7.3 Solutions
315
yield
2ω
2
(l sin θ + a sin φ) = g tan θ
(7)
2ω
2
(l sin θ +
4
3
a sin φ) = g tan φ
(8)
Equation (7) and (8) can be solved to obtain θ and φ.
7.20 Express the Cartesian coordinates in terms of plane polar coordinates (r , θ )
x = r cos θ, y = r sin θ
(1)
˙
x = ˙
r cos θ − r ˙
θ sin θ
(2)
˙
y = ˙
r sin θ + r ˙
θ cos θ
(3)
Square (2) and (3) and add
v
2
= ˙
x
2
+ ˙
y
2
= ˙
r
2
+ r
2 ˙
θ
2
(4)
T =
1
2
mv
2
=
1
2
m(˙ r
2
+ r
2 ˙
θ
2
)
(5)
V = U (r )
(6)
∴ L = T − V =
1
2
m(˙ r
2
+ r
2 ˙
θ
2
) − U (r ) (Lagrangian)
(7)
Generalized momenta:
p k =
∂ L
∂ ˙
q k
, p r =
∂ L
∂ ˙
r
= m ˙
r, p θ =
∂ L
∂ ˙
θ
= mr
2 ˙
θ
(8)
Hamiltonian:
H = T + V =
1
2
m(˙ r
2
+ r
2 ˙
θ
2
) + V (r )
(9)
Conservation of Energy: In general H may contain an explicit time dependence as in some forced systems. We shall therefore write H = H (q, p, t).
Then H varies with time for two reasons: first, because of its explicit dependence on t, second because the variable q and p are themselves functions of
time. Then the total time derivative of H is
dH
dt
=
∂ H
∂t
+
n
β=1
∂ H
∂q β
˙
q β +
n
β=1
∂ H
∂ p β
˙
p β
(10)