Transcript SPH4UI Lecture 1 Notes

SPH3UW: Lecture 1
“Kinematics”
Fundamental Units


How we measure things!
All things in classical mechanics can be expressed in terms of the
fundamental units:




Length: L
Mass : M
Time : T
For example:
 Speed has units of L / T (e.g. miles per hour).
 Force has units of ML / T2 etc... (as you will learn).
Units...

SI (Système International) Units:
 mks: L = meters (m), M = kilograms (kg), T = seconds (s)
 cgs: L = centimeters (cm), M = grams (gm), T = seconds (s)

British Units:
 Inches, feet, miles, pounds, slugs...

We will use mostly SI units, but you may run across some
problems using British units. You should know where to look to
convert back & forth.
Ever heard of Google
Converting between different
systems of units

Useful Conversion factors:
 1 inch
= 2.54 cm
 1m
= 3.28 ft
 1 mile
= 5280 ft
 1 mile
= 1.61 km

Example: convert miles per hour to meters per second:
mi
mi
ft
1 m
1 hr
m
1
 1  5280


 0.447
hr
hr
mi 3.28 ft 3600 s
s
Dimensional Analysis

This is a very important tool to check your work
 It’s also very easy!

Example:
Doing a problem you get the answer distance
d = vt 2 (velocity x time2)
Units on left side = L
Units on right side = L / T x T2 = L x T

Left units and right units don’t match, so answer must be
wrong!!
Dimensional Analysis

The period P of a swinging pendulum depends only on
the length of the pendulum d and the acceleration of
gravity g.
 Which of the following formulas for P could be
correct ?
(a) P = 2
(dg)2
(b)
d
P  2
g
(c)
P  2
d
g
Given: d has units of length (L) and g has units of (L / T 2).
Solution


Realize that the left hand side P has units of
time (T )
Try the first equation
2
(a)
(a)
L 
L4

L 2   4  T
 T  T
P  2  dg 
2
(b)
Not Right!
P  2
d
g
(c) P  2
d
g
Solution


Realize that the left hand side P has units of
time (T )
Try the second equation
(b)
(a)
L
T2 T
L
T2
P  2  dg 
2
Not Right!
(b)
P  2
d
g
(c) P  2
d
g
Solution


Realize that the left hand side P has units of
time (T )
Try the first equation
(c)
(a)
L
 T2 T
L
T2
P  2  dg 
2
(b)
Dude, this is it
P  2
d
g
(c) P  2
d
g
Time to Drop an Apple
Want to ask myself a question: If I drop an Apple from a certain height,
h, what then happens to the time, t, it takes for the apple to fall?
The time t, must be proportional to the height to the power of some
value. The time t may be proportional to the mass of the Apple to the
power of some value. The time t, may be proportional to the
acceleration due to gravity, g to some power


t  Ch m g

1

  L 
T

C
L
M
 
     2
T 
There is no M on Left side:   0
There is no L on Left side:
   0
There is T to power 1 on Left side: 1  2
1
1





Therefore
2
2

Dimensional
Analysis to the
Rescue
1
2
t  Ch m 0 g
C
 h
h
g

1
2
Time for Apple to Drop
t h
Dimensional Analysis tells us, that if we increase the height by a factor
of 100, then to time will increase by the square root of 100, or by a
factor of 10
Let’s verify this by dropping an apple from 3 m then from 1.5 m and
compare the times
 x   y 
e  E     
 x   y 
2
2
2
 2   2 
 1.417 
 

 781   551 
 0.006
2
Since we double the height, the time
should be
2  1.4142 longer.
The experiment shows: 1.417±0.006
781ms
 1.417
551ms
Vectors

A vector is a quantity that involves both
magnitude and direction.
55 km/h [N35E]
 A downward force of 3 Newtons


A scalar is a quantity that does not involve
direction.
55 km/h
 18 cm long

Vector Notation

Vectors are often identified with arrows
in graphics and labeled as follows:
We label a vector with a variable.
This variable is identified as a vector either by an arrow
above itself :
A
Or
By the variable being BOLD:
A
Displacement

