Theoretical mechanics
t)(? 12)( ttt ????????
① Kinematics
② Objects of
kinematics
③ Why we
study it?
④ Relativity
of motion
⑤ Instant of time,
interval of time
⑥ Classification
of motions
Basic concepts of kinematics The science of the relation between a given motion of an
object and time,in which only the geometrical characteristics
of the motion such as trajectories,velocities and accelerations,
are studied,but not the reasons for the motion considered,
① Develops the methods for describing mechanical motions,
② Establishes the relations between the variables related to
motion,
Foundations for the following courses and and direct
applications in engineering practice,
Reference object; frame of reference system,frame of static
reference System,moving frame of reference system,
1) Motion of a particles,
2)Motion of a rigid body,
Introduction
t)(? 12)( ttt ????????
① 运动学
② 运动学研究的对象
③ 运动学学习目的
④ 运动是相对的
⑤ 瞬时,时间间隔
⑥ 运动分类
运动学的一些基本概念
是研究物体在空间位置随时间变化的几何性质的科学 。
(包括,轨迹, 速度,加速度等 )不考虑运动的原因。
① 建立机械运动的描述方法
②建立运动量之间的关系
为后续课打基础及直接运用于工程实际 。
( relativity ):参考体 (物 );参考系 ;静系 ;动系。
1)点的运动 2)刚体的运动
引 言
Theoretical mechanics
§ 6–1 Description of the motion of a particle
using a vector of position
§ 6–2 Description of the motion of a particle
using rectangular coordinates
§ 6–3 Description of the motion of a particle
using natural coordinates
Chapter 6, Kinematics of a Particle
§ 6–1 点的运动矢量分析方法
§ 6–2 点的运动的直角坐标法
§ 6–3 点的运动的自然坐标法
第六章 点的运动学
9
1,Equation of motion,
trajectory (path)
2.Velocity of a particle
3,Acceleration of a particle
OMr ?
rdt rdΔ trΔv
Δ t
????
? 0
lim
rdt rddt vdΔ tvΔa
Δ t
??????
? 2
2
0
lim
§ 6-1 Description of the motion of a particle using a vector of position
10
一,运动方程，轨迹
二,点的速度
三,加速度
OMr ?
rdt rdΔ trΔv
Δ t
????
? 0
lim
rdt rddt vdΔ tvΔa
Δ t
??????
? 2
2
0
lim
§ 6-1 点的运动矢量分析方法
11
1,Equation of motion,
trajectory (path)
2.Velocity of a particle
kzjyixr ???
kdtdzjdtdyidtdxdt rdv ????
kvjvivv zyx ????
2
z
2
y
2
x vvvv ???
v
viv x?? )c o s (
v
v
jv y?
?
)c o s (
v
vkv z?? )c o s (
§ 6-2 Description of the motion of a particle using rectangular coordinates
12
一,运动方程轨迹
二,点的速度
kzjyixr ???
kdtdzjdtdyidtdxdt rdv ????
kvjvivv zyx ????
2
z
2
y
2
x vvvv ???
v
viv x?? )c o s (
v
v
jv y?
?
)c o s (
v
vkv z?? )c o s (
§ 6-2 点的运动的直角坐标法
13
3,Acceleration of a
particle
kajaiak
dt
zd
j
dt
yd
i
dt
xd
k
dt
dv
j
dt
dv
i
dt
dv
dt
vd
a
zyx
zyx
??????
????
2
2
2
2
2
2
zyx aaaa 222 ???? ? )c o s (
a
aia x??
Note,x,y,z are continuous functions of time
?
?
?
?
?
?
?
?
( t )fz
( t )fy
( t )fx
3
2
1
The equation of the trajectory,
F(x,y,z)=0,can be determined by
eliminating the parameter t,
14
三, 加速度,
kajaiak
dt
zd
j
dt
yd
i
dt
xd
k
dt
dv
j
dt
dv
i
dt
dv
dt
vd
a
zyx
zyx
??????
