9 29

Chapter 6,9,10Chapter 6,9,10
Circular Motion, Gravitation,Circular Motion, Gravitation,
Rotation, Bodies inRotation, Bodies in
EquilibriumEquilibrium

Circular MotionCircular Motion
 Ball at the end of a string revolvingBall at the end of a string revolving
 Planets around SunPlanets around Sun
 Moon around EarthMoon around Earth

The RadianThe Radian
 The radian is a unitThe radian is a unit
of angular measureof angular measure
 The radian can beThe radian can be
defined as the arcdefined as the arc
length s along alength s along a
circle divided bycircle divided by
the radius rthe radius r

s
r
θ =
57.3°

More About RadiansMore About Radians
 Comparing degrees and radiansComparing degrees and radians
 Converting from degrees to radiansConverting from degrees to radians
°=
π
°
= 3.57
2
360
rad1
]rees[deg
180
]rad[ θ
°
π
=θ

Angular DisplacementAngular Displacement
 Axis of rotation isAxis of rotation is
the center of thethe center of the
diskdisk
 Need a fixedNeed a fixed
reference linereference line
 During time t, theDuring time t, the
reference linereference line
moves throughmoves through
angle θangle θ

Angular Displacement, cont.Angular Displacement, cont.
 TheThe angular displacementangular displacement is definedis defined
as the angle the object rotatesas the angle the object rotates
through during some time intervalthrough during some time interval

 The unit of angular displacement isThe unit of angular displacement is
the radianthe radian
 Each point on the object undergoesEach point on the object undergoes
the same angular displacementthe same angular displacement
if θθθ −=

Average Angular SpeedAverage Angular Speed
 The averageThe average
angular speed, ω,angular speed, ω,
of a rotating rigidof a rotating rigid
object is the ratioobject is the ratio
of the angularof the angular
displacement todisplacement to
the time intervalthe time interval
tt
if θθθ
ω
−
==

Angular Speed, cont.Angular Speed, cont.
 TheThe instantaneousinstantaneous angular speedangular speed
 Units of angular speed areUnits of angular speed are
radians/secradians/sec
• rad/srad/s
 Speed will be positive if θ isSpeed will be positive if θ is
increasing (counterclockwise)increasing (counterclockwise)
 Speed will be negative if θ isSpeed will be negative if θ is
decreasing (clockwise)decreasing (clockwise)

Average Angular AccelerationAverage Angular Acceleration
 The average angular accelerationThe average angular acceleration
of an object is defined as the ratio ofof an object is defined as the ratio of
the change in the angular speed tothe change in the angular speed to
the time it takes for the object tothe time it takes for the object to
undergo the change:undergo the change:
t
if ωω
α
−
=

Angular Acceleration, contAngular Acceleration, cont
 Units of angular acceleration are rad/s²Units of angular acceleration are rad/s²
 Positive angular accelerations are in thePositive angular accelerations are in the
counterclockwise direction and negativecounterclockwise direction and negative
accelerations are in the clockwise directionaccelerations are in the clockwise direction
 When a rigid object rotates about a fixedWhen a rigid object rotates about a fixed
axis, every portion of the object has theaxis, every portion of the object has the
same angular speed and the same angularsame angular speed and the same angular
accelerationacceleration

Angular Acceleration, finalAngular Acceleration, final
 The sign of the acceleration does notThe sign of the acceleration does not
have to be the same as the sign ofhave to be the same as the sign of
the angular speedthe angular speed
 The instantaneous angularThe instantaneous angular
accelerationacceleration

Analogies Between Linear andAnalogies Between Linear and
Rotational MotionRotational Motion
atvv += 0
( )vvvaverage += 0
2
1
2
00
2
1
attvxx ++=
)(2 0
2
0
2
xx
vv
a
−
−
=
tαωω += 0
( )ωωω += 0
2
1
average
2
00
2
1
tt αωθθ ++=
)(2 0
2
0
2
θθ
ωω
α
−
−
=
Linear Motion with constant
acc.
(x,v,a)
Rotational Motion with fixed
axis
and constant α
(θ,ω,α)

