2-Dimensional Motion
Part I: Projectile Motion
Part II: Circular Motion
First of All, What is 2-D
Motion?
Before, we talked about motion in one
dimension (just the x axis)
Now we are investigating things that can
move it 2 dimensions (x and y)!
1D Motion
x
y
2D Motion
x
2-D Motion Part I:
Projectile Motion
A “projectile” is an object on which
the only force acting is gravity.
– These are things that fly through
the air!
(Examples: footballs, baseballs,
cars flying off cliffs, bullets,
cannon shells, and batman)
– Usually they are thrown/fired/etc.
but are not things like rockets,
which still have something
“pushing” them as they fly
through the air!
As always, gravity will accelerate the
object down.
The Dimensions are
Independent
Since 2D motion is
complicated, it is helpful
to think of the x and y
dimensions separately
when we analyze motion.
This is because the x and
y dimensions are
completely independent.
This lets us use all of our
favorite 1D equations
when solving problems!
Notice that after this idiot
jumps off the cliff his
velocity in the x dimension
does not change.
BUT……His velocity in the
y dimension is increasing
due to gravity.
The same is true for this
cannonball:
Gravity
What causes projectiles to
accelerate?
– Gravity!
In which dimension does
gravity exist?
– The vertical (y)!
The acceleration is always
9.81 m/s2 downward. This is
called “acceleration due to
gravity” or “g”
Since acceleration only
exists in the y dimension
this means it is always zero
for the x dimension.
Horizontal Velocity
Since we have established that acceleration only exists
in the y dimension and the acceleration in the x
dimension is zero, what does this mean about the
velocity of a projectile in the x dimension?
– It is always constant!
This needs to be remembered (by YOU)!
Solving Problems
We are going to take some time to go over
the problem solving process for these
projectile motion problems.
It is a “modified” version of GUESS.
The best part is that you don’t have to
learn any complicated equations… we will
use the ones you
already know from
the last unit! It’s easy!
Givens
Of course, the first step to solving
any problem is to list your givens.
Always begin by drawing a
diagram representing the motion.
And remember to include your
arrows for initial velocity,
acceleration, displacement, and
the positive direction.
Remember, we are separating the
variables into the two dimensions.
You will write out the same givens
table as before, but this time it will
be a different set for each
dimension (x or y)!
+
a
+
v
v
d
d
x
vo =
v=
ax =
t=
x=
y
vo =
v=
ay =
t=
y=
Equations
The equations you will use are the same
ones.
You will just be using the variables from
one dimension (x or y) at a time.
v f  v0  at
x  x0  v0t  12 at 2
2a( x  x0 )  v  v
2
f
2
0
Time
In projectile motion, we
will treat the x and y
variables as completely
different sets.
This means the givens
we use for one set cannot
be used with the other!
However, Time is a
variable that is always
the same for both x and
y.
x
vo = 25 m/s
v = 25 m/s
a = 0 m/s2
t = 4.0 s
x = 190 m
y
vo = 0 m/s
v = 48 m/s
ay = 10.0m/s2
t = 4.0 s
y = 22 m
Sample Problem
A cannon is placed at the edge of a cliff so the barrel is 60 meters above
the ground. The cannon fires a projectile horizontally with a velocity of
95 m/s. How far from the base of the cliff does the projectile land?
As always, we will start by drawing a
diagram of the situation. Remember
to label any givens and include a
direction vector for v, d, a, and +
+
Vx=95m/s
vy
ay
y=
60 m
+
x = ?? m
After you have drawn your picture, fill
in your Givens table.
And of course our unknown variable
is x. Always circle the unknown in the
table.
x
vo =95m/s
v = 95m/s
a = 0m/s2
t= ?
x= ?
y
vo = 0m/s
v= ?
ay = 10.0m/s2
t=?
y = 60 m
Sample Problem
A cannon is placed at the edge of a cliff so the barrel is 60 meters above
the ground. The cannon fires a projectile horizontally with a velocity of
95 m/s. How far from the base of the cliff does the projectile land?
+
Vx=95m/s
vy
dy=
60 m
ay
+
Now determine the equations you will
need to solve the problem.
