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Балаковский институт техники, технологии и управления

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Теоретическая механика

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d23_05

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23 ( 5) http://www.termeh.ru

, .

1 = (t),

1 ( ). 1.

: 1 – 1; 4 – ! 4 r4;

6 – 6 ;– " ;

0 – ,

;

− V0 1 .

# 6 4. . 1 1,4,6

( ).

: = 0.6 , r4 = 0.1`5 , m1 = 1 , m4 = 3 , m6 = 3 , = 16 / ,

( t = 0 ): 0 = 0, 0 = 8 .

$ II . %

1						,
,
	d
		¶T	− ¶T	+	¶ = 0,	(1)

	dt ¶y		¶y		¶

– " ; – " . & "

, " –

" , " 1,4 6: ' 4 6, , ,

ω 4 = ω 6 =		,	(2)
	0 .5 × cos ϕ

ϕ − 6 ( 4) .

																												http://www.termeh.ru
																				co s ϕ = 1 - ϕ 2 +							ϕ 4		- . . .			,
																	2									2 !	4 !
					ω 4				= ω 6					≈			2			.															(3)
					ω 4				= ω 6					≈						.															(3)

	( 4
					J	4		= 0 .5 m								4	r 2 .
						4										4		4
	( 6
					J	6		=			m 6				2	.
					J	6		=			1 2					.
											1 2
&" 1, !														,
					T1		= 0 .5 m										2	.
					T1		= 0 .5 m							1 y				.
&" 2 4, !																							,
T = 0 .5 J ω 2 = 0 .5 m r 2 ×				2 y				2		; T = 0 .5 J ω 2 = 0 .5 × m 6																		2		2 y		2	.
T = 0 .5 J ω 2 = 0 .5 m r 2 ×				2 y						; T = 0 .5 J ω 2 = 0 .5 × m 6																		×		2 y			.

4		4 4	4 4										6									6	6			1 2
4		4 4	4 4										6									6	6

) , "
									2											2			2 y		2		m 6		2		2 y			2

	T = T1 + T4 + T6 = 0 .5 m1 y + 0 .5 m 4 r4 × + 0 .5 ×																													× ;
																												1 2


												4 m					2				m 6			2
			T = m1						+			4 m		4 r4					+		m 6		×	y		.									(4)
			T = m1						+						2				+	3			×	2		.									(4)
																				3				2
	* " ,
																					* ,
, ,
:						= + .																													(5)
	% ",																														1
						= −G1 y = −m1gy.																													(6)

	= 0.5 ( f + λK )2 - 0.5 f 2 = 0.5 λK2 + cf λK = λK ( f + 0.5λK ).
f − ;
λK −		,			.
			λK = −0.25 sinϕ ≈ −0.25 ϕ ≈ −0.5y.
) ,			= -0.5cy ( f - 0.25y).		(7)
+(5) (6) (7) "
			= -m1gy - 0.5cy ( f - 0.25y).		(8)
) ,
¶	,		−m1g − 0.5c ( f − 0.25y) − 0.5c (−0.25) = 0		− m1g − 0.5cf = 0,
¶

¶ =0			f = - 2m1g .	=0
				=0


			c

%" (8) = -m gy - 0.5cy																						-
																			1

* (1):
	d ¶T		=	d		+	4m r2			+	m			m	+	4m r2			+	m	y;
			=		m	+		4	4	+	6	y =		m	+		4	4	+	6	y;
					1								1
					1			2			3		1				2			3
	dt	¶y		dt							3									3

2m1g - 0.25y c

¶¶Ty = 0;

= 0.125cy2 . (9)

¶ = 0.25c .

										http://www.termeh.ru
' (1)		+	4m4r42				m6
' (1)	m1	+			2	+			y + 0.25 y = 0
							3
							3

					0.25
	+				0.25					= 0.
	+			4m4r42				m6		= 0.
			+	4m4r42			+	m6
	m		+				+		3
		1				2			3

y = 0.

(10)

k " ,

y + k

k =

0.25

0.25×1600

400 » 12.06 −1.

4m r2

4×3×0.152

m +

2.75

4 4 +

0.62

- 1:

2 ×3.14

ν =

12.06

» 1.92 ;

T =

2π =

» 0.521 c.

2 ×3.14

12.06

2π

+ (10), 1:

= 1 cos kt + C2kt.

. 1

2 *

= -k 1 sin kt + kC2 cos kt

. + = (t) = (t) t =0

y0 = C1; 0 = kC2 . #, C1 = y0 ; C2 = 0 k . %" 1 2 = (t):

= 0 cos kt + ( 0 k)sin kt.

