制作张昆实 yangtze university

制作张昆实制作张昆实

Yangtze UniversityYangtze University







BilingualBilingual MechanicsMechanics BilingualBilingual MechanicsMechanics Chapter 2

MotionMotion

Chapter 2

MotionMotion

Chapter 2 Chapter 2 MotionMotion Chapter 2 Chapter 2 MotionMotion

2-1 What Is Physics

2-2 Vectors and Scalars

2-3 Multiplying Vectors

2-4 Motion

2-5 Position and Displacement

2-6 Average Velocity and Instantaneous

Velocity

2-7 Acceleration

2-1 What Is Physics

2-2 Vectors and Scalars

2-3 Multiplying Vectors

2-4 Motion

2-5 Position and Displacement

2-6 Average Velocity and Instantaneous

Velocity

2-7 Acceleration

Chapter 2 Chapter 2 MotionMotion Chapter 2 Chapter 2 MotionMotion

2-8 Constant Acceleration: A Special

Case

2-9 Graphical Integration of Motion

Analysis

2-10 Projectile Motion

2-11 Projectile Motion Analyzed

2-12 Uniform Circular Motion

2-13 Relative Motion

2-8 Constant Acceleration: A Special

Case

2-9 Graphical Integration of Motion

Analysis

2-10 Projectile Motion

2-11 Projectile Motion Analyzed

2-12 Uniform Circular Motion

2-13 Relative Motion

2-1 What Is Physics 2-1 What Is Physics

★ All objects in nature are in constant motion, nothing stands still absolutely.

★ All objects in nature are in constant motion, nothing stands still absolutely.

★ In this chapter we learn what is physicswhat is physics through the motion of particles

★ In this chapter we learn what is physicswhat is physics through the motion of particles

MechanicsMechanics KinematicsKinematics

DynamicsDynamics

Deals with the concepts discribing motion, without any reference to force.

Deals with the concepts discribing motion, without any reference to force.

Deals with the effect that forces have on motion . Deals with the effect that forces have on motion .

(chapter 2)(chapter 2)

(chapter 3)(chapter 3)

.

2-2 Vectors and Scalars2-2 Vectors and Scalars

★ Vectors are quantities that have both magnitude and direction and obey the vector law of addition.

★ Vectors are quantities that have both magnitude and direction and obey the vector law of addition.

★ Scalars are quantities that have magnitude only. ★ Scalars are quantities that have magnitude only.

length, time, masslength, time, mass , density, energy ,

temperature etc. are all scalars.

length, time, masslength, time, mass , density, energy ,

temperature etc. are all scalars.

Displacement, velocity, acceleration, Displacement, velocity, acceleration,

force, torque, force, torque, etc. are all vectors.

Displacement, velocity, acceleration, Displacement, velocity, acceleration,

force, torque, force, torque, etc. are all vectors.

.

2-2 Vectors and Scalars2-2 Vectors and Scalars

★Three notations of a vector★Three notations of a vector

a) unit-vector notationa) unit-vector notation

b) component notationb) component notation

c) magnitude-angle notationc) magnitude-angle notation

x ya a i a j

2 2x ya a a

cosxa a sinya a

tan y

x

a

a

a

xa

ya

x

y

o

2-3 Multiplying Vectors2-3 Multiplying Vectors

★ There are three ways in which vectors can be multiplied.

★ There are three ways in which vectors can be multiplied.

Multiplying a Vector by a ScalarMultiplying a Vector by a Scalar

Multiplying a Vector by a VectorMultiplying a Vector by a Vector

s a sa

a s

magnitudemagnitude

(new vector)(new vector)directiondirection

as )0(a s

as 0( )a s

1.1.

2. The Scalar Product2. The Scalar Product a b

3. The Vector Product3. The Vector Product a b

2-3 Multiplying Vectors2-3 Multiplying Vectors

Multiplying a Vector by a VectorMultiplying a Vector by a Vector

2. The Scalar Product ( dot product )2. The Scalar Product ( dot product ) a b

cosa b ab

a b c

3. The Vector Product ( cross product )3. The Vector Product ( cross product ) a b

“a dot b”:“a dot b”:

“a cross b”:“a cross b”:

sinc ab

(scalar)(scalar)

(vector)(vector)

magnitudemagnitude

Direction right hand screw ruleDirection right hand screw rule

c

a

b

c

2-4 Motion2-4 Motion

★ Objects move in one, two or three

dimension(s).

★ Objects move in one, two or three

dimension(s).

★The moving object is either a particle or an object that moves like a particle.