Remember, we will keep the sets of
variables completely separate (except
time)!!!
dx = ?? m
x
vo =95m/s
v = 95m/s
a = 0m/s2
? s
t = 3.5
x= ?
y
vo = 0m/s
v= ?
ay = 10.0m/s2
t = ?3.5 s
y = 60 m
x = vot + ½ at2
This 2is the only equation
x = (95)(3.5) + ½ (0)(3.5)
we can use to solve for
x = 332.5 m
x. But we are missing
time! We need to use the
variables in the ydimension to find time
first.
y = vot + ½ at2
t = √(2y / a)
t = √[(2)(60) / (10.0)] = 3.5 s
Practice Problem
Thelma and Louise drive their car off
the edge of the grand canyon at a
speed of 30 m/s and die 3.4 seconds
later.
(a) how tall is the canyon at that spot?
Answer:
57.8 m
(b) how far away from the bottom of
the canyon do they land?
Answer:
102 m
Which of the following does not
effect the hang time of a
projectile?
A. Angled fired
B. Vertical velocity
C. Horizontal velocity
D. Height it was fired
at.
Angled Projectile Motion
Occasionally, you will
encounter problems
where the projectile is
initially fired at an angle.
– This means that the initial
velocity will have a
horizontal and a vertical
component.
You will need to calculate
these components in
order to fill in your givens
table on each problem.
Vector Components
In order to find the
components of a
vector (like velocity)
you will need to use
those timeless
Trigonometric
Functions.
Vector Components
So we have a
projectile that is fired
at an angle of 30o
with respect to the
horizontal at a
velocity of 50 m/s.
We need to think of
it like the projectile
is fired vertically and
horizontally at the
same time, giving it
both components.
vy
Θ=30o
vx
Vector Components
In order to calculate the
components, we need
to shift vy to make a
right triangle.
Then we can use trig
functions to solve for vy
and vx like they are
sides of a right triangle.
To solve for vx, we will
use cosine because it is
adjacent and we have
the hypotenuse.
vy
Θ=30o
vx
Vector Components
To solve for vx, we will use cosine
because it is the adjacent side and
we have the hypotenuse.
adj
hyp
v
cos   x
V
v x  V cos 
cos  
To solve for vy, use the same
process but with sine.
sin θ 
sin θ 
opp
hyp
vy
V
v y  V sin θ
vy
Θ=30o
vx
Sample Problem
A soccer ball is kicked at an angle of 70o from the horizontal with a velocity of
20m/s. What is the range of the soccer ball? (how far away does it land)
As always, start by drawing a
We
need toincluding
use trig functions
to
diagram,
all vectors.
solveasfor
vix and
viy your
And
always
circle
Beginunknown
to fill in what
you can
variable.
on your givens table
x
y
?
?
vi = 6.84m/s
vi = +18.8m/s
vf = 6.84m/s
?
vf = -18.8m/s
?
2
ax = 0 m/s ay = -10.0 m/s2
t= ?
t= ?
dx = ?
dy = 0 m
a
+
vix = VcosΘ
vix=(20)cos(70)
vix=6.84 m/s
vy
Θ=70o
vx
dx
viy = VsinΘ
viy=(20)sin(70)
viy=18.8 m/s
+
Sample Problem
A soccer ball is kicked at an angle of 70o from the horizontal with a velocity of
20.0m/s. What is the range of the soccer ball? (how far away does it land)
x
y
vi = 6.84 m/s
vf = 6.84 m/s
ax = 0 m/s2
t=?
dx = ?
vi = +18.8 m/s
vf = -18.8 m/s
ay = -10.0m/s2
t=?
dy = 0 m
a
dx = 25.7 m
vf = vi + at
t = (vf – vi) / a
t = (-18.8 – 18.8) / (-10.0)
t = 3.76 s
vy
+
Θ=70o
vx
dx = vixt + ½ at2
dx = (6.84)(t) + ½ (0)t2
dx = (6.84)(3.76) + 0
dx
+
Practice Problem
A cannon shoots a cannon ball at 50° from
the horizontal at a velocity of 300
m/s.
What is the range (dx) of the cannon ball?
Answer:
~8,900 m
Projectile Motion Formulas
viy  v sin 
vix  v cos 
v f  vi  at
d  vi t  1 at 2
2
v f  vi  2ad
2
2
2-D Motion Part II:
Circular Motion
You spin me right round….
Circular Motion
Circular motion is
exactly what is sounds
like: objects moving in
a circular path.
