Chapter 5 Rotation of a Rigid Body
第五章 刚体的转动
§ 5-5 Angular Momentum of a rigid Body
Conservation of Angular Momentum 定轴转动
的刚体的角动量定理 和角动量守恒定律
§ 5-1 Motion of a Rigid body 刚体的平动、转动和定
轴转动
§ 5-2 Torque The Law of Rotation rotational
Inertia 力矩 刚体定轴转动定律 转动惯量
§ 5-3 Applying the Law of rotation 转动定律的应用
§ 5-4 Kinetic Energy and Work in Rotational
Motion 定轴转动的动能定理
1,理解描述刚体定轴转动的基本物理量的定义和性质;
2,理解力矩、转动动能和转动惯量的物理意义;
3,掌握定轴转动的转动定律和角动量定理;
4,掌握定轴转动的角动量守恒定律和机械能守恒定律。
教学基本要求
Now,we consider the motion of a body with a
certain size and shape:
大小和形状!
For simplicity( 为简单 ),we
ignore the deformation(形变 ), In
general,the motion of the body is
more complicated than that
discussed above.
Why?
猫坠楼,但不易受伤!
The body without deformation is called a rigid
body.
(1) translation(平动) ;
(2) rotation(转动)。
Examples,dancing,skating,motion of Earth,motion
of electron in the atom world,……,
Its motion includes usually two types:
铁石心肠
不可塑造
1,The model of a rigid body 刚体模型
Rigid body( 刚体 ), an ideal model
刚体是一种理想模型:刚体是在任何外力作用下任意两点间
均不发生位移,形状大小均不发生改变的物体。
§ 51 Motion of a Rigid Body
刚体的平动、转动和定轴转动
The distance between two points in a
rigid body maintains constant forever,
Thus the body has a perfectly definite,
unchanged shape and size,
2,Translation & Rotation of a rigid body
刚体的平动和转动
(平动:刚体上所有点的运动轨迹都相,
可当作质点来处理 。 )
All particles describe parallel
( 平行 ) paths,and have the
same velocity & acceleration.
Therefore,the motion of any
point of the body can represent
the translational motion of
entire( 整个 ) rigid body.
(1) Translation(平动),
水平飞行
看成质点
All particles describe
circular paths around a line
called the axis( 轴 ) which
may fixed or may changing its
position during the motion.
(2) Rotation:
(转动:刚体上各点都绕某一轴
作圆周运动, 转轴可能是固定
的也可能在运动中改变位置 。 )
圆周运动
Often a rigid body can simultaneously(同时 )
have two kinds of motions.
(3) Translation and Rotation at same time
平动和转动(轴动)
平动和转动(转轴位置变)
定轴转动
3,Rotation of a rigid body around a fixed axis
刚体的定轴转动
Example,A rotation around a fixed axis
轴
Every point of the rigid body
moves in a circle(刚体上各点都
作圆周运 动 ).
( 刚体上各点都绕同一转轴作不同半
径的圆周运 动 )
P
Q
轴
All the points in the rigid body:
? Circular motion with different radii.
?To rotate around an axis.
( 1) Characteristic,all points have the same angular
displacement,angular speed & angular
acceleration.
特点:
各质点均在垂直转轴的
平面内作圆周运动且角
位移, 角速度和角加速
度均相同 。
P
Q
轴
???,,?
We select ( 选择 ) one
arbitrary(任意 ) point P in the
body whose circular motion can
represents the motion of the
entire body and use its circular
plane as a reference to analyze
( 分析 ) the rotation of the
rigid body around the fixed axis.
y
x0
?
V?
r?
P
P
Q
轴
( 2) Kinematics of rotation of a
rigid body around a fixed axis is
the same as the circular motion of
a particle that we discussed in
chapter 1.
Representative(代表) P,
① Angular position ?
② Angular displacement ??
?Angular speed
td
d?? ?
??? tt d
d
d
d 2????Angular acceleration
P
Q
轴
描述刚体定轴转动的物理量
? Angular displacement d?
?Angular speed
td
d?? ?
2tt d
d
d
d 2??? ???Angular acceleration
( 4) Relation between two kinds of quantity
r
s
2
??
??
??
??
nt a ra
rVr
y
x0
?
V?
r?
P
?????,,,
nt a,a,V,s,r ?
§ 5-2 Torque The Law of Rotation Rotational Inertia
力矩 转动定理 转动惯量
1,Torque
Based on(根据 ) the concept of torque of a force
acting on a particle with respect to a fixed point ( in
§ 4-4),we can define the torque of a force acting on a
rigid body.
