Transcript car would

Motion in Two Dimensions
A projectile is an object that has no thrust of its own but is traveling through the air.
A plane is not a projectile because it produces its own thrust to keep it in the air.
When we throw a rock or hit a baseball or shoot a bullet, these objects are projectiles.
A projectile’s path through the air is called its trajectory. Once an object leaves the
device that gives it thrust (a person’s hand, a baseball bat, a gun), the only force acting
on the projectile is gravity (we will ignore air resistance).
Imagine a situation where we fire a gun horizontally along the positive x-axis and at
the exact same time we fire the gun we drop a bullet from the same position. Ignoring
wind resistance, which bullet would hit the ground first?
Even though the gun provides enormous thrust in the horizontal direction, both bullets
have the same force acting in the vertical direction. Gravity is the only force pulling
the two bullets downward. As a result, both bullets would hit the ground at the same
time (note that one bullet would be a great distance down the x-axis when it hits while
the other would still be at the origin.
Fired bullet versus dropped bullet diagram:
+y
+x
Start Position for both bullets (x = 0, y = 2 m)
Ground (or y = 0)
The pathway of a falling object that has some amount of horizontal motion will be a
parabola. The trajectory will be parabolic.
When we have motion in two dimensions, we will break the problem down into two
connected one dimensional problems.
Separate the motion into a vertical component (y-axis) and a
horizontal component (x-axis).
The vertical component will have gravity (g) acting on the object (just like dropping an
object or throwing an object straight up from chapter 3).
The horizontal component will have no net force acting on the object. It will simply be
a constant velocity problem from chapter 2. The object will move in the indicated
horizontal direction at a constant velocity until the time runs out (it impacts some
surface).
The piece that connects the two components of the problem is time. If we know the
time of the motion of the object in one dimension then we know the time for the
motion in the other dimension as well.
Example: a boy throws a ball horizontally at a speed of 25 m/s from a height of 1.8 m.
a) How long does it take for the ball to hit the ground?
b) How far does the ball land from the boy (horizontal distance).
c) What are the horizontal and vertical components of the ball’s velocity just as it hits
the ground?
a) In the vertical dimension, yi = 0, dyi = 1.8 m, dyf = 0, and g = - 9.80 m/s2
1
From chapter 3: dyf = dyi + υyi tf + g tf2
2
Plugging in values and solving for tf gives: 0.606 s = 0.61 s (confirm this yourself)
b) In the horizontal dimension: x = 25 m/s, dxi = 0, and t = 0.61 s
From chapter 2:
dx = x t + dxi
Plugging in values and solving for dx gives: 15.15 m = 15 m (confirm this yourself)
c) The velocity in the horizontal direction is unchanged: x = 25 m/s
From chapter 3:
yf2 = yi2 + 2g(df - di)
Plugging in values and solving for yf gives: 5.9 m/s
See practice problems on page 150.
(confirm this yourself)
How would the previous problem change if the boy threw the ball at an upward angle?
In this type of problem yi ≠ 0. Therefore, we must account for the fact that gravity (g)
first has to bring the upward velocity to a stop and then change the direction to
downward. So the path of the object is a parabola where the object will rise to some
maximum height before falling downward. Note that the velocity is still constant in
the x-direction, so that part of the problem does not change from our previous example.
A ball is launched with initial velocity = i at an angle of o. What is the maximum
height that the ball will reach, how long will the ball stay in the air and how far down
range will the ball travel?
To solve the problem, we need to first break the velocity vector into its components.
From the previous chapter: yi = i (sin ) and xi = i (cos )
From previous work the velocity at any time is:  = i + at
Velocity is constant in the x direction (xi = x), so we need this equation for the y
direction:
y = yi + gt
Note that at the maximum height y = 0, so we can use the equation to solve for the
time it takes to reach the maximum height. In addition since we launched from dyi = 0,
the time it takes to fall from the maximum height will be the same as it took to rise to
the maximum height.
Projectiles launched at a angle continued
Once we know the time it takes to reach the maximum height, we can use the distance
equation to solve for the maximum height.
dy, max = dyi + υyi tf +
1
g tf2
2
The total time the ball is in the air is twice the time it took to reach the highest point.
The downrange distance can be found using the x and the total time the ball is in the
air and the equation :
dx = x t + dxi
To summarize: in order to solve this type of problem we need the following equations:
yi = i (sin )
xi = i (cos )
dy, max = dyi + υyi tf +
1
g tf2
2
y = yi + gt
dx = x t + dxi
Practice: A ball is launched at 4.5 m/s at a 66o angle above horizontal.
a) What is the maximum height of the ball?
b) How long does the ball stay in the air?
c) How far downrange did the ball travel?
yi = i (sin ) = 4.11 m/s
i = 4.5 m/s
 = 66o
x = i (cos ) = 1.83 m/s
y = yi + gt so t = 0.41938 s (time to reach maximum height)
dy, max = dyi + υyi tf +
1
g tf2
2
a)
dy,max = 0.86 m
Total time = twice time to reach max height
b)
c)
2 t = 2(0.41938 s) = 0.83876 s = 0.84 s
dx = x t + dxi
so dx = 1.5349 m = 1.5 m
See practice problems page 152.
Think about what the path of a thrown ball looks like.
We know that the actual pathway through the air depends upon how fast we throw
the ball and the angle at which we throw it. But does the pathway look the same
from all vantage points?
What does the path of the ball look like if we are watching from a position that is
off to the side of the person throwing the ball? (i.e. perpendicular to the x axis)
What would the path of the ball look like if we are watching from a position
directly behind the person throwing the ball? (i.e. on the x axis)
What would the path of the ball look like if we are watching from a position
above the person throwing the ball? (i.e. above the x axis)
Uniform Circular Motion:
When an object travels in a circle with fixed radius at constant velocity, the object
experiences uniform circular motion.
If we define the position vector (r) of an object that is undergoing uniform circular
motion to be an arrow pointing from the center of the circle and stretching to the
object at any given instant, we can use the motion of this vector around the circle to
find the velocity () and acceleration (a) vectors for the object at any time.
When we do this, we find that the direction of the velocity vector for the object is
always perpendicular to the position vector (tangent to the circle).
We also find that the acceleration vector always points toward the center of the
circle. An object whose acceleration vector always points to the center of a
circular pathway is experiencing centripetal acceleration.

