8E (Classical Dynamics)

Let

θ

be the angle from the downward vertical to the hoop's centre, measured at the suspension point, and let

ϕ

be the absolute angle from the downward vertical to the bead's radius from the hoop centre. The bead position is the sum of two vectors of length

a

. Including the hoop's moment of inertia about the pivot gives

T = \frac{m a ^{2}}{2} (3 \dot{θ}^{2} + \dot{ϕ}^{2} + 2 \dot{θ} \dot{ϕ} cos (θ - ϕ)),

while

V = - m g a (2 cos θ + cos ϕ) .

Thus one suitable Lagrangian is

L = \frac{m a ^{2}}{2} (3 \dot{θ}^{2} + \dot{ϕ}^{2} + 2 \dot{θ} \dot{ϕ} cos (θ - ϕ)) + m g a (2 cos θ + cos ϕ) .

b

To quadratic order about

θ = ϕ = 0

T_{2} = \frac{m a ^{2}}{2} (\dot{θ} \dot{ϕ}) (3111) (ϕ ˙ θ ˙),

and

V_{2} = \frac{m g a}{2} (θ ϕ) (2001) (ϕ θ) .

c = ω^{2} a / g

. The characteristic equation is

det (2 - 3 c - c - c 1 - c) = 2 c^{2} - 5 c + 2 = 0,

c = 1/2

2

ω_{1} = \frac{1}{2} \frac{g}{a}, ω_{2} = 2 \frac{g}{a} .

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