решения курсовых (семестровых) работ из задачника / Динамика / d11_12

11 ( 12)

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( 1

2) 3. 1

ix1 = 0.20

2

, ix2 = 0.60 , = 5700 + 50t (H ), = 1500 , g = 9.81 2 , ω10 = 2 ,

t1 = 2 .

% ": ϕ1 = f (t), T , S.

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.1.

" " 1 2

", " 3 ! .

% !

&&$ '

,

,

,

3, ( 1 2 $ ( . 1).

1 G1,

,

$ !

X A , YA , S1 $

N1

2.

2

G2 ,

,

$ ! X B , YB ,

T , "! 3,

S2 $ N2

1.

3 G3 ,

T ′.

, $ .)

&&$ 1

(1)

Jx1ϕ1 = PR1 − S1r1.

)&&$ 2

2:

E

Jx2ϕ2 = M x2 ,

*M xE2 = MixE2 − " ! , 2,

2:

MixE2 = S2 R2 − TR2 − Mc .

( , " S2 , 2

' , T

2 , , $ .)

&&$ 2

(2)

Jx2ϕ2 = S2 R2 − TR2 − Mc .

&&$ 3

m3 y = T′ − G3.

(3)

(1), (2), (3) T′ = T S1 = S2

= S,

&$ ϕ1

(t), ϕ2

(t) ' y(t) − 1, 2 3

.

% &$ "!

ϕ1

R2

=

,

(4)

r1

ϕ2

(5)

y = ϕ2 R2.

, – : T , S, ϕ2.

ϕ1

(4): ϕ1 =

R2

ϕ2

(1):

r1

Jx1

R2

ϕ2

= PR1 − S1r1.

(6)

r

1

+ &&$ (2) , (T′ = T )

(3): = m3 y + G3.

, (5), !

= m3ϕ2 R2 + G3.

= S2 R2

- (2) Jx2ϕ2

− (m3ϕ2 R2 + G3 )R2 − Mc

2

(7)

(Jx2 + m3R2 )ϕ2 = S2 R2 − G3R2 − M c.

+ S(S = S1 = S2 )

(6) (7), (6) R2 , a (7) – r1 :

R2

ϕ2 = PR1R2 − S1r1R2 ,

Jx1 r

2

1

2

= S2r1R2 − G3r1R2 − Mcr1.

(Jx2 + m3R2 )r1ϕ2

) ",

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2

2

R2

+ (Jx2

+ m3R2 )r1

= R1R2 - G3r1R2 - Mcr1,

Jx1

r1

ϕ2

ϕ2 =

PR1R2 − G3r1R2 − Mcr1

.

(8)

R2

J

x1

2

+ (J

x2

+ m R2 )r

r

3

2 1

1

(8) 2 .

, ! (4), (8)

1:

PR R - G r R - M r

(5700 + 50t) × R1 - m3 gr1 - M c

r1

ϕ1 =

×

R

=

R

.

(9)

R

2

3 1

2

2

1

r

2

2

1

2

c

2

Jx1

2

+ (Jx2 + m3R2 )r1

1

Jx1

r1

2

r1

+ Jx2

+ m3r1

R2

*G3 = m3 g, g − .

$ 1 2 "x1 2:

J

x1

= m i2

;

J

x2

= m i

2 .

(10)

1 x1

2 x2

. &(10) (9), :

J

= m i2 = 100 × (0.20

)2

= 8 × 2 ;

J

= m i2

= 200 × 0.602

= 72 × 2.

x1

2

x2

1 x1

2 x2

(5700 + 50t)R1 - m3 gr1 - M c

r1

(5700 + 50t) × 0.30 - 600 × 9.81× 0.20 -1500 × 0.20

R2

ϕ1 =

=

0.60

;

2

0.20 2

r1

2

8 + 36 ×

+ 600 × 0.202

Jx1

+ Jx2

R2

+ m3r1

0.60

ϕ1 = 0.375t + 0.82 .

(11)

+' , :

t = 0

ϕ10 = 0;

ϕ10

= ω10 = 2 .

. " ϕ1 = 0.1875t

2

+ 0.82t

+ C1.

" ϕ = 0.0625t3 + 0.41t2

+ C t + C

.

1

1

2

= C2 ,

% ! ϕ1 = ϕ1(t) ϕ1 = ϕ1(t) t = 0 : ϕ10 = C1; ϕ10

C1 = 2 ;

C2 = 0.

, " 1

ϕ1 = 0.1875t

2

+ 0.82t + 2 .

(12)

+ 2 :

ϕ = 0.0625t3 + 0.41t2 + 2t ( ).

(13)

1

% ",

,

(3): = m3 y + G3

r1

+ g).

(14)

= m3 (ϕ2 R2 )

+ m3 g = m3

ϕ1R2

+ g

= m3 (r1ϕ1

R2

. t1 = 2 c, (11) ,

= m3 (r1 × (0.375t1 + 0.82) + g) = 600 × (0.20 × (0.375 × 2 + 0.82) + 9.81) = 6074.4 .

(15)

# (1):

PR1

- J x1ϕ1

J x1ϕ1 = PR1 - S1r1.

S = S1 =

,

t1 = 2 c, (11),

r1

S =

(5700 + 50t1) × R1 - Jx1 × (0.375t1 + 0.82)

=

(5700+ 50× 2) × 0.30 - 8× (0.375× 2 + 0.82)

= 8637.2

r1

0.20

: ϕ1 = 0.0625t3 + 0.41t2 + 2t ( );

= 6074 ;

S = 8637 .