Motion in Two
Dimensions
Chapter 6
What is Projectile Motion?
Projectile Motion


A projectile can be a football, a bullet, or a drop
of water.
A projectile is any object which once projected
continues in motion by its own inertia and is
influenced only by the downward force of
gravity.
Video
Projectile Motion

Same requirements as free fall, but projectile
motion is two-dimensional motion of an object

Vertical direction


Horizontal direction


Constant acceleration = -g = -9.8m/s2
Constant velocity in x direction
They are INDEPENDENT of each other
Concepts of Projectile
Motion



Horizontal
 Motion of a ball rolling freely along a level
surface
 Horizontal velocity is ALWAYS constant
 Nothing pushes or pulls on it in the
horizontal direction to cause it to accelerate
Vertical
 Motion of a freely falling object
 Force due to gravity
 Vertical component of velocity changes
with time
Parabolic
 Path traced by an object accelerating only
in the vertical direction while moving at
constant horizontal velocity
•The trajectory (or path) of any projectile is a
parabola.
•Gravity acts to influence the vertical motion of the
projectile, thus causing a vertical acceleration.
•The horizontal motion of the projectile is the result of the
tendency of any object in motion to remain in motion at
constant velocity Newton’s First Law
•
The horizontal motion of the cannonball
is the result of its own inertia.
As the cannonball falls, it undergoes a
downward acceleration.
horizontally-launched projectile
Parabola

At a given location on the earth and in the
absence of air resistance, all objects fall with the
same uniform acceleration.

Thus, two objects of different sizes and weights,
dropped from the same height, will hit the
ground at the same time.

Videos click on objects

An object is controlled by two independent motions.

So an object projected horizontally will reach the
ground at the same time as an object dropped vertically.
No matter how large the horizontal velocity is, the
downward pull of gravity is always the same.
•A projectile launched horizontally has no initial vertical velocity.
Therefore, its vertical motion is like that of a dropped object.
Click Picture
Factors Affecting Projectile
Motion

What two factors would affect projectile
motion?
Angle
 Initial velocity

Initial Velocity
Angle
For a particular range less than the maximum
and for a particular launch velocity,
two different launch angles will give that range.
The two angles add to give 900.
450 gives the maximum range.

Initial velocity has a vertical and horizontal
components

Maximum height: is the height of the projectile when

Range, R: is the horizontal distance the projectile

Flight time: is the time the projectile is in the air.
the vertical velocity is zero and the projectile has only
its horizontal velocity component.
travels.
(called hang time in football game)

The range R is the horizontal distance the projectile
has traveled when it returns to its launch height.
Projectile Motion Solutions
PROBLEM-SOLVING STRATEGY Projectile motion problems
MODEL Make simplifying assumptions.
VISUALIZE Use a pictorial representation. Establish a coordinate system with the xaxis horizontal and the y-axis vertical. Show important points in the motion on a sketch.
Define symbols and identify what the problem is trying to find.
SOLVE The acceleration is known: ax  0 and
a y  - g , thus the problem
becomes one of kinematics. The kinematic equations are (THESE ARE NOT NEW –
THEY ARE JUST OUR 1-D EQUATIONS).
ASSESS Check that your result has the correct units, is
reasonable, and answers the question.
Summary of Projectile Motion Equations
Component of Motion
Equation
X Acceleration
ax = 0
Y Acceleration
ay = -g
X Velocity
vxf = vxi = vi cos
Y Velocity
vyf = vyi + ayt = visin+ ayt
X Position
x = xi + vxit
Y Position
y = yi + vyit + 1/2ayt2
v2 = vx2 + vy2 (v is the velocity vector for the projectile)
Example Problem
A Projectile Launched Horizontally
A stone is thrown horizontally at 15 m/s
from the top of a cliff 44 m high..
A. how far from the base of the cliff does
the stone hit the ground?
B. How fast is it moving the instant before it
hits the ground?
•projectile is launched upward at an angle
to the horizontal
Range, R
Example Problem 1
An object is fired from the ground at 100 meters
per second at an angle of 30 degrees with the
horizontal
a)
b)
c)
Calculate the horizontal and vertical components
of the initial velocity
How long does it take to reach highest point?
How far does the object travel in the horizontal
direction?
Example Problem
The Flight of a Ball
 The ball in the photo was launched with an
initial velocity of 4.47 m/s at an angle of 66
above the horizontal.
a. What was the maximum height the ball attained?
b. How long did it take the ball to return to the
launching height?
c. What was its range?
Example: A Home Run
A baseball is hit so that it leaves the
bat making a 300 angle with the
ground. It crosses a low fence at the
boundary of the ballpark 100 m from
home plate at the same height that it
was struck. (Neglect air resistance.)
What was its velocity as it left the
bat?
v0 x  v0 cos  ; v0 y  v0 sin  ;
x1  x0  v0 x (t1  t0 )  (v0 cos  )t1 ;
y1  0  y0  v0 y (t1  t0 )  12 g (t1  t0 ) 2  (v0 sin  )t1  12 gt12 ;
0  (v0 sin  )t1  12 gt12  (v0 sin   12 gt1 )t1;
so t1  0 or t1  2v0 sin  / g;
Therefore, x1  (v0 cos  )t1  (v0 cos  )(2v0 sin  / g );
x1  2v0 2 sin  cos  / g  v0 2 sin 2 / g ;
v0  x1 g / sin 2  (100 m)(9.80 m/s 2 ) / sin(60)  33.6 m/s
Sample Problem 3
The figure illustrates the flight of Emanuel Zacchini over three Ferris wheels,
located as shown and each 18 m high. Zacchini is launched with speed v0 = 26.5
m/s, at an angle  0 = 53° up from the horizontal and with an initial height of 3.0
m above the ground. The net in which he is to land is at the same height.

