Part One Mechanics 力学Part One Mechanics 力学

Chapter 1 Kinematics

( 质点 ) 运动学

第一章 质点运动学第一章 质点运动学 (Kinematics)(Kinematics)§1-1 参考系 质点 Frame of reference particle

§1-2 位置矢量 位移 Position vector and displacement

§1-3 速度 加速度 Velocity and acceleration

§1-4 两类运动学问题 Two types of Problems

§1-6 运动描述的相对性 Relative motion

§1-5 圆周运动及其描述 Circular motion

§1-1 Frame of reference Particle

1. Frame of Reference

!!Choose different objects as the reference frames to describe the motion of a given body, the indications will be different.

Coordinate system:

Cartesian coordinate system( 直角坐标系）

Mathematical reference frame

For describing the motion of a given bodyquantitatively

Nature coordinate system （自然坐标系）

2. Particle

Ignore the size and shape of a body, only think of its mass

Ideal model

Translational motion( 平动）

An object can be simplified a particle when…

Its size << moving size

§1-2 Position Vector and Displacement

P(x,y,z)

z

r

YX

1. Position Vector

Express the position of particle

r

222 zyxrr

Magnitude:

r

xcos

r

zcos

r

ycos

i j

k

kzjyixr

Direction:

In the two dimension:

jtyitxtrr

)()()(

Its two components

)(

)(

tyy

txx

)(xyy Path equation

eliminating t

x

y

o

P

r

)y,x(

(Position function)

(Moving equation)

2. Displacement( 位移 ):

Describe the

change of position

during a given

time interval t:

r

x

y

o

P

1r

1t

2t

2r

r

12 rrr

jyyixxr

)()( 1212 In Cartesian coordinate system:

x y

x

y

o

P

1r

1t

2t

2r

r

Its magnitude

22 )()( yxr

are different.

rrs

,,

s

Note: r

On the condition of limitation: rdds

§1-3 Velocity and Acceleration

Average velocity ：

1.Velocity

12

12

tt

rr

t

rv

Its direction is same with that of r

Average speed( 速率） :t

sv

所用的时间走过的路程

x

y

o

P

1r

1t

2t

2r

r

s

x

y

o

P

r

t

2t

2r

r

v（ Instantaneous 瞬

时） velocity at time t ：

td

rd

t

rv

t

0

lim

In the tangent( 切线） of the path, to point at t

he advance direction.

Direction:

Magnitude ：

vtd

sd

t

rv

t

0lim 速率

jtd

ydi

td

xd

td

rdv

In Cartesian coordinate system:

xv yv

x

y

o

P

r

t

v

xv

yv

22yx vvv

x

y

v

v1tan

2.Acceleration

x

y

o

P1t 1v

2t2v

1v

2v v

Average acceleration

12

12

tt

vv

t

va

Instantaneous acceleration

20lim

td

rd

td

vd

t

va

2

t

In the coordinate system:

jaiajtd

ydi

td

xdj

td

vdi

td

dva yx

22yx

22

22yx aaa

Example 1.1: The of a particle is

r

jtitr 322

where and are constants. Find the velocity and acceleration.

dt

rdv

dt

vda

Solution:

jtit

234

jti

64

Example 1. 2 The position function of a particle in SI unit is x=2t, y=192t2. Calculate （ 1 ） Path function. （ 2 ） Velocity and acceleration at t=1s.

2

2

119 xy

(2) Take time derivation of position function

Solution

2219

2

ty

tx

（ 1）

tv

v

y

x

4

2

(1) Eliminate t from position function

)0( x

(2)

4

0

y

x

a

aTake time derivation of velocity, we have

jitv

42)1( jta

4)1( Substitute t=1s into Eq.(1) and (2)

The magnitude and direction of velocity:

122 47.4 msvvv yx

4.632

4tan 1 (Formed by and +x direction)v

0v

vx

h

0v

a

Example 1-3 Someone stands on a dam, pulls a boat with constant speed

Find: Speed and acceleration of boat at any position x

Take time derivation of it

Solution

222 hrx

dt

drr

dt

dxx 22

0

22

0 vx

hxv

x

rv

3

220

x

hva

Set up a coordinate system shown in Fig.