ExamplesExamples
 78 rev/min=?78 rev/min=?
 A fan turns at a rate of 900 rpmA fan turns at a rate of 900 rpm
 Tangential speed of tips of 20cmTangential speed of tips of 20cm
long blades?long blades?
 Now the fan is uniformly acceleratedNow the fan is uniformly accelerated
to 1200 rpm in 20 sto 1200 rpm in 20 s

Relationship Between Angular andRelationship Between Angular and
Linear QuantitiesLinear Quantities
 DisplacementsDisplacements
 SpeedsSpeeds
 AccelerationsAccelerations
 Every point on theEvery point on the
rotating object hasrotating object has
the same angularthe same angular
motionmotion
 Every point on theEvery point on the
rotating objectrotating object
doesdoes notnot have thehave the
same linear motionsame linear motion
Rx θ=
Rv ω=
Ra α=//

Centripetal AccelerationCentripetal Acceleration
 An object traveling in a circle, evenAn object traveling in a circle, even
though it moves with a constantthough it moves with a constant
speed, will have an accelerationspeed, will have an acceleration
 The centripetal acceleration is due toThe centripetal acceleration is due to
the change in thethe change in the directiondirection of theof the
velocityvelocity

Centripetal Acceleration, cont.Centripetal Acceleration, cont.
 Centripetal refersCentripetal refers
to “center-seeking”to “center-seeking”
 The direction of theThe direction of the
velocity changesvelocity changes
 The acceleration isThe acceleration is
directed toward thedirected toward the
center of the circlecenter of the circle
of motionof motion

Centripetal Acceleration, finalCentripetal Acceleration, final
 The magnitude of the centripetalThe magnitude of the centripetal
acceleration is given byacceleration is given by
• This direction is toward the center of theThis direction is toward the center of the
circlecircle
R
v
a
2
=

Centripetal Acceleration andCentripetal Acceleration and
Angular VelocityAngular Velocity
 The angular velocity and the linearThe angular velocity and the linear
velocity are related (v = ωR)velocity are related (v = ωR)
 The centripetal acceleration can alsoThe centripetal acceleration can also
be related to the angular velocitybe related to the angular velocity
Ra 2
ω=

Forces Causing CentripetalForces Causing Centripetal
AccelerationAcceleration
 Newton’s Second Law says that theNewton’s Second Law says that the
centripetal acceleration is accompanied bycentripetal acceleration is accompanied by
a forcea force
• F = maF = ma ⇒⇒
• FF stands for any force that keeps an objectstands for any force that keeps an object
following a circular pathfollowing a circular path
 Tension in a stringTension in a string
 GravityGravity
 Force of frictionForce of friction
R
v
mF
2
=

ExamplesExamples
 Ball at theBall at the
end ofend of
revolvingrevolving
stringstring
 Fast carFast car
rounding arounding a
curvecurve

More on circular MotionMore on circular Motion
 Length of circumference = 2Length of circumference = 2ππRR
 Period T (time for one completePeriod T (time for one complete
circle)circle)
2
22
)2(
2
τ
π
π
τ
R
R
R
v
a
v
R
==
=
2
2
4
τ
π R
a =

ExampleExample
 200 grams mass revolving in uniform200 grams mass revolving in uniform
circular motion on an horizontalcircular motion on an horizontal
frictionless surface at 2frictionless surface at 2
revolutions/s. What is the force onrevolutions/s. What is the force on
the mass by the string (R=20cm)?the mass by the string (R=20cm)?

Newton’s Law of UniversalNewton’s Law of Universal
GravitationGravitation
 Every particle in the UniverseEvery particle in the Universe
attracts every other particle with aattracts every other particle with a
force that is directly proportional toforce that is directly proportional to
the product of the masses andthe product of the masses and
inversely proportional to the squareinversely proportional to the square
of the distance between them.of the distance between them.
2
21
R
mm
GF =

Universal Gravitation, 2Universal Gravitation, 2
 G is the constant of universalG is the constant of universal
gravitationalgravitational
 G = 6.673 x 10G = 6.673 x 10-11-11
N m² /kg²N m² /kg²
 This is an example of anThis is an example of an inverseinverse
square lawsquare law