= 0 + 812.06sin12.06t » 0.663sin12.06t.

' = (t) ,

β ,

C1 = sin β ,

= cos β.

) = a sin(kt + β ), a =

+ C2 ,

β = arctg(C

)

β = arctg (k 02 0 ).

a =

02 + ( 0

k )2

* , β :

β = arctg (0 8) = 0

a =

0 + (8 12.06)2

≈ 0.663 ,

β = 0 .

: ν = 1.92 ;

T = 0.521 c;

= 0.663 ;

= 0.663sin12.06t.

0.663

sinϕ

≈ 2.21 > 1,

0.5×0.6

0.5

, ,

% ( t = 0 ): 0 = 0,

= 8 = 0.08 .

β = arctg (0 0.088) = 0

) a =

0 + (0.08 12.06)2

≈ 0.00663 ,

β = 0 .

sinϕ =

0.0066333 » 0.022111

» 0.022113 ,

osϕ » 0.99976,

0.5

0.5×0.6

sinϕ

≈ ϕ

6 , (1- cosϕ ) » 1- 0.9998 » 0.0002,

4 .

: ν = 1.92 ;

T = 0.521 c;

= 0.0066 = 6.6 ;

= 0.0066sin12.06t ( ).

http://www.termeh.ru

& "

, " –

" , " 1,4 6: ' 4 6, , ,

ω 4	= ω 6	=								.														(11)

				0 .5 × cos ϕ
				0 .5 × cos ϕ
									co s ϕ = 1 - ϕ 2		+	ϕ 4			- . . .
										2 !		4 !
, ϕ » sin ϕ =					y					co s ϕ » 1 -			ϕ		2	= 1 -					y 2			(12)
							,												2			.
			0 .5				,						2						2			.
			0 .5										2
) ,	ω 4 = ω 6				»							=				×				.				(13)
) ,	ω 4 = ω 6				»					y 2		=		0 .5 2 -					y 2	.				(13)
										y 2				0 .5 2 -					y 2
							0 .5 × 1			- 2
							0 .5 × 1			- 2


( 4																	J	4	= 0 .5 m				r 2 .
																		4					4	4
( 6																	J 6 =				m 6 2			.
( 6																	J 6 =				1 2			.
																					1 2
&" 1, !					,
						2		.
T1 = 0 .5 m1 y								.

&" 2 4, ! ,

T		= 0 .5 J		ω 2	= 0 .5 m	r 2	×		y ×				2	; T		= 0 .5 J		ω 2	= 0 .5 ×	m	6	2	×		y ×
	4		4												6		6
										2		2														2			2
				4		4 4		0 .5		2	- y	2						6		12				0 .5		2	-	y	2
								0 .5			- y									12				0 .5			-	y

) , "

												2						2		y ×				2						m 6 2
T = T1 +			T4 + T6 = 0 .5 m1 y										+ 0 .5 m 4 r4						×					+ 0 .5 ×
T = T1 +			T4 + T6 = 0 .5 m1 y										+ 0 .5 m 4 r4						×	2			2	+ 0 .5 ×
																				0 .5 - y											12
																				0 .5 - y											12
					2		2	+	m 6			4			y	2				3 × 3 × 0 .15			2	× 0 .6			2	+ 3 × 0 .6					4
				3m 4 r4				+	m 6						y					3 × 3 × 0 .15				× 0 .6				+ 3 × 0 .6
T » m1			+										×				=	1 +
				6 (0 .5							)	2								6 × (0 .5 ×										)	2
							2	-	y	2		2			2									0 .6		2		- y	2		2
								-	y						2									0 .6				- y
																		0 .0 76 95						2
																		0 .0 76 95						y
												T »			1 +							×			.
												T »			1 +			(0 .18 - y 2 )2				×		2	.
	* (1):
	d	¶T						0.07695												0.07695								0.3078y
	d	¶T			d			0.07695											+	0.07695								0.3078y
			=			1+							2 2 y				=	1	+		2 2 y +											2 3

	dt ¶y				dt			(0.18 - y						)					(	0.18 - y		)				(0.18 - y )

			y ×		2
×					;
×	0 .5 2 -			y 2	;
	0 .5 2 -			y 2
		2
×	y		;
×			;
	2
	2

y2 ;

2 .

(14)

¶T		0 .1 53 9			∂ = 0.25cy = 0.25×1600 y = 400 y.
	=		2	;
		(0 .18 - y 2 )3
¶y					¶

0.07695

0.1539 y

' (1)

+ 400 = 0.

(15)

(0.18 - y2 )2

(0.18 - y2 )3

/ ! "

, ! . ) ,

! .

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