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PHY6938-01: The Road to the Comps
Classical Mechanics – Quiz
Solution prepared by: Jorge Piekarewicz
September 22, 1999
(a) Given the components of the angular velocity of the rigid body along the body axes,
the Lagrangian for a symmetric rigid body under the influence of gravity becomes:
L =
I
θ
2
φ
2
sin
2
θ) +
I
z
φ cos θ +
ψ)
2
− M gl cos θ. (1)
(b) Since the Lagrangian is independent of φ, ψ, and t, namely,
∂L
∂φ
∂L
∂ψ
∂L
∂t
the canonical momenta conjugate to φ and ψ — as well as the total energy of the
system — are constants of the motion:
p ψ
∂L
ψ
= I
z
φ cos θ +
ψ) = I z
ω z
≡ Ia , (2)
p φ
∂L
φ
= I
φ sin
2
θ + I z
φ cos θ +
ψ) cos θ ≡ Ib , (3)
E =
I
θ
2
φ
2
sin
2
θ) +
I
z
φ cos θ +
ψ)
2
(c) To write an expression for
φ and
ψ in terms of θ one can use Eqs. (2) and (3) to obtain:
φ =
b − a cos θ
sin
2 θ
ψ =
Ia
I
z
(b − a cos θ) cos θ
sin
2 θ
(d) The above expressions may now be inserted into the energy equation which becomes
E
′
≡ E −
I
z
ω
2
z
= constant =
I
θ
2
(θ) , (6)
where we have introduced the “effective” potential through the definition:
Veff (θ) = M gl cos θ +
I
(b − a cos θ)
2
sin
2
θ
(e) For the particular case of a = b we can write the effective potential in terms of x ≡
cos
2 (θ/2) as:
V
eff
(x) = 2M glx +
Ia
2
(
1 − x
x
)
− M gl , (8)
Clearly, the effective potential supports circular orbits, namely, a solution to the equa-
tions of motion of the form θ(t) = θ 0
= constant. This solution is obtained by de-
manding:
V
′
eff
(x) = 2M gl −
Ia
2
2 x
2
= 0 , or x 0
= cos
2
(θ 0
√
Ia
2
4 M gl
Note that under these conditions (a = b) a real solution exists only for:
Ia
2
4 M gl
Moreover, by inspecting the effective potential one concludes that the circular orbit
is stable, V
′′
eff
(x 0
) = Ia
2 /x
3
0
0, so that the frequency of small oscillations may be
extracted from the energy equation [ξ ≡ (θ − θ 0 )]:
E
′′
≡ E
′
− V eff
(θ 0
I
ξ
2
V
′′
eff
(θ 0
)ξ
2
+... , so that (11)
ω s.o.
√
V
′′
eff
(θ 0
)/I. (12)
