fisdas - minggu 05 - gerak melingkar 1

7/30/2019 FisDas - Minggu 05 - Gerak Melingkar 1

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Quick Quiz (15 Minute)

Jika masa mA=4kg, mB=2kg dan gesekan kinetiknyaadalah mkA=0.300 dan mkB=0.400 dan percepatangravitasi sebesar 9.8 m/s2. Tentukan (a) tegangan talipada sistem dan (b) percepatan dari sistem

30O

T


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Circular Motion

Setyawan P. Sakti


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Some Example


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Angular Displacement

Recall for linear motion: displacement, velocity, acceleration

Need similar concepts forobjectsmoving in circle (CD, merry-go-

round, etc.) As before:

need a fixed reference system (line)

use polar coordinate system

tva

trvrrr if

,,


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Every point on the objectundergoes circular motion aboutthe point O

Angles generally need to bemeasured in radians

Note:

r

s

3.572

3601

rad

]degrees[180]rad[

length of arc

radius


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The angular displacementisdefined as the angle the objectrotates through during some

time interval

Every point on the discundergoes the same angulardisplacement in any given timeinterval

if


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Angular Velocity

The average angular velocity(speed), , of a rotating rigidobject is the ratio of theangular displacement to thetime interval

ttt if

if


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Angular Speed

The instantaneousangularvelocity (speed) is defined as thelimit of the average speed as thetime interval approaches zero

Units of angular speed areradians/sec (rad/s)

Angular speed will be positive if is increasing

(counterclockwise)

negative if is decreasing(clockwise)

tt

0lim


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Angular Acceleration

What if object is initially at restand then begins to rotate?

The average angular acceleration,a, of an object is defined as theratio of the change in the angular

speed to the time it takes for theobject to undergo the change:

Units are rad/s

Similarly, instant. angular accel.:

ttt if

if

a

tt

a 0lim


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Notes about angular kinematics:

When a rigid object rotates about a fixed axis, everyportion of the object has the same angular speed andthe same angular acceleration

i.e. ,, and a are not dependent upon r, distance formhub or axis of rotation


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1. Bicycle wheel turns 240 revolutions/min. What is its angular

velocity in radians/second?

secradians1.25secradians8rev1

rads2

sec60

min1

min

rev240

2. If wheel slows down uniformly to rest in 5 seconds, what is the

angular acceleration?

2secrad5

sec5

secrad250

t

if a

Examples:


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Given:

1. Angular velocity:

240 rev/min

2. Time t = 5 s

Find:

1. = ?

3. How many revolution does it turn in those 5 sec?

srevolution102

rev1rad5.62)(

rad5.62sec5secrad52

1sec5secrad25

2

1

2

2

0

a

rev

tt

Recall that for linear motion we had:

Perhaps something similar for angular quantities?

2

02

1attvx

Examples:


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Analogies Between Linear andRotational Motion

Rotational Motion About aFixed Axis with Constant

Acceleration

Linear Motion withConstant Acceleration

ti a

2

2

1

tti a

a 222 i xavv i 222

2

2

1

attvx i

atvv i

R l i hi B A l d


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Relationship Between Angular andLinear Quantities

Displacements

Speeds

Accelerations

r

s

vr

ts

rt

1

or

1

ra a

R l ti hi B t A l d


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Relationship Between Angular andLinear Quantities

Displacements

Speeds

Accelerations

Every point on therotating object has thesame angular motion

Every point on the

rotating object does nothave the same linearmotion

rs

rv

ra a


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Concept

A ladybug sits at the outer edge of a merry-go-round, anda gentleman bug sits halfway between her and the axis ofrotation. The merry-go-round makes a complete revolutiononce each second.The gentleman bugs angular speed is

1. half the ladybugs.2. the same as the ladybugs.3. twice the ladybugs.4. impossible to determine


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Concept

A ladybug sits at the outer edge of a merry-go-round, anda gentleman bug sits halfway between her and the axis ofrotation. The merry-go-round makes a complete revolutiononce each second.The gentleman bugs angular speed is

1. half the ladybugs.2. the same as the ladybugs.3. twice the ladybugs.4. impossible to determine

Note: both insects have an angular speed of 1 rev/s


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Centripetal Acceleration

An object traveling in a circle,even though it moves with aconstant speed, will have anacceleration (since velocity

changes direction) This acceleration is called

centripetal (center-seeking).

