Linear Motion

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 – Measure the anglefrom a carefully drawn diagram.

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!



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





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


What is happening to speed?, velocity?





4 - slow down in negative direction = positive accel.

fin

al

init

ial+ direction

vf = -5 m/s vi = -8 m/s

t = 3 seconds


What is happening to speed?, velocity?






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







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


• 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

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)


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)


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.

10 degrees

Projection angle = 10 degrees

10 degrees30 degrees


10 degrees30 degrees40 degrees


10 degrees30 degrees40 degrees45 degrees


10 degrees30 degrees40 degrees45 degrees60 degrees


10 degrees30 degrees40 degrees45 degrees60 degrees75 degrees


So angle that maximizes Range(optimal) = 45 degrees (or so it appears)

Projection Height

• Projection height = release height - landing height

Effect of Projection Height on Range(when release = 45 degrees)

hrelease = hlanding

hrelease > hlanding

hrelease >> hlanding

R1

R2

R3

h1 < h2 < h3

R1 < R2 < R3

Projection Height and Projection Angleinteract to affect Range

hprojection = 0 R45 > R40 > R30

R30

R40

R45

hreleasehlanding

When hlanding < hrelease

hrelease

hlanding

hprojection > 0 R45 > R40 > R30

BUT differenceb/w R’s is

smaller

When hlanding << hrelease

hrelease

hlanding

hprojection > 0 R40 > R30 > R45

So … as hprojection increasesthe optimal release decreases

It’s possible to have a negative projection height (hrelease < hlanding)

In this case the optimal release is greater than 45 degrees

1009080706050403020100

0

10

20

30

40 10 m/s @ 45 degrees Range ~ 10 m

The effect of Projection Velocity onthe Range of a projectile

1009080706050403020100

0

10

20

30

40

20 m/s @ 45 degrees Range ~ 40 m



1009080706050403020100

0

10

20

30

40




So R v2


•Because R v2, it has the greatest influence on the Range of the projectile

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

Athlete

Distance ofJump Analyzed

(m)

Speed ofTakeoff

(m/s)

OptimumAngle of Takeofffor Given Speed

(deg)

ActualAngle ofTakeoff(deg)

Mike Powell (USA) 8.95 9.8 43.3 23.2Bob Beamon (USA) 8.90 9.6 43.3 24.0Carl Lewis (USA) 8.79 10.0 43.4 18.7Ralph Boston (USA) 8.28 9.5 43.2 19.8Igor Ter-Ovanesian (USSR) 8.19 9.3 43.2 21.2Jesse Owens (USA) 8.13 9.2 43.1 22.0

Elena Belevskaya (USSR) 7.14 8.9 43.0 19.6Heike Dreschler (GDR) 7.13 9.4 43.2 15.6Jackie Joyner-Kersee (USA) 7.12 8.5 42.8 22.1Anisoara Stanciu (Rom) 6.96 8.6 42.9 20.6Vali Ionescu (Rom) 6.81 8.9 43.0 18.9Sue Hearnshaw (GB) 6.75 8.6 42.9 18.9

Actual Angle of Takeoff ~ 17-23 degrees

The Best of the Best

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 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’


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 =


• 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)

• tdown = 2.04 s

• ttotal = 2.32 s

Other Boards (cont.)

1.74/0.38 = 4.64.5 somersaults

2.32/0.38 = 6.16 somersaults

(if you’re crazy!)

Speed of Impact

Know: vi = ay = d = vf =

2.15 m

• 1 m boardd = 2.15 m vf = 6.5 m/s (14.5 mph)

• 3 m boardd = 4.15 m vf = 9.0 m/s (20.1 mph)

• 5 m platformd = 5.4 m vf = 10.3 m/s (23.0 mph)

• 10 m platformd = 10.4 m vf = 14.3 m (32.0 mph)

• 20 m cliffd = 20.4 m vf = 20.0 m/s (44.7 mph)

Linear Motion

Documents

actual distance

accelerationthe position

length of travel

displacementexample

milesntotal distance

dx dydistance distance

y coordinates

xdirection dx