Linear Motion
Angular Motion
TRANSLATION ONLY!
ROTATION ONLY!
Object rotates about a fixed point (axis)
Object maintains angular
orientation (q)
but this point does NOT have to lie
within the object
measured in meters - SI unit
other units - inches, feet, miles,
centimeters, millimeters
Rectilinear motion - if path of one point
on object is a straight line
Curvilinear motion – if path of one
point on object is curved
q
measured in radians – SI Unit
other units – degrees, revolutions
q
Types of Motion
• General
– combination of linear and angular motion
• translation and rotation
B
B
A
A
Kinematics is the study of motion without
regard for the forces causing the motion or …
the description of motion
• there are three basic kinematic variables
– position, velocity and acceleration
• the position of an object is simply its location in
space
– changes in position can be described by distance or
displacement
• the velocity of an object is how fast it is changing
its position
• the acceleration of an object is how fast the
velocity is changing
Acquisition of Position Data
Frame 1
(x1,y1)
(x2,y2)
Y
(x4,y4)
(x5,y5)
(0,0)
(x3,y3)
X
Position data is often
acquired by digitizing the
x and y coordinates from
film or video.
Velocities and
accelerations are
calculated from the
position data.
Position and Displacement (d)
• Position (s) is the location of an object in space
• units: m, cm, km, in, ft, mi
• Displacement (Ds= sf - si) is the change in
position of an object
s2
displacement = d
s1
d
d = s1 – s2
• Problem:
• how do you describe s1 and s2?
• If you put the arrow on graph paper you
describe position with x- & y-coordinates
Y
s2 = (x2,y2)
d
d = s1 – s2
s1 = (x1,y1)
X
d = s 1 – s2
d = (x1,y2) – (x2,y2)
How do you do this?
•Realize that displacement is a vector so you must
determine either the Cartesian or polar coordinates
Y
s2 = (x2,y2)
d
d = s1 – s2
s1 = (x1,y1)
X
•Two choices to describe vector
•Cartesian Coordinates (dx,dy)
•dx = x2 – x2 = distance in the x-direction
•dy = y2 – y1 = distance in the y-direction
•Polar Coordinates (d,q)
•“How far and in which direction”
d  ( x2  x1 ) 2  ( y2  y1 ) 2
q = measured directly from graph
Second problem: Since this movement occurs over time,
displacement (as a vector) does NOT represent changes in
the direction of movement well.
For example – what if s1 represents you at Building A and
s2 represents you at Building B 10 minutes later.
Y
s2
d
d = s1 – s2
s1
X
Assuming this city is like most cities you have to walk up and
down city blocks and not through buildings.
Y
s2
d
s1
X
So your actual route is around the buildings, traveling up and
down city blocks.
Y
s2
d
s1
dy
dx
X
Thus the actual distance you covered is more than displacement represents
distance = the length of your travel in the x-direction (dx) plus
the length of your travel in the y-direction (dy)
BUT since we are only concerned with the length of travel we
don’t distinguish between directions
Y
distance = dx + dy
d
s1
s2
dy
dx
X
Distance ( )
• distance is the length of the path traveled
• it is a scalar - “How far”
• units: same as displacement
dx + dy = distance =
d
dy
dx
Note: use “ ” for length
Example - Distance vs.
Displacement
N
leg 3 = 2 miles
leg 2 = 3 miles
leg 1 = 2 miles
Total DISTANCE Traveled
= 2 miles + 3 miles + 2 miles
= 7 miles
Describing Displacement
N
Describing Displacement
First Method (Cartesian)
3 miles East
4 miles North
(3, 4) miles
put ‘horizontal’ coordinate 1st
put ‘vertical’ coordinate 2nd
Displacement Magnitude
Second Method (Polar)
1st - calculate length of
displacement vector
N
q
3 miles

d  32  4 2

d  25

d  5 miles
Displacement Direction
2nd – Measure the angle
qfrom a carefully drawn
diagram.