????
2
2
2
2
2
2
zyx aaaa 222 ???? ? )c o s (
a
aia x??
[注 ] 这里的 x,y,z 都是时间单位连续函数 。
?
?
?
?
?
?
?
?
( t )fz
( t )fy
( t )fx
3
2
1 当消去参数 t 后,可得到 F(x,y,z)=0
形式的轨迹方程 。
15
§ 6-3 Description of the motion of a particle
using natural coordinates
The position of a moving particle is determined by measuring the displacement
along its path from a reference point,
1,Curvilinear coordinate,natural
coordinate system 1) Equation of motion,S=f (t)
In addition,polar coordinate and column
coordinate can also be used to describe the
motion of a particle,
)(
)(
2
1
tf
tfr
?
?
?
16
§ 6-3 点的运动的自然坐标法
以点的轨迹作为一条曲线形式的坐标轴来确定
动点的位置的方法叫 自然坐标法 。
一,弧坐标,自然轴系 1.弧坐标的运动方程 S=f (t)
补充：极坐标法 (对平面曲线运动时可用 ) )( )(
2
1
tf
tfr
?
?
?
同理可导出柱坐标下的点的运动方程
17
2,velocity of a
particle
2) Natural coordinate system
??
?
?
?
?
?
?
?
?
?
?
??
??
????
????
???
??
??
v
dt
dS
dS
rd
dt
dS
S
r
t
S
t
S
S
r
t
r
v
tt
tt
00
00
limlim
)(limlim
Osculating plane
tangent
Principal normal
Normal plane
18
二,点的速度
2.自然轴系
??
?
?
?
?
?
?
?
?
?
?
??
??
????
????
???
??
??
v
dt
dS
dS
rd
dt
dS
S
r
t
S
t
S
S
r
t
r
v
tt
tt
00
00
limlim
)(limlim
19
① Tangential acceleration
----represents the rate of change
of speed of the particle
dt
τdvτ
dt
Sd
dt
τdvτ
dt
dv)τ(v
dt
d
dt
vda ??????????
2
2
??? ??? 2
2
dt
Sd
dt
dva
?a
3,Acceleration of a particle
② Normal acceleration -----represents the rate
of change of speed direction of the particle
S
v
t
S
S
v
t
v
dt
d
va
t
tt
n
?
??
?
?
?
??
?
???
?
??
0
2
00
lim
)(limlim
?
??
??
?????
)lim(
0
vdtdStS
t
??
? ?
?
?
20
① 切向加速度
----表示速度大小的变化
dt
τdvτ
dt
Sd
dt
τdvτ
dt
dv)τ(v
dt
d
dt
vda ??????????
2
2
??? ??? 2
2
dt
Sd
dt
dva
?a
三,点的加速度
② 法向加速度 -----表示速度方向的变化
S
v
t
S
S
v
t
v
dt
d
va
t
tt
n
?
??
?
?
?
??
?
???
?
??
0
2
00
lim
)(limlim
?
??
??
?????
)lim(
0
vdtdStS
t
??
? ?
?
?
21
The left figure shows that
?
?
?
??
??
??
?
??
?
??
???
1
)
2
2
s i n
(lim2
s i n2
lim||lim
000
??????
??? dS
d
SSS ttt
2s i n22s i n||2|'|||
????????? ????
22s i n,0,0 as
????????? St
??????? h e n c ea n d1||
nvtvaaanva nn ??? ?
22
d
d,??????即
n
n a
aaaa ||a r c t g,22 ?
? ? ???
22
由图可知
?
?
?
??
??
??
?
??
?
??
???
1
)
2
2
s i n
(lim2
s i n2
lim||lim
000
??????
??? dS
d
SSS ttt
2s i n22s i n||2|'|||
????????? ????
22s i n,0,0
?????? ??? St 时当
????? ?? 于是1||
nvtvaaanva nn ??? ?
22
d
d,??????即
n
n a
aaaa ||a r c t g,22 ?
