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# Chapter IV 动量与角动量 Momentum and Angular Momentum - PowerPoint PPT Presentation

Chapter IV 动量与角动量 Momentum and Angular Momentum. 4.1 冲量与动量定理 Impulse and momentum theorem. 4.2 动量守恒定理 Theorem of conservation of momentum. * 火箭飞行原理. * 质心 Center of Mass. * 质心运动定理. 4.3 质点的角动量 Angular momentum of a particle. 4.4 角动量守恒定律 Conservation law of angular momentum.

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Chapter IV 动量与角动量

Momentum and Angular Momentum

4.1 冲量与动量定理

Impulse and momentum theorem

4.2 动量守恒定理

Theorem of conservation of momentum

* 火箭飞行原理

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* 质心 Center of Mass

* 质心运动定理

4.3 质点的角动量

Angular momentum of a particle

4.4 角动量守恒定律

Conservation law of angular momentum

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4.1 冲量与动量定理

momentum

According to Newton’s Law

We define a quantity: 冲量Impulse

I=FΔt to express the accumlation of forces over the time 力的时间积累

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Quantity of process

F 合外力 is a joint external force

The change of quantity of state

This is called impulse-momentum theorem

An impulse is a force applied over time

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F

Integral

form

t

Average

impulse

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Ex1.小球质量0.1kg，初速度为

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y

x

o

Ex2：4.1 一质量m=140g的垒球以v=40m/s的速率沿水平方向飞向击球手，被击后它以相同的速率沿=60º的仰角飞出，求垒球受棒平均打击力。设球、棒接触时间为Δt=1.2ms.

• solution

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F(max)

t

O

0.019s

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m

L

h

dm

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of group particles

F3

F2

F1

Internal force

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a粒子,质量m,初速v1，末速v2; 氧核,M,初速0，末速V。求轰击后速度的大小和方向。

v2

q

v1

b

V

y方向

0=mv2sinq-MVsinb

x方向

mv1=mv2cosq+MVcosb

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Ex4.3 一辆装煤车以v=3m/s的速率从煤斗下面经过， 每秒中落入车厢的煤为Δm=500kg。如果车厢速率不变，应使用多大的牵引力拉车厢？（忽略摩擦）

dm

v

F

m

x

Solution:

t时刻： mv+dm•0=mv

t+Δt时刻：mv+dm•v=(m+dm)v

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- = F Dt

x:

(m+Dm)v

mv

F =vDm/Dt

=3×500 N

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4.2 动量守恒定律

Conservation law of momentum

Newton’s Law

namely：

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Ex4.水银小球m竖直落在水平桌上，分成质量相同的三份,沿桌面运动,其中两等分的速度分别为v1和v2，且相互垂直地散开.试求第三等分的速度大小和方向。

XY面系统动量守恒

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x人地

x车地

Ex5.水平光滑铁轨上有一小车M，长l， 车端站有一人m，人和车原都不动。现人从车的一端走到另一端。问人和车各移动多少距离？

＋相对运动

x车地+x人地=l

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Sol

mv人地 - MV车地= 0

m v人地 dt = M V车地 dt

m x人地=M x车地

x人地 + x车地 = l

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4.6 角动量（动量矩）Angular momentum (moment of momentum)

moment

Angular momentum

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o

r

Fsina

o

F

a

r

F

rsina

1、力矩意义（在转动中）

r 是 质点与o 的连线

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r

o

p

mv

q

mvsinq

r

o

2、角动量意义（在转动中 in rotating）

L方向垂直于r,F相交的平面

p= mv 为动量，

o为惯性参照系中确定的点，

r为o到质点的矢径，

q为p与r的夹角

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4.7 角动量守恒定律

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r

v

q

r0

2、 sina=0

Ex3-15、作匀速直线运动的质点对任一固定

L=m r v sinq

= mvr0=C

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v

q

r

Ex6、开普勒第二定律：行星对太阳的矢径

(万有引力在矢径方向上）

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d

mv

4.16 一质量2200kg的汽车以60km/h的速度沿一平直公路开行。求汽车对公路一侧距公路50m的一点的角动量是多大？对公路上任一点的角动量又是多大？

O是定轴

O’是定轴, L=0

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m1

m3

m2

Ex7:O是定轴在杆a中点，m3以速度v0射入m2点并一起转动。 m1 =m2 =m3

Sol: 根据角动量守恒L始= L末

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4.20 用绳系一小球使之在光滑的水平面上做圆周运动，圆半径为r0，速率为v0。今缓慢地拉下绳的另一端，使圆半径缩至 r时，小球的速率v为多大？

M=0

L前= L后=C

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a为 r和 F的夹角

q为 r 和 v 的夹角

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Assignments:P107 4-2；4-11；4-17

End of Chapter IV

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