★The moving object is either a particle or an object that moves like a particle.

★To describ the motion of an object a

reference frame ( 参考系 ) must be chosen and a coordinate system ( 坐标系 ) must be constructed on it.

★To describ the motion of an object a

reference frame ( 参考系 ) must be chosen and a coordinate system ( 坐标系 ) must be constructed on it.

2-5 Position and Displacement2-5 Position and Displacement

★ Position vector a vector extends fromthe origin of a coordinate system to the particle

● unit vector

● scalar components

★ Position vector a vector extends fromthe origin of a coordinate system to the particle

● unit vector

● scalar components

r * P

x

y

z x

z

y

o

222 z yxr

magnitude ：magnitude ：

r xi y j zk , ,i j k

, ,x y z

(2-25)

★ Position vector

★ Position vector

rr * P

x

y

z x

z

y

o

r

direction ：direction ：

，r

xcos

rzcos，

r

ycos

222 z yxr

magnitude ：magnitude ：


r xi y j zk

★Displacement

★Displacement

2 1r r r

r

x

y2

2r

1r 1

r

o2 1x x

2 1y y

1x 2x

2 2 2

1 1 1

( )

( )

r x i y j z k

x i y j z k

2 1 2 1 2 1( ) ( ) ( )r x x i y y j z z k

r xi y j zk


(2-27)

(2-28)

(2-26)

P20 Sample Problem 2-4P20 Sample Problem 2-4


(2-29)(2-29)

P20 Sample Problem 2-4P20 Sample Problem 2-4A rabbit runs across a parking plot on which a set of coor-dinate axes has been drawn. The coordinates of the rabbit’s position as function of time are given by

A rabbit runs across a parking plot on which a set of coor-dinate axes has been drawn. The coordinates of the rabbit’s position as function of time are given by

20.22 9.1 30y t t

20.31 7.2 28x t t (2-30)(2-30)

With in seconds and and in meters. With in seconds and and in meters. t

15t s

t

2(0.22)(15) (9.1)(15) 30 57y m

2( 0.31)(15) (7.2)(15) 28 66x m

x y(a) At , What is the rabbit’s position vector in unit-vector notation and in magnitude-angle notation?(a) At , What is the rabbit’s position vector in unit-vector notation and in magnitude-angle notation?

15t s r

( ) ( ) ( ) (66 ) (57 )r t x t i y t j m i m j unit-vector notationunit-vector notation

(Answer)(Answer)

SolutionSolution



(a) At , What is the rabbit’s position vector in magnitude-angle notation?(a) At , What is the rabbit’s position vector in magnitude-angle notation?

1 1 57tan tan

66

y m

x m

15t s r

41

87m

magnitude-angle notationmagnitude-angle notation

(Answer)(Answer)

2 2 2 2(66 ) ( 57 )r x y m m



(b) Graph the rabbit’s path for to(b) Graph the rabbit’s path for to 25 .t s0t

magnitude-angle notationmagnitude-angle notation

★ Average Velocity If a particle moves through a displacement in a timeInterval ,Then its

★ Average Velocity If a particle moves through a displacement in a timeInterval ,Then its

avg

rv

t

avg

x y zv i j k

t t t

rt

avgv

tr

Average displacement Velocity time intervalAverage displacement Velocity time interval

2-6 Average Velocity and Instantaneous Velocity2-6 Average Velocity and Instantaneous Velocity

(2-33)

(2-32)

★ Instantaneous Velocity

★ Instantaneous Velocity

0

dlim

dt

r rv

t t

avgv v

dx dy dzv i j k

dt dt dt

drdt

0t

0t 0r

v

The direction of is always tangent to the particle’s path at the particle’s position The direction of is always tangent to the particle’s path at the particle’s position

x y zv v i v j v k


●

●

●

(2-34)

(2-35)

★ Instantaneous Velocity and its components

★ Instantaneous Velocity and its components

x

dxv

dt

y

dyv

dt

z

dzv

dt

Fig.2-18 The velocity of a particle, along with the scalar components of

Fig.2-18 The velocity of a particle, along with the scalar components of v

v


x

y

o

vyv

xv

(2-36)


(2-37)(2-37)


For the rabbit in Sample Problem 2-4, Find the velocity at time , in unit-vector notation and in magnitude-angle notation.

For the rabbit in Sample Problem 2-4, Find the velocity at time , in unit-vector notation and in magnitude-angle notation.

