理論力學(xué)作業(yè)：6-11、思考題6-7§6-2動(dòng)力學(xué)方程動(dòng)力學(xué)方程是研究剛體定點(diǎn)運(yùn)動(dòng)和一般運(yùn)動(dòng)的基本方程2010-5-151理論力學(xué)問(wèn)題的引出應(yīng)用背景2010-5-152理論力學(xué)問(wèn)題的引出問(wèn)題1：已知二

度陀螺的運(yùn)動(dòng)，如何求軸承C、D的約束力？問(wèn)題2：硬盤工作時(shí)，搬動(dòng)計(jì)算機(jī)，為什么易損毀硬盤？2010-5-153理論力學(xué)問(wèn)題的引出筆記本電腦2010-5-154硬盤自動(dòng)保護(hù)裝置理論力學(xué)問(wèn)題的引出計(jì)算機(jī)硬盤結(jié)構(gòu)示意圖2010-5-155定點(diǎn)運(yùn)動(dòng)剛體動(dòng)力學(xué)問(wèn)題：研究力與運(yùn)動(dòng)間的關(guān)系。問(wèn)題：用什么方法建立力與運(yùn)動(dòng)的關(guān)系？理論力學(xué)§6-2動(dòng)力學(xué)方程一、剛體定點(diǎn)運(yùn)動(dòng)的動(dòng)量矩Ox’y’z’為隨體參考系

Oxyz

為慣性參考系剛體對(duì)O點(diǎn)的動(dòng)量矩：M

MLo

r

vdm

r

(

r

)dm

[(r

r

)

(

r

)r

]dmy'xyzx'z'orM上述矢量在不同參考系中可分別表示為：若在Oxyz

參考系中：xi

,yi

,zi

(i

1,2,L);x

,y

,z若在

Ox’y’z’

參考系中：

i',

j',

k

';x

'

,y

'

,z

'是隨時(shí)間t變化的量是隨時(shí)間t變化的量r

xi

yj

zk

or

r

x'i'y'

j'z'k'xi

y

j

zk

or

x'i'y'

j'z'k'2010-5-156理論力學(xué)§6-2

動(dòng)力學(xué)方程Loi

ri

mivi

ri

mi

(

ri

)

mi

ri

(

ri

)

mi

[(ri

ri

)

(

ri

)ri

]y'yzx'z'x

orx'i'y

'

j'z

'k',ri

x'i

i'

y'i

j'

z'i

k'2

22iOi

i

i

iL

m

[(x'

y'

z'

(x'2

y'2

)

]k'i

i

z

'L

m

[(

y'2

z'2

)Oi

i

i

i

x'i

i

y

'

y'

z'

i

i

x

'y

'

y'i

z'i

z

'

]

j'

(x'2

z'2

)i

ii

i

x'i

i

z

'

x'

z'

]i'i

i

y

'

x'

y'

i

m

[x'

z'

i

m

[x'

y'

整個(gè)剛體對(duì)O點(diǎn)的動(dòng)量矩：LO

LOiy

'

i

z

'

i

ix'

ix'

y'

z' )r

])

(用隨體坐標(biāo)系描述定點(diǎn)運(yùn)動(dòng)剛體對(duì)固定點(diǎn)O

的動(dòng)量矩設(shè)：mi

是定點(diǎn)運(yùn)動(dòng)剛體上的一個(gè)質(zhì)點(diǎn)，其對(duì)O

點(diǎn)的動(dòng)量矩：2010-5-157理論力學(xué)§6-2

動(dòng)力學(xué)方程

y'

z'

(x'2

y'2

)

]k'i

i

y

'

i

i

z

'

(x'2

z'2

)

y'

z'

]

j'i

i

y

'

i

i

z

'

x'i

y'i

y

'

x'i

z'i

z

'

]i'L

m

[(

y'2

z'2

)Oi

i

i

i

x'i

i

x'i

i

x'i

m

[x'

z'

i

m

[x'

y'

(x'2

y'2

)

]k'i

i

z

'2

2ii

i

x'i

i

y

'i

iy

'x'

i

i

y

'i

i

z

'2

2

(x'

z'

)

y'i

z'i

z

'

]

j'

x'

y'

x'

z'

]i'

m

[(

y'

z'

)

m

[x'

z'

mi

[x'i

y'i

x'

y'

z'

LO

LOi

ii

iLO

[Jx

'x'

J

x

'y

'y

'

Jx'z

'z'

]i'[Jx

'y

'x'

J

y

'y'

J

y

'z

'z

'

]

j'[Jx'z

'x'

J

y

'z

'y

'

J

y

'z

'

]k'

LOx'i'LOy

'

j'LOz

'k'：定點(diǎn)運(yùn)動(dòng)剛體動(dòng)量矩的最簡(jiǎn)表達(dá)式2010-5-158理論力學(xué)§6-2

動(dòng)力學(xué)方程

Loz

'