 The force thatThe force that
mass 1 exerts onmass 1 exerts on
mass 2 is equalmass 2 is equal
and opposite to theand opposite to the
force mass 2force mass 2
exerts on mass 1exerts on mass 1
 The forces form aThe forces form a
Newton’s third lawNewton’s third law
action-reactionaction-reaction

 The gravitational force exerted by aThe gravitational force exerted by a
uniform sphere on a particle outsideuniform sphere on a particle outside
the sphere is the same as the forcethe sphere is the same as the force
exerted if the entire mass of theexerted if the entire mass of the
sphere were concentrated on itssphere were concentrated on its
centercenter

Gravitation ConstantGravitation Constant
 DeterminedDetermined
experimentallyexperimentally
 Henry CavendishHenry Cavendish
• 17981798
 The light beam andThe light beam and
mirror serve tomirror serve to
amplify the motionamplify the motion

Applications of UniversalApplications of Universal
 Weighing the EarthWeighing the Earth
G
gR
m
R
m
Gg
R
mm
Gmg
R
mm
GFw
E
E
E
E
E
E
E
E
g
2
2
2
2
=
=
=
==
kg106
6380
/8.9take
24
2
×=⇒
=
=
E
E
m
kmR
smg

Applications of UniversalApplications of Universal
 ““g” will vary withg” will vary with
altitudealtitude
2
""
r
m
Gg E
=

Escape SpeedEscape Speed
 The escape speed is the speedThe escape speed is the speed
needed for an object to soar off intoneeded for an object to soar off into
space and not returnspace and not return
 For the earth, vFor the earth, vescesc is about 11.2 km/sis about 11.2 km/s
 Note, v is independent of the mass ofNote, v is independent of the mass of
the objectthe object
E
E
esc
R
Gm
v
2
=

Various Escape SpeedsVarious Escape Speeds
 The escape speedsThe escape speeds
for variousfor various
members of themembers of the
solar systemsolar system
 Escape speed isEscape speed is
one factor thatone factor that
determines adetermines a
planet’splanet’s
atmosphereatmosphere

Motion of SatellitesMotion of Satellites
 Consider onlyConsider only
circular orbitcircular orbit
 Radius of orbit r:Radius of orbit r:
 Gravitational forceGravitational force
is the centripetalis the centripetal
force.force.
hRr E +=
2
2
2
v
r
m
G
r
v
m
r
mm
GmaF
EE
=⇒=⇒=
r
Gm
v
E
=
r

Motion of SatellitesMotion of Satellites
 PeriodPeriod ττ
v
rπ
τ
2
=
EGm
r 23
2π
τ = Kepler’s 3rd
Law
milesmrm
Gs
E
4724
11
106.21023.4106
,1067.6,86400
×=×=⇒×=
×== −
τ

Communications SatelliteCommunications Satellite
 A geosynchronous orbitA geosynchronous orbit
• Remains above the same place on the earthRemains above the same place on the earth
• The period of the satellite will be 24 hrThe period of the satellite will be 24 hr
 r = h + Rr = h + REE
 Still independent of the mass of the satelliteStill independent of the mass of the satellite
milesmrm
Gs
E
4724
11
106.21023.4106
,1067.6,86400
×=×=⇒×=
×== −
τ

Satellites and WeightlessnessSatellites and Weightlessness
 weighting an object in an elevatorweighting an object in an elevator
 Elevator at rest: mgElevator at rest: mg
 Elevator accelerates upward:Elevator accelerates upward:
m(g+a)m(g+a)
 Elevator accelerates downward:Elevator accelerates downward:
m(g+a) with a<0m(g+a) with a<0
 Satellite: a=-g!!Satellite: a=-g!!

Force vs. TorqueForce vs. Torque
 Forces cause accelerationsForces cause accelerations
 Torques cause angular accelerationsTorques cause angular accelerations
 Force and torque are relatedForce and torque are related

TorqueTorque
 The door is free to rotate about an axis through OThe door is free to rotate about an axis through O
 There are three factors that determine theThere are three factors that determine the
effectiveness of the force in opening the door:effectiveness of the force in opening the door:
• TheThe magnitudemagnitude of the forceof the force
• TheThe positionposition of the application of the forceof the application of the force
• TheThe angleangle at which the force is appliedat which the force is applied