? ? ???
23
Teach yourself in the class,
① Write the equations of motion of a particle using column
coordinates,
② What’s the difference between and? Examine
it for rectilinear and curvilinear cases,respectively,dt
vd
dt
dv
is valid for all cases.,the rate of change of the
value of the velocity,is only the tangential component of the
acceleration for the curvilinear case,
a
dt
vd ? a
dt
dv ?
dt
dva ?
?
?
?
?
?
?
?
?
?
)(
)(
)(
3
2
1
tfz
tfr
tf?
Answer,
24
课堂自学,
① 用柱坐标法给出点的运动方程 。
② 与 有何不同?就直线和曲线分别说明 。
dt
vd
dt
dv
(直线,曲线都一样 ),为速度的
大小变化率,在曲线中应为切向加速度 。
a
dt
vd ? a
dt
dv ?
dt
dva ?
?
?
?
?
?
?
?
?
?
)(
)(
)(
3
2
1
tfz
tfr
tf?
柱坐标法方程
25
③ Determine the types of motion of the particle M
in the following cases,<1>,
<2>,
<3>
<4>,
<5>
<6>
<7>
<8>
<9>
<10>
0?na c o n s t a n t??a
???
c o n s t a n t??
0?a
c o n s t a n t,0 ?? va ?
c o n s t a n t,0 ?? naa ?
0?na
0??a
c o n s t.,c o n s t ?? naa ?
(uniform rectilinear motion)
(uniform circular motion)
(uniform rectilinear motion or rest)
(rectilinear motion)
(uniform motion)
(circular motion)
(uniform motion)
(rectilinear motion)
(uniform curvilinear motion)
(uniformly accelerated motion )
c o n s t a n t??
0??a
0?na
26
③ 指出在下列情况下,点 M作何种运动?
<1>,
<2>,
<3>
<4>,
<5>
<6>
<7>
<8>
<9>
<10>
常数??a
常数??
常数?? va,0?
常数??
常数?? naa,0?
常数常数 ?? naa,?
(匀变速直线运动 )
(匀速圆周运动 )
(匀速直线运动或静止 )
(直线运动 )
(匀速运动 )
(圆周运动 )
(匀速运动 )
(直线运动 )
(匀速曲线运动 )
(匀变速曲线运动 )
0?na
0?a
0?na
0??a
0??a
0?na
27
④ The particles are moving along a curved path,draw
the direction of its acceleration in the following cases,
<1> M1 is moving with a uniform velocity;
<2>M2 is moving with a positive acceleration;
<3>M3 is moving with a negative acceleration,
⑤ Examine whether the following motions are possible and
determine their types in the possible cases,
(accelerated motion) (impossible) (uniform curvilinear motion (impossible or change to
accelerated rectilinear motion)
(impossible or change
to decelerated
rectilinear motion)
(impossible) (decelerated curvilinear motion)
28
④ 点作曲线运动,
画出下列情况下点的加速度方向 。
<1>M1点作匀速运动
<2>M2点作加速运动
<3>M3点作减速运动
⑤ 判断下列运动是否可 能出现,若能出现判断是什么运动?
(加速运动 ) (不可能 ) (匀速曲线运动 ) (不可能或改作
直线加速运动 )
(不可能或改作
直线减速运动 )
(不可能 ) (减速曲线运动 )
29
⑥ <1> If the velocity of a particle traveling along a straight line is zero,then
its acceleration must be zero too,
<2>When a particle performs a curvilinear motion,and its velocity has a
constant value is its acceleration zero?
Solution,<1> not always,
If a particle has zero velocity,its acceleration may not be zero.(for example,
at the acme of a upcasted object.)
<2> The acceleration is not zero,A centripetal acceleration always exists for a
particle traveling along a curved path,
⑦ What are the physical meanings of the tangential and the normal
accelerations?
answer,represents the rate of change of the magnitude of the velocity
represents the rate of the change of the direction of the velocity
dt
dva ?