Lox

'

LO

LOx'i'LOy'

j'LOz'k'

[

i

'

j

'

k

'

]

Loy

'

Jx'z'

x'

Jz'

z'

Jy'z'

y'

Jx'z'

Jx'y'Jy'

Jy'z'

x'y'oy'L

JLox'

Jx'剛體對(duì)O點(diǎn)的動(dòng)量矩：(*)Jz'

z'

y'z'

y'

y'J

JLoz'

Jx'y'

Jx'z'

x'

x'y'oy'

LOx

'

J

x

'x

'

J

x

'y

'y

'

J

x

'z

'z

'LOy'

Jx'y'x'

Jy'y'

Jy'z'z'

LOz'

Jx'z'x'

Jy'z'y'

Jy'z'

Jx'Loz'

Jx'z'

Jy'z'Lox'

L

[i'

j'k']L

[i'

j'k']

JO問(wèn)題：式（*）能否進(jìn)一步簡(jiǎn)化？提示：從數(shù)學(xué)角度和物理角度探討如何簡(jiǎn)化。2010-5-159理論力學(xué)§6-2

動(dòng)力學(xué)方程數(shù)學(xué)：對(duì)稱矩陣經(jīng)過(guò)正交變換后可化成對(duì)角矩陣。物理：如果x’y’z’是剛體的慣量主軸Jx'z

'

Jx'y'

Jx'y'0Lo

Jx'x'i'Jy'y'

j'Jz'z'k'

Jz'

z'

0

y'

0

x'

0J

y'0

0Jx'

Jz'

z'

Jy'z'

y'

[i'

j'k']

0

Jx'z'

x'

Jx'z'Jx'y'Jy'Jy'z'x'y'

Jx'L

[i'

j'k']

JO問(wèn)題：LO與是否共線，在什么情況下共線？結(jié)論：當(dāng)且僅當(dāng)剛體繞慣量主軸轉(zhuǎn)動(dòng)時(shí)，Lo與共線。2010-5-1510理論力學(xué)xCAB§6-2

動(dòng)力學(xué)方程，質(zhì)心在AB軸的中點(diǎn)，長(zhǎng)邊為a，短邊為b,AB=2L，求圖示瞬時(shí)均質(zhì)板對(duì)C點(diǎn)和AB軸的動(dòng)量矩。若板的角速度為常量，求板對(duì)質(zhì)心C和對(duì)x軸的動(dòng)量矩及其對(duì)時(shí)間的導(dǎo)數(shù)例：已知：m,a,b,x'y'yx

'y

'LC

Jx'x'i'

J

y

'y

'

j'

J

z

'z

'k'oLx'

cos,y'

sin

z'

01

1Jx'

mb2

,

J

y'

ma212

12x'i'y

'

j'z

'k'12CL

1

m(b2

cos

i'a2

sin

j')

(cos

i'sin

j')1、求板對(duì)質(zhì)心C的動(dòng)量矩2010-5-1511理論力學(xué)§6-212

1

m(a2

sin2

b2

cos2

)問(wèn)題：如何求板對(duì)AB軸的轉(zhuǎn)動(dòng)慣量？xCABx'y'動(dòng)力學(xué)方程yx

'y

'Lx

JxxLx

LC

ioL12CL

1

m(b2

cos

i'a2

sin

j')

i2、求板對(duì)

x

軸的動(dòng)量矩dtd

(i')

i'

sin

k'3、求板對(duì)質(zhì)心C的動(dòng)量矩對(duì)時(shí)間的導(dǎo)數(shù)12

dt

dtCL&

1

m[b2

cos

d

(

i')

a2

sin

d

(

j')]dt

d

(

j')

j'

cos

k'12

1

m

2

cos

sin

[a2

b2

]k'CL&2010-5-1512理論力學(xué)§6-2

動(dòng)力學(xué)方程xCABx'y'yx

'y

'oL12CL

1

m(b2

cos

i'a2

sin

j')

(cos

i'sin

j')若Lo與共線若上式成立，有：sin

cos

(a2

b2)

0L

0C驗(yàn)證：當(dāng)且僅當(dāng)剛體繞慣量主軸轉(zhuǎn)動(dòng)時(shí)，Lo與共線sin

cos

(a2

b2

)

012a2

sinsin12m2

b2

cos

cosm

2則：x

軸為慣量主軸12xyJ

m

sin

cos

(a2

b2

)

0x軸為慣量主軸的充分必要條件2010-5-1513理論力學(xué)§6-2

動(dòng)力學(xué)方程思考題

:若定軸轉(zhuǎn)動(dòng)剛體的動(dòng)量守恒、動(dòng)能守恒，則對(duì)質(zhì)心的動(dòng)量矩（矢量）：

。A：一定守恒；B：一定不守恒；C：不一定守恒。xCABx'y'yx

'y

'Lo12CL

1

m(b2

cos

i'a2

sin

j')

