理论力学_章节答案 10章作业

ω = ω0 −αt
纯滚动满足 vC = rω
t = 5.575s vC = 4.732m/s
纯滚动后 mg
α
∑ maCx = Fx
maC = −FS
(1)
∑ JCα = M C
aC
mρ 2α = FS r (2) aC = rα (3)
FS = 0 aC = 0 α = 0
FS FN
ω = const vC = const
( ) LO = JOω =
JC
+
mR 2
ω
=
3 2
mR 2ω
逆时针
10-5
α
FOy
FOx
解:受力分析 运动分析：突然释放，ω = 0 O定轴转动
aC mg
aC = OC ⋅α = αe
刚体定轴转动动力学方程
J Oα = M O maCx = ∑ Fx
maCy = ∑ Fy
⎡1 ⎢⎣ 2
mr 2
+
永远 纯滚下去．
10-16
解２Ｏ为地面任意固定点．
mg
∑ L&O = M O = 0 LO = const
JCω0
轮对Ｏ的动量矩守恒 mv0 状态１（又滚又滑）
O(固定点) Fd FN
LO1 = LO2
ω = ρ 2ω0 + v0r ρ2 + r2
LO1 = mv0r + mρ 2ω0
状态２（纯滚动）
me2
⎥⎦⎤α
=
mg
sin
30°e
− meα cos 30° = FOx
− meα sin 30° = FOy − mg
⇒ α = 17.3rad/s2 ⇒ FOx = −8.99N ⇒ FOy = 34.0N
10-7
Μ,α1
解：轮1(顺时针)
O1
Ft Fn
J1α1 = M − M f 1 − Ft r1 (1)
轮2：平面运动 mv3
LO2 = (m2v2 )(1.5r )+ JC2ω2
计算略．
−M J2 +
f 2 − mgR mR 2
R
10-12
aDC
解:刚体作平面运动
C2
aC maC = F
aC = 4m/s2
C
C1
α
aC
∑ JCα = M C
JC
=
2⎢⎣⎡112
m l2 2
+
m 2
(CC1
)2
⎤ ⎥⎦
=
0.1
0.1α = F × 0.3
α = 36rad/s2
基点法
avD = avC + avDC
( ) aC
=
Fr2
r2 cosθ − r1 m ρ 2 + r22
10-16
mg
aC
α
vBC v0
vB
Fd
FN
Fd = FN f (4)
解1：刚体作平面运动
C为基点 vvB = vvC + vvBC
vBC = ω0r = 9m/s
vB = −v0 + vBC = 7m/s
∑ maCx = Fx maC = Fd
const
永远 纯滚下去．
10-2
ω
D
v3 v2 vD
v2 = 1.5rω v3 = 3rω
ω2 = v2 /(r / 2) = 3ω
轮3上的D点速度
vD = ω2r = 3ωr
曲柄：定轴转动 轮2：平面运动 轮3：瞬时平移
10-2
JC2ω2
m2 v2
轮3：瞬时平移
LO1 = (mv3 )(3r)
Mf1
轮2(逆时针)
J 2α2 = Ft r2-FT R-M f 2 (2)
Fn
Μf2
Ft
FT
α2
FT
物块
a ma = FT -mg (3)
运动学关系
mg a = Rα2 (4)
α1 = r2 = i (5) α 2 r1
a
=
Mi
− M f 1i J1i2 +
−M J2 +
f 2 − mgR mR 2
mR 2α
=
FT
R
(2)
aC = Rα (3)
FT
=
1 mg 3
10-15
解:刚体作平面运动
mg
α
∑ maCx = Fx maC = F cosθ − FS
(1)
aC
∑ JCα = M C
FS FN
mρ 2α = FS r2-Fr1 (2) aC = r2α (3)
FS
=
F
ρ2
cosθ + r1r2 ρ 2 + r22
(1)
∑ maCy = Fy
∑ JCα = M C
0 = −mg + FN (2)
mρ 2α = Fd r (3)
aC = fg = 0.49m/s2 α = fgr / ρ 2 = 5.1rad/s2
10-16
aC = fg = const
圆心匀加速直线运动
vC = v0 + aCt
α = fgr / ρ 2 = const 轮匀减速转动
aDC = CDα = 3.6 10
avC = 4iv
avDC = −aDC
1 10
iv
+
aDC
3 vj 10
avD = 0.4iv +10.8 vj m/s2
10-13
aC
FT
α
mg
aC
=
2g 3
解:刚体作平面运动
∑ maCy = Fy
ma = mg − FT (1)
∑ JCα = M C
1 2
LO2 = mvC r + mρ 2ω
vC = ωr
vC
=
r
ρ 2ω0 + v0r ρ2 +r2
=
4.732m/s2
10-16
mv − mv0 = Fdt = fmgt
ω = ρ 2ω0 + v0r = const ρ2 + r2
t = 5.575s
vC
=
r
ρ 2ω0 + v0r ρ2 + r2
=
R
Μf2
Fn Ft
也可以应用对定点的动量 矩定理.
dLO dt
=
MO
( ) ω2, α2
d dt
J2ω2 + mvR
Байду номын сангаас
= Ftr2 − mgR− M f 2
v,a J2α2 + maR = Ftr2 − mgR− M f 2 (6)
联立求解(1) (4) (5) (6)
mg
a
=
Mi
− M f 1i J1i2 +
第10章 动量矩定理作业
10-1
LO
=
J Oω
=
1 3
ml 2ω
逆时针
LO
=
J Oω
=
⎡1 ⎢ ⎢⎣12
ml 2
+
m⎜⎛ ⎝
l
⎟⎞
2
⎤
⎥ω
6 ⎠ ⎥⎦
=
1 9
ml 2ω
逆时针
10-1
LO
=
J Oω
=
⎡1 ⎢⎣12
m 2
(2a)2
+
1 3
m 2
(2a)2 ⎥⎦⎤ω
=
5 6
ma 2ω
逆时针
10-1