(a) Does he clear the first Ferris wheel?
SOLUTION
If he reaches his maximum height when he is over the middle Ferris wheel, what is
his clearance above it?
V fy2  Viy2  2ay

At max height Vf = 0
 Viy2
 y 
(vi sin(  )) 2
2a
19.6
Solving for ∆y you get 22.85m
Since he begins 3m off the ground, he clears the Ferris wheel
by (25.85 – 18) = 7.85 m
How far from the cannon should the center
of the net be positioned?



SOLUTION:
V fy  Viy  at
Solve for t at max
height = 2.16 sec
Viy  at  Vi sin 
Total time is 4.32 sec
X  X i  Vxi t  0  (Vi cos  )t

Should be 68.90m away
Sample Problem 2

In Fig. 4-15, a rescue
plane flies at 198 km/h
(= 55.0 m/s) and a
constant elevation of 500
m toward a point directly
over a boating accident
victim struggling in the
water. The pilot wants to
release a rescue capsule
so that it hits the water
very close to the victim.
(a) What should be the angle  of the pilot's
line of sight to the victim when the release is
made?
Solution
  tan
1
x
h
x  x 0  ( v 0 cos 0 ) t
1 2
y  y 0  ( v 0 sin  0 ) t  g t
2
 500 m  (55.0 m / s) (sin 0  ) t 
1
(9.8 m / s 2 ) t 2
2
Solving for t, we find t = ± 10.1 s (take the positive root).
x  0  (55.0 m / s) (cos 0 ) (10.1 s)
x  555.5 m
  tan
1
555.5 m
 48
500 m
(b) As the capsule reaches the water, what

is its velocity v as a magnitude and an
angle?
When the capsule reaches the water,
v x  v 0 cos  0  (55.0 m / s) (cos 0  )  55.0 m / s
v y  v0 sin 0  g t
 (55.0 m / s) (sin 0 )  (9.8 m / s 2 ) (10.1 s)
  99.0 m / s
v  113 m / s and
   61
55 m/s
61°
99 m/s
113 m/s
A rock is thrown upward at an
angle. What happens to the
horizontal component of its
velocity as it rises? (Neglect air
resistance.)
(a) it decreases
(b) it increases
(c) it remains the same
In baseball which path would a
home run most closely
approximate? (Neglect air
resistance.)
(a) hyperbolic
(b) parabolic
(c) ellipse
A projectile is launched upward at an
angle of 75° from the horizontal and
strikes the ground a certain distance
down range. What other angle of launch
at the same launch speed would produce
the same distance? (Neglect air
(a) 45°
resistance.)
(b) 15°
(c) 25°
A horizontally traveling car drives off
of a cliff next to the ocean. At the
same time that the car leaves the cliff a
bystander drops his camera. Which
hits the ocean first? (Neglect air
(a) car
resistance.)
(b) camera
(c) they both hit at the
same time
Uniform Circular Motion
http://www.physicsclassroom.com/mmedia/circmot/circmotTOC.html
32
Circular Motion