The acceleration is directed

toward the center of the circleof motion

C t i t l A l ti d A l


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Centripetal Acceleration and AngularVelocity The angular velocity and the linear

velocity are related (v = r)

The centripetal acceleration canalso be related to the angularvelocity

t

s

r

v

at

v

sr

vv

r

s

v

v

a

but,

rar

va CC

22

or

Thus:

Similar

triangles!


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Total Acceleration

What happens if linearvelocity also changes?

Two-componentacceleration: the centripetal component of

the acceleration is due tochanging direction

the tangential component of theacceleration is due to changingspeed

Total acceleration can befound from thesecomponents:

22

Ct aaa

slowing-down car


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Vector Nature of Angular Quantities

As in the linear case,displacement, velocityand acceleration arevectors:

Assign a positive ornegative direction

A more complete way isby using the right hand

rule Grasp the axis of rotationwith your right hand

Wrap your fingers in thedirection of rotation

Your thumb points in thedirection of

F C i C t i t l


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Forces Causing CentripetalAcceleration Newtons Second Law says that the centripetal

acceleration is accompanied by a force

F stands for any force that keeps an object following acircular path

Force of friction (level and banked curves)

Tension in a string Gravity

rmr

v

mmaF C2

2


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Example1: level curves

Consider a car driving at 20 m/s (~45mph) on a level circular turn ofradius 40.0 m. Assume the carsmass is 1000 kg.

1. What is the magnitude offrictional force experienced bycars tires?

2. What is the minimum coefficientof friction in order for the car tosafely negotiate the turn?


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Example1:

Given:

masses: m=1000 kg

velocity: v=20 m/s

radius: r = 40.0m

Find:

1. f=?

2. m=?

1. Draw a free body diagram, introduce

coordinate frame and consider vertical

and horizontal projections

mgN

mgNFy

0

Nm

sm

kgr

v

mmaf

fmaFx

4

22

100.140

20

1000

2. Use definition of friction force:

02.18.91000

101.0

thus,10

2

4

42

smkg

N

Nr

vmmgf

m

m

Lesson: m for rubber on dry concrete is 1.00!

rubber on wet concrete is 0.2!

driving too fast


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Concept

Is this static or kinetic friction is the car does not slide or skid?

1. Static2. Kinetic


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Example2: banked curves

Consider a car driving at 20 m/s (~45mph) on a 30 banked circularcurve of radius 40.0 m. Assumethe cars mass is 1000 kg.

1. What is the magnitude offrictional force experienced bycars tires?

2. What is the minimum coefficientof friction in order for the car to

safely negotiate the turn?

A component of the normal force adds to thefrictional force to allow higher speeds

rg

v2

tan


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Given:

masses: m=1000 kg

velocity: v=20 m/s

radius: r = 40.0m

angle: a = 30

Find:

1. f=?

2. m=?

1. Draw a free body diagram,

introduce coordinate frame and

consider vertical and

horizontal projections

Nmgr

vmf

mgfr

vmFx

376030sin30cos

30sin30cos

2

2

NmgrvmN

mgNr

vmFy

4

2

2

103.130cos30sin

30cos30sin

2. Use definition of friction force:

28.0

101.3

3760

isminimalthus,

4

s

N

N

N

f

Nf

ss

s

m

mm

Lesson: by increasing angle of banking,

one decreases minimal m or friction with

which one can take curve!