Displacement is an object’s change in position.
Distance is the total length of space traversed
by an object.
1m
6.7m
3m
Start
= 500 m
Finish
5m
Displacement:
 6m   3m
2
2
 6.7m
Distance: 5m  3m  1m  9m
Displacement = 0 m
Distance = 500 m
Vector Addition
A
R
E
B
R
E
C
B A
D
D
E
D
R
C
A
C
B
A + B + C + D + E = Distance
R = Resultant = Displacement
Rectangular Components
Quadrant II
R  A2  B2
R sin  
y
A
R

-x
A opp
sin   
R hyp
B adj
cos   
R hyp
A opp
tan   
B adj
Quadrant I
R cos  
Quadrant III
B
Quadrant IV
-y
x
Vectors...

The components (in a particular coordinate system) of r,
the position vector, are its (x,y,z) coordinates in that
coordinate system
 r = (rx ,ry ,rz ) = (x,y,z)

Consider this in 2-D (since it’s easier to draw):
where r = |r |
 rx = x = r cos 
 ry = y = r sin 
(x,y)
y
arctan( y / x )
r

x
Vectors...

The magnitude (length) of r is found using the
Pythagorean theorem:
r
y
r r
x2  y 2
x

The length of a vector clearly does not depend on its direction.
Vector Example



Vector A = (0,2,1)
Vector B = (3,0,2)
Vector C = (1,-4,2)
What is the resultant vector, D, from adding A+B+C?
(a) (3,5,-1)
(b) (4,-2,5)
(c) (5,-2,4)
Resultant of Two Forces
• force: action of one body on another;
characterized by its point of application,
magnitude, line of action, and sense.
• Experimental evidence shows that the
combined effect of two forces may be
represented by a single resultant force.
• The resultant is equivalent to the
diagonal of a parallelogram which
contains the two forces in adjacent legs.
• Force is a vector quantity.
Vectors
• Vector: parameters possessing magnitude and direction
which add according to the parallelogram law. Examples:
displacements, velocities, accelerations.
• Scalar: parameters possessing magnitude but not
direction. Examples: mass, volume, temperature
P
Q
• Vector classifications:
- Fixed or bound vectors have well defined points of
application that cannot be changed without affecting
an analysis.
- Free vectors may be freely moved in space without
changing their effect on an analysis.
- Sliding vectors may be applied anywhere along their
line of action without affecting an analysis.
P
-P
• Equal vectors have the same magnitude and direction.
• Negative vector of a given vector has the same magnitude
and the opposite direction.
Addition of Vectors
• Trapezoid rule for vector addition
• Triangle rule for vector addition
P
• Law of cosines,
Q
R 2  P 2  Q 2  2 PQ cos B
Q
R  PQ
P
P
Q
• Law of sines,
sin A sin B sin C


Q
R
A
• Vector addition is commutative,
PQ  Q P
-Q
Q
P-Q
P
• Vector subtraction
 
P  Q  P  Q
Addition of Vectors
Q
S
• Addition of three or more vectors through
repeated application of the triangle rule
P
Q
S
P
• The polygon rule for the addition of three or
more vectors.
• Vector addition is associative,



PQ S  PQ  S  P Q S
P
2P
-1.5P

• Multiplication of a vector by a scalar
increases its length by that factor (if scalar
is negative, the direction will also change.)
Resultant of Several Concurrent Forces
• Concurrent forces: set of forces
which all pass through the same
point.
A set of concurrent forces applied
to a particle may be replaced by
a single resultant force which is
the vector sum of the applied
forces.
• Vector force components: two or
more force vectors which,
together, have the same effect as
a single force vector.
Sample Problem
• Graphical solution - construct a
parallelogram with sides in the
same direction as P and Q and
lengths in proportion.
Graphically evaluate the
resultant which is equivalent in
direction and proportional in
magnitude to the diagonal.
Two forces act on a bolt
at A. Determine their
resultant.
• Trigonometric solution - use the
triangle rule for vector addition
in conjunction with the law of
cosines and law of sines to find
the resultant.
Sample Problem Solution
R
Q

