Linear Motion TRANSLATION ONLY! bject maintains angular rientation () sured in meters - SI unit er units - inches, feet, miles, timeters, 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 Angular Motion ROTATION ONLY! Object rotates about a fixed point (axis) measured in radians – SI Unit other units – degrees, revolutions but this point does NOT have to lie within the object
Transcript
Linear MotionTRANSLATION ONLY!
Object maintains angular orientation ()measured in meters - SI unitother units - inches, feet, miles, centimeters, millimeters
Rectilinear motion - if path of one point on object is a straight lineCurvilinear motion – if path of one point on object is curved
Angular MotionROTATION ONLY!
Object rotates about a fixed point (axis)
measured in radians – SI Unitother units – degrees, revolutions
but this point does NOT have to lie within the object
Types of Motion
• General– combination of linear and angular motion
• translation and rotation
A
B
A
B
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
(x2,y2)
(x3,y3)
(x4,y4)
(x5,y5)
(0,0)
Y
X
(x1,y1)
Frame 1Position 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 (s= sf - si) is the change in position of an object
displacement = d d
s1
s2
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
d
s1 = (x1,y1)
s2 = (x2,y2)
d = s1 – s2
X
Y
d
s1 = (x1,y1)
s2 = (x2,y2)
d = s1 – s2
X
Y
d = s1 – 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
•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,)
•“How far and in which direction”2
122
12 )()( yyxxd
= 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.
d
s1
s2
d = s1 – s2
X
Y
Assuming this city is like most cities you have to walk up and down city blocks and not through buildings.
d
s1
s2
X
Y
So your actual route is around the buildings, traveling up and down city blocks.
d
s1
s2
X
Y
dx
dy
Thus the actual distance you covered is more than displacement represents
d
s1
s2
X
Y
dx
dy
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
distance = dx + dy
Distance ( )
• distance is the length of the path traveled
• it is a scalar - “How far”
• units: same as displacement
dx
dy
ddx + dy = distance =
Note: use “ ” for length
Example - Distance vs. Displacement
leg 1 = 2 miles
leg 3 = 2 miles
leg 2 = 3 miles
N
Total DISTANCE Traveled= 2 miles + 3 miles + 2 miles
= 7 miles
Describing Displacement
N
disp
lace
men
t vec
tor
Describing Displacement
First Method (Cartesian)3 miles East4 miles North(3, 4) miles
put ‘horizontal’ coordinate 1st
put ‘vertical’ coordinate 2nd
Displacement Magnitude
N
disp
lace
men
t vec
tor
3 miles
4 m
iles
Second Method (Polar)1st - calculate length ofdisplacement vector
d
d
d miles
3 4
25
5
2 2
Displacement Direction
N
disp
lace
men
t vec
tor
3 miles
4 m
iles
2nd – Calculate the angle using trigonometric relationships
hordverd1tan
1.5334tan 1milesmiles
Displacement Vector(Polar Notation)
N
disp
lace
men
t vec
tor
3 miles
4 m
iles
Describe the displacement vector by its length and direction
d miles 5 531@ .
Average Speed
• speed is a scalar quantity
• it is the rate of change of distance wrt time
• units: same as velocity
distancetime
Speed
What is the average speed of the basketball?
(0,0)
(60,10)
(80,40)
0.5
s
feet 363020 = 22 l ft/s 725.0
36
t
lspeed
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
tdvvelocity
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.
Use the laws of vector algebra.
Example - the pathof the swimmer isdetermined by thevector sum of the swimmer’s velocityand the river current’svelocity.
Adding Velocities
Example:
vswimmer = 2 m/svriver = 0.5 m/s
What is the swimmer’sresultant velocity?