2( 0.31 7.2 28) 0.62 7.2x

dx dv t t t

dt dt

(2-38)(2-38)

15t sv

2(0.22 9.1 30 0.44 9.1y

dy dv t t t

dt dt

unit-vector notationunit-vector notation

(Answer)(Answer)


SolutionSolution

15t s 2.1 / ,xv m s 2.5 /yv m s( 2.1 / ) ( 2.5 / )v m s i m s j

taking derivetives of the components of to get the components oftaking derivetives of the components of to get the components of v

r


Finding the velocity at time , in magnitude-angle notation.Finding the velocity at time , in magnitude-angle notation.

(2-38)(2-38)

15t sv

3.3 /m s

magnitude-angle notation

magnitude-angle notation

(Answer)(Answer)


SolutionSolution2 2x yv v v

1 1 2.5 /tan tan

2.1 /y

x

v m s

v m s

The magnitude of :The magnitude of :v

1tan 1.19 130

2 2( 0.21 / ) ( 0.25 / )m s m s

3.3 /m s

(in the third quadrant)(in the third quadrant)

★ Average AccelerationWhen a particle’s velocitychanges from to in a time interval ,Then its

★ Average AccelerationWhen a particle’s velocitychanges from to in a time interval ,Then its

1v

2v

t

x

y

O

1v

t1

2v

t2

v

1v

2v

yx zavg

vv vva i j k

t t t

t

2 1avg

v v va

t t

2-7 Acceleration2-7 Acceleration

average change in velocityacceleration time intervalaverage change in velocityacceleration time interval

(2-39)

★ Instantaneous acceleration

★ Instantaneous acceleration

0

dlim

dta

t t

v v

0t


(2-40)

(2-41)

xx

dva

dt y

y

dva

dt z

z

dva

dt (2-42)


dd d

d d dyx za i j k

t t t vv v

x y za a i a j a k


For the rabbit in Sample Problem 2-4 and 2-5, Find the acceleration at time , in unit-vector notation and in magnitude-angle notation.

For the rabbit in Sample Problem 2-4 and 2-5, Find the acceleration at time , in unit-vector notation and in magnitude-angle notation.

2( 0.62 7.2) 0.62 /xx

dv da t m s

dt dt

15t sa

2(0.44 9.1) 0.44 /yy

dv da t m s

dt dt

unit-vector notationunit-vector notation

(Answer)(Answer)

SolutionSolution

2 2( 0.62 / ) (0.44 / )a m s i m s j


taking derivetives of the components of to get the components oftaking derivetives of the components of to get the components of a

v


in magnitude-angle notation.in magnitude-angle notation.

SolutionSolution


magnitude-angle notation magnitude-angle notation

(Answer)(Answer)

2 2x ya a a

2 2 2 2( 0.62 / ) (0.44 / )m s m s

20.76 /m s

1 1 0.44 /tan tan 145

6.2 /y

x

a m s

a m s

(in the second quadrant)(in the second quadrant)

The magnitude of :The magnitude of :a

0

0avg

v va a

t

0v v at (2-43)

(2-44)

0

0avg

x xv

t

0 avgx x v t

102 ( )avgv v v (2-45)

10 2avgv v at (2-46)

(2-47)21

0 0 2x x v t at

★ In many cases the acceleration is constant

★ In many cases the acceleration is constant

The average velocityThe average velocity

2-8 Constant Acceleration: A Special Case2-8 Constant Acceleration: A Special Case

Eq. 2-43 and 2-47: are the basic equations for constant acceleration.

Eq. 2-43 and 2-47: are the basic equations for constant acceleration.

Recast this equation:Recast this equation:

SimilarlySimilarly

Substituting 2-43 for :Substituting 2-43 for :

Substituting 2-46 into 2-44 :Substituting 2-46 into 2-44 :

v

0v v at (2-43) (2-47)21

0 0 2x x v t at

2 20 02 ( )v v a x x (2-48)

(2-49)10 02 ( )x x v v t

(2-50)21

0 2x x vt at


Eliminate :Eliminate :t

Eliminate : Eliminate :

a

Eliminate :Eliminate :0v

the basic equations 2-43 and 2-47 for constant acceleration can be combined in three ways to yield additional equations:

the basic equations 2-43 and 2-47 for constant acceleration can be combined in three ways to yield additional equations:

P26 Table 2-1

Eq. 2-43 and 2-47 can be ontained by integration

Eq. 2-43 and 2-47 can be ontained by integration

(2-43)

dv adt

dv adt a dt Indefinite integralIndefinite integral

v at C To determine constant C, let

To determine constant C, let 0t

0v v 0C v

0v v at

dx vdtIndefinite integralIndefinite integral

0( )dx vdt v at dt 21

0 2x v t at C

To determine constant C’, letTo determine constant C’, let 0t

0x x 0C x

(2-47)21

0 0 2x x v t at


Toss an object either up or down (eliminate the air resistance) the object accelerates downward at a certain constant rate: free-fall acceleration

Toss an object either up or down (eliminate the air resistance) the object accelerates downward at a certain constant rate: free-fall acceleration g


The value of varies slightly with latitude and with elevation. The value of varies slightly with latitude and with elevation.

