程序代写代做代考 C 已知:

已知:
Mf11 5,Mf12 0,Mf13 0
Mf 21 0,Mf 22 60,Mf 23 0 Mf31 0,Mf32 0,Mf33 70 J11 4,J12 0,J13 0
J21 0,J22 12,J23 0
J31 0,J32 0,J33 11
ms 40,mh 37,mm 5,mr 5,mb 3,mo 0,Bh 490,B0,g9.81
r0.1,rr1 0.3,rm2 0,rm3 0
ro1  0,ro2  0,ro3  0,rb1  0.7,rb2  0,rb3  0
r 0,r 0,r 0 hb1 hb2 hb3
KD0 2,KD 45,KL0 0,KL 132.5
KM1 60,KM2 50,KM3 100
K 20
Kq1 20,Kq2 60,Kq3 20 变量:
  r t 00000000 ,,,,,,,m1均为时间的函数,且 , , , ,
 0 0, 0 0, 0 0,rm1 0 0.04 函数关系:
ud  2 2 2  dt Vuvw
  1 coscos sin 0
d  w   v  arctan   cossin cos 0  dt  u 2 

d v  w  d t      a r c t a n  V 
sin 01 3 


coscos cossin sin R  sin cos 0 
wb  sincos sinsin cos 
  cos cos
coscos sinsinsin
cossin sinsincos ba 
cos sin
sin  sincos
S
DK K2V2
 coscos 
sinsin cossincos  D0 D
sincos cossinsin 
SF  K V 2 
LK KV2  L0 L
M K V2KV2 w1M1 q11
M K K V2K V2 w2M0M2 q22
M K V2KV2 w3M3 q33
X D YS R SF
  ba wb   Z  L 
 M1  Mw1 
M S R M   2 ba wb w2
M  M   3  w3
rr2 rsin rr3 rcos

0 XT
 r T T
  F0Y
h1  rrr 
1   BmgZ
1 h h2   r 
 h h  0  0
T
 h3   r T
   
F 0 S ucos

  r1 rrSr
2   ba r
m g u sin

2 r  bar2  r 
rr  0 T
 
 r3   rT
    FS ucos
  r1 rSr
 3  ba  r    usin
3  ba r2   r
r 
r3
  rT  m   m1
 0  uT F  F  0 S 0 
r r r  S r  j4 mbam2
 rT   m1
j4ba  m g 0 
  r    m3
 m    uT
   m
FS0 rSr
 5  ba    0 
5  ba m2  r 
   
  m3
  rT
 0 T
  o1 rrS r
   F  0 6
6o   ba o2   r 
 mg o
 o3   r T
 0 T F 0 
   b1 7 rrSr
  mgB

7bbab2 
q6  T6 M1sinM2 sincosM3 coscos q7  T7 0
qr T0 8m8
q2 
T 0 T2 0 T3 0
q3 
b b
r   b3
q 
1 1
q 
q4 T4 1
j q i
5 52 3
Qi 
7 j1
rj
Fjq Ti i18
j
T M
T M cosM sin


    
T  a h     
  
 
 0 0  
cos 
1 0 sin 
T
   0 sin sinsincoscos   0 cos sincos 
h 

    

  
  
0 sin 
sincos cossin   
     T  
0 sin 
cos cos    
   h 1   T  
  1 
s i n 
h h2  
0
 0 cos sincos 
     0 sin coscos
 h3    rrrT
  r1  r1 r1 aaS r 2S r S r
 r h  b a  r 2  b a  r 2  b a  r 2    r r r
   r3   r3   r3 
  0 0T       r h Sba 0Sba 0
   
 r1 T  0T     S 0
 ba    
 r  r2 h r3 
 rrrT   m1  m1 m1
  aaS r 2S r S r
m h 
 bam2 bam2 bam2
 I0 1 0
 m3   m3  M f 11
 m3 
M f 12 M f 13 

  
rrr
1 0
  f f21 f22 f23
0 0 D 0 00
0
001 MMM 
M M
  f31 f32 f33
M M  0 r r rT 
0 00  
 h3 h2  s1   rr0rrrSr rr0r
h  h3 h1 s s2   bahb2 s  s3 s1
r r 0 r r r r 0 h2 h1  s3hb3 s2 s1 
0h3h2  0 
0 r3 r2  0 
h h3 h1 0
r r3 r1 0
r2 r1  J06 0 J06 0 JJJ
h2 h1 
200 200
sfsf
0 0 5.5 0 0 5.5 
0
f  
r T hb1
0rr  s3 s2

65  g 7 7
ba r ba r  m m I
g g g 0
G q

g 0 44
0 a a/q i18 rr i
r2r2rrrr r2r2 rr rr  m2m3m1m2m1m3 r2 r3 r1r2 r1r3
Jmrrr2r2rr Jmrrr2r2rr
m mm1m2 m1 m3 m2m3 rr rr r2r2
r
r r1r2  r r
r1 r3 r r
r2r3  r2 r2 
m1m3 m2m3 m1 m2  g 1 1  m s I  M f
 r1r3
r2r3
r1 r2
g mrDT 12 sh f
g    m I  M   r 13 s fhs
g mrD 21 sh f
g S JST S J ST S J ST 22 ba s ba ba m ba ba f ba
g TS J J ST TS J ST 23 hbasfbarbamba

 g   S  J  J  S T   S J T S T 
g31mIM r
g44 mrI
g S JST 55 ba r ba
g TS JST 56 r ba r ba
g S JTST
sf
T
 11 12
g g g 0
0 0 0
0 0 0 0
0aTa/q hhi
 21
22 32
23 0
0 0 0 0
0 /q hh i
g g
0
 31
 0 0
i 00 

32 ba s f bah bambar
sh
13
0/q hh i
0 0 
0 g 0 65
0 /q rr i
 0 0 
0 0 0 g   a   a /  q   77 m  m i
恒等式:
Qi G i18 q
0gg 55 56
0 /q rr i
i
求:利用 Matlab 编程,采用 Runge-Kutta 方法,数值求解在t  3000s 时间范围内, , , ,,,随时间t的变化规律,并绘出图像。
要求提供所有代码,每条代码都要加注释。工期 7 天,结果正确才交付(有论文结果作比对)。