 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2013/h/1/a/solution

Let

s

be distance along the rod from its fixed end. The position of that material point is

s n (t)

, where

n = (sin θ cos ϕ, sin θ sin ϕ, - cos θ)

s \overset{n}{˙}

, and direct differentiation gives

∣ \overset{n}{˙} ∣^{2} = \dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2} .

The uniform linear mass density is

M / a

. Integrating the elemental kinetic energy therefore proves, rather than assumes, the rod formula:

T = \frac{1}{2} \int_{0}^{a} s^{2} ∣ \overset{n}{˙} ∣^{2} \frac{M}{a} d s = \frac{M a ^{2}}{6} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) .

Equivalently, its moment of inertia about a perpendicular axis through the pivot is

M a^{2} /3

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