Does a force make a rigid body rotate?
depending on its magnitude and
acting position,Torque
( 力矩 ) ! F?F?
FrM ??? ??
FdFs i nFrM ??? ?r ?
r?
F?
?
?r
z
P
F?
轴
The torque with respect to z axis is
the Z-component of,Mz,
briefly labeled as M
M?
⑴ If the force acting on a rigid
body is located in the plane
perpendicular to the axis,the
torque with respect to point o is
defined as
F?
⑵ If the force is not placed in the plane perpendicular to
the axis,we can resolve into ( in the plane)and
( at right angle to the plane),Obviously( 显然 ),only
contributes to (有贡献 ) the torque,
1F? 2
F?
1F?
F?
P
F?
轴
F?
2F?
1F?
dFFs i nrFM ???? ??? r ?
⑶ The resultant Toque
合力矩
???? ??? ?? MMM
???? ???? dFdFM z
P
1F
?
轴
2F
?
The torque have only two possible directions,
Counterclockwise(反时针),positive
Clockwise(顺时针), negative
说明:
与转轴垂直但通过转轴的力对转
动不产生力矩; (Why?)
与转轴平行的力对转轴不产生力
矩; (Why?)
刚体内各质点间内力对转轴不产
生力矩。 (why?)
2,The Law of rotation 转动定理 (very important)
Our method is to treat( 处
理 ) the rigid body as a
collection ( 集体 ) of
particles with same angular
acceleration and to apply
( 应用 ) Newton’s second
law to all particles,then
add up(相加 ) all equations,
P
F?
轴
方法 — 对组成刚体的所有质点用牛顿第二定律, 再相
加 。
Apply Newton’s second law to the ith particle,
法向,2 ???
iiiniinin rmamfF ???
??? iiitiitit rmamfF ???
切向,
To ith particle(第 i个粒子),we introduce
Fi— the external force on the ith particle
iF?
fi — the internal force on the ith particle
if
?
?
?
?
ij
iji ff
?? iff
?
fif
?
i
j
??? 2iiiitiiitiit rmramrfrF ???
Add all of equations for all particles
?? )rm(rfrF iiiitiit ??? ? ??
(5-11)
Considering
??? iit rf (内力成对出现)
and the resultant torque of the external forces about the
axis is
?? iit rFM
Multiplying(乘) both sides of second equation by,
we have
ir
Thus (5-11) becomes
? )( 2ii rmM ? ??
(5-12)
The sum( 求和 ) on the right size is defined as the
rotational inertia( 转动惯量 ) of the body with respect
to the axis and labeled as I (转动惯量定义 )
2ii rmI ? ??
(5-13)
( 注意:有的书用 J表示 )
Equation (5-12) becomes
?IM ?
(5-14)
转动定理
I
M
??
(5-14‘)
or rewrite it as
P
F?
轴
I
M
??
(5-14‘)
Equation (5-14) is called as the law of rotation,转动定理
表明角加速度与力矩成正比, 与转动惯量成反比 。
P
F?
轴
Eq.(5-14) has exactly the
same form as that for
acceleration of a particles,by
which the rotation of a rigid
body is governed( 控制 ),
Therefore,it is often regarded
as( 称为 ) Newton’s second
law for rotation.
I
M
??
3,Calculation of rotational inertia(转动惯量的计算 ):
The rotational inertial I is not a unique(唯一 ) property
of the body but depends on the axis,Rotational inertial of
a body is determined by:
(1) the total mass of the body(总质量) ;
(2) the mass distribution of the body(质量的分布) ;
(3) the position of the axis(转轴的位置),
axis
axis
axis
三个刚体, 质量相
等, 因转轴位置或
质量分布不同, 转
动惯量不相等;
We can use the
following example to
show above
conclusion(结论 )
The rotational inertial of a body can be found by
?Experimental method;
? Calculation method.
o
M
m x
高度与时间
质量离散分布的物体,
? ?? 2ii rmI ( 5-15)
(会计算简单分离物体的 I)
If the mass distribution of a body is known,its
rotational inertial may be calculated as follows:
1r
3r
2r
1m
2m
3m
axis
Example:
????????? ??? rmrmrmI
质量连续分布的物体,
( 记住:棒, 圆盘和圆柱体的 I)
?? mrI d2 ( 5-16)
线积分
?
面积分
?