Then the position of boat depend by

o

y

r

h

xx0v

r

=-v0

=v

§ 1-4 Two Types Problems in Kinematics

(2) Given acceleration(or velocity) and initial con

dition, find the velocity and position by me

ans of integration method

积分法 .

There are two kinds of problems to be solved:

(1) Given position function, find the velocity and

acceleration by using derivation method

微分法 .

See the examples above.

Solution

ktdtv

dv

2Separate variables

tv

v

ktdtv

dv

02

0

Integrate in both side of “=”

Example1.4

tkva 2 (k is constant) and its speed is att=0.

0vFind ?)( tv

An object moves along a straight line. Its Acce. is

tkva 2dt

dv

20

0

2

2

tkv

vv

Example1.5 A particle moves in a plane with an Acce. , Whe

n t=0, its Vel. is at a initial point (0,0). Find its velocity at time t and path equation.

jga

jViVV

sincos

Solution jga

, we can obtain:

dtjgVd

jgtViVVjgtV

)sin(cos 000

VjgtV

Using , we havejsinVicosVV

000

From

Integrate

Using jdt

dyi

dt

dxV

and the initial condition(0,0), we have

tVx cos0

20 2

1sin gttVy

22

0

2

cos2

1

v

gxtgxy

§1-5 Circular MotionMany of circular motions in our world

Take any point on path as origin point of coordinate system

The position of a particle depend on the path length from O to P

Two coordinate axes are on the moving particle.

1. The nature coordinate system （自然坐标系）

o

P

S= S(t) S

n

:tangential direction

:normal direction

n Both are unit vectors.

1 n

Position ： )(tSS

Velocity ： vv

2. Description of circular motion with nature coordinate system

dt

ds

Acce.dt

vda

v

dt

d

dt

dv

dt

dv

=?

tdt

dt

0lim

AB

ntt

ˆlim0

AvBv

B

A

n̂

n̂

nR

v ˆ

a n

R

v

dt

dv ˆ2

R

van

2

---- Normal accelerationChanges the direction of the velocity.

dt

dvat ---- tangential acceleration

Changes the magnitude of the velocity.

3. General curvilinear motion on a plane

R

dt

dvat

--Radius of curvature at

any point on the curve

2van

o

),( yxP

.constR

4. Angular variables in circular motion

Angular position Position function

)(t Angular displacement

Angular velocity

td

d

Angular acceleration 2td

d

td

d 2

－ rad － rad.s-1

－ rad.s-2

Units:

Counterclockwise( 反时针） : positive directio

n

Clockwise （顺时针） : negative directio

n

Two directions:

o.constR

Special Example

Uniform circular motion

t

const

0

,0

Circular motion with constant angular Acce.

)(2

2

1

,

020

2

200

0

tt

t

const

5. Relation between linear & angular variables

2,

,

ra ra

rV

rS rS

nt

),(),,( oraav nt

r

请自己推导！

§1-6 Relative motion1. Relative motion

2.Relativity of the description about a motion

'0 rrr 0r

:K 相对 K 的牵连坐标

K

K’

0r

r

'r

o

'o

A particle is moving in the space.

K’ moves with respect to K

Measure particle in K:

)()()( tatvtr

)(')(')(' tatvtr Measure particle in K’:

P

'vuv

'0 aaa

dt

rd

dt

rd

dt

rd

0

v

'v

u

： K 测得质点速度 --- 绝对速度

： K 测得质点速度 --- 相对速度

： K 相对 K 速度 --- 牵连速度

dt

vd

dt

ud

dt

vd

P

ox

y

zo x

y

z

),,(

),,(

zyx

zyx

u

Assuming that O and O’ coincide at t=0 and moves along the x-axis at speed of u, we have

zyxo

tt

zz

yy

utxx

tt

zz

yy

utxx

Galilean transformations( 伽利略变换）