!
v 0 =??
= 35 0 = 3.3 km = 9.4 km
v
0
x
cos 0
t t
v
0 sin 0
v
2 0 sin 2 0 2
gh g
Using a lot of algebra and tricks this equation becomes:
v
0
x
cos 0 2
x g
tan 0
h
255 .
5
m s
v 0 =255 m/s = 35 0 t= ?
= 3.3 km = 9.4 km Next, I would like to find the time of flight. We can rearrange this equation to solve for time:
x
v
0
x t
t
x v
0
x
v
0
x
cos 0 9400 255
m s m
cos 35 0 45
s
v 0 =255 m/s = 35 0 t= 45 s = 3.3 km = 9.4 km Finally how would air resistance change our initial velocity? We expect the air to provide resistance but no appreciable lift to the rock, so we would need a greater launching speed to reach the same target.