?
?
2v
a n ?
30
⑥ <1>点作直线运动时,若其速度为零,其加速度也为零
<2>点作曲线运动时,若其速度大小不变,加速度是否一定为零
答,<1>不一定,
速度为零时加速度不一定为零 (自由落体上抛到顶点时 )
<2> 加速度不一定为零，只要点作曲线运动，就有向心加速度
⑦ 切向加速度和法向加速度的物理意义？
答,表示速度大小的变化
表示速度方向的变化
dt
dva ??
?
2v
a n ?
31
⑧ Point M moves along a helical curve from the outside to the inside and its
distance from the origin is proportional to time,Does its acceleration become
bigger or smaller? Does it move more quickly or more slowly?
⑨ Show is directed to the direction of the principal
normal,i.e,dt
d?
?? ?dtd
dt
d
dt
d
dt
d
dt
d
dt
)d( ???????????? ??????????????? 02 0 0 1 ?
Proof,
c o n s t
???
??
? b
dt
dS
v
abtS?
????
222,,0 baabva
dt
dva
nn ?????? ?
The radius of curvature ? becomes smaller as M moves from
the outside to the inside and,consequently,its acceleration
becomes bigger and bigger,Since v =const it moves with a
constant velocity,
solution,
32
⑧ 点 M沿着螺线自外向内运动，它走过的弧长与时间的一次方成
正比，问点的加速度是越来越大，还是越来越小？点是越跑越快，
还是越跑越慢？
⑨ 证明 是沿着主法线方向,即 。
dt
d?
?? ?dtd
dt
d
dt
d
dt
d
dt
d
dt
)d( ???????????? ??????????????? 02 0 0 1 ?
证明,
常数???
??
? b
dt
dSv
abtS
?
????
222,,0 baabva
dt
dva
nn ?????? ?
由于点由外向内运动,曲率半径 ? 越来越小,所以加速度
越来越大 。而速度 v =常数,故点运动快慢不变。
解,
33
34
t)(? 12)( ttt ????????
① Kinematics
② Objects of
kinematics
③ Why we
study it?
④ Relativity
of motion
⑤ Instant of time,
interval of time
⑥ Classification
of motions
Basic concepts of kinematics The science of the relation between a given motion of an
object and time,in which only the geometrical characteristics
of the motion such as trajectories,velocities and accelerations,
are studied,but not the reasons for the motion considered,
① Develops the methods for describing mechanical motions,
② Establishes the relations between the variables related to
motion,
Foundations for the following courses and and direct
applications in engineering practice,
Reference object; frame of reference system,frame of static
reference System,moving frame of reference system,
1) Motion of a particles,
2)Motion of a rigid body,
Introduction
t)(? 12)( ttt ????????
① 运动学
② 运动学研究的对象
③ 运动学学习目的
④ 运动是相对的
⑤ 瞬时,时间间隔
⑥ 运动分类
运动学的一些基本概念
是研究物体在空间位置随时间变化的几何性质的科学 。
(包括,轨迹, 速度,加速度等 )不考虑运动的原因。
① 建立机械运动的描述方法
②建立运动量之间的关系
为后续课打基础及直接运用于工程实际 。
( relativity ):参考体 (物 );参考系 ;静系 ;动系。
1)点的运动 2)刚体的运动
引 言
Theoretical mechanics
§ 6–1 Description of the motion of a particle
using a vector of position
§ 6–2 Description of the motion of a particle
using rectangular coordinates
§ 6–3 Description of the motion of a particle
using natural coordinates
Chapter 6, Kinematics of a Particle
§ 6–1 点的运动矢量分析方法
§ 6–2 点的运动的直角坐标法
§ 6–3 点的运动的自然坐标法
第六章 点的运动学
9
1,Equation of motion,
trajectory (path)
2.Velocity of a particle
3,Acceleration of a particle
OMr ?
rdt rdΔ trΔv
Δ t
????
