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普通物理

Lecture 5

Law of Motion

Law of Motion

(2)

Sir Isaac Newton

Sir Isaac Newton

Sir Isaac Newton

Sir Isaac Newton

1642

1642 –

– 1727

1727

Formulated basic

Formulated basic

concepts and laws of

concepts and laws of

mechanics

mechanics

Universal Gravitation

Universal Gravitation

Calculus

Calculus

Calculus

Calculus

(3)

Newton's Law of Motion

Newton's Law of Motion

i i

i

i

i i

i

i

The branch of physics involving the motion of

The branch of physics involving the motion of

an object and the relationship between that

an object and the relationship between that

motion and other physics concepts

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

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

、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

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

生某一運動所需的力

33

生某一運動所需的力。

生某一運動所需的力。

(4)

Newton's Law of Motion

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

(5)

Contents

Newton’s First Law

Newton’s First Law

牛頓第一運動定律

牛頓第一運動定律

Newton s First Law

Newton s First Law

牛頓第一運動定律

牛頓第一運動定律

Force

Force

Newton’s Second Law

Newton’s Second Law

牛頓第二運動定律

牛頓第二運動定律

Inertia & Mass

Inertia & Mass

慣性與質量

慣性與質量

Newton’s Third Law

Newton’s Third Law

牛頓第三運動定律

牛頓第三運動定律

Inertial Reference Frames

Inertial Reference Frames

慣性參考體

慣性參考體

Inertial Reference Frames

Inertial Reference Frames

慣性參考體

慣性參考體

Application of Newton’s Laws

Application of Newton’s Laws

牛頓運動定律的應用

牛頓運動定律的應用

Free Body Diagrams

Free Body Diagrams

自由體圖

自由體圖

Friction

Friction

摩擦力

摩擦力

Drag force and terminal Speed

Drag force and terminal Speed

拖曳力與終端速率

拖曳力與終端速率

Uniform Circular Motion/ Centripetal force Speed

Uniform Circular Motion/ Centripetal force Speed

55

Uniform Circular Motion/ Centripetal force Speed

Uniform Circular Motion/ Centripetal force Speed

等速率圓周運動與向心力

等速率圓周運動與向心力

(6)

Newton’s First Law

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

(7)

Newton’s First Law

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

(8)

Force

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

(9)

Force

Force

External 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

(10)

Inertia & Mass

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

(11)

Inertia & Mass

Inertia & Mass

F

mo

Apply F on a second body of unknown mass

m

X

which results in an acceleration

a

X

.

The

ao

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

aX mX

o

o

X

X

o

a

a

m

m

m

m

o

a

X

a

X

m

a

a

11 11

(12)

Newton’s Second Law

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 FF 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 FFFF     

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

(13)

Newton’s Second Law

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

Fnet

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

(14)

Newton’s Second Law

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

m

Expressed by Newton’s Law of Universal Gravitation:

2 2 1 g

r

m

m

G

F

(15)

Newton’s Second Law

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 gm = 1.7 m/s2) , the mass does not

(16)

Newton’s Second Law

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

(17)

Newton’s Second Law

Newton’s Second Law

Tension:

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

(18)

Newton’s Third Law

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

(19)

Newton’s Third Law

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

(20)

Inertial Reference Frames

Inertial Reference Frames

慣性參考體

慣性參考體

慣性參考體

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

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

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。

The earth rotates about its axis once every 24 hours and thus it is accelerating with respect to an inertial reference frame. Thus we are making an approximation when we consider the earth to be an inertial reference frame. This approximation is excellent for most small scale phenomena Nevertheless for large scale phenomena small scale phenomena. Nevertheless for large scale phenomena

(21)

Application of Newton’s Laws

Application of Newton’s Laws

pp

pp

Free body Diagrams

Frictional force between two objects

The drag force exerted by a fluid on an object

moving through the fluid

moving through the fluid

Uniform circular motion

21 21

(22)

Free Body Diagrams

Free Body Diagrams

y

y

g

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 Fapp is exerted.

