решения курсовых (семестровых) работ из задачника / Динамика / d23_01

,

= (t)

1

.

d

¶T

¶T

= - ¶ ;(1)

×

-

dt

¶

¶

¶

–

&

;

" –

&

.

2, 6 #

:

υ1 = ; ω2 = r2 ; ω6 = ;

υ6′ =

( 2 6 # :

2

= m × r2 2

;

6

= m × 2 3

2 2

6

T = 1 × m ×υ 2

=

1 × m × 2

T2 = 1 × 2 ×ω22 =

1 × m2 × r22 ×

2

= m2

× 2

;

2

1

2

1 1

2

1

;

2

2

2

4

r2

T =

1

× ×ω

2

=

1

×

m × 2

×

2

=

1

× m ×

2

T =

1

× m

×υ

2

=

1

× m ×

2

6

;

;

2

2

2

6

2

2

6

6

6

2

6

6′

6

6′

6

T = T + T + 2T + T

= 1 × m × 2 +

1 × m × 2

+ 2 ×

1 × m × 2 + 1 × m × 2

=

1

2

6

6′

2

1

4

2

6

6

2

6

= 1 × m + 1 × m + 5 × m

× 2;

2

1

2

2

3

6

'

& ,

, 1 $,

, :

= I + II

"

& ,

#

#,

I = -G1 × + 2 × G6 × h6 + G6 × h6′ ;

h6 ,h6′ -

h =

-

× cosϕ

=

×(1- cosϕ

)

6

6

6

2

2

2

h6′ = - × cosϕ6 = × (1- cosϕ6 )

ϕ

=

6

;

%

cosϕ = 1-ϕ 2 2! +ϕ 4 4!...

, :

ϕ 2

2

2

ϕ 2

2

2

h =

×

6 =

×

=

;

h

= ×

6 = ×

=

;

6

2 2 2 2

× 2

4 ×

6′

2

2 × 2

2 ×

= -G × + 2 × G

×

2

+ G

×

2

= -G × + G

×

2

I

2 ×

1

6

4 ×

6

1

6

"

&

% #

:

× ( f . + λ

)2

× f 2

II =

-

.

,

2

2

$,

λ

=

1 ×

;

2

×

f .

+

2

× f

2

1

2

.

2

2

II =

-

=

×

f . + f . ×

+

×

-

2

2

2

4

-

× f 2 .

=

1

× f . × × +

1

× × 2 ;

2

8

2

=

+

= -G × + G

×

2

+

1 × f

× × +

1

× × 2 ;

I

II

8

1

6

2

.

∂

= 0

,

-G +

1 × f

.

× = 0,( )

.

1

2

∂

$ ( )

,

i

= 0

O

2

( 3).

i

= -Po ×

+ G1 × = 0; P = × f

;

2

O

o

.

2

- × f . ×

+ G1 × = 0;

-G1 +

1

× f . × = 0;

2

2

,

& :

2

1

G

C

= G6

×

+

8

×

×

2 =

6 +

× 2 ;

8

' (1) :

d

×

¶T

=

m + 1 × m

+

5 × m

× ;

∂T = 0;

2

6

dt

¶

1

2

3

¶

¶

=

2 × G

+

C

,

¶

6

4

×

(1) :

m +

1 × m

2

+

5 × m

6

× +

2 × G6 +

C

× = 0;

1

2

3

4

2 × G6

+

C

+

4

× = 0;

m +

1 × m

2

+ 5

× m

6

1

2

3

2

&%% $, : + 2 × = 0,(2)

2 × 3 × 9,81

4000

2 × G6

+

C

+

=

4

=

0,5

4

=

m +

1 × m

2

+ 5 × m

6

1 + 1 × 2 + 5 × 3

1

2

3

2

3

=

117,72 + 1000

= 12,64 1

7

http://www.termeh.ru/

"

= 2π

= 2 × 3,14 = 0,497

12,64

* (2),

1:

= 1 × cos ×t + 2 × sin ×t .

)

C1 C2

:

= - × 1 × sin ×t + × 2 × cos ×t

. * = (t) = (t) t = 0 :

= C1 ; = × C2

C =

= 0,001( )

C2

=

=

5,0

= 0,396

( )

+,

;

12,64

.

1

" &

C1 C2 = (t) :

= 0,001× cos12,64 ×t + 0,396 × sin12,64 ×t

$ = (t)

,

, C1 = × sin β , C2 = × cos β .

= × sin ( ×t + β )

,

=

2

+ 2

,

1

2