P
R  98 N   35
• Graphical solution - construct a
parallelogram with sides in the same
direction as P and Q and lengths in
proportion. Graphically evaluate the
resultant which is equivalent in
direction and proportional in
magnitude to the diagonal.
Sample Problem Solution
• Trigonometric solution
From the Law of Cosines,
R 2  P 2  Q 2  2 PQ cos B
  40N    60N   2  40N  60N  cos155
2
2
R  97.73N
From the Law of Sines,
sin A sin B

Q
R
sin A  sin B
Q
R
 sin155
A  15.04
  20  A
 35.04
60N
97.73N
Sample Problem
• Find a graphical solution by applying the
Parallelogram Rule for vector addition.
The parallelogram has sides in the
directions of the two ropes and a
diagonal in the direction of the barge
axis and length proportional to 5000 N.
A barge is pulled by two
• Find a trigonometric solution by
tugboats. If the resultant of the
applying the Triangle Rule for vector
forces exerted by the tugboats
addition. With the magnitude and
is 5000 N directed along the
direction of the resultant known and the
axis of the barge, determine
directions of the other two sides parallel
to the ropes given, apply the Law of
a) the tension in each of the
Sines to find the rope tensions.
o
ropes for α = 45 ,
Sample Problem
• Graphical solution Parallelogram Rule with known
resultant direction and
magnitude, known directions for
sides.
T1  3700 N T2  2600 N
T1
30
5000N
45
• Trigonometric solution Triangle Rule with Law of Sines
30
45
T2
T1
T2
5000 N


sin 45 sin 30 sin105
5000N
45
T2
30
105
T1
T1  3700 N T2  2600 N
Rectangular Components of a
Force: Unit Vectors
• May resolve a force vector into perpendicular
components so that the resulting parallelogram is
a rectangle. Fx and Fy are referred to as
rectangular vector components
• Define perpendicular unit vectors iˆ and ˆj which
are parallel to the x and y axes.
• Vector components may be expressed as
products of the unit vectors with the scalar
magnitudes of the vector components.
F  Fxiˆ  Fy ˆj
Fx and Fy are referred to as the scalar
components of Fx and Fy

F
Addition of Forces by Summing Components
• Wish to find the resultant of 3 or more
concurrent forces,
P
Py j
S
Sy j
Qxi Px i
S xi
Qy j
Q
R  PQ S
• Resolve each force into rectangular
components
Rxi  Ry j  Pxi  Py j  Qxi  Qy j  S xi  S y j
  Px  Qx  S x  i   Py  Qy  S y  j
Ry j
R

Rxi
• The scalar components of the resultant are
equal to the sum of the corresponding
scalar components of the given forces.
Ry  Py  Qy  S y
Rx  Px  Qx  S x
  Fx
  Fy
• To find the resultant magnitude and
direction,
R R R
2
x
2
y
  tan
1
Ry
Rx
Sample Problem
Plan:
Four forces act on bolt A as
shown. Determine the
resultant of the force on the
bolt.
• Resolve each force into
rectangular components.
• Determine the components of
the resultant by adding the
corresponding force
components.
• Calculate the magnitude and
direction of the resultant.
Sample Problem Solution
• Resolve each force into rectangular
components.
force mag
F1 150
F2
80
F3 110
F4 100
x  comp
129.9
27.4
0
96.6
Rx  199.1
y  comp
75.0
75.2
110.0
25.9
Ry  14.3
• Determine the components of the
resultant by adding the
corresponding force components.
• Calculate the magnitude and direction.
R  199.1 14.3
2
2
tan  
14.3 N
199.1N
R  199.6N
  4.1
Equilibrium of a Particle
• When the resultant of all forces acting on a particle is zero, the
particle is in equilibrium.
• Newton’s First Law: If the resultant force on a particle is zero, the particle
will remain at rest or will continue at constant speed in a straight line.
• Particle acted upon by
two forces:
- equal magnitude
- same line of action
- opposite sense
• Particle acted upon by three or more
forces:
- graphical solution yields a closed
polygon
R  F  0
- algebraic solution
 Fx  0  Fy  0
Free-Body Diagrams
TAB
50
Space Diagram: A sketch
showing the physical
conditions of the problem.
A
TAC
30
736N
Free-Body Diagram: A sketch
showing only the forces on the
selected particle.
Sample Problem
Plan of Attack:
• Construct a free-body diagram
for the particle at the junction
of the rope and cable.
In a ship-unloading operation, a
3500-lb automobile is supported
by a cable. A rope is tied to the
cable and pulled to center the
automobile over its intended
position. What is the tension in
the rope?
• Apply the conditions for
equilibrium by creating a
closed polygon from the forces
applied to the particle.
• Apply trigonometric relations
to determine the unknown
force magnitudes.
Sample Problem
SOLUTION:
• Construct a free-body diagram
for the particle at A.
• Apply the conditions for
equilibrium in the horizontal
and vertical directions.
TAB
horizontal
2
A
 cos  88  TAB  cos  30  TAC  0
TABy  Tcar  TACy  0
30
TAC
3500lb
TABx  TACx  0
Vertical cos  2 T  3500lb  sin  30 T  0
AB
AC
Sample Problem
 cos 88TAB  cos 30TAC  0
cos  2TAB  3500lb  sin 30TAC  0
TAC
cos 88 TAB