Torque, contTorque, cont
 Torque,Torque, ττ, is the tendency of a force, is the tendency of a force
to rotate an object about some axisto rotate an object about some axis
 ττ is the torqueis the torque
 F is the forceF is the force
• symbol is the Greek tausymbol is the Greek tau
 l is the length of lever arml is the length of lever arm
 SI unit is NSI unit is N..
mm
 Work done by torque W=Work done by torque W=τθτθ
Fl=τ

Direction of TorqueDirection of Torque
 If the turning tendency of the forceIf the turning tendency of the force
is counterclockwise, the torque willis counterclockwise, the torque will
be positivebe positive
 If the turning tendency isIf the turning tendency is
clockwise, the torque will beclockwise, the torque will be
negativenegative

Multiple TorquesMultiple Torques
 When two or more torques are actingWhen two or more torques are acting
on an object, the torques are addedon an object, the torques are added
 If the net torque is zero, the object’sIf the net torque is zero, the object’s
rate of rotation doesn’t changerate of rotation doesn’t change

Torque and EquilibriumTorque and Equilibrium
 First Condition of EquilibriumFirst Condition of Equilibrium
 The net external force must be zeroThe net external force must be zero
• This is a necessary, but not sufficient,This is a necessary, but not sufficient,
condition to ensure that an object is incondition to ensure that an object is in
complete mechanical equilibriumcomplete mechanical equilibrium
• This is a statement of translational equilibriumThis is a statement of translational equilibrium
0
0 0x y
or
and
Σ =
Σ = Σ =
F
F F
r
r r

Torque and Equilibrium, contTorque and Equilibrium, cont
 To ensure mechanical equilibrium,To ensure mechanical equilibrium,
you need to ensure rotationalyou need to ensure rotational
equilibrium as well as translationalequilibrium as well as translational
 The Second Condition of EquilibriumThe Second Condition of Equilibrium
statesstates
• The net external torque must be zeroThe net external torque must be zero
∑ = 0τ

Equilibrium ExampleEquilibrium Example
 The woman, mass m,The woman, mass m,
sits on the left end ofsits on the left end of
the see-sawthe see-saw
 The man, mass M, sitsThe man, mass M, sits
where the see-saw willwhere the see-saw will
be balancedbe balanced
 Apply the SecondApply the Second
Condition ofCondition of
Equilibrium and solveEquilibrium and solve
for the unknownfor the unknown
distance, xdistance, x

Moment of InertiaMoment of Inertia
 The angular acceleration is inverselyThe angular acceleration is inversely
proportional to the analogy of theproportional to the analogy of the
mass in a rotating systemmass in a rotating system
 This mass analog is called theThis mass analog is called the
moment of inertia,moment of inertia, I, of the objectI, of the object
• SI units are kg mSI units are kg m22
2
I mr≡ Σ

Newton’s Second Law for aNewton’s Second Law for a
Rotating ObjectRotating Object
 The angular acceleration is directlyThe angular acceleration is directly
proportional to the net torqueproportional to the net torque
 The angular acceleration is inverselyThe angular acceleration is inversely
proportional to the moment of inertiaproportional to the moment of inertia
of the objectof the object
Iτ αΣ =

More About Moment of InertiaMore About Moment of Inertia
 There is a major difference betweenThere is a major difference between
moment of inertia and mass: themoment of inertia and mass: the
moment of inertia depends on themoment of inertia depends on the
quantity of matterquantity of matter and itsand its
distributiondistribution in the rigid object.in the rigid object.
 The moment of inertia also dependsThe moment of inertia also depends
upon the location of the axis ofupon the location of the axis of
rotationrotation

Moment of Inertia of a UniformMoment of Inertia of a Uniform
RingRing
 Image the hoop isImage the hoop is
divided into adivided into a
number of smallnumber of small
segments, msegments, m11 ……
 These segmentsThese segments
are equidistantare equidistant
from the axisfrom the axis
2 2
i iI m r MR= Σ =

Other Moments of InertiaOther Moments of Inertia

ExampleExample
Wheel of radius R=20 cm andWheel of radius R=20 cm and
I=30kg·m². Force F=40N actsI=30kg·m². Force F=40N acts
along the edge of the wheel.along the edge of the wheel.
1.1. Angular acceleration?Angular acceleration?
2.2. Angular speed 4s after startingAngular speed 4s after starting
from rest?from rest?
3.3. Number of revolutions for the 4s?Number of revolutions for the 4s?
4.4. Work done on the wheel?Work done on the wheel?

