中国石油大学(华东)理论力学zuoye8综

y1
y: sinθva=vB cosθ +vA’B
θ& = − v A ' B /( l / cos θ ) = − 0 .559 rad/s x: 0=-vr sinθ+vA’Bcosθ+vB vr =7.715cm/s n τ 2.加速度 aan a=ar+ak+aB+aA’B+aA’B H +a τ
n y1: sinθaa=ar - aA’Bn
vA’B A ω O vA’B=16.29cm/s vB
n
ar vr
an a
τ aA’B aA’B l
√√ 0
?√ √√ 0
√√
ar=20.68cm /s2 B
x
?√
P
θ
C
aτe e.x.综 e.x.综-8 解:一、AB: A—基点,C—动点 :
ar ak vr D aA y n 1.速度 vC=vA +vCA vC=vCA=vA=0.5lω0 B aBA O ?√ √√ ?√ φ n ae vBA ve aAC 0 vCA n ωC=vC/R=0.5lω0/R nω τ aA n aBA vA A ωAB=vCA/CA=0.5ω0 n τ τ vA 2.加速度 aC=aAn+aA+aCA+aCA O1 θ C vC aC
l 2 l 3 ω 0 cot 60 ° + ω 02 cot 60 ° = l ω 02 , 2 4 4
τ ε AB = a CA / AC =
3 2 ω0 4
x: a -ak=0.5a - aτ e???√ √√ ?√ √√
n BA τ e
√√
√√
2
√√
√√
τ
n A
a =1.063lω0,
εO1 =
& θ& = −
3ω sin (θ + ωt ) O1 ω ω 3 cos θ + 3 sin (θ + ωt ) 4r θ & & 3θ 2 sin θ − 3 (ω + θ ) 2 cos(θ + ωt )
2.求
& & vC = vE = xE (θ = 30°, ωt = 0, θ = −ω / 4) = −1.335rω & & aC = aE = &&E (θ = 30°, ωt = 0,θ = −ω / 4,θ& = −0.2164ω 2 ) = −0.769rω 2 x
y e.x.综 e.x.综-5 解:A—动点,H—动系 : & v B = x , v A ' B = θ&l / cos θ 1.速度 va=vr+vB +vA’B ak va va=rω, √√ ?√ √√ ?√
AB = l 2 + x2 = 29.155cm,sinθ = x l = 0.5144 cosθ = , = 0.8575 AB AB
xE = 4r tan θ
y
.3.求aAr 3.求 3.
4r r cos ωt yA = − cosθ sin θ & & a Ar = &&A (θ = 30°, ωt = 0,θ = −ω / 4,θ& = −0.2164ω 2 ) = 0.2792rω 2 y
z x: -cos30ºaC=aAcos60º aC=0, εC=aC/R=0 τ vA x aAC τ y: 0=an cos60º+anCAcos60º -aCAsin60º A
?√ √√n 0 √√
n aCA,
?√
τ a CA =
二、O1D—动系,B—动点, A—基点: x: ve=vAcos30º=0.433lω0 1.速度 va=ve+vr=vA +vBA v 0.433lω0 ?√ ?√ √√ √√ ωO1 = e = = 0.375ω0 O1 B 2l tan 30° z: vr=vBA+vAsin30º=0.75lω0 n τ n τ n τ 2.加速度 aa=ar+ae+ae+ak=aA+aA+aBA+aBA
x1 e.x.8e.x.8-25 解:一速度分析：1. :一速度分析：1.EF—动系,O2—动点 y x M va=vr+ve ve=vO’2=-vr (1) n C 0 ?√ ?? aAE ar1 E 2.EF—动系,A—动点, O’2—基点 2. n vA’O’2 va1 aAO’2 va1=vr1+vO’2 +vA’O’2 AO’2=r/sinθ=2r √√ ?√ ?√ ?√ ω2=vA’O’2/AO’2 aa1O1 ω vr1 aE x: sin30ºva1=vAO’2=rω/2, , =ω/4 aE 4r vE y: cos30ºva=vO’2+vr1, (2) A θ vO’2 vA’E E—基点 va1=vr1+vE +vA’E ak1 √√ ?√ ? √ √√ vC=vE = ak ve x: sin30ºva1=cos30ºvE-vA’E, aτAO’2 1.335rω ae vr O2 A’E=4r/cosθ-2r=2.619r τ aAE x1: va1=cos30ºvr1-vA’Esin30º, vr1=1.533 rω ar F 由(2) vr=ve=-0.6667 rω 二加速度分:1. :1.EF—动系,O2—动点 二加速度分:1. ae=-ar-ak=aO’2 (3) aa=ar+ak +ae 2..EF—动系,A—动点, O’2—基点 2.. 0 ?√ √√ ? ?
ae 17 3lω0 = = 0.375 02 ω O1B 32
n τ n τ aa1=ar1+ak1+aO’2+ aAO’2+aAO’2=ar1+ak1-ar-ak+ aAO’2+aAO’2 τ √√ x: cos30ºaa1=ak1-ak-aA’O’2, ?√ √√ ?√ √√ √√ ?√ τ ε2=aτA’O’2/AO’2=-0.2164ω2 aA’O’2=-0.4329rω2 n τ E—基点 aa1=ar1+ak1+aE+ aAE+aAE √√ ?√ √ ?√ √√ √√ aE=0.7692rω2 x: cos30ºaa1=ak1-aEcos30º-aτ AE, y: -sin30ºaa1=ar1-sin30ºaE+an AE, ar1=-0.2792rω 2
方法2—给定法 O1O2 = r cot 30° = 3r 解:1.求ε2, ω2 :1.求 C 由余弦定理 3r sin θ = r cos(θ + ωt ) 两边微分，解得
θ& = −
M
x E A ωt
3 cos θ + 3 sin (θ + ωt ) O2 & = − ω , ε = θ& = − 3 3 ω 2 & θ = 30°, ωt = 0 → ω2 = θ 2 4 8 F .2.求vC, aC 2.求