Lectures - Classical mechanics

8/10/2019 Lectures - Classical mechanics

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Classical Mechanics

Nguyen Thi Hong VanINSTITUTE OF PHYSICS, HANOI

December 3, 2013

Nguyen Thi Hong Van


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Contents

1 Chapter 1:

Introduction to classical mechanics Motion in one dimensionMotion in three dimension

2 Chapter 2: Forces and Newtons laws of motion

3

Chapter 3: Work and energy 4 Chapter 4: Momentum conservation

5 Chapter 5: Circular motion

6 Chapter 6: Rotational motion

7 Chapter 7: Angular momentum

8 Chapter 8: Law of gravity

9 Chapter 9: Fluids mechanics

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Evaluation

Attendance/Attitude: 5 %

Class exercise(s): 10 %

Mid-term test: 25 %

Final exam: 60 %

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Major sources

Major sources

1 Classical Mechanics: an introductory course , Richard Fitzpatrick(professor of Physics, the University of Texas at Austin)

2 R. Resnick, D. Halliday, and K.S. Krane, Fourth edition, Vol. 1(John Wiley & Sons, New York NY, 1992)

3 G.R. Fowles, Third edition (Holt, Rinehart, & Winston, New York

NY, 1977).

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Chapter 1: Overview of Classical Mechanics (CM)

CM is the study of the motion of bodies in accordance with the generalprincipals rst enunciated by Issac Newton (1687)

CM has many applications in different areas of science such asastronomy , Chemistry , Geology , and Engineering

Types of motion studied in CM:

Translational motion : a body shifts from one point in space toanother

Rotational motion : an extended body changes orientation, withrespect to other bodies in space, without changing position

Oscillatory motion : continually repeats in time with a xed periodCircular motion : a body executes a circular orbit about anotherxed body.

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Chapter 1: Units of measurement (1)

mks units (Meter, Kilogram, Second)

Three fundamental quantities which are subject to measurement used inCM:Interval in space lengthQuantities of inertia , or mass , processed by various bodiesInterval in time

SI system (Systemme International) mks units SI system is a standard units for basis quantities. It was establishedin 1960 by an international committeemks system is for length (meter), mass (kilogram) and time (second)

Derived quantities:

Velocity: [v ] = ms 1

Acceleration: [a] = ms 2

angular momentum: [ p ] = kgms 1

mks units of these quantities are the corresponding combination of themks units of length, mass and time.

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Chapter 1: Units of measurement (2)

Standard prexes A set of standard prexes is a modication of the mks units of length, mass and time which is devised to deal more easily with very small and very largequantities (the motion of molecules and the motion of stars in the Galaxy).

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Chapter 1: Uncertainty and signicant gures (2)

Signicant gures

Zero may NOT be signicant gures: those used to position the decimalpoint such as: 0.03 (1 signicant gure) and 0.0075 (2 signicant gures).Zero may be signicant gures: when zeros come after other digits.However there is the possibility of mis-interpretation should usescientic notation to indicate the number of signicant gures:

1500 1.5 103

if there are 2 SFs1500 1.50 103 if there are 3 SFs1500 1.500 103 if there are 4 SFs0.00023 2.30 10 4 if there are 3 SFs.

When numbers are added or subtracted, the number of decimal places in

the result should equal the smallest number of decimal places of any termin the sum:123 + 5.35 = 1281.002 - 0.998 = 0.004

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Chapter 1: Dimensional analysis

All laws of physics are dimensionally consistent:

A = B [A] = [B ]

Application of dimensional analysis: to check the forms of simplelaws of physics.

Nguyen Thi Hong Van

h d ( )


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Chapter 1: Motion in one dimension (1) Displacement

Velocity Velocity is the rate of change of displacement with timeAverage velocity:

v = x t

Instantaneous velocity:v = lim t 0

x t

Average speed: the ratio of the total distance it travels to the total time ittakes to travel that distance.Instantaneous speed = |v |

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Ch 1 M i i di i (3)


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Chapter 1: Motion in one dimension (3) Motion with constant velocity

Problem: given v = const , nd displacement x with time t ?Answer:

v = dx / dt dx = v .dt x = v .t + C C =? : t = 0 x = x 0 (displacement of the object at t = 0 or initialdisplacement). x (t ) = x 0 + v .t

Problem: Plot the graph of displacement versus time for a body movingwith constant velocity?Conclusion with motion with constant velocity:

The graph of x versus t is a straight line.v = x / t gradient of the graph/straight line.a = d 2 x / dt 2 = 0 v = const : consistent with the denition!x increases v > 0 OR object moving to the right,x decreases v < 0 OR object moving to the left,x = const v = 0 OR object remains at rest.