体积分
?
dVordsord ??? ??dm
§ 5- 3 Applying the Law of rotation
转动定律的应用( important!!!!)
基本步骤
( 1) 隔离法选择研究对象;
( 2) 受力分析和运动情况分析;
( 3) 对质点用牛顿定理, 对刚体用转动定理;
( 4) 建立角量与线量的关系, 求解方程;
( 5) 结果分析及讨论 。
Example 5-3,一个质量为 M,半径为 R的定滑轮(当作均匀圆盘)
上面绕有细绳。绳的一端固定在滑轮边上,另一端挂一质量为 m的
物体而下垂。忽略轴处摩擦,求物体 m由静止下落 h高度时的速度
和此时滑轮的角速度 (定滑轮的转动惯量 )。
221 MRI ?
解,(1)研究对象:滑轮和物体 m;
T
T
mg
o
M
m x
(2)受力分析如图:
滑轮,T,Mg,和轴的支持力,
只有 T产生力矩 ( why?),顺 时
针转动;
物体,mg和 T,向下运动
对物体 m:
?? 221 MRIRT ??
maTmg ??
?Ra ?
( 3) 对滑轮:
( 4) 以上三式联立, 可得物体下落的加速度和速度:
gMm ma 2?? Mm m g hahV ??? 2 42
这时滑轮转动的角速度为
Mm
m g h
RR
V
??? 2
41?
滑轮和物体的运动学关系为,
o
M
m x
Example 5-4:质量 M=1.1kg,半径 =0.6m的匀质圆盘, 可
绕通过其中心且垂直于盘面的水平光滑固定轴转动 。 圆盘
边缘绕有轻的柔绳, 下端挂一质量 m=1.0kg的物体, 如图
所示, 起初在圆盘上加一恒力矩使物体以速率 V0=0.6m/s
上升, 如撤去所加的恒力矩, 问经历多少时间圆盘开始作
反向转动 ( ) 。 221 MrI ?
解,( 1) 研究对象:物体和圆盘;
( 2) 受力分析如图, 设逆时针
方向为转动的正向, 角加速度为
?,物体向下的加速度为 a。
mg
T
T
( 4) 解上面的方程组:
Imr
rm g
?? 2?
( 5) 圆盘作匀加速转动, 故有:
t??? ?? 0
其中
r
V 0
0 ???
,代如数据, 令 ?=0可求得反转时间 。
请同学求出反转所需时间 !!
(3)列方程:
?
?
?
?
?
?
??
?
?
?
ra
maTmg
ITr
mg
T
T
Example 5-5,As shown in below figure,the body A is
connected to the body B by the light rope which is
through two uniform solid cylinder( 圆柱体 ) with a
mass m and a radius r,The body A has a mass of m and
the mass of B is 2m.There is not relative motion between
the rope and cylinder,Find the tension force between the
solid cylinders with, ?
?
?? mrI
m
A B
2m
m,r m,r
T=?
解,( 1) 研究对象,A,B和两圆柱体;
( 2) 受离分析如图,TA
A B
mg 2mg TA
TB
TTB
A向上运动, 有加
速度 aA;B 向下运
动, 加速度 aB,圆
柱体顺时针转动 。
( 3) 可有下列方程:
?
?
?
?
?
?
?
?
?
??
??
??
??
??
?
?
?
raa
maTT
IrTT
IrTT
maTmg
BA
AA
A
B
BB
)(
)(
22
3
11 mgT ?
( 5) 解方程组可得
Example 5-6:如图, 长为 L质量为 m的匀质棒, 可绕其通
过端点的光滑轴在竖直平面内转动 。 求棒从水平位置转
到图中位置的角加速度 ( ) 。
231 mLI ?
?
mg
解,( 1) 研究对象:棒;
( 2) 受重力作用, 可
证明重力对转轴的力矩
为:
?c o sLmgM 21?
( 3) 由转动定理, 可得:
L
c o sg
mL
c o sLmg
?
?
?
?
?
?
?
?
?
??
?
类似的问题:
?
mg
请同学求出角加速度 ?
§ 5-4 Kinetic energy and work in rotational motion
1,Kinetic Energy of Rotation
A moving body has the kinetic
energy,What is the kinetic
energy of a rotating body?
Windmill( 风车 )
唐吉,可德(?)
风力发电, 水轮机等
How can we get the the expression( 表达式 ) of kinetic
energy of a rigid body in rotational motion? — Treat the
body as a collection of particles.