Uniform circular motion is the motion of an
object in a circle with a constant or uniform
speed.
As an object moves in a circle, it is accelerating
inward due to its change in direction.
Circular Motion Vocabulary











r = radius
m= mass
v = velocity
Fc = centripetal force
FT = tension force
(sometimes written as T, not to be confused with
the T for period)
Ff = friction force
t = time
T = period = “sec / rev”
linear (tangential) velocity = 2πr/T “m / s”
frequency (f) = “rev / sec” = 1 / T
ac = centripetal acceleration
Rotation → axis inside object
(Earth rotates around its axis)
Revolution → axis outside object
(Earth revolves around the Sun)
Rotational speed → rot. speed = # of rev / time (rev/s)
Period
Linear velocity →
→
T
=
time / # of rev (s/rev)
v
= 2πr / T
(m/s)
What affects the speed of a ball tied to a
string moving in a circular path?
Tension, mass, radius
Linear = Polar = Angular = Rotational
2πr
360º
2π rad
1 Rev
Convert rotational speed to linear velocity:
Example: The spinning ball (in the above picture) has a
string radius of 0.5 m with the tube. There are 5
revolutions in 2.5 seconds (2 rev/s).
Convert to linear velocity …
Example: The spinning ball (in the
picture at left) has a string radius of
0.5 m with the tube. The ball
completes 5 revolutions in 2.5 seconds.
(Rotational velocity)
r = 0.5 m
t = 2.5 seconds for 5 revolutions = T (period)
T = time/ # of rev = 2.5 s/ 5 rev = 0.5 s/rev
Convert rotational speed of 5 rev / 2.5 s to linear
speed in m/s. Remember T = 1 / rot. speed
Linear velocity: v = 2 π r / T
v = (2 π 0.5 / 0.5) = 6.28 m/s
Uniform Circular Motion: Period
Object repeatedly
finds itself back where
it started.
The time it takes to
travel one “cycle” is
the “period”.
distance = rate  time
distance 2r
time =

rate
v
2r
T=
v
38
Quantifying Acceleration: Magnitude
v v  t

v
r
2
v t
v 
r
v v
a

t
r
2
Centripetal Acceleration
39


Speed of an object moving in circle is:
v = 2r /T
Substituting that for v,
(2r / T ) 4 r
ac 
 2
r
T
2
2
•The net force (aka inward or centripetal force) is
directed towards the center of the circle.
•Without such an inward force, an object would
continue in a straight line due to its inertia.

Newton’s Second Law Fnet = m×ac
mv 2
Fnet 
r
2
4 r
Fnet  m(
T
2
)
Sources of net forces on some objects
•For Earth circling the sun, the force is the sun’s
gravitational force on Earth.
•For a stopper swinging, the force is the tension in
the chain attached to the stopper.
•For a car turning around a bend, the inward force
is the frictional force of the road on the tires.
Applying Newton’s 2nd Law:
F  ma
mv
F
r
2
Centripetal Force
Always points toward center of circle.
(Always changing direction!)
Centripetal force is the magnitude of the force
required to maintain uniform circular motion.
42
Direction of Centripetal Force,
Acceleration and Velocity
With a centripetal
force, an object in
motion follows uniform
circular motion.
Without a centripetal
force, an object in
motion continues along
a straight-line path.
43
Direction of Centripetal Force,
Acceleration and Velocity
44
What if velocity decreases?
45
What if mass decreases?
46
What if radius decreases?
47
What provides the centripetal force?
• Tension
• Gravity
• Friction
• Normal Force
Centripetal force is NOT a new “force”. It is simply a
way of quantifying the magnitude of the force
required to maintain a certain speed around a circular
path of a certain radius.
48
Relationship Between Variables of Uniform
Circular Motion
Suppose two identical objects go around in
horizontal circles of identical diameter but one
object goes around the circle twice as fast as the
other. The force required to keep the faster object
on the circular path is
The answer is E. As the
A. the same as
velocity increases the
B. one fourth of
centripetal force required to
maintain the circle increases
C. half of
as the square of the speed.
D. twice
E. four times
the force required to keep the slower object on the path.49
Relationship Between Variables of Uniform
Circular Motion
Suppose two identical objects go around in
horizontal circles with the same speed. The
diameter of one circle is half of the diameter of
the other. The force required to keep the object
on the smaller circular path is
A.
the same as
The answer is D. The centripetal force needed
B.
one fourth of to maintain the circular motion of an object is
inversely proportional to the radius of the circle.
C.
half of
Everybody knows that it is harder to navigate a
D.
twice
sharp turn than a wide turn.
E.
four times
the force required to keep the object on the larger
path.
50
Relationship Between Variables of Uniform
Circular Motion
Suppose two identical objects go around in horizontal circles of
identical diameter and speed but one object has twice the
mass of the other. The force required to keep the more
massive object on the circular path is
A.
the same as
B.
one fourth of
Answer: D.The mass is directly
C.
half of
proportional to centripetal force.
D.
twice
E.
four times
51
Tension Can Yield a Centripetal Acceleration:
If the person doubles the
speed of the airplane,
what happens to the
tension in the cable?
mv
F = ma 
r
2
Doubling the speed, quadruples the force (i.e.
tension) required to keep the plane in uniform circular
52
motion.
Friction Can Yield a Centripetal Acceleration:
53
Car Traveling Around a Circular Track
Friction provides the centripetal acceleration
54
Friction Can Yield a Centripetal Acceleration
FN
fs
W
What is the maximum
speed that a car can use
around a curve of radius
“r”?
Force
X
Y
W
0
-mg
FN
0
FN
fs
-sFN
0
Sum
ma
0
55
Maximum Velocity
F
y
 0   mg  FN
FN  mg
F
x
 ma   
max
s
mv 2
max

