质点的转动Chapter 10&11
Chapter 10 & 11
Rotational Motion
Angular Quantities
Torque
Angular Momentum
Conservation of Angular Momentum
质点的转动Chapter 10&11
10-1,Angular Quantities (P235-237)
r
s
Unit of?,radian( rad) 弧度?
o
sr
x
s
We measure the angular position? of this line
relative to a fixed direction (x).
1.Angular position? p235:
质点的转动Chapter 10&11
2.Angular Displacement(角位移 ):
● Whereis positive for counterclockwise（逆时针） rotation and negative for clockwise（顺时针）
rotation.
Always,? =?(t)
a body that rotates about a rotation axis,changing its
angular position from?1 to?2,undergoes an angular
displacement:
12
质点的转动Chapter 10&11
x
y
o
)( tP?
0P
r?
ttt d
dlim
0
3.Angular Velocity p236:
The magnitude of the body’s angular
velocity is the angular speed.
方向：满足右手定则,沿刚体转动方向右旋大拇指指向 。
Right-hand-rule,When the fingers of the right hand
are curled around the rotation axis and point in the
direction of the rotation,then the thumb points in the
direction of,(p241)?
质点的转动Chapter 10&11
ttt d
dl i m
0
4.Angular Acceleration p237:
The limit of the ratio as time interval
approaches zero.
质点的转动Chapter 10&11
rdtddtds
rv
or
5,Relation of Linear and Angular Variables p237
The distance along a circular arc:
Differentiating above equation with
respect to time (r again holds constant)
rdtddtdv
The linear speed (with r held constant):
rs m e a s u r e ),( r a d i a n
质点的转动Chapter 10&11
The tangential component at,
rt
The radial component an:
2
2
rv
r
va
n
质点的转动Chapter 10&11
ter
v
r?
te
v?
2
n
t
ra
ra
ta
na
n
2
t erera
a?
Example 10-2 page 238
质点的转动Chapter 10&11
10-2,Kinematics Equations for Uniformly Accelerated
Rotational Motion(p238)
Equations of Motion for Constant Linear Acceleration
are analogous to the ones of Constant Angular
Acceleration:
绕 定轴作匀变速转动质点 匀变速直线运动
at 0vv
2
2100 attxx v
)(2 0202 xxa vv
t 0
)(2 0202
2
2100 tt
质点的转动Chapter 10&11
1,Torque on an axis (对轴的力矩 ):
11-3 Angular Momentum of a particle (P277-278)
s i nRF?
where is called moment arm —— the distance?r
FR
Its SI units,m?N
The torque about a given axis is defined as:
or
from the axis of rotation
that is perpendicular to the
line of action of the force.
RF
质点的转动Chapter 10&11
The torque acting on a particle relative to a fixed
point O is a vector quantity defined as:
Fr
2,Torque about a point:
where is the position vector of a particle
relative to O.
r?
质点的转动Chapter 10&11
Example 11-1,page 277
质点的转动Chapter 10&11
vmP
The linear momentum of a particle:
3,Angular momentum of a Particle(p277)
The Angular momentum relative to a fixed point：
)( vrmPrl
Magnitude：
s i nm v rll
Direction,right-hand rule
l
m
O r?
v?rv
m v rll
A particle moves in a circle
whose relative to the central Ol?
质点的转动Chapter 10&11
v?
v mrprL v
r?
L?
L?
r?
x y
z
o
m
质量为 的质点以速度在空间运动，某时刻相对原点
O 的位矢为,质点相对于原点的角动量
m
r?
v?
s invrmL?大小的方向符合右手法则,L?
质点的角动量
L?
r?
p?
m
o
质点以角速度 作半径为 的圆运动，相对圆心的角动量
r
2mrL?
质点的转动Chapter 10&11
Supplementary Explanation:
(1) It is a vector quantity (J.s);
(2) It is not same relative to different point.
(3) The angular momentum of particle relative
to some point in the motion of straight line.
)( vrmPrl
m v dm v rlls i n?
大小：
O
m
r?
d
v
Direction:
质点的转动Chapter 10&11
Solution:
A particle moves in the xOy-plane,Write down
its angular momentum in terms of its
coordinates and its components of velocity,
vrmprl
)(or xyz yvxvml
, jyixr, jvivv yx
)( kyvkxvm xy
EXAMPLE:
质点的转动Chapter 10&11
In linear motion,we have Newton’s
second law for a single particle:,dt
pdF
n et
one can guess for its angular form,
dt
ld
n e t
The time rate of change of angular momentum of
a particle is equal to the net torque applied to it.