? 0
lim
rdt rddt vdΔ tvΔa
Δ t
??????
? 2
2
0
lim
§ 6-1 Description of the motion of a particle using a vector of position
10
一,运动方程，轨迹
二,点的速度
三,加速度
OMr ?
rdt rdΔ trΔv
Δ t
????
? 0
lim
rdt rddt vdΔ tvΔa
Δ t
??????
? 2
2
0
lim
§ 6-1 点的运动矢量分析方法
11
1,Equation of motion,
trajectory (path)
2.Velocity of a particle
kzjyixr ???
kdtdzjdtdyidtdxdt rdv ????
kvjvivv zyx ????
2
z
2
y
2
x vvvv ???
v
viv x?? )c o s (
v
v
jv y?
?
)c o s (
v
vkv z?? )c o s (
§ 6-2 Description of the motion of a particle using rectangular coordinates
12
一,运动方程轨迹
二,点的速度
kzjyixr ???
kdtdzjdtdyidtdxdt rdv ????
kvjvivv zyx ????
2
z
2
y
2
x vvvv ???
v
viv x?? )c o s (
v
v
jv y?
?
)c o s (
v
vkv z?? )c o s (
§ 6-2 点的运动的直角坐标法
13
3,Acceleration of a
particle
kajaiak
dt
zd
j
dt
yd
i
dt
xd
k
dt
dv
j
dt
dv
i
dt
dv
dt
vd
a
zyx
zyx
??????
????
2
2
2
2
2
2
zyx aaaa 222 ???? ? )c o s (
a
aia x??
Note,x,y,z are continuous functions of time
?
?
?
?
?
?
?
?
( t )fz
( t )fy
( t )fx
3
2
1
The equation of the trajectory,
F(x,y,z)=0,can be determined by
eliminating the parameter t,
14
三, 加速度,
kajaiak
dt
zd
j
dt
yd
i
dt
xd
k
dt
dv
j
dt
dv
i
dt
dv
dt
vd
a
zyx
zyx
??????
????
2
2
2
2
2
2
zyx aaaa 222 ???? ? )c o s (
a
aia x??
[注 ] 这里的 x,y,z 都是时间单位连续函数 。
?
?
?
?
?
?
?
?
( t )fz
( t )fy
( t )fx
3
2
1 当消去参数 t 后,可得到 F(x,y,z)=0
形式的轨迹方程 。
15
§ 6-3 Description of the motion of a particle
using natural coordinates
The position of a moving particle is determined by measuring the displacement
along its path from a reference point,
1,Curvilinear coordinate,natural
coordinate system 1) Equation of motion,S=f (t)
In addition,polar coordinate and column
coordinate can also be used to describe the
motion of a particle,
)(
)(
2
1
tf
tfr
?
?
?
16
§ 6-3 点的运动的自然坐标法
以点的轨迹作为一条曲线形式的坐标轴来确定
动点的位置的方法叫 自然坐标法 。
一,弧坐标,自然轴系 1.弧坐标的运动方程 S=f (t)
补充：极坐标法 (对平面曲线运动时可用 ) )( )(
2
1
tf
tfr
?
?
?
同理可导出柱坐标下的点的运动方程
17
2,velocity of a
particle
2) Natural coordinate system
??
?
?
?
?
?
?
?
?
?
?
??
??
????
????
???
??
??
v
dt
dS
dS
rd
dt
dS
S
r
t
S
t
S
S
r
t
r
v
tt
tt
00
00
limlim
)(limlim
Osculating plane
tangent
Principal normal
Normal plane
18
二,点的速度
2.自然轴系
??
?
?
?
?
?
?
?
?
?
?
??
??
????
????
???
??
??
v
dt
dS
dS
rd
dt
dS
S
r
t
S
t
S
S
r
t
r
v
tt
tt
00
00
limlim
)(limlim
19
① Tangential acceleration
----represents the rate of change
of speed of the particle
dt
τdvτ
dt
Sd
dt
τdvτ
dt
dv)τ(v
dt
d
dt
vda ??????????