N
q  53.1o
q
3 miles
Displacement Vector
(Polar Notation)
Describe the displacement
vector by its length and
direction
N

d  5 miles @ 531
.
q
3 miles
Average Speed
• speed is a scalar quantity
• it is the rate of change of distance wrt time
• units: same as velocity
Speed  distance
time
What is the average speed of the
basketball?
(80,40)
(60,10)
(0,0)
l=
20  30  36 feet
2
2
l
36
speed  
 72 ft/s
t
0.5
Average Velocity (v)
• rate of change of displacement wrt time
• velocity is a vector quantity
– “How fast and in which direction”
• units: m/s, km/hr, mi/hr, ft/s

 Dd
velocity v 
Dt
NOTE: displacement (d)
is a vector so must obey
rules of vector algebra
when computing velocity.
•When two velocities act on an object you find the net
or resultant effect by adding the velocities.
•Because velocity is a vector you can’t simply add the
numbers.
•Instead – you must use vector algebra to add the
velocities.
In this example the boat
is propelled to the right
by its motor while the
river’s current carries it
towards the top of the
picture. This describes 2
velocities
Other examples of velocities
that can be added together
include the wind direction
when flying.
Adding Velocities
Use the laws of
vector algebra.
Example - the path
of the swimmer is
determined by the
vector sum of the
swimmer’s velocity
and the river current’s
velocity.
Example:
vswimmer = 2 m/s
vriver = 0.5 m/s
What is the swimmer’s
resultant velocity?
Example - Solution
vR = (2 m/s)2 + (0.5 m/s)2
vR = 2.06 m/s
q = 14
50 m
0.5 m/s
2 m/s
vR
Average Speed and Velocity
• average speed has a greater magnitude than
average velocity unless there are no
direction changes associated with travel
• in sports
– average speed is often more important than
average velocity
1996 Olympic Marathon
Men
2:12:36
Josia Thugwane - RSA
Women
2:26:05
Fatuma Roba - ETH
Distance
26 miles + 385 yards
26 miles * 1.61 km/mile
= 41.86 km
385 yards * 0.915 m/yd
= 352 m
Total = 41.86 km + .35 km
= 42.21 km
Average Speed & the Marathon
• marathon example (cont.)
t = 2:12:36
t=2 hrs (3600s/1 hr) + 12 min (60 s/ 1min) + 36 s
= 7,956 s
t = 2:26:05
= 8,765 s
Average Speed and the Marathon
• average speed = distance/time
speed = 42,210m/7956 s
= 5.3 m/s
speed = 42,210/8765 s
= 4.8 m/s
average velocity???
Average vs. Instantaneous
• average velocity is not very meaningful in
athletic events where many changes in
direction occur
• e.g. marathon

– start and end in same place so

v  0 ???
d 0
Instantaneous Values
• instantaneous velocity (v) is very
important
– specifies how fast and in what direction one is
moving at one particular point in time
– magnitude of instantaneous velocity is exactly
the same as instantaneous speed
Average vs. Instantaneous Speed
14
1991 World Championships - Tokyo
12
speed (m/s)
10
8
Lewis
Burrell
Mitchell
Lewis Avg
Burrell Avg
Mitchell Avg
6
4
2
0
0
2
4
6
time (s)
8
10
Average Acceleration (a)
• rate of change of velocity with respect to time
– “How fast the velocity is changing”
• acceleration is a vector quantity
• units: m/s/s or m/s2 , ms2, ft/s/s
Dv v  v
acceleration  a  
Dt t  t
f
f
i
i
Average Acceleration
velocity (m/s)
v0.0 = 0 m/s
v2.5 = 5 m/s
v5.0 = 0 m/s
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
1
2
time (s)
3
4
5
velocity (m/s)
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
1
2
time (s)
3
4
5
1st interval
a0.0  2.5 
v v
2.5
0.0 
2.50
m
5m

0
s
s
 2.0 m2
2.5s
s
Note: velocity is positive and acceleration is positive.