*** Note: For free fall ***

(2) The free-fall acceleration, being downward, is now negative!

(2) The free-fall acceleration, being downward, is now negative!

29.8 /a g m s

(1) The directions of motion are now along a verticle axis with the positive direction of upward.(1) The directions of motion are now along a verticle axis with the positive direction of upward.y y

g

0v v at (2-43)

(2-47)

210 0 2x x v t at

2 20 02 ( )v v a x x

(2-48)

Horizontal motion Vertical motion Horizontal motion Vertical motion

0v v gt (2-43)

210 0 2y y v t gt

(2-47)


2 20 02 ( )v v g y y

(2-48)

2-9 Graphical Integration of Motion Analysis2-9 Graphical Integration of Motion Analysis

Using a graph of an object, the object’s can be found by integrating on the graph:Using a graph of an object, the object’s can be found by integrating on the graph:

a tv

a dv dt1

01 0

t

tv v adt

1

0

t

tadt Area between acceleration curve

and time axis, from toArea between acceleration curve

and time axis, from to0t 1t

1t0t

Area

t

a When the acceleration curve is above the time axis, the area is positive;When the acceleration curve is above the time axis, the area is positive;

When the acceleration curve is below the time axis, the area is negative;When the acceleration curve is below the time axis, the area is negative;

(definite integral)(definite integral)

2-9 Graphical Integration of Motion Analysis2-9 Graphical Integration of Motion Analysis

Using a graph of an object, the object’s can be found by integrating on the graph:Using a graph of an object, the object’s can be found by integrating on the graph:

v tx

v dx dt1

01 0

t

tx x vdt

1

0

t

tvdt Area between velocity curve

and time axis, from toArea between velocity curve

and time axis, from to0t 1t

1t0t

Area

t

v When the velocity curve is above the time axis, the area is positive;When the velocity curve is above the time axis, the area is positive;

(definite integral)(definite integral)

When the velocity curve is belowthe time axis, the area is negative;When the velocity curve is belowthe time axis, the area is negative;

★ Projectile Motion : A particle moves in a vertical planewith some initial velocity and an angle with respect to the horizontal axis but its acceleration is always the free-fall acceleration , which is downward.

★ Projectile Motion : A particle moves in a vertical planewith some initial velocity and an angle with respect to the horizontal axis but its acceleration is always the free-fall acceleration , which is downward.

00v

g

0d

y

0v

xv

yv v

xv

yv vx

o0xv

0yv

0

0 0 0cosxv v 0 0 0sinyv v 0 0 0x yv v i v j

(2-58)

(2-59)

2-10 Projectile Motion2-10 Projectile Motion

Fig. 2-29 The projectile bullet always hits the falling coconutFig. 2-29 The projectile bullet always hits the falling coconut

2-10 Projectile Motion2-10 Projectile Motion

★ In projectile motion , the horizontal motion andthe vertical motion are independent of each other.

★ In projectile motion , the horizontal motion andthe vertical motion are independent of each other.

0xa

xv

yv

v

xv

yv

v0dx

y

o

0v

x0v

y0v

0 0 0 0( cos )xx x v t v t (2-60)

0xa ★ the horizontal motion:★ the horizontal motion:

2-11 Projectile Motion Analyzed2-11 Projectile Motion Analyzed

★ the vertical motion:

★ the vertical motion:

0xa xv

yv

v

xv

yv

v0dx

y

o

0v

x0v

y0v

210 0 2yy y v t gt

(2-61)

ya g

210 0 2( sin )v t gt

0 0sinyv v gt 2 2

0 0 0( sin ) 2 ( )yv v g y y (2-62)

(2-63)


★ The Equation of the Path★ The Equation of the Path

0 0 0 0( cos )xx x v t v t 21

0 0 2yy y v t gt 210 0 2( sin )v t gt (2-61)

(2-60)

Let Solving Eq.2-60 for t and

Substituting into Eq.2-61:

Let Solving Eq.2-60 for t and

Substituting into Eq.2-61:0 0,x 0 0y

This is the equation of a parabola, so the path (trajectory) is parabolic.This is the equation of a parabola, so the path (trajectory) is parabolic.