4,Relation between Angular Momentum and
Torque:(p278)
The rotational equivalent of N-II law for a particle.
质点的转动Chapter 10&11
Proof,From the definition
differentiating both sides of the equation we have
0
netF?
So,we have,
n e tFrdt
ld
dt vdrvdt rdmdt ld
or,
dt
ld
n t
)( vrml
)( dt vdmrvvm
The time rate of change of angular momentum of
a particle is equal to the net torque applied to it.
质点的转动Chapter 10&11
力矩的时间累积效应 冲量矩、角动量、
角动量定理,
力的时间累积效应 冲量、动量、动量定理,
质点的转动Chapter 10&11
d
d,
d
d
t
LF
t
p
p
t
r
t
prpr
tt
L
d
d
d
d)(
d
d
d
d
t
LM
d
d
作用于质点的合力对 参考点 O
的力矩,等于质点对该点 O 的 角动量 随时间的 变化率,
Fr
t
pr
t
L
d
d
d
d
0,
d
d p
t
r vv
prL
质点的角动量定理质点的转动Chapter 10&11
If the angular momenta of individual particles
change with time,then we have
.
1
,
1
n
i
in et
n
i
i
dt
ld
dt
Ld
dt
ld i
in e t
,? is the net torque acting on the ith particle.
n
i
in llllL
1
21
For a system of particles,total angular momentum
of the system is the (vector) sum of the angular
momentum of the individual particles,L
l?
11-4 Angular Momentum for a system P278
质点的转动Chapter 10&11
Total net torque acting on all particles of system
The time rate of change of the total angular
momentum of a particle’s system (or a rigid body)
equals resultant external torque on the system.
)(~ dtPdF
Angular Momentum Principle
(质点系角动量定理 )
dt
LdFr
n e tn e t
质点的转动Chapter 10&11
例 1 一半径为 R 的光滑圆环置于竖直平面内,一质量为 m 的小球穿在圆环上,并可在圆环上滑动,小球开始时静止于圆环上的点 A (该点在通过环心 O 的水平面上 ),
然后从 A 点开始下滑,设小球与圆环间的摩擦略去不计,求小球滑到点 B 时对环心 O 的角动量和角速度,
解 小球受重力和支持力作用,支持力的力矩为零,
重力矩垂直纸面向里由质点的角动量定理
c osm gRM?
t
Lm g R
d
dc o s
质点的转动Chapter 10&11
t
Lm g R
d
dc o s
tm gRL dc osd
考虑到
2,dd mRmRLt v
dco sd 32 gRmLL?
得由题设条件积分上式
0320 dc o sd gRmLLL
2123 )s in2(?gmRL? 21)s in2(
R
g?
2mRL
质点的转动Chapter 10&11
If net external torque acting on particle’s system
is zero,the total angular momentum of the system
remains conserved,合外力矩为零，质点系总角动量守恒 !
,0If CLn et
11-7 Conservation of Angular Momentum P284
质点所受对参考点 O 的合力矩为零时，质点对该参考点 O 的角动量为一恒矢量,
LM,0
恒矢量质点的角动量守恒定律质点的转动Chapter 10&11
—– each planet moves so that a line from the sun
to the planet sweeps out equal areas in equal
period of time行星对太阳的径矢在相等的时间内扫过相等的面积,
EXAMPLE 11-6:When the planet moves in an ellipse,
the area it sweeps out is:
)s i n)((21?dtvrdA?
s i n21 vrdtdA?
Here,the torque about the sun is zero,then
L=constant,So Kepler’ 2nd laws can be got:
Lmvrmm 2 1s i n2 1
质点的转动Chapter 10&11
v1 r
1 r2
O
v2
F?
A cord makes a ball,mass m,move in a circle (r1
v1) on the surface of frictionless table,(1) When
one pulls the cord downward slowly and makes
the radius equal to r2,v2=? (2) The work done by
the during r1?r2.F?
2211 rmvrmv?
)(
2
1
12 r
rvv?
12 vv?
