 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iv/4/b/solution

Write

c = k \cdot e

. Since

s = ℓ (k - e)

, the total torque simplifies to

τ = (κ ℓ^{2} - m g h) e \land k .

e

gives

C \overset{n}{˙} = 0

: both torque terms are perpendicular to

e

, as are the other terms on the left. Hence

n = constant .

Taking the dot product of

\dot{L}_{Q} = τ

with the fixed vector

k

gives

\overset{p}{˙} = k \cdot \dot{L}_{Q} = 0, p = k \cdot L_{Q} = constant .

These are conservation of the axial spin and of the vertical component of angular momentum.

T = \frac{1}{2} ω \cdot L_{Q} = \frac{A}{2} ∣ \dot{e} ∣^{2} + \frac{C}{2} n^{2}, V = m g h c + \frac{κ}{2} ∣ s ∣^{2} = m g h c + κ ℓ^{2} (1 - c) .

\overset{n}{˙} = 0

, with

e \land \dot{e}

. The cross product identity

(e \land u) \cdot (e \land v) = u \cdot v - (e \cdot u) (e \cdot v)

gives

A \ddot{e} \cdot \dot{e} = (κ ℓ^{2} - m g h) k \cdot \dot{e} .

\dot{T} = (κ ℓ^{2} - m g h) \overset{c}{˙}, \dot{V} = (m g h - κ ℓ^{2}) \overset{c}{˙},

E = T + V = \frac{1}{2} ω \cdot L_{Q} + m g h k \cdot e + \frac{κ}{2} ∣ s ∣^{2} = constant .

The applied force is conservative even if

κ

is negative: its potential energy is

κ ∣ s ∣^{2} /2

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