Example - Solution
50 m
2 m/s
0.5 m/s
vR
vR = (2 m/s)2 + (0.5 m/s)2
vR = 2.06 m/s
= 14
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
Distance26 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/timespeed = 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
d 0
???0v
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
0
2
4
6
8
10
12
14
0 2 4 6 8 10
time (s)
sp
ee
d (
m/s
)
Lewis
Burrell
Mitchell
Lewis Avg
Burrell Avg
Mitchell Avg
1991 World Championships - Tokyo
Average vs. Instantaneous Speed
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 , ms, ft/s/s
acceleration at
vt t
i
f i
v vf
00.5
11.5
22.5
33.5
44.5
5
0 1 2 3 4 5time (s)
velo
city
(m
/s)
v0.0 = 0 m/sv2.5 = 5 m/sv5.0 = 0 m/s
Average Acceleration
1st interval
20.05.2
5.20.00.2
5.2
05
05.2 sm
ssm
smvv
a
Note: velocity is positive and acceleration is positive.
00.5
11.5
22.5
33.5
44.5
5
0 1 2 3 4 5time (s)
velo
city
(m
/s)
2nd interval
Note: velocity is positive but acceleration is negative.
00.5
11.5
22.5
33.5
44.5
5
0 1 2 3 4 5time (s)
velo
city
(m
/s)
25.20.5
0.55.20.2
5.2
50
5.20.5 sm
sm
sm
smvv
a
whole interval
2
0.00.5
0.00.50.50.0 0
05
00
sm
sssm
sm
tt
vva
00.5
11.5
22.5
33.5
44.5
5
0 1 2 3 4 5time (s)
velo
city
(m
/s)
Six Cases of Acceleration
1 - speed up in positive direction = positive accel.
a
+ direction
vf = 8 m/svi = 5 m/s
t = 3 seconds
fin
al
init
ial
Calculate average acceleration!
Six Cases of Acceleration
1 - speed up in positive direction = positive accel.
2 - slow down in positive direction = negative accel.
a
+ direction
vi = 8 m/s vf = 5 m/s
t = 3 seconds
fin
al
init
ial
Calculate average acceleration!
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.
+ direction
vf = -8 m/s vi = -5 m/s
a
t = 3 seconds
fin
al
init
ial
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.
fin
al
init
ial+ direction
vf = -5 m/s vi = -8 m/s
t = 3 seconds
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.
vf = -1 m/svi = +1 m/s
a
+ direction t = 3 seconds
init
ial
fin
al
Calculate average acceleration!
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.
vf = -1 m/svi = +1 m/s
a
+ direction t = 3 seconds
init
ial
fin
al
Calculate average acceleration!
• 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.
Human Response to Sustained g’s
Data primarily from: Bioastronautics Data Book, second edition, 1973, NASA)
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
Relationships Betweens, v, & a
• v is the rate of change of s wrt time
• a is the rate of change of v wrt time
• consider a graph of s vs. time– s on vertical axis– time on horizontal axis– rate of change is interpreted as the slope
Slope of a Curve
• “Slope” = number which describes the steepness of a line
– rise/run
– Note: this is the definition for the tangent of , opposite / adjacent
Changes in the slope
• positive slope– up and to the right
• negative slope– down and to the right
• quick change– very steep slope
• slow change– very flat slope
0
10
20
30
40
50
60
0 5 10 15 20 25 30
Time (s)
Position
slope = rise
run =
21-8
10 -5 =
p
t
The slope of the position by time curve is the velocity.
Note that this is the average velocity during the period from 5 seconds to 10 seconds.
0
10
20
30
40
50
60
0 5 10 15 20 25 30
Time (s)
Position
The tangent of a curve is the instantaneous slope at a single point. This slope represents the instantaneous velocity.
0
10
20
30
40
50
60
0 5 10 15 20 25 30
positive s
lope
change in directionslope = 0
negative slope
• the instantaneous velocity (v) curve is the plot of how the slope of the s vs. t curve changes
•a similar relationship exists between a and v
Relationship of s, v, & a
s
v
a
Where is velocity greatest?
Where is acceleration greatest?
Where is velocity closest to zero?What is happening to the position curve at this point?
Where is acceleration negative?What is happening to the velocity curve at this point?