cos  30
sin  30 TAC  3500lb
TAB 
cos  2
TAB
2
TAC
A
30
cos  88   sin  30  TAC  3500lb 



cos  30  
cos  2 

 0.02016015TAC  141.12105lb
0.979839TAC  141.12105lb
TAC  144lb
3500lb
Sample Problem
• Solve for the unknown force
magnitudes using Sine Law.
TAB
TAC
3500lb


sin120 sin 2 sin 58
TAB  3570lb
TAC  144lb
TAC
58
120
TAB
3500lb
2
Sometimes the Sine Law / Cosine
Law is faster than component
vectors. Intuition should tell you
which is best.
Sample Problem
PLAN OF ATTACK:
• Choosing the hull as the free body,
draw a free-body diagram.
It is desired to determine the drag
force at a given speed on a
prototype sailboat hull. A model is
placed in a test channel and three
cables are used to align its bow on
the channel centerline. For a given
speed, the tension is 40 lb in cable
AB and 60 lb in cable AE.
Determine the drag force exerted
on the hull and the tension in cable
AC.
• Express the condition for
equilibrium for the hull by writing
that the sum of all forces must be
zero.
• Resolve the vector equilibrium
equation into two component
equations. Solve for the two
unknown cable tensions.
Sample Problem
SOLUTION:
• Choosing the hull as the free
body, draw a free-body diagram.
1.5 ft
7 ft
tan


 0.375


tan   
 1.75
4
ft
4 ft
  20.56
  60.25
TAC
TAB=40
60.26
69.44
A
FD
• Express the condition for
equilibrium for the hull by
writing that the sum of all forces
must be zero.
TAE=60
R  TAB  TAC  TAE  FD  0
Sample Problem
TAC cos  69.44
TAB
40sin  60.26
TAC
TAC sin  69.44
40cos  60.26
A
FD
• Resolve the vector equilibrium
equation into two component
equations. Solve for the two
unknown cable tensions.
TAE  TAC sin  69.44   40cos  60.26 
60  0.936305TAC  19.842597
TAE=60
TAC  42.889
FD  TAC cos  69.44   40sin  60.26 
 0.351188TAC  34.73142
   0.351188  42.889   34.73142
 19.66 lb
Sample Problem
This equation is satisfied only if all
vectors when combined, complete a
closed loop.
Rectangular Components in Space
• The vector F
is contained in
the plane
OBAC.
• Resolve Fh into
• Resolve F into
horizontal and vertical rectangular
components
components.
Fh  F sin  y
Fy  F cos y
Fx  Fh cos 
 F sin  y cos 
Fy  Fh sin 
 F sin  y sin 
Rectangular Components in Space
• With the angles between F and
the axes, Fx  F cos x Fy  F cos y Fz  F cos z
F  Fxi  Fy j  Fz k

