理论力学(胡运康)第七章作业答案

aC
aτ e
a
n a
aC 方向投影 :
n τ aa cos 60° + aτ sin 60 ° = a a e + aC
2 2 v 600 n aa = a = = 1200mm/s 2 OA 300
aen
α1
2 aτ = α ⋅ AB = 1000 3 mm/s e 1
aC = 2ω 1 vr = 2 × 3 × 300 = 1800mm/s 2
第7章 习题解答
1
7-1求轮边缘处水流对轮的vr
ve vr va
【解】 动点: M，动系: 轮
r r r va = ve + v r
va = 15 m s
nπ ve = R ⋅ = 6.28m s 30
x : va sin60 o = ve + vrx
⇒ v rx = 6.7 m s
y : − va cos60 o = 0 + vry
300
解 动点A，动系：BC
r r r va = ve + vr
ve = ω1 ⋅ AB = 3 OA = 300 3mm/s 3
va = ve / cos 30° = 600mm/s vr = va sin 30° = 300mm/s
12
aτ a ar
300
r n rτ r n r τ r r aa + aa = ae + ae + ar + aC
ω ω1 = ve / O1 D = 2
18
a
n ae
t e
r n r n rτ r r aa = ae + ae + ar + aC
n aa = ω2R
n aa
aC
ar
1 a = O1Dω = Rω 2 2 ω aC = 2ω1 vr = 2 Rω = Rω2 2
n e 2 1
τ n a cos 60 ° = − a cos 60 ° − a aC 方向投影 a e e sin 60° + aC 1 τ ⇒ ae = Rω 2 2 3
vr
n aa
va
300 300
ve
1 2va cos 30 = ve⇒ va = 0.116m s 2 va ⇒ ωOB = = 0.33 rad s l
aa = ar + ae
y
300
t aa
n t cos60° + aa cos30° = 0 向y轴投影: aa
n 2 aa = l ⋅ ωOB = 0.0385m/s2
3 2 α1 = a / O1 D = ω 12
τ e
19
aτ a = α 1 ⋅ AB = 5 × 0.15 = 0.75
ωBC
a
ar
α2
n e
a
n a
aC
aC = 2ω2 vr = 0.3456
aC 方向投影
4 τ3 a + a a = aτ e + aC 5 5
n a
⇒ aτ e = 0.5844
α 2 = a / DB = 2.338rad/s
17
α AB
ve = 2va = 0.4 m s
⇒ ω AB vB 0.4 = = = 2 rad s AB 0.2
ห้องสมุดไป่ตู้
aa = ar + aet + aen
a
ar
n e
aa a
300
t e
n t 向y轴投影: aa = a e cos30° + ae cos60°
n 2 ae = AB ⋅ ωAB = 0.8m/s2
15
7-20 AB=150 mm， ω1=2rad/s，α1 =5rad/s2。杆BC穿过套筒，套 筒绕D转。AD=200 mm，此时AD⊥AB。求杆BC的ωBC 、αBC 。
vr va
ωBC
5 4 3
解: 动点B，动系：套筒D
r r r va = ve + vr
ve
va = ω1 ⋅ AB = 2 × 0.15 = 0.3
n aa
8 2 4 a = ω GE GE = ω (1.5r ) = ω r 3 3
2
2
y
8 2 a = 3a = 3ω r 3
τ a n a
α GE
a 16 = = 3ω 2 = 3.079ω 2 1.5r 9
7
τ a
7-14 已知：匀vr =0.5m/s，匀ω =6 rad/s 。r1=300 mm。 r2=100 mm。求销子通过点B和点D时的绝对加速度。
10
r n r τ r n rτ r r aa + aa = ae + ae + ar + aC
ar 垂直方向y轴投影：
n n τ aa cos 30° + aτ sin 30 ° = − a sin 30 ° + a a e e cos 30° + aC n aa = ω 2 AB = 1.6m/s 2 2 aτ = α ⋅ AB = 2 m/s a
D
θ
O
α AB
n a 向 r 方向投影，
aC aτ e
4 τ 3 n 4 n 3 τ aa − aa = ae + ae + arn − aC 5 5 5 5 n 2 aa = AB ⋅ ω AB = 0.1× 0.122 = 0.144
a
n a
A
θ θ
n ae
n ae = OA⋅ ω 2 = 0.04× 42 = 0.64
arn
a
τ a
a
τ r
aτ e = OA ⋅ α = 0.04 × 2 = 0.08 2 2 v 0 . 2 arn = r = = 0.8 r 0.05
aC = 2ω e vr = 2 × 4 × 0.2 = 1.6
2 ⇒ aτ = − 0 . 192 m/s a
2 α AB = a τ a / AB = – 0.192/0.1= –1.92 rad/s
n ae = ω12 ⋅ BC = 0.7111m/s2
y
aC
n a aτ a e
aτ a
ar
300
α1 ω1
n ae
aC = 2ω1 ⋅ vr = 2.844m/s2
2 ⇒ aτ = − 0 . 1192 m/s e
aτ α1 = e = −0.3974rad/s 2 BC
11
7-18 OA=300 mm，ω1=3rad/s，α1 =10rad/s2，求：aA。 va vr ve
θ
ω AB
r r r va = ve + v r
ve = OA ⋅ ω = 0.04 × 4 = 0.16m/s 3 va = ve = 0.12 m/s 4
ω AB
ve
A
θ
vr
va = = 0.12 / 0.1 = 1.2 rad/s AB
va
vr =
5 ve = 0.2 m/s 4
14
r n rτ r n rτ r n rτ r aa + aa = ae + ae + ar + ar + aC
3 ve = va = 0.18 5
⇒ ωBC = ve 0.18 = = 0.72rad/s BC 0.25
4 vr = va = 0.24 5
16
t ae t aa
r n r τ r n rτ r r aa + aa = ae + ae + ar + aC
n aa = ABω12 = 0.15 × 2 2 = 0.6
E
t ⇒ ae = −0.386m/s 2⇒ α AB
t aB = = −1.93m/s2 AB 4
7-8 尖劈匀v=0.2 m/s ,ϕ=300，l=200 3 mm。求θ = ϕ 时，ωΟB、 αOB。 （平动） 【解】 动点: B，动系:尖劈
ωOB
r r r va = ve + v r
o
αOB
⇒ v ry = −7.5 m s
r v r vr = 6.7 i − 7.5 j m ⋅ s -1
2
7-2 OA匀ω 1= 2rad/s，CB=200mm， 求图示位置CB的ωCB。 【解】 动点: B，动系: OA
va ve
ωBC
r r r va = vr + ve
vr
ve = 400 × ω1 = 800 mm s
r r r va = v e + v r
ve va
600
ve = vB = rω va = 2ve = 2 rω
⇒ ωGE va 4 = = ω 1.5r 3
6
vr
ωGE
r n rτ r r aa + aa = ae + ar
向y轴投影: aD
aτ a ae
α GE
ar
n a
n aa cos 30° − aτ a sin 30° = 0
ve va = = 1600mm s sin30
⇒ ωCB
va = = 8 rad s 200
3
7-7 v=0. 2m/s， a= 0.5m/s2 ，AB=CD=200mm，求此时ωAB、αAB。 【解】 动点: E，动系: BC杆
ω AB
vB vr
va
ve
r r r va = ve + v r
（平动）
va ve
300
vr
解 动点B，动系：CDE