Nguyen Thi Hong Van

Ch t 1 M ti i di i (4)


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Chapter 1: Motion in one dimension (4) Motion with constant acceleration

Prob.: a = const v =?; x =?

a = dv / dt dv = a .dt v (t ) = v 0 + a .t straight line

dx = v .dt x = x 0 + v 0 .t + a.t 2 / 2 parabola.

Conclusion:S = x x 0 = v 0 .t + a .t 2 / 2 the net distance traveled after t

secondsv = v 0 + a.t v 2 = v 20 + 2 .a .S

Free-fall under gravity Galileo: all bodies in free-fall close to the Earth surface accelerate

vertically downward with the same acceleration: g = 9 .81m / s 2

???

t = 2hg Nguyen Thi Hong Van

Ch t 1 M ti i th di i (1)


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Chapter 1: Motion in three dimensions (1)

To deal with 3-dimensional motion, it is necessary to set up a suitablecoordinate system. Vectors

Cartesian coordinates

Cartesian coordinate system consistsof three mutually perpendicular axes.

Vector and scalar quantities

Scalar quantity is specied by a single value with an appropriate unitand has no direction such as temperature, volume, mass, timeA vector quantity has both magnitude and direction. Examples:displacement,velocity, accelerator, force .

Nguyen Thi Hong Van



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Chapter 1: Motion in three dimensions (2) Some properties of vector

Equality of two vectors:

A = B |

A| = |

B | &

A and

B point in the same direction.

Vector additionSum of two vectors is the vector down from the tail of the rstvector to the tip of the last vector.Parallelogram rule of addition A + B = B + AAssociative law of addition: A + ( B + C ) = ( A + B ) + C

Negative of a vector: A + ( A) = 0 have the same magnitude butpoint in opposite direction.Subtracting vector: A B = A + ( B )

Components of a vector and unit vectors

Components of a vector are the projections of that vector alongcoordinate axes: A = ( Ax , Ay , Az ).?? Can the components of a vector ever be greater than the magnitude of the vector?

hint: Ax = | A| .sin.cos ; Ay = | A| .sin sin ; Az = | A| .cos Nguyen Thi Hong Van



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Chapter 1: Motion in three dimensions (3) Components of a vector and unit vectors

Unit vectors

a unit vector is a dimensionless vector having a magnitude of exactly1. It is used to specify a given direction and have no other physicalsignicance:

i x ; j y ; k z .| i | = | j | = | k | = 1 i . j = i . k = j . k = 0 A = Ax . i + Ay . j + Az . k .

Vector magnitude: A = A2x + A2y + A2z Scalar multiplication: S = . r , is a number/scalar

S = (x , y , z ) = ( x , y , z ): multiplying all the components of the original.Scalar product of two vectors:

A. B = Ax .B x + Ay .B y + Az .B z

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Chapter 1: Motion in three dimensions (4) Vector displacement, vector velocity and vector acceleration

Vector displacement: r (t ) = [ x (t ), y (t ), z (t )]

Vector velocity:

v (t ) = lim t 0 r (t + t ) r (t )

t =

d r dt

v x = dx

dt

; v y = dy

dt

; v z = dz

dt Vector acceleration:

a(t ) = lim t 0 v (t + t ) v (t )

t =

d v dt

= d 2 r dt 2

ax =

dv x dt =

d 2 x

dt 2

ay = dv y

dt =

d 2 y dt 2

a z = dv z

dt =

d 2 z dt 2

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Chapter 1: Motion in three dimensions (5) Motion with constant velocity

r (t ) = r 0 + v .t

The trajectory of the object is parallel to v a straight-line

Motion with constant acceleration

r (t ) = r 0 + v 0 .t + a .t 2 / 2

S = v 0 .t + a .t 2

/ 2 v (t ) = v 0 + a .t

v (t )2 = v 20 + 2 a . S

Projectile motion Given: a = (0 , 0, g ) ; v 0 = v 0 .cos , 0, v 0 .sin?? Find horizontal range R = v 20 .sin2/ g R max = v 20 / g when = 45 o

?? Find altitude of the projectile h = v 20 .sin 2 / 2g hmax = v 20 / 2g when = 90 o

Nguyen Thi Hong Van


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Chapter 2: Forces and Newtons laws of motion


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Chapter 2: Forces and Newton s laws of motion Newtons rst law of motion

If the motion of a given body is not disturbed by external inuences then

that body moves with constant velocity r = r 0 + v .t r 0 , v ; constant vectors v = 0 the body simply remains at rest.

