普通物理－Lecture4－Law of Motion

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普通物理

Lecture 4

Law of Motion

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Sir Isaac Newton



1642

1642 –

– 1727

1727



Formulated basic

concepts and laws of

mechanics



Universal Gravitation



Calculus



Calculus



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Newton's Law of Motion

i i

i

i i

i

The branch of physics involving the motion of

an object and the relationship between that

motion and other physics concepts



Kinematics

(

(運動學

運動學):

):

description of motion



Kinematics

(

(運動學

運動學):

):

description of motion



Kinematics

(

(運動學

運動學):

):

description of motion

是在討論物體運動的狀況，不考慮產生運動的原因，只研究

是在討論物體運動的狀況，不考慮產生運動的原因，只研究位

移、速度、加速度及時間之間的關係。

之間的關係。



Kinematics

(

(運動學

運動學):

):

description of motion



Dynamics

(

(運動力學

運動力學):

):

cause of the motion

移、速度、加速度及時間之間的關係。

之間的關係。



Dynamics

(

(運動力學

運動力學):

):

cause of the motion

研究作用於物體的

研究作用於物體的力

力、物體

、物體質量

質量及物體

及物體運動狀況

運動狀況之間所存在的

之間所存在的

關係。可以用來預測某些已知力所造成的運動，或是決定要產

生某一運動所需的力

33

生某一運動所需的力。

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Newton's Law of Motion

In this chapter we will introduce

Newton’s three laws of

In this chapter we will introduce Newton s three laws of

motion

which is at the heart of classical mechanics.

Newton’s laws describe physical phenomena of

a vast range

Newton s laws describe physical phenomena of

a vast range

.

Ex:

the motion of stars and planets

Exceptions:

p



When the speed of objects approaches

(1% or more) the speed of

light in vacuum (c = 3×10

8

_m/s)

_{In this case we must use}

_Einstein’s

light in vacuum (c 3×10 m/s).

In this case we must use Einstein s

special theory of relativity (1905)



When the objects under study become very small (e g

electrons

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

Newton’s First Law

牛頓第一運動定律



Newton s First Law

牛頓第一運動定律



Force

力



Newton’s Second Law

牛頓第二運動定律



Inertia & Mass

慣性與質量



Newton’s Third Law

牛頓第三運動定律



Inertial Reference Frames

慣性參考體



Inertial Reference Frames

慣性參考體



Application of Newton’s Laws

牛頓運動定律的應用



Free Body Diagrams

自由體圖



Friction

摩擦力



Drag force and terminal Speed

拖曳力與終端速率



Uniform Circular Motion/ Centripetal force Speed

55



Uniform Circular Motion/ Centripetal force Speed

等速率圓周運動與向心力

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Newton’s First Law

Before Newton



A force was required in order to keep an object moving at

constant

velocity



An object was in its “natural state” when it was at rest.

After Newton



F i ti

i d t b

f

d i t d

d



Friction

was recognized to be a force and introduced

For example:

 Slide an object on a floor with an initial speed, very soon the object will

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Newton’s First Law

An object continues in a state of rest

or in a state of motion at a

constant speed

along a straight line unless compelled to

constant speed

along a straight line, unless compelled to

change that state by a net force.

If no force

acts on a body, the body’s velocity cannot change;

that is the body cannot accelerate

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Force

A push or pull exerted on an object We can define a force exerted on an A push or pull exerted on an object. We can define a force exerted on an object quantitatively by measuring the acceleration it causes

We place an object of mass m = 1 kg on a frictionless surface and measure

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Force

External Force

Any force that results from the interaction between the object and its

i

environment

Internal Force

Forces that originate within the object itself, and they

cannot change the

object’s velocity

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Inertia & Mass

慣性與質量

Inertia

慣性與質量

Inertia

The natural tendency of an object to remain at rest or in motion at a

Mass

y

j

constant speed along a straight line.

Mass

An

intrinsic

characteristic of a body that automatically comes with the

existence of the body A

measure of the resistance

of an object to

existence of the body. A

measure of the resistance

of an object to

changes in its motion due to a force

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Inertia & Mass

F

m_o

Apply F on a second body of unknown mass

m

_X

which results in an acceleration

a

_X

.