2
2
??? ??? 2
2
dt
Sd
dt
dva
?a
3,Acceleration of a particle
② Normal acceleration -----represents the rate
of change of speed direction of the particle
S
v
t
S
S
v
t
v
dt
d
va
t
tt
n
?
??
?
?
?
??
?
???
?
??
0
2
00
lim
)(limlim
?
??
??
?????
)lim(
0
vdtdStS
t
??
? ?
?
?
20
① 切向加速度
----表示速度大小的变化
dt
τdvτ
dt
Sd
dt
τdvτ
dt
dv)τ(v
dt
d
dt
vda ??????????
2
2
??? ??? 2
2
dt
Sd
dt
dva
?a
三,点的加速度
② 法向加速度 -----表示速度方向的变化
S
v
t
S
S
v
t
v
dt
d
va
t
tt
n
?
??
?
?
?
??
?
???
?
??
0
2
00
lim
)(limlim
?
??
??
?????
)lim(
0
vdtdStS
t
??
? ?
?
?
21
The left figure shows that
?
?
?
??
??
??
?
??
?
??
???
1
)
2
2
s i n
(lim2
s i n2
lim||lim
000
??????
??? dS
d
SSS ttt
2s i n22s i n||2|'|||
????????? ????
22s i n,0,0 as
????????? St
??????? h e n c ea n d1||
nvtvaaanva nn ??? ?
22
d
d,??????即
n
n a
aaaa ||a r c t g,22 ?
? ? ???
22
由图可知
?
?
?
??
??
??
?
??
?
??
???
1
)
2
2
s i n
(lim2
s i n2
lim||lim
000
??????
??? dS
d
SSS ttt
2s i n22s i n||2|'|||
????????? ????
22s i n,0,0
?????? ??? St 时当
????? ?? 于是1||
nvtvaaanva nn ??? ?
22
d
d,??????即
n
n a
aaaa ||a r c t g,22 ?
? ? ???
23
Teach yourself in the class,
① Write the equations of motion of a particle using column
coordinates,
② What’s the difference between and? Examine
it for rectilinear and curvilinear cases,respectively,dt
vd
dt
dv
is valid for all cases.,the rate of change of the
value of the velocity,is only the tangential component of the
acceleration for the curvilinear case,
a
dt
vd ? a
dt
dv ?
dt
dva ?
?
?
?
?
?
?
?
?
?
)(
)(
)(
3
2
1
tfz
tfr
tf?
Answer,
24
课堂自学,
① 用柱坐标法给出点的运动方程 。
② 与 有何不同?就直线和曲线分别说明 。
dt
vd
dt
dv
(直线,曲线都一样 ),为速度的
大小变化率,在曲线中应为切向加速度 。
a
dt
vd ? a
dt
dv ?
dt
dva ?
?
?
?
?
?
?
?
?
?
)(
)(
)(
3
2
1
tfz
tfr
tf?
柱坐标法方程
25
③ Determine the types of motion of the particle M
in the following cases,<1>,
<2>,
<3>
<4>,
<5>
<6>
<7>
<8>
<9>
<10>
0?na c o n s t a n t??a
???
c o n s t a n t??
0?a
c o n s t a n t,0 ?? va ?
c o n s t a n t,0 ?? naa ?
0?na
0??a
c o n s t.,c o n s t ?? naa ?
(uniform rectilinear motion)
(uniform circular motion)
(uniform rectilinear motion or rest)
(rectilinear motion)
(uniform motion)
(circular motion)
(uniform motion)
(rectilinear motion)
(uniform curvilinear motion)
(uniformly accelerated motion )
c o n s t a n t??
0??a
0?na
26
③ 指出在下列情况下,点 M作何种运动?
<1>,
<2>,
<3>
<4>,
<5>
<6>
<7>
<8>
<9>
<10>
常数??a
常数??
常数?? va,0?
常数??
常数?? naa,0?