We are going to
investigate how
objects rotate and
revolve, and why they
do this…
Rotation vs. Revolution
Before we get started lets make a
quick differentiation.
When we say “Rotation,” we will
be talking about something
spinning.
– The axis of rotation is inside the
object.
When we say “Revolution” we
will be talking about something
going around something else.
– The axis of rotation is outside the
object.
Angular Velocity
For an object in circular
motion, it makes sense to
describe its velocity in angular
terms.
Meaning: we will describe the
velocity of an object in motion
by how many degrees (or
radians) it covers in a certain
amount of time.
The symbol for angular
velocity is a lower-case
omega: ω
Also, the units for Θ must be in
radians.
The equation for angular velocity:
ΔΘ
ω
t
Radians?!?
If you’re in Pre-Calculus (which
you should be) then you have
already been introduced to the
idea of radians. But here is the
basic idea:
– One full revolution of a circle
(360o) is equal to 2π Radians.
– So if you have degrees and
want to convert to radians the
equation is:
2
360
For any problem where you
need to find the angular
velocity, be sure that your units
are in radians!
 rad   deg
Sample Problem
A child on a merry-go-round completes
one rotation in 60 seconds. What is
the child’s angular velocity?
ω = ∆Θ / t
ω = (2π) / (60)
ω = 0.10 rad/s
∆Θ is 2π because he
completes one rotation
Tangential Velocity
Suppose we have 2 horses on a
carosel. The black horse is 1 meter
from the center and the white horse
is 2 meters from the center.
– Which horse has a greater angular
velocity?
They have the same! They will each
cover a full rotation (360o) in the same
amount of time.
– Which horse “feels” like they are going
“faster”?
The white one!
– The white horse is going faster
because it has a greater Tangential
Velocity.
It covers a greater distance
(circumference) in the same amount of
time.
Tangential Velocity
Hopefully you learned in math
what a “Tangent” line is.
– (Basically it’s a straight line that
touches one point on a curve.)
When we talk about the
tangential velocity of an object
in motion we mean its actual
velocity (distance/time).
Tangential velocity can be
found using the angular
velocity and radius.
It can also be found if you
know the distance
(circumference) traveled and
time.
The effect of tangential velocity
can hurt!
vT  r
Sample Problem
A student twirls a rock on a string above their head. The rock
completes 2 revolutions in 4.50 seconds and the length of the string
is 0.750 meters. What is the tangential velocity of the rock?
As always, start by listing Givens and Unknowns:
t = 4.50 s
∆Θ = 2 revs = 4π radians
vT = rω
r = 0.750 meters.
vT = (0.75)(?)
vT = ?
vT = (0.75)(2.79)
vT = 2.09 m/s
The equation we will need to use is:
vT = rω
But there is a problem.. We don’t know ω!
So we must use ω = ∆Θ/t
ω = ∆Θ/t
ω = (4π)/(4.5)
ω = 2.79 rad/s
Question…
We know that it is possible for an
object to rotate in a circle with a
constant speed, but can an object
rotate with a constant velocity?
– The answer is NO! But why?
– Because the direction is constantly
changing. Remember: velocity is a
vector and has a magnitude and
direction!
What is it called when something
has a change in velocity?
– Acceleration! In this case we refer to
it as Centripetal Acceleration.
Centripetal Acceleration
Centripetal Acceleration is what causes
objects to travel in a circle.
“Centripetal” means “center pointing”
As the object is moving, the
acceleration pushes toward the axis of
rotation (center) and causes it to
change direction and travel in a circle.
The equation for Centripetal
Acceleration is:
vT
ac 
r
2
Sample Problem
A car is doing donuts out in the parking lot and
is traveling in a perfect circle with a radius of
15 meters at a constant speed of 18 m/s.
What is the magnitude of the centripetal
acceleration acting on the car?
Givens:
vT = 18 m/s
r = 15 m
ac = ?
ac = vT2/r
ac = (18)2 / 15
ac = 21.6 m/s2
“g” Forces
“g” Forces are used to describe
acceleration in terms of how it compares
to acceleration due to gravity (g) which is
10m/s2.
When undergoing centripetal acceleration,
people can undergo many g’s!
UCM Formulas
2
  deg
360
ΔΘ
ω
t
vT  r
2
 rad
vT
ac 
r