(cos

i'sin

j')2010-5-15142010-5-1515理論力學(xué)§6-2

動(dòng)力學(xué)方程例：求質(zhì)量為m半徑為R的均質(zhì)圓盤對(duì)O點(diǎn)的動(dòng)量矩。1

,

2Lo

Jx'x'i'

J

y'

y'

j'

Jz'z'k'2

4y

'z

'

x'J

1

mR2

,

J

J

1

mR2z'x'y'21z

'

1y'

2x'

2

sin

,

cos411

2

2

2o

sin

i'

cos

j'2

k')L

mR

(41

2

12

k

2

k')L

mR

(O另一種計(jì)算方O

ri

mivaiLO

ri

mivei

ri

mivri

ω1

ω2

x'i'y

'

j'z

'k'理論力學(xué)§6-2動(dòng)力學(xué)方程二、剛體定點(diǎn)運(yùn)動(dòng)的動(dòng)力學(xué)方程

MO

(F

)dtLO

Jx'x'i'

J

y

'y

'

j'

J

z

'z

'k'dLO

(e)

My

'

y

'J

x'&x'

(Jx

'

J

z

'

)x'z'y

'

J

z

'&z

'

(J

y

'

J

x'

)x'y

'

Mz

'

(J

z

'

J

y

'

)y

'z

'

Mx'

J

&xyzx'oz'y'注：隨體軸為慣量主軸2010-5-1516理論力學(xué)§6-2

動(dòng)力學(xué)方程例：已知：m,a,b,

，質(zhì)心C在AB軸的中點(diǎn)，長(zhǎng)邊為a，短邊為b,

AB=2L，

求圖示瞬時(shí)軸承A、B的約束力。x'y'xCFAyyaAz'A

bFBz'F

Bmg問(wèn)題1：如果板不轉(zhuǎn)動(dòng)，如何求約束力？C

C

Mma

FCL&rAy

By2F

F

1

mgFAz'

FBz'

0

F

0

MC

0問(wèn)題2：如果板轉(zhuǎn)動(dòng)，如何求約束力？FByQa

0C 0

FC

C

ML&r問(wèn)題：有幾個(gè)獨(dú)立的方程？2010-5-1517理論力學(xué)§6-2動(dòng)力學(xué)方程J

z'&z'

(J

y'

Jx'

)x'y'

M

z'

J

y'&y'

(Jx

'

Jz'

)x'z'

M

y'

J

x'&x'

(J

z'

J

y'

)y'z'

M

x'

24By

Ay1

m(b2

a2

)2

sin2

F

L

F

Lz'x'

y'

cos

,

sin

,

0x'y'xCBFAyFBymgyabAFAz'FBz'x'y'FAz'

L

cos

FBz'

L

cos

0

0

0

0

0

Fx

Fy

Fz&x'

0,&y'

0,&z'

0FAz'

L

sin

FBz'

L

sin

0

0

FAy

FBy

mg

0FAz'

FBz'

0AC=CB=L2010-5-1518理論力學(xué)§6-2

動(dòng)力學(xué)方程用動(dòng)量矩定理解釋附加動(dòng)反力產(chǎn)生的原因L&C

MC

(F

)x'y'xyCABx'y

'LC

1

ma2

sin

j'12對(duì)C點(diǎn)的動(dòng)量矩矢量的大小不變，并且始終位于板內(nèi)12LC

1

mb2

cos

i'CL&的方向垂直于屏幕向內(nèi)FAyFByA、B處約束力對(duì)C點(diǎn)之矩也應(yīng)垂直于屏幕向內(nèi)。由此得到：FAy>FBy2Az'Bz'

048Lm

(a

2

b2

)2

sin

2F

1

mg

By2F

F48LFAy

1

mg

m

(a2

b2

)2

sin

2即：FAy>FBy2010-5-1519理論力學(xué)§6-2

動(dòng)力學(xué)方程問(wèn)題1：如果圓盤不轉(zhuǎn)動(dòng)，如何求約束力？Cz

Dz2F

F

1

mgFCy

FDy

0

MO

0F

0FCzFDz例：求支架C,

D的約束力。已知：m，R

，CD=2L

1

,

21,2大小為常量maC

FL&O

MOQaC

0問(wèn)題2：如果圓盤轉(zhuǎn)動(dòng)，如何求約束力？ 0

FL&O

MO2010-5-1520理論力學(xué)§6-2

動(dòng)力學(xué)方程41y

'

J

mR221x'z

'J

mR2

,

Jx'

2

sin

,y'

2

cosz'x'y'21J

z'&z'

(J

y'

Jx'

)x'y'

M

z'

J

y'&y'

(Jx

'

Jz'

)x'z'

M

y'

J

x'&x'

(J

z'

J

y'

)y'z'

M

x'

z

'

1&z

'

0&x'

21

cos,&