Rotational Kinetic EnergyRotational Kinetic Energy
 An object rotating about some axisAn object rotating about some axis
with an angular speed, ω, haswith an angular speed, ω, has
rotational kinetic energy KErotational kinetic energy KErr==½Iω½Iω22
 Energy concepts can be useful forEnergy concepts can be useful for
simplifying the analysis of rotationalsimplifying the analysis of rotational
motionmotion
 Units (rad/s)!!Units (rad/s)!!

Total Energy of a SystemTotal Energy of a System
 Conservation of Mechanical EnergyConservation of Mechanical Energy
• Remember, this is for conservativeRemember, this is for conservative
forces, no dissipative forces such asforces, no dissipative forces such as
friction can be presentfriction can be present
• Potential energies of any otherPotential energies of any other
conservative forces could be addedconservative forces could be added
( ) ( )t r g i t r g fKE KE PE KE KE PE+ + = + +

Rolling down inclineRolling down incline
 Energy conservationEnergy conservation
 Linear velocity and angular speed areLinear velocity and angular speed are
related v=Rrelated v=Rωω
 Smaller I, bigger v, faster!!Smaller I, bigger v, faster!!
22
2
1
2
1
ωImvmgh +=
2
2
2
2
2
)(
2
1
)(
2
1
2
1
v
R
I
mv
R
I
mvmgh +=+=

Work-Energy in a RotatingWork-Energy in a Rotating
SystemSystem
 In the case where there areIn the case where there are
dissipative forces such as friction,dissipative forces such as friction,
use the generalized Work-Energyuse the generalized Work-Energy
Theorem instead of Conservation ofTheorem instead of Conservation of
EnergyEnergy
 (KE(KEtt+KE+KERR+PE)+PE)ii++W=(KEW=(KEtt+KE+KERR+PE)+PE)ff

Angular MomentumAngular Momentum
 Similarly to the relationship betweenSimilarly to the relationship between
force and momentum in a linearforce and momentum in a linear
system, we can show thesystem, we can show the
relationship between torque andrelationship between torque and
angular momentumangular momentum
 Angular momentum is defined asAngular momentum is defined as
• L = I ωL = I ω
• andand L
t
τ
∆
Σ =
∆

Angular Momentum, contAngular Momentum, cont
 If the net torque is zero, the angularIf the net torque is zero, the angular
momentum remains constantmomentum remains constant
 Conservation of Angular MomentumConservation of Angular Momentum
states: The angular momentum of astates: The angular momentum of a
system is conserved when the netsystem is conserved when the net
external torque acting on theexternal torque acting on the
systems is zero.systems is zero.
• That is, whenThat is, when
0, i f i i f fL L or I Iτ ω ωΣ = = =

Conservation Rules, SummaryConservation Rules, Summary
 In an isolated system, the followingIn an isolated system, the following
quantities are conserved:quantities are conserved:
• Mechanical energyMechanical energy
• Linear momentumLinear momentum
• Angular momentumAngular momentum

Conservation of AngularConservation of Angular
Momentum, ExampleMomentum, Example
 With hands andWith hands and
feet drawn closerfeet drawn closer
to the body, theto the body, the
skater’s angularskater’s angular
speed increasesspeed increases
• L is conserved, IL is conserved, I
decreases,decreases, ωω
increasesincreases

ExampleExample
A 500 grams uniform sphere of 7.0 cmA 500 grams uniform sphere of 7.0 cm
radius spins at 30 rev/s on an axisradius spins at 30 rev/s on an axis
through its center.through its center.
 Moment of inertiaMoment of inertia
 Rotational kinetic energyRotational kinetic energy
 Angular momentumAngular momentum

ExampleExample
Find work done to open 30Find work done to open 30°° a 1m widea 1m wide
door with a steady force of 0.9N atdoor with a steady force of 0.9N at
right angle to the surface of theright angle to the surface of the
door.door.