2

0 20

(tan )2( cos )

gxy x

0v(2-64)(trajectory)(trajectory)

The Horizontal Range: the horizontal distance theProjectile has traveled when it returns to its initial(launch) height.

The Horizontal Range: the horizontal distance theProjectile has traveled when it returns to its initial(launch) height.

0 0 0( cos )R x x v t

00 y y 210 0 2( sin )v t gt (2-61)

(2-60)

Eliminating t between these two equations yieldsEliminating t between these two equations yields2 20 0

0 0 0

2sin cos sin 2

v vR

g g (2-65)

The horizontal distance R is maximum for a launch angle of The horizontal distance R is maximum for a launch angle of 045

0sin 2 1 002 90 0

0 45


The Effects of the Air (Air resistance):In vacuum: thepath(trajectory) is parabolic.The Effects of the Air (Air resistance):In vacuum: thepath(trajectory) is parabolic.

x

y

o0d

In Air In vacuum

d

In air: the horizontal range, the maximum

height of the path are much less.

In air: the horizontal range, the maximum

height of the path are much less. Air resistance force:

density of air, the cross section area of the projectile, mainly the velocity of the body.

Air resistance force:

density of air, the cross section area of the projectile, mainly the velocity of the body.

200 / ,v m s 2f v


Sample Problem 2-8 (P32)Sample Problem 2-8 (P32)

Figure 2-32 shows a pirate ship 560m from a fort defending the harbor entrance of an island. A defense canno, located at sea level, fires balls at initial speed .

Figure 2-32 shows a pirate ship 560m from a fort defending the harbor entrance of an island. A defense canno, located at sea level, fires balls at initial speed .0 82 /v m s


(a) at what angle from the horizontal must a ball be fired to hit the ship?

(a) at what angle from the horizontal must a ball be fired to hit the ship?

0

(b) How far should the pirate ship be from the canno if it is to be beyond the maximum range of the cannoballs?




(a) at what angle from the horizontal must a ball be fired to hit the ship? (a) at what angle from the horizontal must a ball be fired to hit the ship?

0

SolutionSolution The fired cannoball is a projectileThe fired cannoball is a projectile

(2-65)

0 82 /v m s 560R mWe know:We know:20

0sin 2v

Rg

(2-65)

Which gives usWhich gives us

1 10 2

0

2 sin sin 0.816gR

v

0 02 54.7 2 125.3and

0 027 63and

Solution: The maximum range corresponds to an elevation angle ofSolution: The maximum range corresponds to an elevation angle of




0 45

20

0sin 2v

Rg

(2-65)


2

2

(82 / )

9.8 /

m s

m s

686 690m m

★ Uniform circular Motion : A particle travel around a circle or a circular arc at constant (uniform) speed. The velocity changes only in direction, there are still an acceleration– centripetal acceleration.

★ Uniform circular Motion : A particle travel around a circle or a circular arc at constant (uniform) speed. The velocity changes only in direction, there are still an acceleration– centripetal acceleration.

2va

r (2-70)

(2-71)2 r

Tv

Period: the time for going around a circle exactly once(circumference of the circle)

Period: the time for going around a circle exactly once(circumference of the circle)

2-12 Uniform Circular Motion2-12 Uniform Circular Motion

Fig.2-34

2-13 Relative Motion2-13 Relative Motion

In three dimensions:Two observers are watchinga moving particle P from theorigins of frames A and B,while B moves at , Thecorresponding axes of frame

A and B remain parallel position vecter: B to A :

P to A :

P to B :(4-41)

Frame A

y

P

Frame B

y

x

x

d

dtd

dt

Frame A

y

P

Frame B

y

x

x

(4-41)

d

dt

Take the time derivativeTake the time derivative

Get :

(4-42)

Take the time derivativeTake the time derivative

Since:Since: (4-43)

The acceleration of the particle measured from frames A and B are the same! The acceleration of the particle measured from frames A and B are the same!


(4-38)

d

dtd

dt

d

dt

(4-42)

Since:

(4-43)

The acceleration of the particle measured from frames A and B are the same! The acceleration of the particle measured from frames A and B are the same!

in Two or Three Dimensions in One Dimension

(4-41)

d

dtd

dt

d

dt

(4-39)

Since:

(4-40)


制作 张昆实 yangtze university

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制作张昆实 yangtze university