a
r
See page 153 for more
details about the
relationship between
r, , and a.
Centripetal Acceleration : ac
ac =
Where:
2
𝐫
ac is acceleration for an object undergoing uniform circular motion
 is the velocity of the object
r is the radius of the circle that the object is traveling
Note that the centripetal acceleration is always directed toward the center of the
circle that the object is traveling.
From geometry: circumference of a circle is
2pr
The period (T) of an object traveling in a circle is the time it takes to travel around
the circumference.
Dividing the circumference by the period will give the velocity of the object as it
travels around the circle.
 =
𝟐π𝐫 2
T
Centripetal acceleration continued: ac
ac =
2
 =
𝐫
𝟐π𝐫 2
T
If we substitute the velocity of an object around a circle in terms of circumference
and period we get:
ac =
(𝟐π𝒓)2
T𝟐𝐫
=
𝟒π2𝒓
T𝟐
Since the object is being accelerated toward the center of the circle, a force must be
acting in that direction. We call this force the net centripetal force.
Remember that F = ma, so Fc = mac which is known as Newton’s Second Law for
Circular Motion
Where
Fc is the net centripetal force
m is the mass of the object traveling around the circle
ac is the centripetal acceleration
For centripetal acceleration problems, there needs to be something that causes the
net centripetal force. If an object is in orbit around another object, then the force of
gravity is creating the net centripetal force. If an object is tethered to the center of
the circle, then the tension on the tether is causing the net centripetal force.
In previous problems we used x and y as references for direction. However for
circular motion problems it will be convenient to use “c” as the direction toward
the center of the circle and “tang” which represents a direction tangential to the
circle.
A 2.5 kg ball is attached to a 1.35 m cable. The ball swings around a horizontal
circle in 3.54 s. Find the tension force on the ball.
m = 2.5 kg, r = 1.35 m, T = 3.54 s,Click
FT =to? see solution
+tang
+c

r
FT
ac =
𝟒π2r
T𝟐
=
𝟒(𝟑.𝟏𝟒𝟏𝟔)2(𝟏.𝟑𝟓 𝒎)
(3.54 s)𝟐
= 4.25 m/s2
Fc = mac = (2.5 kg)(4.25 m/s2) = 10.6 kgm/s2 = 11 N
See example page 155 and practice problems on 156.
A nonexistent force: Centrifugal force
Many people misuse the phrase centrifugal force. When an object is moving
around a circular path, there is an inward force (directed toward the center called
centripetal force), but there is no actual force acting outward away from the center.

ac
r
Look at the diagram to the left. If the string were to be
suddenly cut, the ball would not fly directly out away
from the center. Instead it would fly away in the
direction of the velocity vector which is actually in a
direction perpendicular to the string at the moment the
string is cut.
When the car we are riding in turns a sharp corner at high speed, we feel like we
are being “pushed” to the side. The reality is that we are trying to travel in a
straight line and our body happens to run into the door of the car as it turns in front
of us.
Relative Velocity
If two objects are moving, then we enter the area where the idea of relative
velocities become important.
Think about driving on the highway. If two cars are traveling North one at 65 mi/h
the other at 55 mi/h, what would that look like to people in the two cars?
The person in the slow car would see another car pulling away from them at 10 mi/h
while the person in the fast car would see the other car falling behind at 10 mi/h.
How would this change if two cars are traveling one going North at 65 mi/h and the
other going South at 55 mi/h, what would that look like to people in the two cars?
Now the people in both cars would see a car approaching at 120 mi/h.
What would both of these situations look like to an observer standing on the side of
the road?
In many problems, the velocity we use to solve the problem depends upon our frame
of reference (said another way, the velocity is relative to how we look at it).
Relative Velocity continued
Object a ( a sailor working on the aircraft carrier) is walking North at 3 mi/h along
the deck of an aircraft carrier. Object b (an aircraft carrier) is moving North in the
water at 25 mi/h. Object c (a person sitting on a nearby pier) watches the aircraft
carrier go by.
How fast does the sailor appear to be moving to the person on the pier?
a/b = 3 mi/h
b/c = 25 mi/h
a/c = ?
Relative Velocity: a/b + b/c = a/c so a/c = 3 mi/h + 25 mi/h = 28 mi/h
What happens if the sailor decides to turn around and jog 5 mi/h to the South?
Relative Velocity continued
Object a ( a sailor working on the aircraft carrier) is walking East at 5.00 mi/h across
the deck of an aircraft carrier. Object b (an aircraft carrier) is moving North in the
water at 15.00 mi/h. Object c (a person sitting on bridge above the aircraft carrier)
watches the aircraft carrier go under her.
What does the sailor’s motion look like to the person on the bridge?
North (+y)
Click to see solution
East (+x)
From Geometry, the length of the diagonal (red arrow) is (15.002 + 5.002)1/2 = 15.8
From Geometry, the angle the sailor appears to move is tan-1(15.00/5.00) = 71.6o
To the observer on the bridge, the sailor will appear to be moving at 15.8 mi/h in a
direction 71.6o North of East.
See example on page 158 and practice problems on page 159.