The

a_o

m

X

which results in an acceleration

a

X

.

The

ratio of the accelerations is inversely

proportional to the ratio of the masses

F

p p

a_X m_X

o

X

o

a

m

_o



a

_X





a

_X

m

a

11 11

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Newton’s Second Law



F

m

a

When a net external force acts on an object of mass , the

acceleration that results is directly proportional to the net force

and has a magnitude that is inversely proportional to the mass

Note: If several forces act on a body (say F F and F ) the net force F

(say , , and ) the net force is defined as: i i th t f A B C net net A B C F F F F F F  F  F  F     

i.e. is the vector sum of , , and ne C t A B F F F  F

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Newton’s Second Law

The results of the discussions on the relations between the net force

F

The results of the discussions on the relations between the net force F

_net

applied on an object of mass m

and the resulting acceleration a can be

summarized in the following statement known as:

g

“Newton’s second

law”

The net force on a body is equal to the product of

the body’s mass and its acceleration

F_net

the body s mass and its acceleration

a m

F



_net



ma



F



ma

13 13 , net x x

F



ma

F

_{net y}_,



ma

_y

F

_{net z}_,



ma

_z

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Newton’s Second Law

The force that the earth exerts on any object (in the picture

The Gravitational Force

The force that the earth exerts on any object (in the picture a cantaloupe) It is directed towards the center of the earth. Its magnitude is given by Newton’s second law.

y



Mutual force of attraction between any two objects

m



Expressed by Newton’s Law of Universal Gravitation:

2 2 1 g

r

m

G

F



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Newton’s Second Law

Weight

 The magnitude of the gravitational force acting on an object of mass m

near the Earth’s surface is called the weight w of the object

We g

g j

 w = m g is a special case of Newton’s Second Law

 g is the acceleration due to gravity, can also be found from the Law of

U i l G i i

W

Universal Gravitation

 The weight of a body is defined as the magnitude of the force required to prevent the body from falling freely

y g _W ,

0

net y y

F



ma



W



mg

 

W



mg

_mg 15 15

Note: The weight of an object is not its mass. If the object is moved to a location where the acceleration of gravity is different (e.g. the moon where g_m = 1.7 m/s2_{) , the}_{mass does not} change but the weight does.

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Newton’s Second Law

Contact Forces:

Forces act between two objects that are in contact. The contact forces have two components. One is acting along the normal to the contact surface (normal

force) and a second component that is acting parallel to the contact surface

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Newton’s Second Law

Tension:

The force exerted by a rope or a cable attached to an object

Ch t i ti

always directed along the rope

Characteristics

the same value along the rope

always pulling the object

Neglect rope mass compared to the mass of the

Assumptions

the same value along the rope

Neglect rope mass compared to the mass of the object it pulls

Rope does not stretch

17 17

Rope does not stretch

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Newton’s Third Law

Wh

t

b di i t

t b

ti

f

h th

th

When two bodies interact by exerting forces on each other, the

forces are equal in magnitude and opposite in direction

For example consider a book leaning against a bookcase. We label the force exerted the book the case. Using the same convention we label on by the force

BC CB F F   exerted the cason e by the book.

CB

Newton's third law can be written as: The book together with the bookcase are known as

a

BC CB

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Newton’s Third Law

A second example is shown in the picture

h l f Th hi d l

f

i

to the left. The third –law force pair

consists of the earth and a cantaloupe.

U i

h

i

b

Using the same convention as above we

can express Newton’s third law as:

F



 