Newtons second law of motion

Momentum: p = m. v Second law: d p dt = f net inuence/force

m = const f = m . a

Newtons third law of motion

f ab : force exerted on (a) by (b) f ab : force exerted on (b) by (a)

Newtons third law: f ab = f baNguyen Thi Hong Van


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Chapter 2: Everyday forces


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Chapter 2: Everyday forces

Gravitational force: f g = m. g The normal force/normal reaction and weight:

Weight: the weight W of a body is the magnitude of the downwardforce it exerts on any object which supports it.Normal force: under the inuence of the weight of the object,obeying the Newtons third law, the surface of the supporting objectwill act back on the surface of the object this called normalreaction or normal force

Normal force is a force exerted by one surface on another i adirection perpendicular to the surface of contact.

The magnitude of the weight = the normal force .Exercise:

Find weight of an object resting on a table?

Find weight of an object of mass m resting on the oor of anelevator moving with acceleration a?

Relation between mass and weight : W = m.g in case of accelerator of the frame = 0 and f g = m g is perpendicular to the surface.Friction: f = .R n , where : coefficient of friction and R n is normalreaction of the surface.

Nguyen Thi Hong Van

Chapter 3: Work and energy


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Chapter 3: Work and energy Energy Potential energy; kinetic energy; electrical energy; thermal energy;chemical energy; nuclear energy.

An example of energy conservation: energy conservation during free-fall

v 22 v 21 = 2 a . S ; a = g ; S = h2 h1

v 22 v

21 2 g . S

mv 22

2

mv 212

= mgh 2 + mgh 1

12

mv 22 + mgh 2 = 12

mv 21 + mgh 1

Total energy E of the mass m is conserved: E = mv 2 / 2 + mghK = mv 2 / 2 kinetic energyU = mgh potential energyE conserved K = U

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Work W = f . S = | f | .| S | .cos : assuming that the force doesnt vary withposition.

Work-Energy theorem : ?? prove:

K = W

using v 2 = v 20 + 2 . a . S

Find back the energy conservation during free-fall using work-energy

theorem?

Nguyen Thi Hong Van

Chapter 3: Work


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pWork of the force varies with position f = f ( r )

W =N

i =1

f i . r i = limN N

i =1

f i . r i =B

A

f ( r ).d r line-integral

Work performed by a 1-dimensional force: From W =x B

x A f (x )dx ,prove: W = K ?f = m

d 2 x dt 2

W =

x B

x A

md 2 x dt 2

.dx =

t B

t A

md 2 x dt 2

.dx dt

.dt =

t B

t A

d dt

m2

dx dt

2

x (t A) = x A , x (t B ) = x B , v A = dx dt t = t A

, v B = dx dt t = t B

W = 12

m .v 2B 12

mv 2A = K

The object is acted by many forces: W i =B

A f i ( r ).d r W = i W i = i

B

A f i ( r ).d r =B

A f i ( r ) .d r =B

A f ( r ).d r ; f ( r ) =

i

f i ( r )

Nguyen Thi Hong Van

Chapter 3: Conservative and non-conservative force-eld


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p

A non-uniform force-eld f ( r )acts upon an object which movesalong curved trajectory (frompoint A to point B) labeled bypath 1 and path 2.

Work W 1 performed by f ( r ) along path 1: W 1 =path 1

A B f .d r Work W 1 performed by f ( r ) along path 2: W 2 =

path 2

A B

f .d r

If W 1 = W 2 f ( r ) is a conservative force-eld : the line-integral doesntdepend on the path taken between the end points; Ex: gravitational force.If W 1 = W 2 f ( r ) is a non-conservative force-eld : the line-integraldepends both on the end points A, B and the path taken between them;Ex: frictional force.

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Chapter 3: Potential energy


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p gy

Consider a body moving in a conservative force-eld f ( r ). Pick arbitrarilysome point O in this eld.Dene:

U (R ) = R

O f .d r (3.1)(3.1) U (O ) = 0U (R ) dened by (3.1) is unique in conservative force-eld since inthis eld, the work of the eld depends only on the x points butNOT the path taken between them.(3.1) makes no sense in non-conservative force eld since there willbe an innite number of different values corresponding to the innitenumber of different paths the body coild take between O and R.

