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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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Angular Displacement
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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Angular Displacement
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.jpg7/30/2019 FisDas - Minggu 05 - Gerak Melingkar 1
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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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