0
1
2
3
4
5
6
7
0 0.5 1 1.5 2time (s)
sp
ee
d (
m/s
)
Lewis
Burrell
Mitchell
1991 World Championships - Tokyo
Who had the largest accelerationat the beginning of the race?
11
11.2
11.4
11.6
11.8
12
12.2
8 8.5 9 9.5 10
time (s)
sp
ee
d (
m/s
)
Lewis
Burrell
Mitchell
1991 World Championships - Tokyo
Describe the accelerations at the end of the race.
Steps to determining v vs. t curve from s vs. t curve
(1) draw a set of axes (v vs t) directly under the s vs. t curve
(2) locate all minimums, maximums, asymptotes, and inflection points
(3) plot zero value points for each corresponding min, max or asym
(4) plot mins or maxes for each inflection point
negative slope
start negative but get closer to zero
but flattening out
minimum = zero slope
must cross time axis (i.e. v=0)
positive slopebut becoming steeper
Start at zero and increase
positive slope
but becoming steeper
start out flat
slope stops becoming steeper and begins to flatten out
This is known as an inflection pointand corresponds to a relative maximum
on velocity vs. time curve
slope stays +just not as steep
positive slope
but becoming flatter
start out steep
slope flattens out as much asit is going to another inflection point
corresponds to a relative minimumthen slope becomes steeper
positive slope
continues to become steeper
start out steep
Region 1 – negative slope so negative velocity
Region 2 – positive slope so positive velocity but inflection point where slope maxes out
Region 3 – positive slope so positive velocity but inflection point where slope is minimized
Region 4 – positive slope so positive velocity, no special points so velocity continues to rise
The above diagram shows a typical zero-g maneuver. However, the maneuver can be modified to provide any level of g-force less than one g. Some typical g-levels used on different tests and the corresponding time for each maneuver are as follows:
•Negative-g (-0.1 g): Approximately 15 seconds •Zero-g: Approximately 25 seconds •Lunar-g (one-sixth g): Approximately 40 seconds •Martian-g (one-third g): Approximately 30 seconds
NASA’s KC135A“The Vomit Comet”
s
v
a
inf max inf
Quantitative determination of v and a from s
or How to calculate v and a from sFrame Time Pos. (m) Vel. (m/s) Acc. (m/s/s)
1 0.00 0.00
2 0.10 0.59 5.90 -23.00
3 0.20 0.95 3.60 -31.00
4 0.30 1.00 0.50 -10.00
5 0.40 0.95 -0.50 -31.00
6 0.50 0.59 -3.60
t
va ,
t
sv
t = 0.10 s
Calculating v and a from s
Note: the velocity values do not occur at the same time as the position and acceleration values and you lose data at the starting and ending frames.
Frame Time Pos. (m) Vel. (m/s) Acc. (m/s/s)
1 0.00 0.00
2 0.10 0.59 5.90 -23.00
3 0.20 0.95 3.60 -31.00
4 0.30 1.00 0.50 -10.00
5 0.40 0.95 -0.50 -31.00
6 0.50 0.59 -3.60
dt2
xxv 1i1i
xi
dt
xxv 12
1x
dt
xxv 1nn
xn
First Central Difference Method
Last (nth) frame of data
First frame of data
xi is the ith frame of horizontal position data
vxi is the ith frame of horizontal velocity data
dt is the time interval between frames
First Central Difference Method
Frame Time Pos. (m) Vel. (m/s) Acc. (m/s/s)
1 0.00 0.00 5.90 -11.50
2 0.10 0.59 4.75 -19.25
3 0.20 0.95 2.05 -23.75
4 0.30 1.00 0.00 -20.50
5 0.40 0.95 -2.05 -18.00
6 0.50 0.59 -3.60 -15.50
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 groundvertical velocity then decreases due to
gravity
v (m/s)
initial velocity (positive)
velocity decreases
top of the jump ...the body changes direction so velocity is zero
v (m/s)
initial velocity (positive)
velocity decreases
velocity =0
on the way down ...the jumper’s velocity decreases, it becomes negative but the magnitude gets larger - speed increases
v (m/s)
initial velocity (positive)
velocity decreases
velocity =0
final velocity (negative)
velocity decreases
v (m/s)
the change in velocity over timeis linear so we say thechange in velocity is constantThis constant acceleration is
= -9.8 m/s2
This is the rate at which any airborne
object will accelerate.