 F cos  xi  cos y j  cos z k

 F
  cos x i  cos y j  cos z k

of F
•  is a unit vector along the line
action of F
and cosx , cos y , and cosz are the
direction cosines for F
Rectangular Components in
Space
Direction of the force is defined by the location
of two points, M  x1, y1, z1  and N  x2 , y2 , z2 
d  vector joining M and N
 d xi  d y j  d z k
d x  x2  x1 d y  y2  y1 d z  z2  z1
F  F
1
  d xi  d y j  d z k
d
Fd
Fd
Fd
Fx  x Fy  y Fz  z
d
d
d

d is the length of the vector F

Sample Problem
PLAN of ATTACK:
The tension in the guy wire is
2500 N. Determine:
a) components Fx, Fy, Fz of the
force acting on the bolt at A,
b) the angles x, y, z defining
the direction of the force
• Based on the relative
locations of the points A and
B, determine the unit vector
pointing from A towards B.
• Apply the unit vector to
determine the components of
the force acting on A.
• Noting that the components of
the unit vector are the
direction cosines for the
vector, calculate the
corresponding angles.
Sample Problem
SOLUTION:
• Determine the unit vector pointing from A towards B.
AB   40m  i  80m  j   30m  k
AB 
 40m   80m    30m 
2
2
 94.3 m
 40 
 80 
 30 
 
i  
 j 
k
 94.3   94.3 
 94.3 
 0.424 i  0.848 j  0.318k
• Determine the components of the force.
F  F

  2500 N  0.424 i  0.848 j  0.318k
  1060 N  i   2120 N  j   795 N  k

2
Sample Problem
• Noting that the components of the unit vector are the direction
cosines for the vector, calculate the corresponding angles.
  cos x i  cos y j  cos z k
 0.424 i  0.848 j  0.318k
 x  115.1
 y  32.0
 z  71.5
Motion in 1 dimension

In 1-D, we usually write position as x(t).

Since it’s in 1-D, all we need to indicate direction is + or .

Displacement in a time t = t2 - t1 is x = x(t2) - x(t1) = x2 - x1
x
x
some particle’s trajectory
in 1-D
x
2
x
1
t1
t
t2
t
1-D kinematics


Velocity v is the “rate of change of position”
Average velocity vav in the time t = t2 - t1 is:
vav 
x(t2 )  x(t1 ) x

t2  t1
t
x
x
trajectory
x
2
Vav = slope of line connecting x1 and x2.
x
1
t1
t2
t
t
1-D kinematics...


Consider limit t1 t2
Instantaneous velocity v is defined as:
v (t ) 
dx (t )
dt
so v(t2) = slope of line tangent to path at t2.
x
x
x
2
x
1
t1
t2
t
t
1-D kinematics...


Acceleration a is the “rate of change of velocity”
Average acceleration aav in the time t = t2 - t1 is:
v(t2 )  v(t1 ) v
aav 

t2  t1
t

And instantaneous acceleration a is defined as:
dv(t ) d 2 x(t )
a(t ) 

dt
dt 2
Calculus way of saying t
gets very very small
dx (t )
using v (t ) 
dt
Recap

If the position x is known as a function of time, then we can
find both velocity v and acceleration a as a function of
time!
x
x  x( t )
dx
dt
dv
d 2x
a 

dt
dt 2
v 
v
a
Calculus (don’t worry you will
understand this in next year.)
t
t
t
More 1-D kinematics


We saw that v = dx / dt
In “calculus” language we would write dx = v dt, which we
can integrate to obtain:
t2
x(t2 )  x(t1 )   v(t )dt
t1

Graphically, this is adding up lots of small rectangles:
v(t)
+ +...+
= displacement
t
Recap

So for constant acceleration we
find:
x
1
x  x0  v 0 t  at 2
2
v  v 0  at
v
t
a  const
a
t
t
Motion in One Dimension
Question

When throwing a ball straight up, which of the following is
true about its velocity v and its acceleration a at the
highest point in its path?
(a) Both v = 0 and a = 0.
(b) v  0, but a = 0.
(c) v = 0, but a  0.
y
Solution

Going up the ball has positive velocity, while coming down
it has negative velocity. At the top the velocity is
momentarily zero.
x

Since the velocity is
continually changing there must
be some acceleration.
 In fact the acceleration is caused
by gravity (g = 9.81 m/s2).
 (more on gravity in a few lectures)

The answer is (c) v = 0, but a  0.
v
t
t
a
t
Recap:
This is just
for constant
acceleration!