ExampleExample
A turntable is a uniform disk of metalA turntable is a uniform disk of metal
of mass 1.5 kg and radius 13 cm.of mass 1.5 kg and radius 13 cm.
What torque is required to drive theWhat torque is required to drive the
turntable so that it accelerates at aturntable so that it accelerates at a
constant rate from 0 to 33.3 rpm inconstant rate from 0 to 33.3 rpm in
2 seconds?2 seconds?

Center of GravityCenter of Gravity
 The force of gravity acting on anThe force of gravity acting on an
object must be consideredobject must be considered
 In finding the torque produced byIn finding the torque produced by
the force of gravity, all of the weightthe force of gravity, all of the weight
of the object can be considered to beof the object can be considered to be
concentrated at a single pointconcentrated at a single point

Calculating the Center ofCalculating the Center of
GravityGravity
 The object isThe object is
divided up into adivided up into a
large number oflarge number of
very small particlesvery small particles
of weight (mg)of weight (mg)
 Each particle willEach particle will
have a set ofhave a set of
coordinatescoordinates
indicating itsindicating its
location (x,y)location (x,y)

Calculating the Center ofCalculating the Center of
Gravity, cont.Gravity, cont.
 We wish to locate the point ofWe wish to locate the point of
application of theapplication of the single forcesingle force whosewhose
magnitude is equal to the weight ofmagnitude is equal to the weight of
the object, and whose effect on thethe object, and whose effect on the
rotation is the same as all therotation is the same as all the
individual particles.individual particles.
 This point is called theThis point is called the center ofcenter of
gravitygravity of the objectof the object

Coordinates of the Center ofCoordinates of the Center of
GravityGravity
 The coordinates of the center ofThe coordinates of the center of
gravity can be foundgravity can be found
i i i i
cg cg
i i
m x m y
x and y
m m
Σ Σ
= =
Σ Σ

Center of Gravity of a UniformCenter of Gravity of a Uniform
ObjectObject
 The center of gravity of aThe center of gravity of a
homogenous, symmetric body musthomogenous, symmetric body must
lie on the axis of symmetry.lie on the axis of symmetry.
 Often, the center of gravity of suchOften, the center of gravity of such
an object is thean object is the geometricgeometric center ofcenter of
the object.the object.

ExampleExample
Find the center of mass (gravity) ofFind the center of mass (gravity) of
these masses: 3kg (0,1), 2kg (0,0)these masses: 3kg (0,1), 2kg (0,0)
And 1kg (2,0)And 1kg (2,0)

ExampleExample
Find the center of mass (gravity) ofFind the center of mass (gravity) of
the dumbbell, 4 kg and 2 kg with athe dumbbell, 4 kg and 2 kg with a
4m long 3kg rod.4m long 3kg rod.

Torque, reviewTorque, review
 ττ is the torqueis the torque
 FF is the forceis the force
• symbol is the Greek tausymbol is the Greek tau
 ll is the length of lever armis the length of lever arm
 SI unit is NSI unit is N..
mm
Fl=τ

ExampleExample
A 2 m by 2 m square metal plateA 2 m by 2 m square metal plate
rotates about its center. Calculaterotates about its center. Calculate
the torque of all five forces each withthe torque of all five forces each with
magnitude 50N.magnitude 50N.

Torque and EquilibriumTorque and Equilibrium
 First Condition of EquilibriumFirst Condition of Equilibrium
 The net external force must be zeroThe net external force must be zero
0
0 0x y
or
and
Σ =
Σ = Σ =
F
F F
r
r r
The Second Condition of EquilibriumThe Second Condition of Equilibrium
statesstates
•The net external torque must be zeroThe net external torque must be zero
∑ = 0τ

ExampleExample
The system is in equilibrium. CalculateThe system is in equilibrium. Calculate
W and find the tension in the ropeW and find the tension in the rope
(T).(T).

ExampleExample
A 160 N boy stands on a 600 NA 160 N boy stands on a 600 N
concrete beam in equilibrium withconcrete beam in equilibrium with
two end supports. If he stands onetwo end supports. If he stands one
quarter the length from one support,quarter the length from one support,
what are the forces exerted on thewhat are the forces exerted on the
beam by the two supports?beam by the two supports?

9 29

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