F is central force,and v
r,L=mvr =constant
Solution,(1)
质点的转动Chapter 10&11
Homework,P266:9,15;
P267:20
P292,11
P292,16
Chapter 10 & 11
Rotational Motion
Angular Quantities
Torque
Angular Momentum
Conservation of Angular Momentum
质点的转动Chapter 10&11
10-1,Angular Quantities (P235-237)
r
s
Unit of?,radian( rad) 弧度?
o
sr
x
s
We measure the angular position? of this line
relative to a fixed direction (x).
1.Angular position? p235:
质点的转动Chapter 10&11
2.Angular Displacement(角位移 ):
● Whereis positive for counterclockwise（逆时针） rotation and negative for clockwise（顺时针）
rotation.
Always,? =?(t)
a body that rotates about a rotation axis,changing its
angular position from?1 to?2,undergoes an angular
displacement:
12
质点的转动Chapter 10&11
x
y
o
)( tP?
0P
r?
ttt d
dlim
0
3.Angular Velocity p236:
The magnitude of the body’s angular
velocity is the angular speed.
方向：满足右手定则,沿刚体转动方向右旋大拇指指向 。
Right-hand-rule,When the fingers of the right hand
are curled around the rotation axis and point in the
direction of the rotation,then the thumb points in the
direction of,(p241)?
质点的转动Chapter 10&11
ttt d
dl i m
0
4.Angular Acceleration p237:
The limit of the ratio as time interval
approaches zero.
质点的转动Chapter 10&11
rdtddtds
rv
or
5,Relation of Linear and Angular Variables p237
The distance along a circular arc:
Differentiating above equation with
respect to time (r again holds constant)
rdtddtdv
The linear speed (with r held constant):
rs m e a s u r e ),( r a d i a n
质点的转动Chapter 10&11
The tangential component at,
rt
The radial component an:
2
2
rv
r
va
n
质点的转动Chapter 10&11
ter
v
r?
te
v?
2
n
t
ra
ra
ta
na
n
2
t erera
a?
Example 10-2 page 238
质点的转动Chapter 10&11
10-2,Kinematics Equations for Uniformly Accelerated
Rotational Motion(p238)
Equations of Motion for Constant Linear Acceleration
are analogous to the ones of Constant Angular
Acceleration:
绕 定轴作匀变速转动质点 匀变速直线运动
at 0vv
2
2100 attxx v
)(2 0202 xxa vv
t 0
)(2 0202
2
2100 tt
质点的转动Chapter 10&11
1,Torque on an axis (对轴的力矩 ):
11-3 Angular Momentum of a particle (P277-278)
s i nRF?
where is called moment arm —— the distance?r
FR
Its SI units,m?N
The torque about a given axis is defined as:
or
from the axis of rotation
that is perpendicular to the
line of action of the force.
RF
质点的转动Chapter 10&11
The torque acting on a particle relative to a fixed
point O is a vector quantity defined as:
Fr
2,Torque about a point:
where is the position vector of a particle
relative to O.
r?
质点的转动Chapter 10&11
Example 11-1,page 277
质点的转动Chapter 10&11
vmP
The linear momentum of a particle:
3,Angular momentum of a Particle(p277)
The Angular momentum relative to a fixed point：
)( vrmPrl
Magnitude：
s i nm v rll
Direction,right-hand rule
l
m
O r?
v?rv
m v rll
A particle moves in a circle
whose relative to the central Ol?
质点的转动Chapter 10&11
v?
v mrprL v
r?
L?
L?
r?
x y
z
o
m
质量为 的质点以速度在空间运动，某时刻相对原点
O 的位矢为,质点相对于原点的角动量
m
r?
v?
s invrmL?大小的方向符合右手法则,L?
质点的角动量
L?
r?
p?
m
o
质点以角速度 作半径为 的圆运动，相对圆心的角动量
r
2mrL?
质点的转动Chapter 10&11
Supplementary Explanation:
(1) It is a vector quantity (J.s);
(2) It is not same relative to different point.
(3) The angular momentum of particle relative
to some point in the motion of straight line.
)( vrmPrl
m v dm v rlls i n?
大小：
O
m
r?
d
v
Direction:
质点的转动Chapter 10&11
Solution:
A particle moves in the xOy-plane,Write down
its angular momentum in terms of its
coordinates and its components of velocity,
vrmprl
)(or xyz yvxvml
, jyixr, jvivv yx
)( kyvkxvm xy
EXAMPLE:
质点的转动Chapter 10&11
In linear motion,we have Newton’s
second law for a single particle:,dt
n et
one can guess for its angular form,
dt
ld
n e t
The time rate of change of angular momentum of
a particle is equal to the net torque applied to it.