Consider the object move between A and BFrom the work-energy theorem:

K =B

A f .d r =B

A f .d r =B

O f .d r A

O f .d r = U E = K (A) + U (A) = K (B ) + U (B ) = K + U = constant U represents some form of potential energy

Nguyen Thi Hong Van

Chapter 3: Some conclusions about potential energy


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p p gy

Possible to associate a potential energy with any conservativeforce-eld.

Any force-eld for which we can dene a potential energy mustnecessary be conservative.

The concept of potential energy is meaning-less in anon-conservative force-eld.

Potential energy dene to within an arbitrary additive constant sowe can choose an arbitrary point at which U = 0.

Potential energy is not a property of the body BUT a property of the force-eld within which the body moves.

Nguyen Thi Hong Van

Chapter 3: Hookes law


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Force: f = kx

Work: W =x B

x A

f (x )dx = k x B

x A

x .dx = [kx 2B / 2 kx 2A/ 2]

Potential energy: U (x B ) U (x A) = x B

x A f (x ).dx = kx 2B / 2 kx 2A/ 2 U (x ) = kx 2 / 2

f = dU dx

In three-dimensional; f =

U x

, U y

, U z

Example in gravitational force-eld: U = mgz f = (0 , 0, mg )Total energy of the mass: E = K + U

Power

P = dW

dt = f . v

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Chapter 4: Conservation of momentum: Multi-component systems


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Consider a system of N mutually interacting point mass objects whichmove in 3-dimensionsInternal force: f ij = f ji ; i , j = 1 , N , i = j External force: F i

Second law of Newton for object i : m i r i = j = i

j =1 , N f ij + F i Summing over all objects:

i =1 , N

m i r i = j = i

i , j =1 , N

f ij +i =1 , N

F i =i =1 , N

F i

M r cm = F ; M =i =1 , N

m i , F =i =1 , N

F i ; r cm =

N

i =1

m i r i

N

i =1 m i If F = 0 r cm =

N

i =1 m i r i /N

i =1 m i = const OR P =N

i =1 m i r i = const In general case: d P / dt = F

Nguyen Thi Hong Van

Chapter 4: Collisions in 1-dimension


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m 1 and m2 collide in1-dimensionignore zero net external force inthe direction of moving

Momentum conservation gives

m 1 v i 1 + m2 v i 2 = m 1 v 1 f + m2 v f 2 (1)

In 1-dimention: m1 v

i 1

+ m2 v

i 2

= m 1 v

f 1

+ m2 v

f 2

There are many types of collision

Elastic collision: Total kinetic energy of the two colliding objects isconserved

1

2m 1 v 2i 1 +

1

2m 2 v 2i 2 =

1

2m 1 v 2f 1 +

1

2m 2 v 2f 2

Inelastic collision: some fraction of the initial kinetic energy of thecolliding objects is usually converted into some other form of energy.Totally inelastic collision: two objects stick together after the collision v f 1 = v f 2

Nguyen Thi Hong Van

Chapter 4: Collisions in 1-dimension: Elastic collision


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In the laboratory frame:

v cm = m1 v i 1 + m2 v i 2

m 1 + m2=

m1 v f 1 + m2 v f 2m 1 + m2

(3)

In the center of mass frame: v = v v cm

v i 1 = m2

m 1 + m2(v i 2 v i 1 ); v i 2 =

m1m 1 + m2

(v i 2 v i 1 )

v f 1 = m2

m 1 + m2(v f 2 v f 1 ); v f 2 =

m1m 1 + m2

(v f 2 v f 1 ) (4)

p i 1 = p i 2 = (v i 2 v i 1 ); p f 1 = p f 2 = (v f 2 v f 1 ); = m1 m 2m 1 + m2

(5)

p i 1 + p i 2 = p f 1 + p f 2 = 0 trivial !This result is valid also for inelastic collision!The center of mass kinetic energy conservation equation:

p i 12

2m 1+

p i 22

2m 2=

p f 12

2m 1+

p f 22

2m 2(6)

(5)&(6) v i 2 v i 1 = (v f 2 v f 1 ) (7)