Example 2:


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Example 3: Horizontal Circle

The horizontal component ofthe tension causes thecentripetal acceleration

tangaC

E l 4


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Example 4:

7 m49 m
http://www.gaydayatbuschgardens.com/rollercoaster.jpghttp://www.gaydayatbuschgardens.com/rollercoaster.jpghttp://www.gaydayatbuschgardens.com/rollercoaster.jpg


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THE VIPER

(Six Flags Over Magic Mountain)188 feet high

70 mph

One of the largestlooping roller coasters

in the world

Roller Coaster


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Top Thrill Dragster(Cedar Point)

420 feet tall

120 mph0 to 120 mph in 4 sec

free-falls back to Earth,reaching a speed of 120

mph for the second time

$ 25 million

Which way do you perceive the


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Perceived acceleration

Which way do you perceive theacceleration for circular motion?


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FORCE


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FPerceived acceleration

Circular motion:force is toward center of circle

Note: the kid in the middle feels no acceleration!


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look at the passengers!

NASAs vomit comet


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What has to be true about theacceleration a of the vomitcomet (if the apparent weightof the passengers is zero)?

A. a points upward

B. a points downward

C. a points towards the

center of their

circular trajectory

Clicker Question


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You better believe in conservation of energy if you want tolive!

Roller coasters


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Most people find this sudden

reduction in apparent weightterrifying.

Note: this is not uniform circular motion!

Forces in Accelerating Reference


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Forces in Accelerating ReferenceFrames

Distinguish real forces from fictitious forces

Centrifugal force is a fictitious force

Real forces always represent interactionsbetween objects


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Newtons Law of Universal Gravitation

Every particle in the Universe attracts every otherparticle with a force that is directly proportional tothe product of the masses and inverselyproportional to the square of the distancebetween them.

2

21

r

mm

GF

G is the universal gravitational constant

G = 6.673 x 10-11 N m /kg

This is an example of an inverse square law


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Gravitation Constant

Determined experimentally

Henry Cavendish

1798

The light beam and mirror serve toamplify the motion

E l


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Example:

Question: Calculate gravitational attraction between two students 1

meter apart

Nmkgkg

kg

mN

r

mm

GF7

22

211

2

21

102.41

9070

1067.6

Extremely small

Compare:

NmgF 686

Applications of Universal Gravitation 1: Mass of


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Applications of Universal Gravitation 1: Mass ofthe Earth

Use an example of anobject close to thesurface of the earth

r ~ REGgRM EE

2

Applications of Universal Gravitation 2:


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Applications of Universal Gravitation 2:Acceleration Due to Gravity

g will vary with altitude

2r

MGg E

mgrMGm

rmMGF EE

22

Gravitational Potential


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Gravitational PotentialEnergy

PE = mgy is valid onlynear the earths surface

For objects high above

the earths surface, analternate expression isneeded

Zero reference level isinfinitely far from theearth

r

mMGPE E


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Escape Speed

The escape speed is the speed needed for an objectto soar off into space and not return

For the earth, vesc is about 11.2 km/s

Note, v is independent of the mass of the object

E

Eesc

R

GMv

2


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Keplers Laws

All planets move in elliptical orbits withthe Sun at one of the focal points.

A line drawn from the Sun to any planet

sweeps out equal areas in equal timeintervals.

The square of the orbital period of any

planet is proportional to cube of theaverage distance from the Sun to theplanet.


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Keplers Laws, cont.

Based on observations made by Brahe Newton later demonstrated that these laws

were consequences of the gravitational force

between any two objects together withNewtons laws of motion

K l Fi L


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Keplers First Law

All planets move inelliptical orbits withthe Sun at onefocus.

Any object bound toanother by aninverse square lawwill move in anelliptical path

Second focus isempty

K l S d L


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Keplers Second Law

A line drawn from theSun to any planet willsweep out equal areasin equal times

Area from A to B and C toD are the same

K l Thi d L


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Keplers Third Law

The square of the orbital period of anyplanet is proportional to cube of the averagedistance from the Sun to the planet.

For orbit around the Sun, KS = 2.97x10-19 s2/m3

K is independent of the mass of the planet

32KrT

K l Thi d L li ti


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Keplers Third Law application

Mass of the Sun or othercelestial body that hassomething orbiting it

Assuming a circular orbit

is a good approximation

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