velocity (m/s)
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
1
2
time (s)
3
4
5
2nd interval
v v
m 5 m
0
s  2.0 m
a2.5  5.0  5.0 2.5  s
2
5.0 2.5 2.5 m
s
s
Note: velocity is positive but acceleration is negative.
velocity (m/s)
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0
1
2
time (s)
3
4
whole interval
m m
v5.0 v0.0 0 m

0
s 0
a0.0  5.0 
 s
t5.0 t0.0
5s 0s
s2
5
Six Cases of Acceleration
1 - speed up in positive direction = positive accel.
+ direction
t = 3 seconds
a
vi = 5 m/s
Calculate average acceleration!
vf = 8 m/s
Six Cases of Acceleration
1 - speed up in positive direction = positive accel.
2 - slow down in positive direction = negative accel.
+ direction
t = 3 seconds
a
vi = 8 m/s
Calculate average acceleration!
vf = 5 m/s
Six Cases of Acceleration
1 - speed up in positive direction = positive accel.
2 - slow down in positive direction = negative accel.
3 - speed up in negative direction = negative accel.
t = 3 seconds
+ direction
a
vf = -8 m/s
Calculate average acceleration!
What is happening to speed?, velocity?
vi = -5 m/s
Six Cases of Acceleration
1 - speed up in positive direction = positive accel.
2 - slow down in positive direction = negative accel.
3 - speed up in negative direction = negative accel.
4 - slow down in negative direction = positive accel.
+ direction
vf = -5 m/s
t = 3 seconds
vi = -8 m/s
Calculate average acceleration!
What is happening to speed?, velocity?
Six Cases of Acceleration
1 - speed up in positive direction = positive accel.
2 - slow down in positive direction = negative accel.
3 - speed up in negative direction = negative accel.
4 - slow down in negative direction = positive accel.
5 - reverse directions from pos to neg = negative accel.
t = 3 seconds
+ direction
a
vi = +1 m/s
Calculate average acceleration!
vf = -1 m/s
Six Cases of Acceleration
1 - speed up in positive direction = positive accel.
2 - slow down in positive direction = negative accel.
3 - speed up in negative direction = negative accel.
4 - slow down in negative direction = positive accel.
5 - reverse directions from pos to neg = negative accel.
6 - reverse directions from neg to pos = positive accel.
+ direction
t = 3 seconds
a
vi = +1 m/s
Calculate average acceleration!
vf = -1 m/s
In certain activities people
experience + & - accelerations.
By standardizing these
accelerations to the normal
acceleration on earth (-9.8
m/s/s) you get an idea of how
much force they are
experiencing
Human Response to
Sustained g’s
•
6-9 Gs: "Increased chest pain and pressure;
breathing difficult, with shallow respiration
from position of nearly full inspiration; further
reduction in peripheral vision, increased
blurring, occasional tunneling, great
concentration to maintain focus; occasional
lacrimation; body, legs, and arms cannot be
lifted at 8 G; head cannot be lifted at 9 G."
•
9-12 Gs: "Breathing difficulty severe;
increased chest pain; marked fatigue; loss of
peripheral vision, diminution of central acuity,
lacrimation."
•
15 Gs: "Extreme difficulty in breathing and
speaking; severe vise-like chest pain; loss of
tactile sensation; recurrent complete loss of
vision.
Data primarily from: Bioastronautics Data
Book, second edition, 1973, NASA)
What influences the shape of the path that an object follows
when it is airborne?
Gravity makes it return to earth
(i.e., fall).
Any initial horizontal velocity will
make it move either forward or
backward.
When both of these influences are
present the object always follows a
parabolic path.
