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UNIVERSITY PHYSICS 1. Chapter 5 Rotation of a Rigid Body 第五章 刚体的转动. §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. - PowerPoint PPT Presentation
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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 ig
nore the deformation( 形变) . In
general, the motion of the body is
more complicated than that discus
sed 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 sa
me velocity & acceleration. Ther
efore,the motion of any point of t
he body can represent the transl
ational motion of entire ( 整个) rigid body.
(1) Translation (平动) :
水平飞行
看成质点
All particles describe circula
r paths around a line called the
axis (轴) which may fixed o
r may changing its position dur
ing 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 arbitr
ary( 任意) point P in the body
whose circular motion can repres
ents the motion of the entire bod
y and use its circular plane as a r
eference to analyze (分析) th
e rotation of the rigid body arou
nd the fixed axis.
y
x0
V
r P
P
Q
轴
( 2 ) Kinematics of rotation of a
rigid body around a fixed axis is th
e same as the circular motion of a
particle that we discussed in chapt
er 1.
Representative( 代表) P :
① Angular position
② Angular displacement
Angular speed td
d
tt d
d
d
d 2Angular 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
s2
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 act
ing on a particle with respect to a fixed point ( in §4-
4), we can define the torque of a force acting on a rigi
d body.
Does a force make a rigid body rotate?
depending on its magnitude and acting position. Torque(力矩) ! F
F
FrM
FdFsinFrM 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
2F
1FF
P
F
轴
F
2F
1F
dFFsinrFM 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?)
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 colle
ction (集体) of particles w
ith same angular acceleratio
n and to apply (应用) Ne
wton’s second law to all part
icles , then add up( 相加) al
l 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 particleiF
fi —the internal force on the ith particleif
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 haveir
Thus (5-11) becomes
)( 2iirmM (5-12)
The sum (求和) on the right size is defined as the
rotational inertia (转动惯量) of the body with respec
t to the axis and labeled as I ( 转动惯量定义 )
2iirmI (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 sam
e form as that for acceleration o
f a particles, by which the rotati
on of a rigid body is governed
( 控制) . Therefore, it is often r
egarded 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 ) propert
y of the body but depends on the axis. Rotational inertial o
f 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 followi
ng example to show a
bove 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
3r2r
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。
( 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 )。2
2
1MRI
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 :
2
2
1MRIRT
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
mghahV
2
42
这时滑轮转动的角速度为Mm
mgh
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( )。 2
2
1MrI
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
rmg
2
( 5 )圆盘作匀加速转动,故有:t 0
其中r
V00 ,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 co
nnected to the body B by the light rope which is through
two uniform solid cylinder (圆柱体) with a mass m a
nd 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 cy
linders with .
mrI
mA 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 2mgTA
TB
TTBA 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
11mgT
( 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( )。2
3
1mLI
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:
cosLmgM2
1
( 3 )由转动定理,可得:
L
cosg
mL
cosLmg
类似的问题:
mg
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 kineti
c energy of a rigid body in rotational motion ? — Treat th
e body as a collection of particles.
iik VmE
iiiik rmVmE
(2) Take the sum of kinetic energies o
ver all the particles that make up the
body:
(1) Write the ith particle’s Kinetic energy as
axis
ir im
iV
ii rV i
iirmI 2
That is
(5-16)2k Iω
2
1E
( 转动动能 )
( 对比质点平动动能 ) 2k mV
2
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 rFscosFsFW t
Fig. 5-14
axisir
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
axisir
F
d
Instantaneously power
M
tM
WP
d
d
d
d
t(5-20)
(2) Kinetic energy theorem of rotation 转动动能定理
Rewrite the element work
dd MW
Fig. 5-14
axisir
F
dt
IIM d
d
d
d
d
tI
dIt
tI d
d
d
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 r
otation. This is kinetic theorem of rotation.
末动能 初动能
if II 2
1 f
d
i
IWkikf 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 MghE
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:
ttanconsI 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:
d2
1dd cosmgLMW
mgLmgLM2
1d
2
1dW 2
0
2
0
cos
两边积分得:
22
2
10
2
1
2
1 IImgL
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
ImgLM sin2
1外
mg
2
3
1mLI
sinL
g
2
3
( 2 )利用机械能守恒求角速度:取 o 点的重力势能为零,则
cosmgLImgL2
1
2
1
2
1 2
)cos( 12
3
L
g
mg
§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)
IωL axis
ir im
iV
ωndIL, a对刚体: are about the same axis. This implies t
hat
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
tI
t
I(
t
L
dd
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 0
d
d
t
LWe have:
constant IL 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 bo
dy or a system , the angular momentum will remains const
ant. 如果系统不受外力矩的作用,系统的角动量将保持不变。
constant iii IL (5-24)
Examples the spin of the earth; the example in ① ②Fig.5-16; the example in Fig.5-17 and Fig 5-19.③
(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
20
22
2
1
2
1MRmRMR )(
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 。2
3
1mLI
解:显然,这一系统的角动量守恒,则:
)( 22 23
1mLmLmLVmLV
L
V
7
6
mLVmVr
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:
OA
dxx
l
mrI d2
12dd
32
2
220
lxxmrI
l
l
有
ml 将 代入上式,得: 20 12
1mlI
( 2 )当转轴通过棒的一端 A并与棒垂直时
X ’O’
23
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
R
rdr
O
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
hRdrhrIIR 4
0
3
2
12d
hR
m2
代入得 2
2
1mRI