For constant acceleration:
1
x  x0  v 0 t  at 2
2
v  v 0  at
a  const

From which we know:
v 2  v 02  2a(x  x0 )
v av
1
 (v 0  v)
2
Recap:

For constant acceleration:
1 2
x  x0  v0t  at
2
Problem Solving Tips:




Read Carefully!
 Before you start work on a problem, read the problem
statement thoroughly. Make sure you understand what
information is given, what is asked for, and the meaning of
all the terms used in stating the problem.
Using what you are given, set up the algebra for the problem and
solve for your answer algebraically
 Invent symbols for quantities you know as needed
 Don’t plug in numbers until the end
Watch your units !
 Always check the units of your answer, and carry the units
along with your formula during the calculation.
Understand the limits !
 Many equations we use are special cases of more general
laws. Understanding how they are derived will help you
recognize their limitations (for example, constant
acceleration).
1-D Free-Fall


This is a nice example of constant acceleration (gravity):
In this case, acceleration is caused by the force of gravity:
 Usually pick y-axis “upward”
y
 Acceleration of gravity is “down”:
ay   g
v y = v 0y - gt
t
v
t
1 2
y  y0  v0 y t  g t
2
a
y
ay =  g
t
Gravity facts:
Penny
& feather

g does not depend on the nature of the material!
 Galileo (1564-1642) figured this out without fancy clocks &
rulers!
 On the surface of the earth, gravity acts to give a constant
acceleration

demo - feather & penny in vacuum

Nominally, g = 9.81 m/s2



At the equator
g = 9.78 m/s2
At the North pole g = 9.83 m/s2
More on gravity in a few lectures!
Gravity facts:




Actually, gravity is a “fundamental force”.
 Other fundamental forces: electric force, strong and weak
forces
It’s a force between two objects, like me and the earth. or earth
and moon, or sun and Neptune, etc
Gravitational Force is proportional to product of masses:
 F(1 acting on 2) proportional to M1 times M2
 F(2 acting on 1) proportional to M1 times M2 too!
Proportional to 1/r2
 r is the separation of the 2 masses
 For gravity on surface of earth, r = radius of earth
 Example of Gauss’s Law (more on this later)
At the surface of earth gravitational force attracts “m” toward the
center of the earth, is approximately constant and equal to mg.
The number g=9.81 m/s2 contains the effect of Mearth and rearth.
Question:

The pilot of a hovering helicopter
drops a lead brick from a height
of 1000 m. How long does it
take to reach the ground and
how fast is it moving when it gets
there? (neglect air resistance)
1000 m
Solution:

First choose coordinate system.
 Origin and y-direction.

Next write down position equation:
y  y 0  v 0y t

1 2
gt
2
1000 m
Realize that v0y = 0.
1 2
y  y0  gt
2
y
y=0
Solution:

Solve for time t when y = 0 given
that y0 = 1000 m.
1
y  y0 - gt 2
2
t

2 y0
2 1000m

 14.3s
2
g
9.81 m s
y0 = 1000 m
Recall:
vy2 - v02y  2a( y - y0 )

Solve for vy:
y
v y   2 gy0
 140 m / s
y=0
1D Free Fall

(a)
Alice and Bob are standing at the top of a cliff of
height H. Both throw a ball with initial speed v0, Alice
straight down and Bob straight up. The speed of the
balls when they hit the ground are vA and vB
respectively. Which of the following is true:
vA < vB
(b) vA = vB
Alice v0
(c) vA > vB
Bob
v0
H
vA
vB
1D Free fall

Since the motion up and back down is symmetric, intuition
should tell you that v = v0
 We can prove that your intuition is correct:
2
2
v

v
0  2( g )  H  H   0
Equation:
Bob
v0
H
v = v0
This looks just like Bill threw the
ball down with speed v0, so
the speed at the bottom should
be the same as Alice’s ball.
y=0
Does motion in one direction affect
motion in an orthogonal direction?