  s mg
r
2
max
v  s  g  r
v  smax  g  r
mg
Force
X
Y
W
0
-mg
FN
0
FN
Fs
-sFN
0
Sum
ma
0
56
Centripetal Force: Question
A car travels at a constant
speed around two curves.
Where is the car most likely to
skid? Why?
mv
F = ma 
r
2
Smaller radius: larger force
required to keep it in uniform
circular motion.
57
Centripetal vs Centrifugal

Centripetal: This is the force needed to make
something move in a circle.



The force could actually be a number of things such as:
friction, gravity, tension in a rope or any combination.
Centripetal force is a name for a real force that has the role of
making something move in a circle. This force is always
directed towards the center of the circle of motion.
Centrifugal: This is a fictitious force needed to make a
non-inertial (accelerating) reference frame seem like it is
not accelerating. This fake force is what it “seems” like
pushes you away from the center of the circle of
motion. It’s actually due to your inertia.
Banked Road problem
FN
ө
ө
Fg
ac
Ff
Banked Curves
Q: Why exit ramps in highways are banked?
A: To increase the centripetal force for the higher exit speed.
60
The Normal Force Can Yield a Centripetal Acceleration:
Engineers have learned to “bank” curves so that
cars can safely travel around the curve without
relying on friction at all to supply the centripetal
acceleration.
How many forces are
acting on the car (assuming
no friction)?
61
The Normal Force as a Centripetal Force:
Two: Gravity and Normal
Force
X
Y
W
0
mg
FN
Sum
FNsin FNcos
mac
0
62
The Normal Force as a Centripetal Force:
F
y
  mg  FN cos  0
mg
FN 
cos
mv 2
 Fx  FN sin   ma  r
mg
mv 2
 sin  
cos
r
v2
tan  
gr
63
The Normal Force and Centripetal Acceleration:
How to bank a curve…
2
v
tan  
gr
…so that you don’t rely on friction at all!!
64
Vertical Circular Motion
65
1. Tied to a post and moving in a circle at
constant speed on a frictionless horizontal
surface. Coming straight out of the paper.
FN
FT
ac
Fg
ΣFx = mac
-FT = m(-ac)
FT = mac
ΣFy = ma = 0
FN - Fg = 0
FN = Fg
2. Tied to point A by a string. Moving in a horizontal
circle at constant speed. Not resting on a solid
surface. No Friction. Coming straight out of paper.
A
FT
ac
Fg
ΣFx = mac
-(FTcosθ) = m(-ac)
FTcosθ = mac
ΣFy = ma = 0
FTsinθ - Fg = 0
FTsinθ = Fg
3. Riding on a horizontal disk that is rotating at
constant speed about its vertical axis. Friction
prevents rock from sliding. Rock is moving straight
out of the paper.
FN
Ff
ac
Fg
ΣFx = mac
-Ff = m(-ac)
Ff = mac
ΣFy = ma = 0
FN - Fg = 0
FN = Fg
4. Stuck by friction against the inside wall of a drum
rotating about its vertical axis at constant speed.
Rock is moving straight out of the paper.
Ff
FN
ac
ΣFx = mac
-FN = m(-ac)
FN = mac
Fg
ΣFy = ma = 0
Ff - Fg = 0
Ff = Fg
5. Swinging on a rope, at the bottom of a
vertical circle.
FT
ac
Fg
ΣFx = ma
ΣFy = mac
FT - Fg = mac
FT = Fg + mac
6. Swinging on a rope, at the top of a vertical
circle.
ac
Fg FT
ΣFx = ma
ΣFy = mac
-(FT + Fg) = m(-ac)
FT + Fg = mac
A ball held by a string is coasting around in a
large horizontal circle. The string is then pulled in
so the ball coasts in a smaller circle. When it is
coasting in the smaller circle its speed is …
(Assume tension and mass stay constant)
a) greater
b) less
c) Unchanged
Explain.
Problem #1
If the radius of a circle is 1.5 m and it
takes 1.3 seconds for a mass to swing
around it (1 rev).
a) What is the speed of the mass?
b) Find the tension if the mass is 2 kg.
s = 7.25 m/s
FT = 70.1 N
Problem #2
A 1200 kg car traveling at 8 m/s is
turning a corner with a 9 m radius.
a)
b)
How large a force is needed to keep the
car on the road?
b) Find the coefficient of friction.
Ff = 8533.3 N
μ = .726
Problem #3
A car travels around a circular flat track
with a speed of 20 m/s. The coefficient
of friction between the tires and the
road is 0.25. Calculate the minimum
radius needed to keep the car on the
track.
r = 163.27 m
What speed must a 1.5 kg pendulum bob swing in
the circular path of the accompanying figure if
the supporting cord is 1.2 m long and  is 30?
Also find the tension in the cord.
ө
Answer:
v = 1.84 m/sec;
T = 16.97 N
Bill the Cat, tied to a rope, is twirled around in a
vertical circle. Draw the free-body diagram for Bill in
the positions shown. Then sum the X and Y forces.
ΣFy = mac
FT + mg = mv²/r
FT = mv²/r - mg
ac
FT = m ((v²/r) - g)
ΣFy = mac
ac
mg
FT
FT
FT - mg = mv²/r
FT = mv²/r + mg
FT = m ((v²/r) + g)
mg
Minimum velocity needed for an object to
continue moving in a vertical circle. Any less
velocity and the object will fall.
At this point, FT = 0, so…
ΣFy = mac
FT + mg = mv²/r
0 + mg = mv²/r
g = v2/r
rg = v2 or, vc = rg
Practice
1. A race car travels around a flat, circular track with a radius of 180 m.
….The coefficient of friction between the tires and the pavement is 1.5.
a. Draw all forces on the car, including direction of acceleration.
b. Write the summation and net force equations (x & y).
c. Calculate the maximum velocity the car can go and stay on the track.
20°
2. A ball is swung in a horizontal circle.
The 1.5 m long string makes a 20° to
the horizontal.
a. Draw all forces on the picture, including direction of acceleration.
b. Write the summation and net force equations (x & y)
c. Determine the velocity of the ball?
Practice #3
3. A loop-the-loop rollercoaster has a radius of
20 m. Draw a FBD (at the top of the loop)
showing all forces and calculate the minimum
velocity the roller coaster must have in order to
stay on the track.
v = __________
Example Problem
Uniform Circular Motion
A 13-g bucket is attached to a 0.93-m string. The
bucket is swung in a horizontal circle, making
one revolution in 1.18 s. Find the tension force
exerted by the string on the bucket.
G-Forces