F



F

_CE



F

_EC

19 19

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Inertial Reference Frames

慣性參考體

In

Newtonian physics

and

special relativity

, an inertial frame of

reference (or

Galilean

reference frame) is a

frame of reference

in

which

Newton

's

first law of motion

applies: an object moves at a

which

Newton

s

first law of motion

applies: an object moves at a

constant velocity unless acted on by an external

force

. All inertial

frames are in a state of constant, rectilinear motion with respect to one

,

p

another; they are not

accelerating

(in the sense of

proper acceleration

that would be detected by an

accelerometer

). Measurements in one

inertial frame can be converted to measurements in another by a

simple transformation (the

Galilean transformation

in Newtonian

physics and the

Lorentz transformation

in special relativity) In

general

physics and the

Lorentz transformation

in special relativity). In

general

relativity

, an inertial reference frame is only an approximation that

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Inertial Reference Frames

慣性參考體



牛頓認為宇宙中存在一個與任何物體均無相互作用，而且永遠

靜止的空間，稱作絕對空間(absolute space)，所有的物體均

在此絕對空間中運動。



凡是相對於絕對空間靜止，或作等速率直線運動(且不旋轉)的

參考體，即稱之為慣性參考體 (inertial reference frame)。



牛頓運動定律及其所推展出的力學定理，僅成立於慣性參考體

中。

中

21 21

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Application of Newton’s Laws

pp

Free body Diagrams



Frictional force between two objects



The drag force exerted by a fluid on an object

moving through the fluid

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Free Body Diagrams

y

g

Among the many parts of a given problem we choose one which we call the “system”

Among the many parts of a given problem we choose one which we call the system .

Then we choose axes and enter all the forces that are acting on the system and omitting those acting on objects that were not included in the system.

An example is given in the figure below. This is a problem that involves two blocks labeled "A" and "B" on which an external force F_app is exerted.

We have the following "system" choices:

a. System = block A + block B. The only horizontal force is F_app

  b. . There are now two horizontal forces: and c. . The only horizontal force

System = block A System = block B is app AB BA F F F    23 23

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Free Body Diagrams

y

g

1. Choose the system to be

studied

2. Make a simple sketch of the

system

3. Choose a convenient

coordinate system

4. Identify all the forces that

act on the system. Label

them on the diagram

5. Apply Newton’s laws of

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Friction

25 25

The free body diagrams for frames a-d show the existence of a new force which balances the force with which we push the crate. This force is called the static frictional force.

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Friction

When F reaches a certain limit the crate “breaks away” and accelerates to the left. Once the crate starts moving the force

Once the crate starts moving the force opposing its motion is called the kinetic frictional force , . Thus if we wish the crate to move with constant speed we must decrease F so that it

balances ƒ_k (frame f) In frame (g) we plot balances ƒ_k (frame f). In frame (g) we plot ƒ versus time t

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Friction

F_N

Properties of friction:

F

F_N

Properties of friction:

The frictional force is acting between two dry un-lubricated surfaces in contact

mg

y

Property 1.

If the two surfaces do not move with respect to each other, then the static frictional force balances the applied force .

Property 2.

s,max s N

f





F

ope y .

The magnitude ƒ_s of the static friction is not constant but varies from 0 to a maximum value ƒ_{s, max} = μ_SF_N The constant μ_s is known as the coefficient

f t ti f i ti If F d ƒ th t t t t lid

27 27

of static friction. If F exceeds ƒ_s,maxthe crate starts to slide

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Friction

Property3.

Once the crate starts to move the frictional force is known as kinetic friction. Its magnitude is constant and is given by the equation: ƒ_k= μ_kF_N

μ is known as the coefficient of kinetic friction

μ_Kis known as the coefficient of kinetic friction.

ƒ

_k

< ƒ

_{s max}

F F_N

ƒ

_k

ƒ

_s,max

Note 1:

The static and kinetic friction acts parallel to the surfaces in contact The direction

mg

the surfaces in contact. The direction

opposes the direction of motion (for kinetic friction) or of attempted motion (in the case of static friction)

Note 2: Note 2:

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Example 1:

29 29

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Drag force and terminal Speed

g

p

拖曳力與終端速率

When an object moves through a fluid (gas or liquid) it experiences an When an object moves through a fluid (gas or liquid) it experiences an

opposing force known as “drag”. Under certain conditions (the moving object must be blunt and must move fast so as the flow of the liquid is turbulent) the magnitude of the drag force is given by the expression:

1

₂

1

2 D



C Av



2

C : drag Coefficient (拖曳係數) d it f th di fl id (流體密度)



_{: density of the surrounding fluid (流體密度)}

A : effective cross sectional area of the moving object (物體有效截面積)

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Drag force and terminal Speed

g

p

拖曳力與終端速率

Consider an object (a cat of mass m in this case) start moving in air Consider an object (a cat of mass m in this case) start moving in air.