(1)&(7) v f 1 = m1 m 2m 1 + m2

v i 1 + 2m 2m 1 + m2

v i 2 ; v f 2 = 2m 1m 1 + m2

v i 1 m1 m 2m 1 + m2

v i 2

Nguyen Thi Hong Van

Chapter 4: Totally inelastic collisions


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We have: v f = v f 1 = v f 2

m 1 v i 1 + m2 v i 2 = m 1 v f 1 + m2 v f 2 = ( m 1 + m2 )v f

v f = m1 v i 1 + m2 v i 2

m 1 + m2= v cm

two objects remain stationary in the center of mass velocity frame.Suppose that v i 2 = 0

v f = m1

m 1 + m2v i 1

The fractional loss in kinetic energy of the system due to the collisionis

f = K i K f

K i =

m1 v 2i 1 (m 1 + m2 )v 2f

m 1 v 2i 1=

m2m 1 + m2

If m2 m 1 the loss in energy is small

If m2 m 1 the loss is almost 100 %

Nguyen Thi Hong Van

Chapter 4: Collision in 2-dimensions


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Suppose v i 2 = 0; m1 , m 2 , v i 1 knownCollision is NOT head-on

Unknown: v

f 1

,v

f 2

, 1

, 2

m 1 v i 1 = m 1 v f 1 + m2 v f 2 (1)(1)/ox:m 1 v i 1 = m 1 v f 1 cos 1 + m2 v f 2 cos 2(1)/oy: m1 v f 1 sin 1 = m 2 v f 2 sin 2

Elastic collision :m 1 v 2i 1 / 2 = m 1 v 2f 1 / 2 + m2 v

2f 2 / 2 need

one more condition!

Totally inelastic colliding :Unknown: f , v f m 1 v i 1 + m2 v i 2 = ( m 1 + m2 ) v f (2)(2)/ox:m 1 v i 1 + m2 v i 2 cos i = ( m 1 + m2 )v f cos f (2)/oy: m2 v i 2 sin i = ( m 1 + m2 )v f sinf

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Chapter 5: Circular motion


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An object executes a circular orbit of radius r with uniform tangential speed

v with | v | doesnt change.t = 0 : = 0 (t ) = .t angular velocity of the object =const in case of uniformly rotating

arc length: S = 2360 o .r .(o )

Dene radian unit for angle: (rad ) = 2360 o (o )

S = r . S = r . arc length in an interval time t v = S / t & = / t v = r .Repetition period of the circular motion (time to execite a completecircle): T = 2 /Repetition frequency: f = 1 / T = / (2 ); [f ] = Hz cycles per second

Nguyen Thi Hong Van

Chapter 5: Centripetal acceleration


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An object executing a circular orbit of radius r with uniform tangentialspeed v : magnitude | v | is constant while its direction is continuouslychanging exist an acceleration !Object moves from P Q between time t , t + t , rotate an angle of during the time interval t

From the diagram one can see: v is directed towards the center of the circle v = 2 v sin(/ 2) is small sin

v

= v

.a = v / t = v ./ t = v . = v 2 / r = r .2

Centripetal force: f = ma = mv 2 / r

Nguyen Thi Hong Van

Chapter 5: The conical pendulum


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P + T = m a

oy: mg + T cos = 0 T cos = mg ox: T sin = ma = m2 r tan = 2 r / g ; also tan = r / h 2 r / g = r / h =

g / h

h = l cos = g / l cos Example:

l = 0 .2m , = 30 o

= 9.810.2 cos 30o = 7 .53 rad / s

f = 2

= 1 .20 Hz

Nguyen Thi Hong Van

Chapter 5: Non-uniform circular motion (1)


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The vertical pendulum : an example

of non-uniform circular motionObject mass m attached to the end of a light rigid rod, or light string, of length r at = 0 v

Find v in order for the object toexecute a complete vertical circle?

mv 2 / 2 = mv 2 / 2 + mgr (1 cos ) (1)v 2 = v 2 2gr (1 cos ) (2)

Newton law at B: mv 2

/ r = T mg cos T = mv 2 / r + mg cos T = m [v 2 2gr (1 cos )]/ r + mg cos = mv 2 / r + mg (3cos 2)In case of string :T 0 mv 2 / r + mg (3cos 2) 0 with v 2 > 5rg In case of rigid rod: v 2 0 (2) v 2 4gr

Nguyen Thi Hong Van

Chapter 5: Non-uniform circular motion (2)


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Motion on curved surfaces

Object mass m slides without frictionaround the inside of the hoop?? Condition for the object to executea complete vertical circle? Find v?