常数常数 ?? naa,?
(匀变速直线运动 )
(匀速圆周运动 )
(匀速直线运动或静止 )
(直线运动 )
(匀速运动 )
(圆周运动 )
(匀速运动 )
(直线运动 )
(匀速曲线运动 )
(匀变速曲线运动 )
0?na
0?a
0?na
0??a
0??a
0?na
27
④ The particles are moving along a curved path,draw
the direction of its acceleration in the following cases,
<1> M1 is moving with a uniform velocity;
<2>M2 is moving with a positive acceleration;
<3>M3 is moving with a negative acceleration,
⑤ Examine whether the following motions are possible and
determine their types in the possible cases,
(accelerated motion) (impossible) (uniform curvilinear motion (impossible or change to
accelerated rectilinear motion)
(impossible or change
to decelerated
rectilinear motion)
(impossible) (decelerated curvilinear motion)
28
④ 点作曲线运动,
画出下列情况下点的加速度方向 。
<1>M1点作匀速运动
<2>M2点作加速运动
<3>M3点作减速运动
⑤ 判断下列运动是否可 能出现,若能出现判断是什么运动?
(加速运动 ) (不可能 ) (匀速曲线运动 ) (不可能或改作
直线加速运动 )
(不可能或改作
直线减速运动 )
(不可能 ) (减速曲线运动 )
29
⑥ <1> If the velocity of a particle traveling along a straight line is zero,then
its acceleration must be zero too,
<2>When a particle performs a curvilinear motion,and its velocity has a
constant value is its acceleration zero?
Solution,<1> not always,
If a particle has zero velocity,its acceleration may not be zero.(for example,
at the acme of a upcasted object.)
<2> The acceleration is not zero,A centripetal acceleration always exists for a
particle traveling along a curved path,
⑦ What are the physical meanings of the tangential and the normal
accelerations?
answer,represents the rate of change of the magnitude of the velocity
represents the rate of the change of the direction of the velocity
dt
dva ?
?
?
2v
a n ?
30
⑥ <1>点作直线运动时,若其速度为零,其加速度也为零
<2>点作曲线运动时,若其速度大小不变,加速度是否一定为零
答,<1>不一定,
速度为零时加速度不一定为零 (自由落体上抛到顶点时 )
<2> 加速度不一定为零，只要点作曲线运动，就有向心加速度
⑦ 切向加速度和法向加速度的物理意义？
答,表示速度大小的变化
表示速度方向的变化
dt
dva ??
?
2v
a n ?
31
⑧ Point M moves along a helical curve from the outside to the inside and its
distance from the origin is proportional to time,Does its acceleration become
bigger or smaller? Does it move more quickly or more slowly?
⑨ Show is directed to the direction of the principal
normal,i.e,dt
d?
?? ?dtd
dt
d
dt
d
dt
d
dt
d
dt
)d( ???????????? ??????????????? 02 0 0 1 ?
Proof,
c o n s t
???
??
? b
dt
dS
v
abtS?
????
222,,0 baabva
dt
dva
nn ?????? ?
The radius of curvature ? becomes smaller as M moves from
the outside to the inside and,consequently,its acceleration
becomes bigger and bigger,Since v =const it moves with a
constant velocity,
solution,
32
⑧ 点 M沿着螺线自外向内运动，它走过的弧长与时间的一次方成
正比，问点的加速度是越来越大，还是越来越小？点是越跑越快，
还是越跑越慢？
⑨ 证明 是沿着主法线方向,即 。
dt
d?
?? ?dtd
dt
d
dt
d
dt
d
dt
d
dt
)d( ???????????? ??????????????? 02 0 0 1 ?
证明,
常数???
??
? b
dt
dSv
abtS
?
????
222,,0 baabva
dt
dva
nn ?????? ?
由于点由外向内运动,曲率半径 ? 越来越小,所以加速度
越来越大 。而速度 v =常数,故点运动快慢不变。
解,
33
34
