Part One Mechanics 力学 Part One Mechanics 力学. Chapter 1 Kinematics ( 质点 ) 运动学.
-
Upload
gervais-simmons -
Category
Documents
-
view
352 -
download
16
Transcript of Part One Mechanics 力学 Part One Mechanics 力学. Chapter 1 Kinematics ( 质点 ) 运动学.
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)