Energy conservation:12

mv 2 = 12

mv 2 + mgr (1 cos )

v 2 = v 2 2gr (1 cos )Newton second law at B:

mv 2

r = N mg cos

N 0 v 2 5gr independent on the mass of the train!Nguyen Thi Hong Van

Chapter 6: Rotational motion (1)


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Rigid body rotation

Consider a rigid body executingpure rotational motion arroundan axis (the locus of all pointsinside the body which remainstationary as the body rotates).Instantaneous angular velocityof the body (t ):

= lim t 0 t =

d dt

If the body rotates rogidly should be the same for all points P lyingwithin the body.

Rotational speed v (tangential speed): v = .Angular acceleration: (t ) = d / dt = d 2 / dt 2

(t ) = 0 + . t (t ) = 0 + 0 .t + . t 2 / 2Vector quantity in the rotational motion: = x .x + y .y + z .z direction along the axis of the rotation.

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Chapter 6: Rotational motion (2)


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The vector product

Scalar product: a . b = ax .b x + ay .b y + az .b z = | a | .| b | . cos( a , b )

Vector product: a b = c = ( ay b z a z b y , a z b x ax b z , ax b y ay b x )

| c | = | a | .| b | . sin( a , b ) a b = b a

Example: Velocity: v = r

Center of mass

r cm = 1M

N

i =1m i r i ; M =

N

i =1m i ; mi = ( r i ).V i

If the object under consideration is continuous: r cm = 1M . r .dV ; dV = dxdydz If the mass density is constant: r cm = 1V

r .dV

Nguyen Thi Hong Van

Chapter 6: Moment of inertia (1)


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Consider an extended object made of N elementsTotal kinetic energy:

K =N

i =1

1

2m i v

2i

Suppose that the motion of the object consist of rigid rotation at angularvelocity : v i = r i ; = . k

K =N

i =1

1

2m i | k r i | 2 .2 =

1

2I 2

where I =N

i =1

m i | k r i | 2 =N

i =1

m i 2i Moment of inertia of the object

Reminder: in the translational motion: K = Mv 2 / 2

Moment of inertia plays the same role as mass in the translational motion !For continuous object: I = 2 dV ; = | k r | perpendiculardistance from the axis of rotation.For constant density object: I = M 2 dV / dV depends not only onthe size and shape of the object but also on the location of the axis aboutwhich the object is rotating .

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Chapter 6: Moment of inertia (3)


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The moment of inertia of a thin rod of mass M and length about an axispassing through the centre of the rod and perpendicular to its length is

I = 1

12M 2

The moment of inertia of a thin rectangular sheet of mass M anddimensions a and b about a perpendicular axis passing through the centreof the sheet is

I = 1

12M (a 2 + b 2 )

The moment of inertia of a solid cylinder of mass M and radius b aboutthe cylindrical axis is

I = 12

Mb 2

The moment of inertia of a thin spherical shell of mass M and radius b

about a diameter isI =

23

Mb 2

The moment of inertia of a solid sphere of mass M and radius b about adiameter is

I = 25

Mb 2

Nguyen Thi Hong Van

Chapter 6: Torque


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f . cos is canceled out by the reactionat the pivotM v = f sin; v = b . Mb = f . sin I = f .b . sin ; where I = Mb 2

I = Torque = f .b . sin

is a vector quantity:

= I d dt

= r f

Nguyen Thi Hong Van

Chapter 6: Power and work


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Power : P = f . v , with v = r

P = f .( r ) = .( r f ) = .

Work :dW dt

= P = . ; = d dt

dW

dt = .

d

dt dW = .d

W = .d For the case of translational motion, we have

W = f .d r (torque) in the rotaional motion, is an anology of f (force)in the translational motion !

Nguyen Thi Hong Van


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Chapter 7: Angular momentum


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Angular momentum of a point particle

Find a similar quantity as p in rotational motion: p = m v ; d p / dt = f

= r p | | = | r | .| p | .sin ( r , p )

d dt

= d r dt

p + r d p dt

= p p

m + r f =

Angular momentum of an extended object

L =N

i =1 m i r i v i For rotating object: v i = r i ; = . k

L = N

i =1 m i r i ( k r i )Nguyen Thi Hong Van

Chapter 7: Angular momentum of an extended objectC f l h bj i i


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Component of L along the objects rotation axis:

Lk = k . L =

N

i =1m

i k .[ r

i ( k r

i )]

= N

i =1 m i k .[( k r i ) r i ] = N

i =1 m i ( k r i ).( r i k )=

N

i =1 m i | k r i |2 = .I k

Lx = I x x ; Ly = I y y ; Lz = I z z I x , I y , I z are the moment of inertia of the object about x , y , z axis

Assume that the origin of the coordinate system lies on the objects axisof rotation.x , y , z must be aligned along the so-called principal axes of the object(objects main symmetry axes).

