同濟(jì)大學(xué)理論力學(xué)孫杰習(xí)題解答練習(xí)冊3

理論力學(xué)習(xí)題解答練習(xí)冊動(dòng)力學(xué)

2P37

習(xí)題：1AOmm1m2aM動(dòng)量矩定理：wvvLO

=

JOw

+(

m1v)r+(m2v)r

21=(m

+m1

+m2)rv

SMO(Fi

)=

M–

m1gr

+

m2gr其中：v=

rw

JO

=

mr221dtdLO=SMO(Fi

)21(m

+m1

+m2)ra

=

M–

m1gr

+

m2gr∴M

=2.89kN·m對系統(tǒng)：?。篻=9.8m/s2dtdLO=SMO(Fi)研究對象m1gm2gP37

習(xí)題：2FⅠ

OGⅡ

OPGwFOxFOyFOxFOySMO(Fi

)=

Fr剛體定軸轉(zhuǎn)動(dòng)微分方程：JOa

=SMO

(Fi

)∴JOFra

=

JOPr=

質(zhì)心運(yùn)動(dòng)定理：SFx

=

0

FOx

=

0

SFy

=

0

FOy

-

G

-

F

=0∴FOy=

G

+

F=

G

+

Pa動(dòng)量矩定理：LO

=

JOw

+

vr

gP=

(JO

+

r2)w

gPSMO(Fi

)=

PrdtdLO=SMO(Fi

)(JO

+

r2)a

=

PrgPPra

=

JO

+

r2gPJOg

+

Pr2Pr=g

質(zhì)心運(yùn)動(dòng)定理：SFx

=

0

FOx

=

0

FOy

–

G

–

F

=

-

agPFOy=

G

+

JOg

+

Pr2PJOgv=

rw

wavSFy

=

-

agPJOP38

習(xí)題：4AaCmgFaC剛體平面運(yùn)動(dòng)微分方程：S

Fx

=

0

S

Fy

=maCSMC

(Fi

)=JCaaCx

=

0

mg

–

F

=

maC

aC

=raFr

=

JCa其中：JC

=

mr221∴3r2ga

=32aC

=g31F

=mg平面運(yùn)動(dòng)P39

習(xí)題：1qCAaaCmgqFTFFNxy剛體平面運(yùn)動(dòng)微分方程：S

Fx

=

0

S

Fy

=maCSMC

(Fi

)=JCa

FN

–

mgcosq

=

0

FN

=

mgcosqF

=fFN=fmgcosqmgsinq

–

FT

–

F

=

maCFT

=

mg(sinq

–

fcosq)

–

mra(FT

–

F

)r

=JCa

aC

=ra

其中：JC

=

mr221∴3r2ga

=(sinq

–

2fcosq)32=(sinq

–

2fcosq)gaC

=ra

FT

=

mra

+

fmgcosq21平面運(yùn)動(dòng)

圖!摩擦力：f?

f

：動(dòng)摩擦系數(shù)P40

習(xí)題：3AOhlSMA

(Fi

)=JAaA正方形板：A相對于質(zhì)心的動(dòng)量矩定理

mgaA

=0SMA

(Fi

)=0∴整體：相對于轉(zhuǎn)軸O的動(dòng)量矩定理FOxFOyFOxFOymgmgdtdLO=SMO(Fi

)SMO(Fi

)=

mgl+

mgl21LO

=

JOwO

+

mvAl=

mlvA

34=

mgl23mlaA

3431JO

=

ml

2wO

=lvA=

mgl23∴aA=

g89質(zhì)心運(yùn)動(dòng)定理：S

Fx

=

0

FOx

=

0

aAaCaC=

g169FOy

–

2mg=

–

3maCS

Fy

=

–

maA

–

maCFOy=

mg165wOS

Fy

=

–

2maEaEP40

習(xí)題：4qACBaCxaCyFAaAmga剛體平面運(yùn)動(dòng)微分方程：S

Fx

=maCx

S

Fy

=maCySMC

(Fi

)=JCa

運(yùn)動(dòng)學(xué)：aCx=aA

–

a2√32laCy=

aA21FA

=

m()aA

–

a2√32l21×

=(ml

2)a1212FA2lmg

–

FA

=

maA212√3FA

=

mla31aA

=

la94√3∴a

=

7l6√3gaA

=

g78FA

=

mg72√3=

12.12

rad/s2=

11.2

m/s2=

14.55

NaCAtaCA

=

at2l平面運(yùn)動(dòng)