Airborne Motion
• say a person jumps up into
the air
• motion is influenced only
by gravity while the
person is in the air
• the CM will follow a
parabolic path
on the way up ...
initially vertical velocity is high when
the body leaves the ground
vertical velocity then decreases due to
gravity
initial velocity (positive)
velocity decreases
v (m/s)
top of the jump ...
the body changes direction
so velocity is zero
initial velocity (positive)
velocity decreases
v (m/s)
velocity =0
on the way down ...
the jumper’s velocity decreases,
it becomes negative but the magnitude
gets larger - speed increases
initial velocity (positive)
velocity decreases
v (m/s)
velocity =0
velocity decreases
final velocity (negative)
Airborne motion is
UNIFORMLY ACCELERATED MOTION
v (m/s)
the change in velocity over time
is linear so we say the
change in velocity is constant
This constant acceleration is
= -9.8
m/s2
This is the rate at which any
airborne object will accelerate.
Projectile Motion: A special case of
uniformly accelerated motion
If air resistance is negligible then only gravity
affects the path (or trajectory) of a projectile.
This path is a parabola.
Horizontal and vertical components of velocity
are independent.
Vertical velocity decreases at a constant rate
due to the influence of gravity.
Vertical velocity = 0
Positive velocity
gets smaller
Negative velocity
gets larger
Horizontal velocity will remain constant.
3 Primary Factors
Affecting Trajectory
• Projection angle
aka release angle or take-off angle
• Projection height
aka relative height
= release height - landing height)
• Projection velocity
aka release velocity or take-off velocity
Projection Angle
• The optimal angle of
projection is dependent
on the goal of the
activity.
• For maximal height the
optimal angle is 90o.
• For maximal distance
the optimal angle is 45o.
• Optimal angle changes
if projection height is not
equal to 0.
Projection angle = 10 degrees
10 degrees
Projection angle = 30 degrees
10 degrees
30 degrees
Projection angle = 40 degrees
10 degrees
30 degrees
40 degrees
Projection angle = 45 degrees
10 degrees
30 degrees
40 degrees
45 degrees
Projection angle = 60 degrees
10 degrees
30 degrees
40 degrees
45 degrees
60 degrees
Projection angle = 75 degrees
10 degrees
30 degrees
40 degrees
45 degrees
60 degrees
75 degrees
So angle that maximizes Range
(qoptimal) = 45 degrees (or so it appears)
Projection Height
• Projection height = release height - landing height
Effect of Projection Height on Range
(when qrelease = 45 degrees)
h1 < h2 < h3
R1 < R2 < R3
hrelease = hlanding
R1
hrelease > hlanding
hrelease >> hlanding
R2
R3
Projection Height and Projection Angle
interact to affect Range
R45 > R40 > R30
hprojection = 0
hlanding
hrelease
R30
R40
R45
When hlanding < hrelease
hprojection > 0
hrelease
R45 > R40 > R30
BUT difference
b/w R’s is
smaller
hlanding
When hlanding << hrelease
hprojection > 0
R40 > R30 > R45
hrelease
So … as hprojection increases
the optimal qrelease decreases
hlanding
It’s possible to have a negative
projection height (hrelease < hlanding)
In this case the optimal qrelease is
greater than 45 degrees
The effect of Projection Velocity on
the Range of a projectile
40
Range ~ 10 m
10 m/s @ 45 degrees
30
20
10
0
0
10
20
30
40
50
60
70
80
90
100
The effect of Projection Velocity on
the Range of a projectile
40
30
10 m/s @ 45 degrees
Range ~ 10 m
20 m/s @ 45 degrees
Range ~ 40 m
20
10
0
0
10
20
30
40
50
60
70
80
90
100
The effect of Projection Velocity on
the Range of a projectile
40
30
10 m/s @ 45 degrees
Range ~ 10 m
20 m/s @ 45 degrees
Range ~ 40 m
30 m/s @ 45 degrees
Range ~ 90 m
20
So R a v2
10
0
0
10
20
30
40
50
60
70
80
90
•Because R a v2, it has the greatest influence
on the Range of the projectile
100
Long Jump
• What is the optimum angle of takeoff for
long jumpers?