For example, does motion in the y-direction affect motion in
the x-direction?
It depends….
For simple forces, like gravitational and electric forces, NO
For more complicated forces/situations, like magnetism, YES
In any case, vectors are the mathematical objects that we
need to use to describe the motion
Vectors have
 Magnitude
 Units (like meters, Newtons, Volts/meter,
meter/sec2…)
 Direction
Recall Vectors:

In 1 dimension, we could specify direction with a + or - sign.
For example, in the previous problem ay = -g etc.

In 2 or 3 dimensions, we need more than a sign to specify the
direction of something:

To illustrate this, consider the position vector r in 2 dimensions.
Example: Where is Waterloo?
 Choose origin at Toronto
 Choose coordinates of
distance (km), and
direction (N,S,E,W)
 In this case r is a vector that
points 120 km north.
Waterloo
r
Toronto
A vector is a quantity with a magnitude and a direction
2-D Kinematics

Most 3-D problems can be reduced to 2-D problems when
acceleration is constant:
 Choose y axis to be along direction of acceleration
 Choose x axis to be along the “other” direction of
motion

Example: Throwing a baseball (neglecting air resistance)
 Acceleration is constant (gravity)
 Choose y axis up: ay = -g
 Choose x axis along the ground in the direction of the
throw
“x” and “y” components of motion are independent.
A man on a train tosses a ball straight up in the air.
View this from two reference frames:
Reference frame
on the moving train.

Reference frame
on the ground.
Problem:

David Eckstein clobbers a fastball toward center-field. The
ball is hit 1 m (yo ) above the plate, and its initial velocity is
36.5 m/s (v ) at an angle of 30o () above horizontal. The
center-field wall is 113 m (D) from the plate and is 3 m (h)
high.


What time does the ball reach the fence?
Does David get a home run?
v
y0

h
D
Problem...




Choose y axis up.
Choose x axis along the ground in the direction of the hit.
Choose the origin (0,0) to be at the plate.
Say that the ball is hit at t = 0, x(0) = x0 = 0. y(0) = y0 = 1m

Equations of motion are:
vx = v0x
x = vxt
vy = v0y - gt
y = y0 + v0y t - 1/ 2 gt2
Problem...

Use geometry to figure out v0x and v0y :
y
g
v
y0
Find
and

v0x = |v| cos .
v0y = |v| sin .
v0y
v0x
remember, we were told that  = 30 deg
x
Problem...




The time to reach the wall is: t = D / vx (easy!)
We have an equation that tell us y(t) = y0 + v0y t + a t2/ 2
So, we’re done....now we just plug in the numbers: a = -g
Find:
 vx = 36.5 cos(30) m/s = 31.6 m/s
 vy = 36.5 sin(30) m/s = 18.25 m/s
 t = (113 m) / (31.6 m/s) = 3.58 s
 y(t) = (1.0 m) + (18.25 m/s)(3.58 s)
- (0.5)(9.8 m/s2)(3.58 s)2
= (1.0 + 65.3 - 62.8) m = 3.5 m

Since the wall is 3 m high, Eckstein gets the homer!!
Thinking deeper: Can you figure out what angle gives the longest fly ball?
To keep things simple, assume y0 = 0, and go from there…
Motion in 2D


Two footballs are thrown from the same point on a flat field.
Both are thrown at an angle of 30o above the horizontal. Ball 2
has twice the initial speed of ball 1. If ball 1 is caught a
distance D1 from the thrower, how far away from the thrower
D2 will the receiver of ball 2 be when he catches it?
Assume the receiver and QB are the same height
(a) D2 = 2D1
(b) D2 = 4D1
(c) D2 = 8D1
Solution