4,Relation between Angular Momentum and
Torque:(p278)
The rotational equivalent of N-II law for a particle.
质点的转动Chapter 10&11
Proof,From the definition
differentiating both sides of the equation we have
0
netF?
So,we have,
n e tFrdt
ld
dt vdrvdt rdmdt ld
or,
dt
ld
n t
)( vrml
)( dt vdmrvvm
The time rate of change of angular momentum of
a particle is equal to the net torque applied to it.
质点的转动Chapter 10&11
力矩的时间累积效应 冲量矩、角动量、
角动量定理,
力的时间累积效应 冲量、动量、动量定理,
质点的转动Chapter 10&11
d
d,
d
d
t
LF
t
p
p
t
r
t
prpr
tt
L
d
d
d
d)(
d
d
d
d
t
LM
d
d
作用于质点的合力对 参考点 O
的力矩,等于质点对该点 O 的 角动量 随时间的 变化率,
Fr
t
pr
t
L
d
d
d
d
0,
d
d p
t
r vv
prL
质点的角动量定理质点的转动Chapter 10&11
If the angular momenta of individual particles
change with time,then we have
.
1
,
1
n
i
in et
n
i
i
dt
ld
dt
Ld
dt
ld i
in e t
,? is the net torque acting on the ith particle.
n
i
in llllL
1
21
For a system of particles,total angular momentum
of the system is the (vector) sum of the angular
momentum of the individual particles,L
l?
11-4 Angular Momentum for a system P278
质点的转动Chapter 10&11
Total net torque acting on all particles of system
The time rate of change of the total angular
momentum of a particle’s system (or a rigid body)
equals resultant external torque on the system.
)(~ dtPdF
Angular Momentum Principle
(质点系角动量定理 )
dt
LdFr
n e tn e t
质点的转动Chapter 10&11
例 1 一半径为 R 的光滑圆环置于竖直平面内,一质量为 m 的小球穿在圆环上,并可在圆环上滑动,小球开始时静止于圆环上的点 A (该点在通过环心 O 的水平面上 ),
然后从 A 点开始下滑,设小球与圆环间的摩擦略去不计,求小球滑到点 B 时对环心 O 的角动量和角速度,
解 小球受重力和支持力作用,支持力的力矩为零,
重力矩垂直纸面向里由质点的角动量定理
c osm gRM?
t
Lm g R
d
dc o s
质点的转动Chapter 10&11
t
Lm g R
d
dc o s
tm gRL dc osd
考虑到
2,dd mRmRLt v
dco sd 32 gRmLL?
得由题设条件积分上式
0320 dc o sd gRmLLL
2123 )s in2(?gmRL? 21)s in2(
R
g?
2mRL
质点的转动Chapter 10&11
If net external torque acting on particle’s system
is zero,the total angular momentum of the system
remains conserved,合外力矩为零，质点系总角动量守恒 !
,0If CLn et
11-7 Conservation of Angular Momentum P284
质点所受对参考点 O 的合力矩为零时，质点对该参考点 O 的角动量为一恒矢量,
LM,0
恒矢量质点的角动量守恒定律质点的转动Chapter 10&11
—– each planet moves so that a line from the sun
to the planet sweeps out equal areas in equal
period of time行星对太阳的径矢在相等的时间内扫过相等的面积,
EXAMPLE 11-6:When the planet moves in an ellipse,
the area it sweeps out is:
)s i n)((21?dtvrdA?
s i n21 vrdtdA?
Here,the torque about the sun is zero,then
L=constant,So Kepler’ 2nd laws can be got:
Lmvrmm 2 1s i n2 1
质点的转动Chapter 10&11
v1 r
1 r2
O
v2
F?
A cord makes a ball,mass m,move in a circle (r1
v1) on the surface of frictionless table,(1) When
one pulls the cord downward slowly and makes
the radius equal to r2,v2=? (2) The work done by
the during r1?r2.F?
2211 rmvrmv?
)(
2
1
12 r
rvv?
12 vv?
F is central force,and v
r,L=mvr =constant
Solution,(1)
质点的转动Chapter 10&11
Homework,P266:9,15;
P267:20
P292,11
P292,16