Projection height > 0 (take-off height > landing height)
Optimum Angle should be slightly less than 45 degrees
research shows that it should be 42-43 degrees
The Best of the Best
Athlete
Mike Powell (USA)
Bob Beamon (USA)
Carl Lewis (USA)
Ralph Boston (USA)
Igor Ter-Ovanesian (USSR)
Jesse Owens (USA)
Distance of
Jump Analyzed
(m)
8.95
8.90
8.79
8.28
8.19
8.13
Speed of
Takeoff
(m/s)
9.8
9.6
10.0
9.5
9.3
9.2
Optimum
Angle of Takeoff
for Given Speed
(deg)
43.3
43.3
43.4
43.2
43.2
43.1
Elena Belevskaya (USSR)
Heike Dreschler (GDR)
Jackie Joyner-Kersee (USA)
Anisoara Stanciu (Rom)
Vali Ionescu (Rom)
Sue Hearnshaw (GB)
7.14
7.13
7.12
6.96
6.81
6.75
8.9
9.4
8.5
8.6
8.9
8.6
43.0
43.2
42.8
42.9
43.0
42.9
Actual
Angle of
Takeoff
(deg)
23.2
24.0
18.7
19.8
21.2
22.0
19.6
15.6
22.1
20.6
18.9
18.9
Actual Angle of Takeoff ~ 17-23 degrees
Long Jump
• when a jumper is moving at 10 m/s
– the foot is not on the ground long enough
to generate a large takeoff angle
– so jumpers maintain speed and live with a
low takeoff angle
• v is the most important factor in
projectile motion
VALUES FOR HYPOTHETICAL JUMPS
UNDER DIFFERENT CONDITIONS
Variable
Speed of
Takeoff
Angle of
Takeoff
Values for
Actual Jump
(1)
Speed of
Takeoff
Increased 5%
(2)
Angle of
Takeoff
Increased 5%
(3)
Relative Height
of Takeoff
Increased 5%
(4)
8.90 m/s
9.35 m/s
8.90 m/s
8.90 m/s
20
20
21
20
Relative
Ht of
Takeoff
0.45 m
0.45 m
0.45 m
0.47 m
Horizontal
Range
6.23 m
6.77 m
6.39 m
6.27 m
0.54 m
0.16 m
0.04 m
7.54 m
7.16 m
7.04 m
Change in
Horiz
Range
Distance
of Jump
--
7.00 m
Suppose a zookeeper must shoot the banana from the banana cannon to the
monkey who hangs from the limb of a tree. This particular monkey has a habit of
dropping from the tree the moment that the banana leaves the muzzle of the
cannon. If the monkey lets go of the tree the moment that the banana is fired, will
the banana hit the monkey?
Banana’s
gravity-free
path
If there is no gravity then the monkey floats AND
you throw directly at the monkey, then the path of the
banana will be a straight line (the “gravity-free path”).
Since this path will cross the point where the monkey
floats the monkey can catch and eat the banana!
When you take gravity into consideration
you STILL aim at the monkey!
Banana’s
Gravity free path
Monkey’s
Gravity free path is
“floating” at height
of limb
Fall thru same height
YEAH! It works! Since both banana and monkey experience
the same acceleration each will fall equal amounts below their
gravity-free path. Thus, the banana hits the monkey.
What happens when you throw the banana slower?
Monkey’s
Gravity free path is
“floating” at height
of limb
Banana’s
Gravity free path
Fall thru same height
As long as you aim at the monkey he will still catch it. The only
difference is that the monkey will fall farther before he catches it
because it takes longer to travel the necessary horizontal distance.