The distance a ball will go is simply
x = (horizontal speed) x (time in air) = v0x t

To figure out “time in air”, consider the
1
y

y

v
t

g
0
0y
equation for the height of the ball:
2

When the ball is caught, y = y0
1

t  v0 y  g
2

1
v0 y t  g t 2  0
2
t2

t 0

two
solutions
t2
v0 y
g
t 0
(time of catch)
(time of throw)
Solution
v0 y

So the time spent in the air is proportional to v0yt: 2

Since the angles are the same, both v0y and v0x for ball 2
are twice those of ball 1.
v0,2
ball 1
v0x ,1

v0,1
g
v0y ,2
ball 2
v0y ,1
v0x ,2
Ball 2 is in the air twice as long as ball 1, but it also has twice
the horizontal speed, so it will go 4 times as far!!
Projectile Motion
As you can see, it can become difficult to solve problems
that involve motion in both the x and y axis. Lucky for you,
people from all over the world have had the same
difficulties. Therefore a complete set of equations have
been created that will help solve these problems. These are
known as ballistic formulas. They assume launch height and
landing height are the same.
Range
Distance:
vi2 sin  2 
d x 
g
2vi sin  
Travel Time: t 
g
Height
Maximum:
h
 vi sin   
2g
2
Time to top:
vi sin  
t 
g
Motion in 2D Again


Two footballs are thrown from the same point on a flat field.
Both are thrown at an angle of 30o above the horizontal. Ball 2
has twice the initial speed of ball 1. If ball 1 is caught a
distance D1 from the thrower, how far away from the thrower
D2 will the receiver of ball 2 be when he catches it?
Assume the receiver and QB are the same height
(a) D2 = 2D1
(b) D2 = 4D1
2vo  sin  2 

d x 
g
(c) D2 = 8D1
2
vi2 sin  2 
4
g
vi2 sin  2 
d x 
g
Example
A golfer hits a golf ball so that it leaves the club with an
initial speed v0=37.0 m/s at an initial angle of 0=53.1o.
a) Determine the position of the ball when t=2.00s.
b) Determine when the ball reaches the highest point
of its flight and find its height, h, at this point.
c) Determine its horizontal range, R.
a)
v0 x  v0 cos  0    37.0m / s  cos  53.1   22.2m / s
v0 y  v0 sin  0    37.0m / s  sin  53.1   29.6m / s
The x-distance: x  v0xt   22.2m / s  2.00s   44.4m
1 2
1
2
2
y

v
t

gt

29.6
m
/
s
2.00
s

9.80
m
/
s
2
.0
0
s
 39.6m







0y
The y-distance:
2
2
Example
A golfer hits a golf ball so that it leaves the club with an
initial speed v0=37.0 m/s at an initial angle of 0=53.1o.
a) Determine the position of the ball when t=2.00s.
b) Determine when the ball reaches the highest point
of its flight and find its height, h, at this point.
c) Determine its horizontal range, R.
v sin   

h
2
i
2g

  37.0m / s  sin  53.1 
2  9.80m / s
2
2

 44.7m
vi sin  
g
 37.0m / s  sin  53.1  3.02s

9.80m / s 2
t 
vi2 sin  2 
d x 
g

 37.0m / s 
 134m
2
sin  2  53.1  
9.80m / s 2
Shooting the Monkey
(tranquilizer gun)

Where does the zookeeper
aim if he wants to hit the monkey?
( He knows the monkey will
let go as soon as he shoots ! )
Shooting the Monkey...

If there were no gravity, simply aim
at the monkey
r =v0t
r = r0
Shooting the Monkey...

With gravity, still aim at the monkey!
r = v0 t -
1/
2g
t2
r = r0 - 1 / 2 g t 2
Dart hits the
monkey!
Recap:
Shooting the monkey...
x = v0 t
y = -1/2 g t2

This may be easier to think about.
It’s exactly the same idea!!
They both have the same Vy(t) in this case
x = x0
y = -1/2 g t2
Feeding the Monkey
Kinematics Flash Review