المساحة الحرة
01-08-2011, 01:11 PM
تحية طيبة..
الرجاء المساعده لوجود الإختبار غداً
الحل موجود ولكن مطلوب تفاصيل الحل..
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Problem 2. Double Atwood’s Machine
Let replace one of the weights in the simple Atwood machine by a second simple Atwood machine. Then we obtain the system which is known as a compound, or a double Atwood machine that is shown in Figure from the Set of Problems. For the motion of this system there are two degrees of freedom: one is the freedom of mass 1 (and the attached movable pulley) to move up and down about the fixed pulley, and the second one is the freedom of mass 2 (and the attached mass 3) to move up and down about the movable pulley. In general, to describe the configuration of the system we need to have 12 coordinates (3 for each mass m1, m2, m3 plus 3 for a movable pulley). Thus there must be 12-2=10 constraints. 8 of those constraints limits the motion of all the components of the machine to only a single direction. To formulate the remaining two constraints we assume for simplicity that the pulleys are massless, and their radii are small compared with the lengths of the constraining strings l and l0. Then the constraints can be written in a simplified form as
(xp + x1) − l = 0 and (2x1 + x2 + x3 ) − (2l + l0)= 0, (5)
where xi and xp are the vertical positions of the masses and movable pulley relative to the center of the fixed pulley (Note that the second constraint is coming from the formula (x2 −xp)+(x3 −xp)=l0and by using the first constraint in the form xp = l−x1).
Let choose the generalized coordinates x and x0, as shown in Figure. Then the
kinetic and potential energies as well as the resultant Lagrangian can be written down as follows:
12 1
0 2 10 2
T = 2 m1x˙
+ 2 m2(−x˙+ x˙ )
+ 2 m3 (−x˙− x˙ )
V = −m1gx − m2g(l − x + x0)− m3g(l − x + l0− x0)
L = 1m1 x˙2 + 1m2(−x˙+ x˙0)2+ 1m3(x˙+ x˙0)2+ (m1 − m2 − m3)gx + (m2 − m3)gx0+ const
2 2 2
(6)
Lagrange’s equations
,
=
( d ∂L ∂L
dt ∂x˙ ∂x
d ∂L ∂L
(7)
yield
dt ∂x˙ 0 =∂x0
m1x¨ + m2 (x¨ − x¨0)+ m3(x¨+ x¨0)= (m1 − m2 − m3 )g m2(−x¨+ x¨0)+ m3(x¨+ x¨0)= (m2 − m3 )g
(8)
The accelerations can be found from an algebraic solution of this system of equations as
x¨=
m1m3 − 4m2m3 + m1 m2
m1m3 + 4m2 m3 + m1 m2
g (9)
and
x¨0=
2m1(m2 − m3)
m1m3 + 4m2m3 + m1m2
g (10)
2
We can see if m1 = m and m2 = m3 = m
than x¨ = 0 and x¨0= 0.
الرجاء المساعده لوجود الإختبار غداً
الحل موجود ولكن مطلوب تفاصيل الحل..
.
.
Problem 2. Double Atwood’s Machine
Let replace one of the weights in the simple Atwood machine by a second simple Atwood machine. Then we obtain the system which is known as a compound, or a double Atwood machine that is shown in Figure from the Set of Problems. For the motion of this system there are two degrees of freedom: one is the freedom of mass 1 (and the attached movable pulley) to move up and down about the fixed pulley, and the second one is the freedom of mass 2 (and the attached mass 3) to move up and down about the movable pulley. In general, to describe the configuration of the system we need to have 12 coordinates (3 for each mass m1, m2, m3 plus 3 for a movable pulley). Thus there must be 12-2=10 constraints. 8 of those constraints limits the motion of all the components of the machine to only a single direction. To formulate the remaining two constraints we assume for simplicity that the pulleys are massless, and their radii are small compared with the lengths of the constraining strings l and l0. Then the constraints can be written in a simplified form as
(xp + x1) − l = 0 and (2x1 + x2 + x3 ) − (2l + l0)= 0, (5)
where xi and xp are the vertical positions of the masses and movable pulley relative to the center of the fixed pulley (Note that the second constraint is coming from the formula (x2 −xp)+(x3 −xp)=l0and by using the first constraint in the form xp = l−x1).
Let choose the generalized coordinates x and x0, as shown in Figure. Then the
kinetic and potential energies as well as the resultant Lagrangian can be written down as follows:
12 1
0 2 10 2
T = 2 m1x˙
+ 2 m2(−x˙+ x˙ )
+ 2 m3 (−x˙− x˙ )
V = −m1gx − m2g(l − x + x0)− m3g(l − x + l0− x0)
L = 1m1 x˙2 + 1m2(−x˙+ x˙0)2+ 1m3(x˙+ x˙0)2+ (m1 − m2 − m3)gx + (m2 − m3)gx0+ const
2 2 2
(6)
Lagrange’s equations
,
=
( d ∂L ∂L
dt ∂x˙ ∂x
d ∂L ∂L
(7)
yield
dt ∂x˙ 0 =∂x0
m1x¨ + m2 (x¨ − x¨0)+ m3(x¨+ x¨0)= (m1 − m2 − m3 )g m2(−x¨+ x¨0)+ m3(x¨+ x¨0)= (m2 − m3 )g
(8)
The accelerations can be found from an algebraic solution of this system of equations as
x¨=
m1m3 − 4m2m3 + m1 m2
m1m3 + 4m2 m3 + m1 m2
g (9)
and
x¨0=
2m1(m2 − m3)
m1m3 + 4m2m3 + m1m2
g (10)
2
We can see if m1 = m and m2 = m3 = m
than x¨ = 0 and x¨0= 0.