理论力学a-版本刚体平面运动planar motion of rigid

BS§9-4

用基点法求平面图形内各点的加速度BAaBAan

AB

2

AB

aABAaaB

aAaBAnBAα

ASB平面图形内各点的加速度BABAa

a

a

aBAaAaBAaBaAanaBAθBA

ABBAaA

a

a

anBA采用平移系、均为绝对量例12OaO

vOaCxCaCyaByaBx

BaOvOOCCv

2ORa

BOOv

2 O

)

2Ra

(a

a

2O

a

nBCOaCCOOBBOO

a

annBOaO

vOaOaCxCaCyaByaBx

BaOaCvOaＢOCC讨论一：若轮心O速度为匀速，则轮缘上各点加速度有什么特点？BOB

a

n

O二：R

vOR

aOOBvOaOCAa

nBatB三：行星轮机构如图，系杆匀角速度为，求行星轮上B点的加速度。OAr2PA瞬心vA

(r1

r2)vAr1(r1

r2

)A

1

2AvAa

ABAa

a

a

n

aBA

BAAanBAr1aAr1ddt

A

a

an

2

(r

r

)A

0BA2Aa

nBABAa

a

an

BAA

BA

BA

0aA四：若系杆角速度，角加速度，则aB?OAr2r1AvAAa

nAannBAaAvA

(r1

r2)vAr1(r1

r2

)A

2A1

2an

(r

r

)Ar1A

r1

r2APddtAat12

(r

r

)BAatAa

tAa

tBa

a

n

a

tA

A

a

n

a

tBA

BAPBA2A

BABAA

BAa

naBy

an

an

A

BABxA

BAa

at

at

axy例9-13AO’BOABACrO

vA2BABv

BC

(2

2)r

rOO

'

BO

'

B2

r

vB

rO

O2CvBvAABO’BAO’B

OAB

(1

2)rnBABABABa

anαABaAaAαaBBAAa

nn

BAABBannBBaaxyBBABABAa

a

n

a

a

a

nBO2

O

2222OaB

O’BαAaAaABABABanBaαB2Oa

A

r22OBa

nr

12BABAan

a

a12nBAABa

a

a22r2

r2

O

O22Or1

2

ABaBAAB

2O2AB

AO’BαaABAnABa

nBaαxy

aABBABABAa

a

n

a

a

a

n202

vOBABv

BC

'

20

2

2vO

24(cm/s)例14vOOO

aABθO

vO

3

(rad/s)r

2O

aOvA

AC

O 2

rO

2vO2

(rad/s)5ABOAC

'

20

vA

2

v

3C’COαOvAAvBB

9

(rad/s2

)

ABr

4aOAαO(*)Axa

aAy

a

a

an

O

AO

AOa

a

anAxOAOxOyO2O

a

r

18

18

36(cm/s2

)Aya

aAOO

r

1

8

(

c

m

/

s

2

)aOAOnAOOθBABAxAy

3

2

(rad/s)5a

36(cm/s2

)a

18

(cm

/s2

)a

a

a

a

anB

Ax

Ay

BA

BA122AxAyBAa

B(

a

a

)

a

nAy2AB 2

AB

AxaB

aAx

aAy 2

an

a

aBAaB

74.36(cm/s

2

)BBAaanαABA

aAxaAyaAxaAyBBAAann

BAAaaBB已知曲柄OA匀转速绕O轴转动，圆柱沿水平地面作无滑动的滚动，圆柱与DE间也没有滑动。已知OA=r，AB=2r，圆柱半径为0.5r。求该曲柄OA处于铅直位置瞬时及OAB成一直线时物体DE的速度vDE和加速度aDE。OvAvBvvDErr已知OA

=r，AB=2r，圆柱直径dB=r。求(1)OA铅直时物体DE的速度vDE和加速度aDE。PDE

K已知 ，OA

=

r，AB=2r，圆柱半径0.5r。求(1)

OA铅直时物体DE的速度vDE和加速度aDE。OaBraABAnAaAaDE2 3

r23？BAaaaaBABA已求得KaBaBaKBxyaDE3

r23KBan圆柱的角加速度由基点法已知OA

=r，AB=2r，圆柱半径0.5r。求(2)OAB成一直线时物体DE的速度vDE和加速度aDE。Ov

AaABAanBAaBDE2aB

3r2avBAvDE

BABABa

aAaa

n§9-5运动学综合应用举例轮C在水平面纯滚动，轮缘上固定销钉B，此销钉在摇杆OA的槽内滑动，并带动摇杆绕O轴转动。已知轮的半径为r，在图示位置时AO是轮的切线，轮心匀速运动，速度为v，摇杆与水平面的交角θ=60°。求摇杆的角速度和角加速度。vrvev60O

ve

vOB

2rvaBv60再求摇杆的角加速度

0aCaeea

nO2rn

2e

O3v24ra

OB

CO

r3v22ra

2

v

varr将(*)式向轴投影，得3

v26

rO

而得2r已求得

v

，

v

3

vaaB例15解法1：AClv1lABAC

v

vABOBC1vrvaveOACB0

a

a

n

a

a

(*)e

erCe4l

3v2rv

v3v2aC

2evr

4l4lae

3

v

2eC0

a

a/3

AB

ae

3

v

2

2lAO

4l3

3

v

24

l

2ααABABeavrnaeaCarr解法2：OACBlxyx

A

l

c

o

t

Als

i

n

2

v

x

l

v

sin

2

l

v

sin

2v2l2sin2

sin

24

lAB

3

v3

v

28

l

2

AB

3例16O解:OO1CO

Dve

0.25

3.6.18

(rad/s)1vAvrvDave1O

COa

a

a

aBABAaAaABABAABaBBA0anABAa

2OA

5

(

/s2O(

)a

aa(

2

)O

DaAa

DDAAαABaeea

naCaraAaCa

a

a

a

aa

e

e

r

CaD

a110.07

e

a

τO

D

OD

5.457

7

7

.

9

5

7

(

r

a

d

/

s

2

)nerCa

aAae

a

a30°ADAeCaτDA

轴投影3

12

2a

aτ

aττ1aDAae

a

A

2

aC2ADAO1D

r

323aτ12a

2

v2

1

5

3

12.759

2

6

.

1

8

6

0.25

5.457

(

m

/

s

2

)2O1DDAeaea

naCar

OD30°aA例17vA

10aA

103

mm/s,3

mm/s2。ABED

aBvA

C

aAvB求速度解：BDaBvA

C

aAAvBEv

B

=

v

A

+

v

B’A

+

v

r(

3

)v

B

=

v

A

+

v

B’A

+

v

r(

3

)AB’B30°vB

'

A

30 3

mm/s0.866 (

r

a

d

/

s

)2A

B

A

E

v

B

'

A3

vA

vBvB’AvrAEvAABCAE

ra

2

v

103

17.32mm/s2(

5

)B

'Aa

a

a

a

nB

'

A B

'

AB

'

A2AEan

AB

45

mm/s2aAaAarB’aa

=

aBa

CαAE求加速度anaB

'AABB'AABAa

a

aB

'

A

a

n

a

aB

'

A

r

C

和

轴投影30°ABaAaAarB’AB

Aa

a

=

a

Ba

CαAEBa

cos

30

a

sin

30

aτ

aA

B'A

Ca

sin

30