Eqns of Constant Acceleration Motion
ECAM’s
Eqn 1
Eqn 3
Eqn 2
Eqn 4
v f  vi  at
d  (vi  v f )t
1
d  vi t  21 at 2
v 2f  vi2  2ad
2
d = displacement
(d = sf – si)
vi = initial velocity
vf = final velocity
a = acceleration
t = time
Remember: when there is no change
in direction then displacement and
distance are the same thing so …
Often times it is useful to consider
these equations being applied
separately for x- and y-directions
Eqns of Constant Acceleration Motion
ECAM’s
Eqn 1
Eqn 3
Eqn 2
Eqn 4
v f  vi  at
d  vi t  21 at 2
d  (vi  v f )t
v 2f  vi2  2ad
1
2
Eqn
d
vi
vf
a
t
1
2
3
4




















Remember that d = sf - si
ECAM Examples
• Example problem
– a cyclist passes the midpoint of a race moving at a
speed of 10 m/s
– she accelerates at an average rate of 3 m/s/s for 3 s
– how fast is she moving at the end of this period?
ECAM Examples
An object falls 10 meters from the top of a tower.
What is the contact velocity and how much time
does it take to reach the ground?
Steps:
1. Draw a picture.
2. List values for any parameters that are given.
3. Find equations in which all of the variables are
known except the one that you are trying to find.
4. Substitute values for variable and solve.
Example
A runner starts from rest, uniformly accelerates at 3 m/s2
for 3 seconds, then runs at a constant velocity for 5
seconds, then accelerates in the negative direction at -2
m/s2 for 2 seconds. How far does the runner travel during
this 10 second period?
Diving Example
Can the diver
successfully complete
a 2.5 somersault?
PROBLEM DESCRIPTION
It is given that it takes a
minimum of 0.95 s to perform a
2.5 somersault.
+
2m
Only consider the
vertical component.
0.85 m
1m
0.85 m
ventry = ?
1st: find time to reach peak of dive (tup)
+
2m
0.85 m
si =
sf =
d=
ay =
vf =
vi =
t=
1m
Which equation should you use?
tup = tup’
Step 1 (cont.)
+
2m
0.85 m
si =
sf =
d=
ay =
vf =
vi =
t=
1m
Which equation should you use?
2nd: Find time from peak of flight to time of impact with water
2.0 m
1.0 m
si =
sf =
d=
ay =
vf =
vi =
t=
0.85 m
Which equation should you use?
How many somersaults can the diver complete off of other
boards if it take .38 s per somersault?
• 3-m springboard (CM reaches 5M above water)
• tup = 0.48 s
• tdown = 0.92 s
• ttotal = 1.40 s
1.4/0.38 = 3.7
3.5 somersaults
• 5-m platform (CM reaches 1.25 m above platform)
• tup = 0.28 s (raise CM only 0.4 m above initial pos)
• tdown = 0.95 s
1.33/0.38 = 3.5
• ttotal = 1.33 s
3.5 somersaults
(if perfect)
Other Boards (cont.)
• 10-m platform (CM reaches 1.25m above platform)
• tup = 0.28 s
• tdown = 1.46 s
• ttotal = 1.74 s
1.74/0.38 = 4.6
4.5 somersaults
• 20-m cliff (CM reaches 1.25 m above cliff)
• tup = 0.28 s (raise CM only 0.4 m above initial pos)
• tdown = 2.04 s
2.32/0.38 = 6.1
• ttotal = 2.32 s
6 somersaults
(if you’re crazy!)
Speed of Impact
Know:
vi =
ay =
d=
vf =
2.15 m
• 1 m board
d = 2.15 m
vf = 6.5 m/s
(14.5 mph)
vf = 9.0 m/s
(20.1 mph)
• 3 m board
d = 4.15 m
• 5 m platform
d = 5.4 m
vf = 10.3 m/s (23.0 mph)
• 10 m platform
d = 10.4 m
vf = 14.3 m
(32.0 mph)
• 20 m cliff
d = 20.4 m
vf = 20.0 m/s (44.7 mph)