L = I x x .x + I y y .y + I z z .z

Nguyen Thi Hong Van

Chapter 7: Angular momentum of a multi-component sys.C id t i ti g f N t ll i t ti g i t ti l


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Consider a system consisting of N mutually interacting point particles: f ij = f ji : third law of Newton

Assume the internal force acting within the system are central forces:

f ij ( r i r j )Equation of motion of the i th particle: p i =

N

j =1 , j = i f ij + F i r i p i =

N

j =1 , j = i

( r i f ij ) + ( r i F i ) ()

? We have: r i p i = d ( r i p i )

dt =

d Ldt

Total angular momentum of the system: L =N

i =1

( r i p i )

Summing (*) for all particles:

d Ldt

=i = j

i , j =1 , N

( r i f ij ) +N

i =1

( r i F i )=N

i =1

( r i F i ) ()

net external torque acting on the systemNguyen Thi Hong Van

Chapter 7: Conservation of angular momentum


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At (**), consider the system is isolated

= 0

d Ldt

= 0 :

The total angular momentum of the system is a conserved quantity.

Nguyen Thi Hong Van

Chapter 8: The law of gravityNewtons law of universal gravitation:


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Newton s law of universal gravitation:Every particle in the Universe attracts every other particle with a forcethat is directly proportional to the product of their masses and inverselyproportional to the square of the distance between them.

F 12 = F 21 = G m1 m 2

r 2 r 12 , (1)

where

G = 6 .673 10 11

N .m2

/ Kg 2

universal gravitational constantmeasured by experiment.m 1 , m2 masses of particle 1 andparticle 2,r = | r 2 r 1 | distance between twoparticles,r 12 = ( r 2 r 1 )/ | r 2 r 1 | unit vectorwhich is directed from particle 1 toparticle 2.

Nguyen Thi Hong Van

Chapter 8: The law of gravity


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Properties of the gravitational force given at (1) : the gravitational forceexerted by a nite-size, spherically symmetric mass distribution on a particle

outside the distribution is the same as if the entire mass of the distributionwere concentrated at the center .

Nguyen Thi Hong Van


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Chapter 8: Keplers laws


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1 All planets move in elliptical orbits with the Sun at one focal point.

2 The radius vector drawn from the Sun to a planet sweeps out equal areasin equal time intervals.

3 The square of the orbital period of any planet is proportional to the cubeof the semimajor axis of the elliptical orbit.

Nguyen Thi Hong Van

Chapter 8: The law of gravity and the motion of planetsKeplers Third Law


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Kepler s Third LawConsider a planet of mass M p moving around the Sun of mass M S in a circularorbit, as shown in the gure. Because the gravitational force exerted by the

Sun on the planet is a radially directed force that keeps the planet moving in acircle, we can apply Newtons second law to the planet:GM S M p

r 2 =

M p v 2

r

GM S

r 2 =

(2 r / T )2

r

T 2 = 4 2

GM S r 3 = K S r

3 (4)

where K S is a constant given by

K S = 4 2

GM S = 2 .97 10

19

s 2

/ m3

Equation (4) is Keplers third law . It can be shown that the law is also validfor elliptical orbits if we replace r with the length of the semimajor axis a. Notethat the constant of proportionality K S is independent of the mass of theplanet. Therefore, Equation (4) is valid for any planet .

Nguyen Thi Hong Van

Energy considerations in planetary and satellite motionConsider a body of mass m moving with a speed v in the vicinity of a massive


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Consider a body of mass m moving with a speed v in the vicinity of a massivebody of mass M , where M m . The system might be a planet moving aroundthe Sun, a satellite in orbit around the Earth, or a comet making a one-time

yby of the Sun.

E = K + U

E = 12

mv 2 GMm

r Newtons second law applied to m:

GMmr 2

= ma = mv 2

r

E = GMm

2r

(5) is the total energy of the two-body system. The absolute value of E is alsoequal to the binding energy of the system, because this amount of energy mustbe provided to the system to move the two masses innitely far apart.