G-forces are used for explaining the relative effects of
centripetal acceleration that a rider feels while on a
roller coaster.

The greater the centripetal acceleration, the greater
the G-forces felt by the passengers.


A force of 1 G is the usual force of the Earth’s gravitational
pull that a person feels when they are at rest on the Earth’s
surface; it is a person’s normal weight.
When a person feels weightless, as in free fall or in
space, they are experiencing 0 G’s.


When the roller coaster train is going down a hill, the
passengers usually undergo somewhere between 0
and 1 G.
G-Force Continues

If the top of the hill is curved more narrowly than a
parabola, the passengers will experience negative G’s as
they rise above the seat and get pushed down by the lap
bar. This is because gravity and the passengers’ inertia
would have them fall in a parabolic arc.

G-forces greater than 1 can be felt at the bottom of hills as
the train changes direction. In this case the train is
pushing up on the passengers with more than the force of
gravity because it is changing their direction of movement
from down to up. G-forces that are felt when changing
direction horizontally are called lateral G’s. Lateral G’s
can be converted into normal G-forces by banking turns.
Inertia and the Right Hand Turn
•You are a passenger in a car which is making a right-hand turn. As the car
begins to take the turn to the right, you often feel as though you are sliding to
the left. The car is turning to the right due to the inward force, yet you feel as
though you are being forced leftward or outward. In actuality, the car is
beginning its turning motion (to the right) while you continue in a straight line
path.
•Observe in the animation that the passenger (in blue) continues in a straightline motion for a short period of time after the car begins to make its turn.
•An inward net force is required to make a turn in a circle. In the absence of
any net force, an object in motion (such as the passenger) continues in motion
in a straight line at constant speed. This is Newton's first law of motion.
The End!
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