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7 Lagrangian and Hamiltonian Mechanics
Solving (9) and (10)
¨
x =
g sin θ
1 − m cos 2 θ/(M + m)
=
(M + m)g sin θ
M + m sin
2
θ
(11)
¨
x
= −
g sin θ cos θ
(M + m)/m − cos 2 θ
= −
mg sin θ cos θ
M + m sin
2
θ
(12)
which are in agreement with the equations of prob. (2.14) derived from Newtonian mechanics.
7.9 T =
1
2
m( ˙
x
2
+ ω
2 x
2
)
(1)
because the velocity of the bead on the wire is at right angle to the linear velocity of the wire:
V = mgx sin ωt
( 2 )
because ωt = θ , where θ is the angle made by the wire with the horizontal at
time t, and x sin θ is the height above the horizontal position:
L =
1
2
m( ˙
x
2
+ ω
2 x
2
) − mgx sin ωt
(3)
∂ L
∂ ˙
x
= m ˙
x,
∂ L
∂ x
= mω
2 x − mg sin ωt
(4)
Lagrange’s equation
d
dt
∂ L
∂ ˙
x
−
∂ L
∂ x
(5)
then becomes
¨
x − ω
2 x + g sin ωt = 0 (equation of motion)
(6)
which has the solution
x = Ae
ωt
+ Be
−ωt
+
g
2ω 2 sin ωt
( 7 )
where A and B are constants of integration which are determined from initial
conditions.
At t = 0, x = 0 and ˙
x = 0
( 8 )
Further ˙
x = ω(Ae
ωt
− Be
−ωt
) +
g
2ω
cos ωt
( 9 )