Nguyen Thi Hong Van

Chapter 9: Fluid MechanicsA uid is a collection of molecules that are randomly arranged and held


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together by weak cohesive forces and by forces exerted by the walls of acontainer. Both liquids and gases are uids. Pressure

Fluids do not sustain shearing stresses or tensile stresses

The only stress that can be exerted on an object submerged in a uidis one that tends to compress the object

the force exerted by a uid on an object is always perpendicular to thesurfaces of the object.The pressure P in a uid:

P = F A

, F is the magnitude of the force exerted on a surface area A

The pressure at a specic point:

P = dF dA

.

The pressure is a scalar quantity and its SI unit is Pascal ( Pa ):

1 Pa = 1 N / m2

. Nguyen Thi Hong Van

Chapter 9: Variation of pressure with depth (1)Consider a uid of density at rest and open to the atmosphere


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y p pAssume that is constant (the uid is incompressible)Select a sample of the liquid contained within an imaginary cylinder of cross-sectionalarea A extending from the surface to a depth hThe pressure exerted by the outside liquid on the bottom face of the cylinder is P The pressure exerted on the top face of the cylinder is the atmospheric pressure P 0 The upward force exerted by the outside uid on the bottom of the cylinder is PA The downward force exerted by the atmosphere on the top is P 0 A

The mass of liquid in the cylinder:M = V = Ah .The weight of the liquid in the cylinderMg = Ahg .Because the cylinder is in equilibrium, we havePA P 0 A Mg = 0 , PA P 0 A Ahg = 0

P = P 0 + gh . (1)Nguyen Thi Hong Van

Chapter 9: Variation of pressure with depth (2)


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The pressure P at a depth h below the surface of a liquid open to theatmosphere is greater than atmospheric pressure by an amount gh .

We usually take atmospheric pressure to be:

P 0 = 1 .00 atm = 1 .013 105 Pa .

(1) the pressure is the same at all points having the same depth,independent of the shape of the container.

Nguyen Thi Hong Van

Chapter 9: Buoyant force and Archimedess principle (1)The upward force exerted by water on any immersed object is called a


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p y y jbuoyant force .Archimedess principle : the magnitude of the buoyant force alwaysequals the weight of the uid displaced by the object .The buoyant force acts vertically upward through the point that was thecenter of gravity of the displaced uid.

Forces acting on a totally submerged object

Consider an object is totally submerged in a uid of density f The magnitude of the upward buoyant force is B = f V 0 g , where V 0 isthe volume of the object.Weight of the object: F g = Mg = 0 V 0 g (the object has a mass M anddensity 0 )

The net force exerted on the object: B F g = ( f 0 )V 0 g . If 0 < f F g < B , and the unconstrained object acceleratesupward. If 0 > f F g > B , and the unsupported object sinks.

Nguyen Thi Hong Van

Chapter 9: Buoyant force and Archimedess principle (2)


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Forces acting on a oating Object Consider an object of volume V 0 in static equilibrium oating on a

uid (only partially submerged)

The upward buoyant force is balanced by the downwardgravitational force acting on the object.

The buoyant force: B = f V f g , where V f is the volume of the uiddisplaced by the object.

From the condition of equilibrium of the object we have B = F g , orf V f g = 0 V 0 g .

0f =

V f V o

Nguyen Thi Hong Van

Chapter 9: Fluid dynamic


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Because the motion of real uids is very complex and not fullyunderstood, we make some simplifying assumptions in our approach. Inour model of an ideal uid, we make the following four assumptions:

1 The uid is non-viscous : internal friction is neglected. An

object moving through the uid experiences no viscous force.

2

The ow is steady : the velocity of the uid at each pointremains constant.

3 The uid is incompressible : the density of an incompressibleuid is constant.

4 The ow is irrotational : the uid has no angular momentum

about any point. If a small paddle wheel placed anywhere in theuid does not rotate about the wheels center of mass, then the owis irrotational.

Nguyen Thi Hong Van

Chapter 9: Streamlines and the equation of continuityThe path taken by a uid particle


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under steady ow is called astreamline . The velocity of the

particle is always tangent to thestreamline, as shown in the gure. Aset of streamlines form a tube of ow.Equation of continuity : the productof the area and the uid speed at allpoints along the pipe is a constant for

an incompressible uid.

A1 v 1 = A2 v 2 = constant

Bernoullis equation : in laminarow, the sum of the pressure ( P ),kinetic energy per unit volume ( 12 v

2 ),and gravitational potential energy perunit volume (gy ) has the same valueat all points along a streamline.

Nguyen Thi Hong Van

Lectures - Classical mechanics

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Transcript of Lectures - Classical mechanics