We have the following "system" choices:

a. System = block A + block B. The only horizontal force is Fapp

  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)

Free Body Diagrams

Free Body Diagrams

y

y

g

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

23 23

(24)

Friction

Friction

(25)

Friction

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

25 25

(26)

Friction

Friction

FN

Properties of friction:

F

FN

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 ƒs of the static friction is not constant but varies from 0 to a maximum value ƒs, max = μSFN The constant μs is known as the coefficient

f t ti f i ti If F d ƒ th t t t t lid of static friction. If F exceeds ƒs,max the crate starts to slide

(27)

Friction

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: ƒk = μk FN

μ is known as the coefficient of kinetic friction

μK is known as the coefficient of kinetic friction.

ƒ

k

< ƒ

s max F

FN

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:

27 27

Note 2:

(28)
(29)

Drag force and terminal Speed

Drag force and terminal Speed

g

g

p

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) v : object’s speed (速率)

29 29 v : object s speed (速率)

(30)

Drag force and terminal Speed

Drag force and terminal Speed

g

g

p

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 vt

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

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

1

2

1

2

t

D

C Av

mg

2

2

t

mg

v

t

(31)

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

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

31 31

(32)

Uniform Circular Motion/

Uniform Circular Motion/

Centripetal force

Centripetal force

Centripetal force

Centripetal force

The direction of the acceleration vector

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

mv

(33)

Uniform Circular Motion/

Uniform Circular Motion/

Centripetal force

Centripetal force

Centripetal force

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

33 33 ynet

r

(34)

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 FL downward on the car. Assume μs 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 FL?

(b) If , when v=90 m/s, is the car possible to run on the ceiling?FLv2

(35)

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 FL downward on the car. Assume μs 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 FL?

(b) If , when v=90 m/s, is the car possible to run on the ceiling?FLv2

35 35

(36)

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 μs=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 FN 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: mg s mv v Rg v Rg R

    

(37)

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

not fall

37 37

(38)

Assignment 4

Assignment 4

Due Date: 10/28/2009

Due Date: 10/28/2009

Due Date: 10/28/2009

Due Date: 10/28/2009

Assignment 4 Assignment 4 Assignment 4 Assignment 4 1.

The figure shows acceleration-versus-force graphs for two objects pulled by rubber bands What is the mass ratio m /m ?

pulled by rubber bands. What is the mass ratio m1/m2 ?

2.

The period of a pendulum is proportional to the square root of its length. A 2.0-m-long pendulum has a period of 3.0 s. What is the period of a 3.0-m-long pendulum?

3.

It’s a snowy day and you’re pulling a friend along a level road on a sled. You’ve both been taking physics, so she asks what you think the coefficient of friction between the sled and the snow is. physics, so she asks what you think the coefficient of friction between the sled and the snow is.

(39)

Assignment 4 Assignment 4

44.

A ball is shot from a compressed-air gun at twice its terminal speed.

a. What is the ball’s initial acceleration, as a multiple of g, if it is shot straight up? b. What is the ball’s initial acceleration, as a multiple of g, if it is shot straight down? c. Draw a plausible velocity-versus-time graph for the ball that is shot straight down. 5.

A 100 kg basketball player can leap straight up in tile air to a height of 80 cm as shown in the figure A 100 kg basketball player can leap straight up in tile air to a height of 80 cm, as shown in the figure below. You can understand how by analyzing the situation as follows:

a. The player bends his legs until the upper part of his body has dropped by 60 cm, then he begins his jump. Draw separate free-body diagrams for the player and for the floor as he is jumping, but before his feet leave the ground.

b. Is there a net force on the player as he jumps (before his feet leave the ground)? How can that be? Explain.

c With what speed must the player leave the ground to reach a height of 80 c. With what speed must the player leave the ground to reach a height of 80

cm?

d. What was his acceleration, assumed to be constant, as he jumped?

e. Suppose the player jumps while standing on a bathroom scale that reads in newtons. What does the scale read before he jumps, as he is jumping, and after his feet leave the ground?

39 39

(40)

Assignment 4 Assignment 4

6. 6.

In an amusement park ride II called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in the figure. a. Suppose the ring rotates once every 4.5 s. If a rider’s mass is 55 kg, with how much force does the

ring push on her at the top of the ride? At the bottom?

b. What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

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