?
?
??
iik VmE ??
? ? ??? ?????????? ??? ?? ??? iiiik rmVmE
(2) Take the sum of kinetic energies
over all the particles that make up the
body:
(1) Write the ith particle’s Kinetic energy as
axis
ir
im?
iV
?
?ii rV ? ? ?? i ii rmI
2
That is
(5-16)
2
k Iω2
1E ?
(转动动能 )
(对比质点平动动能 ) 2
k mV2
1E ?
axis
?
I
2,Work done by torque Kinetic theorem of rotation
(1) Work done by torque
When a torque acts on a
rigid body,the rigid body
starts to rotate with an
angular acceleration so that it
store the kinetic energy,This
fact shows that the torque
have done work on the rigid
body.
In the following,for
simplicity,we consider only a
force applied at point P of
a rigid body rotating about a
fixed axis through o as shown
in Figure.
F?
Supposing that the rigid body rotates through a
element angular displacement,the work done by the
force is
??? ddddd M????? rFsc o sFsFW t??
Fig,5-14
axis i
r?
F?
?d
The total work done during a finite angular
displacement is then
??? ? ?dW M
(5-18)
In the special case of M is a constant
??? ?? MMMW ??? ?? ?? dd
(5-19)
Fig,5-14
axis i
r?
F?
?d
Instantaneously power
?? MtMWP ??? dddd t
(5-20)
(2) Kinetic energy theorem of rotation 转动动能定理
Rewrite the element work
?? ?dd MW
Fig,5-14
axis i
r?
F?
?d
tIIM d
d ?? ??
?? ddd tI
???? dIttI ?? ddd
td?
When the angular speed changes from to,the
work done by the torque is
fi ??
(5-21)
Equation (5-21) indicates that the work done by torque
equals to the increment ( 增量 ) of kinetic energy of
rotation,This is kinetic theorem of rotation.
末动能 初动能
?? ?
?
?
if II ?? 2
1 ? ?? f d ?
?
??
i
IW kikf EE ??
3.potential energy of weight 刚体的重力势能
?? ???? iiiiP hmgghmE
Mhmmhmh iiiiiC ??? ?????
设势能零点在 x-axis,hc
为质心到势能零点的距
离,
y
?mi
C
hi
hC
o x
M
CP M g hE ?
如刚体在重力矩作用下转动, 计入 刚
体的重力势能 后, 如满足守恒条件, 即
其它力矩作功为零或无其它力矩, 机械
能守恒定律:
tt a nc o n sI ??
?
? ? 势能?
Example 5-7 质量为 m,长为 L的均质细杆可绕水平光滑轴 O在竖直
平面内转动 。 若使杆从水平位置开始由静止释放, 试求杆转至铅垂
位置时的角速度 。
?
mg
L解:可利用动能定理来求解。
当杆的位置由 ? 转到 ? +d? 时,
重力矩所做元功为:
??? d21dd ??? c o sm g LMW
m g Lm g LM 21d21dW 2
0
2
0
???? ??
??
??? c o s
两边积分得:
22
2
10
2
1
2
1 ?? IIm g L ???
L
g3??
( 事实上, 这就是机械能量守恒 )
L
?
由定轴转动动能定理有:
Example 5-8:一长为 L,质量为 m的匀质细杆竖直放
置,其下端与一固定铰链 o相连,并可绕其转动,当其
受到微小扰动时,细杆将在重力的作用下由静止开始
绕铰链 o转动,试计算细杆转到与铅直线呈 角时的
角加速度和角速度,
?
解,受力分析如图
取任一状态,由转动定律
?? Im g LM ?? s i n21外
2
3
1 mLI ?
?? s i nLg23?
( 2) 利用机械能守恒求角速度:
取 o点的重力势能为零, 则
?? c o sm gLIm gL 212121 2 ??
)co s( ?? ?? 1
2
3
L
g
§ 5-5 Angular Momentum( 角动量 ) of a Rigid
body Conservation of Angular momentum角动量守
恒定律
1,Angular Momentum of a Rigid body
As shown in the figure,the
angular momentum of the rigid
body about the fixed axis with
an angular speed ? is defined as
IωL ? (5-22)
轴 ?
刚体对定轴的角动量
Note,based on the concept of angular momentum of a
particle with respect to a fixed point we learned in § 4-4,
a rigid body is treated as a collection of particles to lead
to (see the text)
I ωL ?
axis
ir
im?
iV
?