Airborne motion isUNIFORMLY ACCELERATED MOTION
This path is a parabola.
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.
Horizontal and vertical components of velocityare independent.
Vertical velocity decreases at a constant ratedue to the influence of gravity.
Positive velocitygets smaller
Vertical velocity = 0
Negative velocitygets larger
Vertical velocity = 0
Horizontal velocity will remain constant.
• Projection angle aka release angle or take-off angle
• Projection height aka relative height
= release height - landing height)
• Projection velocityaka release velocity or take-off velocity
3 Primary FactorsAffecting Trajectory
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.
• 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 Speed of Angle of Relative Height
Values for Takeoff Takeoff of TakeoffActual Jump Increased 5% Increased 5% Increased 5%
Variable (1) (2) (3) (4)Speed ofTakeoff 8.90 m/s 9.35 m/s 8.90 m/s 8.90 m/s
Angle of 20 20 21 20Takeoff
Relative 0.45 m 0.45 m 0.45 m 0.47 mHt ofTakeoff
Horizontal 6.23 m 6.77 m 6.39 m 6.27 mRange
Change in -- 0.54 m 0.16 m 0.04 mHorizRange
Distance 7.00 m 7.54 m 7.16 m 7.04 mof Jump
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?
If there is no gravity then the monkey floats ANDyou 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!
Banana’s gravity-free path
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.
When you take gravity into consideration you STILL aim at the monkey!
Monkey’sGravity free path is “floating” at height of limb
Banana’s Gravity free path
Fall thru same height
Banana’s Gravity free path
Monkey’sGravity free path is “floating” at height of limb Fall thru same height
What happens when you throw the banana slower?
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’sEqn 1 Eqn 3
Eqn 2 Eqn 4
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
v v atf i
d v v ti f 1
2( )
d v t ati 12
2
v v adf i2 2 2
d = displacement(d = sf – si)vi = initial velocityvf = final velocitya = accelerationt = time
Eqns of Constant Acceleration Motion
ECAM’sEqn 1 Eqn 3
Eqn 2 Eqn 4
Remember that d = sf - si
v v atf i
d v v ti f 1
2( )
d v t ati 12
2
v v adf i2 2 2
Eqn d vi vf a t
1 2 3 4
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?
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.
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?
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?
Example
Diving Example
Can the diver successfully completea 2.5 somersault?
PROBLEM DESCRIPTION
+
1 m
0.85 m
2 m
0.85 m
ventry = ?
Only consider thevertical component.
It is given that it takes a minimum of 0.95 s to perform a 2.5 somersault.
+
1 m
0.85 m
2 m
t up
1st: find time to reach peak of dive (tup)
ay =vf =vi =t =
si = sf = d =
Which equation should you use?
+
1 m
0.85 m
2 m
tup ’
Step 1 (cont.) tup = tup’
Which equation should you use?
ay =vf =vi =t =
si = sf = d =
2nd: Find time from peak of flight to time of impact with water
2.0 m
1.0 m0.85 m
ay =vf =vi =t =
si = sf = d =
Which equation should you use?
• 3-m springboard (CM reaches 5M above water)
• tup = 0.48 s
• tdown = 0.92 s
• ttotal = 1.40 s
• 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
• ttotal = 1.33 s
How many somersaults can the diver complete off of other boards if it take .38 s per somersault?
1.4/0.38 = 3.73.5 somersaults
1.33/0.38 = 3.53.5 somersaults
(if perfect)
• 10-m platform (CM reaches 1.25m above platform)
• tup = 0.28 s
• tdown = 1.46 s
• ttotal = 1.74 s
• 20-m cliff (CM reaches 1.25 m above cliff)
• tup = 0.28 s (raise CM only 0.4 m above initial pos)