補(bǔ)充方程

P41

習(xí)題：1OkD=(1

–

)k

√221W=k(d12

–

d22)

d2=2r

–

l0

d1=r

–

l0

√2=–

20.7J

ABdW

P41

習(xí)題：2ABPMSW

=

∫

Mdj

–

Fd

S

0

4p

=

∫

(bj

+

hj

2)dj

–

fP

×4pr

0

4p

=8bp2

+

hp

3

–

4

fPrp364Fd

=

f

PS

=4pr其中：M=

bj

+

hj

2

P41

習(xí)題：3OCCAwwJO

=

JO桿

+

JO盤31=

ml

221+

mR

2+

ml

22435=

ml

221T=

JOw

24835=

ml

2w

2vC=v0R–

rrw

=R–

rv0v0vC21T=

mvC2

+

JCw

221其中：

JC

=

mr

2=

2(R

–

r

)2

(r2

+

r2)mv02JOT

(b)鼓輪為mJA

＝

JC

+

mr2AvC、vO

w～P42

習(xí)題：5ABCPPvAvCPwhAC桿平面運(yùn)動(dòng)：瞬心P

121JP

=

ml

2+m()2

2l21TAC

=

JPw

2w

=hvA31=

ml

2ml2=

vA

26h2T=

2TACPl2=

vA

23gh221T=

JPw

2

×2T

：與開始時(shí)靜止無關(guān)vA

w～P43

習(xí)題：2wvAvCaAMOACBqs

T

–

T0

=

SW動(dòng)能定理：21T=mvA2

+

JOw2

+

mvC22121w

=r2vA=r1vCJO

=

mr

2

=

(

r22+

r2+r12)vA

2

2r22mSW

=

Mj

+

mgs

–

mg×sCsinq

代入：dtd：m(r22+r2+r12)

[

M

+

mg(r2

–r1sinq]r2aA

=

∴=

[+

mg(1

–

sinq)]

s

r2Mr2r1(

r22+

r2+r12)vA

2

2r22m–

T0

=

[+

mg(1

–

sinq)]s

r2Mr2r1(

r22+

r2+r12)2vAaA

2r22m–

0

=

[+

mg(1

–

sinq)]vA

r2Mr2r1動(dòng)能定理JO

=mr

2

TO

≠

0

求導(dǎo)

a

(對象)

sCjP44

習(xí)題：4ACBP2P3P1vBvAwAwCT2

–

T1

=

SW動(dòng)能定理：xSW

=

P3x

aAvB

=vA=RwA

=rwC

(摩擦力不作功)

T1

=

021T2

=m1vA2

+

JAwA2212121+

JCwC2

+

m3vB23P1

+P2

+2P3

4g=

vA2

12JA

=

m1R

2

12JC

=

m2r

2

代入：vA2

=

x

3P1

+P2

+2P3

4P3g∴2vAaA

=

vB

3P1

+P2

+2P3

4P3gvA

=√x

√3P1

+P2

+2P3

√4P3gdtd：aA

=3P1

+P2

+2P3

2P3gP45

習(xí)題：2Ak12l5lBCD5lwT2

+V2

=

T1

+V1

機(jī)械能守恒定律：取初瞬時(shí)為零勢能位置

T1

=

0

V1

=

021T2

=

JBw

2625=

ml

2w

2其中：=

ml

2325JB

=

(

5l

)2×22m3121V2=k[(

2l

)2

–

(

8l

)2]

=–

30kl

2

代入：mw2625=30k

∴w=√36k√5m機(jī)械能守恒定理T2、JB

～

md?求：角加速度?

(零位置)

P46

習(xí)題：3ABCDEm1m2m3vCvAwCwBT2

+

V2

=

T1

+

V1

機(jī)械能守恒定律：取初瞬時(shí)為零勢能位置

T1

=

0

V1

=

0sV2

=–

m1gs

21T2

=

m1vA2

+

JBwB2

+

JPwC22121PJB

=

m2r

2

其中：wC

=wB=rvC=RvA32JP

=

m3r

2

2m1R2+2m2r2+3m3r2

4r2=

vC2

代入：=m1gs

2m1R2+2m2r2+3m3r2

4r2vC2

=m1gvA

2m1R2+2m2r2+3m3r2

4r22vCaC

vC

=2r√s

√m1g√2m1R2+2m2r2+3m3r2

dtd：aC=

g

2m1R2+2m2r2+3m3r2

2m1RraCABCDEm1m2m3vCvAwCwBsPJB=m2r2wC=wB=rvC=RvA32JP

=

m3r

2

vC

=2r√s

√m1g√2m1R2+

2m2r2+

3m3r2

aC=

g

2m1R2+

2m2r2+

3m3r2

2m1Rr物塊A：Am1gFaCaA

=aCrR=

g

2m1R2+

2m2r2+

3m3r2

2m1R2m1g–F=m1aA∴F

=m1g

2m1R2+

2m2r2+

3m3r2

2m2r2+3m3r2

aAJB=m2r2aAP4俱6習(xí)題議：4qABm1m2vAvBwAT2–T1=SW動(dòng)能辭定理：21T2

=