Initially D = 0. As the cat accelerates D increases and at a certain speed v_t

D = mg At this point the net force and thus the acceleration become zero

and the cat moves with constant speed v_t known the the terminal speed

1 ₂

1

2 t

D



C Av





mg

2

t

mg

v



31 31 t

C A



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Example 2:

The terminal speed of a sky diver is 160 km/h in the spread-eagle position and 310 km/h in the nosedive position Assume that the diver’s drag

and 310 km/h in the nosedive position. Assume that the diver s drag

coefficient C does not change from one position to the other, find the ratio of the effective cross-sectional area A in the slower position to that in the faster position? faster position?

2



v

C

mg

A



1

2



A

v

C



75 .

3 )

160 /

310 (

)

(

2



2





A

_slow

v

fast

v

)

(

)

(

slow fast

v

A

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Uniform Circular Motion/

Centripetal force

The direction of the acceleration vector

always points towards the center of

rotation C (thus the name centripetal) Its

(

p

)

magnitude is constant:

2 vv

a

r



r

Apply Newton’s law to analyze uniform circular motion we conclude

that the net force in the direction that points towards C

2

mv

33 33

mv

F

r



Centripetal Force

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Uniform Circular Motion/

Centripetal force

Recipe

.

C

r

Recipe



Draw the force diagram for the

m

x

y

g

object



Choose one of the coordinate axes

_m

v

Choose one of the coordinate axes

(the y-axis in this diagram) to point

towards the orbit center C



Determine

ynet

F



Set

2 ynet

mv

F

_ynet



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Example 3:

A racing car with m=600 kg travels on a flat track in a circular arc of radius R= 100 m Because of the shape of the car and the wings on it the

R 100 m. Because of the shape of the car and the wings on it, the passing air exerts a negative lift F_L downward on the car. Assume μ_s between the tires and the track is 0.75.

(a) If the car is on the verge of sliding out of the turn when its speed is 28 6 (a) If the car is on the verge of sliding out of the turn when its speed is 28.6

m/s, what is the magnitude of F_L?

(b) If , when v=90 m/s, is the car possible to run on the ceiling?F_L  v2

(a) The frictional force fs is the centripetal force

35 35

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Example 4:

A racing car with m=600 kg travels on a flat track in a circular arc of radius R= 100 m Because of the shape of the car and the wings on it the

R 100 m. Because of the shape of the car and the wings on it, the passing air exerts a negative lift F_L downward on the car. Assume μ_s between the tires and the track is 0.75.

(a) If the car is on the verge of sliding out of the turn when its speed is 28 6 (a) If the car is on the verge of sliding out of the turn when its speed is 28.6

m/s, what is the magnitude of F_L?

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Example 5:

The Rotor is a large hollow cylinder of radius R=2.5m that is rotated rapidly around its central axis with a speed v. A rider of mass m stands on the Rotor floor with his/her back against the Rotor wall. Cylinder and rider begin to turn. When the y g speed v reaches some predetermined value, the Rotor floor abruptly falls away. The rider does not fall but instead

remains pinned against the Rotor wall. The coefficient of remains pinned against the Rotor wall. The coefficient of static friction μ_s=0.5 between the Rotor wall and the rider is given. What is the minimum speed v ?

2

( 1) mv

F F

The normal reaction F_N is the centripetal force.

, , = (eqs.1) , 0 , (eqs.2) x net N y net s s s N s N F F ma R F f mg f



F mg



F         2 2 min

If we combine eqs.1 and eqs.2 we get: _s

s s mv Rg Rg mg v v R



     37 37 s s



s m v 7.0 / 5 . 0 8 . 9 5 . 2 min   

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Example 6:

In a 1901 circus performance Allo Diavolo introduced the stunt of riding a bicycle in a looping-the-loop The loop is a stunt of riding a bicycle in a looping the loop. The loop is a circle of radius R. We are asked to calculate the minimum speed v that Diavolo should have at the top of the loop and not fall