ωndIL,a
对刚体,are about the same axis,This implies
that
IωL ?
may be negative or positive depending on the choice( 选
择 ) of direction of rotation:
Counterclockwise,positive
Clockwise, negative
Take time derivative of equation (5-22),we have
??? I
t
I
t
I(
t
L ???
d
d
d
d
d
d )
Hence:
t
LM
d
d?
(5-23) 转动定理的另一种形式
It means that the net torque acting on a rigid body
equals the time rate of change of the body’s angular
momentum.
M
2,Conservation of Angular momentum角动
量守恒定律
If no external torque acts on the body or the
system,that is
0M ?
0dd ?tL
We have:
constant ? ?? I L 角动量守恒定理
For a system,(here i represents ith rigid body)
which means that if no external torque acting on a rigid
body or a system,the angular momentum will remains
constant,如果系统不受外力矩的作用, 系统的角动量将
保持不变 。
c o n s t a n t ?? ?? iii IL ?
(5-24)
Examples ① the spin of the earth; ② the example in
Fig.5-16; ③ the example in Fig.5-17 and Fig 5-19.
(机械能守恒吗?)
(2) For a system 对系统
Example 4-9 一个 质量为 M,半径为 R的水平均匀圆盘可绕过中心
的光滑竖直轴转动, 在盘缘上站着一个质量为 m的人 。 初时, 他们
以角速度 ?0作匀速转动, 后人从 盘缘走到圆盘的中心处, 求人在中
心时圆盘的角速度 。
解:选人 +盘为系统, 重力和压力对
转轴的力矩为零, 故系统角动量守恒 。
设人到达圆盘中心时圆盘的角速度为
?,则
?? 2022 2121 MRmRMR ?? )(
0
2 ??
M
mM ??
Example 4-10:如图所示, 一匀质细杆可绕通过上端与杆垂直
的水平光滑轴 o转动, 初始为静止悬挂, 现有一个小球自左方
水平打击细杆, 设小球与细杆之间为非弹性碰撞, 则在碰撞过
程中, 对细杆和小球这一系统:
( A) 只有机械能守恒;
( B) 只有动量守恒;
( C) 只有对转轴 o的角动量守恒;
( D) 机械能, 动量和角动量守恒;
小球
细杆
Example 4-11:光滑的水平桌面上, 有一长为 2L的, 质量为 m
的匀质细杆, 可绕其中点且垂直于杆竖直轴自由转动, 起初杆
静止, 有两个质量均为 m的小球, 各自沿桌面正对着杆的一端,
在垂直于杆长的方向上, 以相同的速率 V相向运动, 如图所示 。
当两小球同时与杆的两端点发生完全非弹性碰撞后, 就与杆粘
在一起转动, 求这一系统碰撞后的转动角速度 ( 杆的转动惯量
为 。
231 mLI ?
解:显然, 这一系统的角
动量守恒, 则:
?)( 22 231 mLmLm L Vm L V ???
L
V
7
6??
m L VmVr ??
Example 5,1 求质量为 m,长为 l的均匀细棒对下面转轴的转动惯量:
1)转轴通过棒的中心并和棒垂直; 2) 转轴通过棒的一端并和
棒垂直 。 ( 课本 P116)
解,1) 在棒上离轴 x处, 取一长度元 dx( 如图所示 ), 如果棒的
质量线密度为 ?,则长度元的质量为 dm=?dx,根据转动惯量计算公
式:
O
A
d xx
l
?? mrI d2
12dd
32
2
22
0
lxxmrI l
l
?? ??? ??
?
有
ml ??将 代入上式,得,2
0 12
1 mlI ?
( 2)当转轴通过棒的一端 A并与棒垂直时
X ’O’
2
3
0
22
3
1
3
'd' d
' ml
lxxmrII l
oA ????? ??
??
Example 5-2 求质量为 m、半径为 R、厚为 h的均质圆盘对通过盘心
并与盘面垂直的轴的转动惯量。(课本 P116)
解:如图所示, 将圆盘看成许多薄圆环组成 。 取任一半径为 r,宽
度为 dr的薄圆环, 此薄圆环的转动惯量为
mrI dd 2?
其中,dm为薄圆环的质量 。 以 ?表示圆
盘的质量体密度, 则有
rrhVm d2dd ??? ???
rhrI d2d 3 ???
???? ???? ?? hRdrhrII R 4
0
3
2
12d
hR
m
2?? ?
代入得
2
2
1 mRI ?