 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2009/h/h1/i/solution

For the first particle, the derivative of its angular momentum about the origin is its torque:

\frac{d}{d t} (m_{1} q_{1} \times \dot{q}_{1}) = q_{1} \times (- \frac{f ^{'} ( r )}{r} (q_{1} - q_{2})) = \frac{f ^{'} ( r )}{r} q_{1} \times q_{2} .

This need not vanish. For example take

f (r) = r^{2} /2

and instantaneous positions

q_{1} = (1, 0, 0)

q_{2} = (0, 1, 0)

; the torque is

(0, 0, 1)

. Therefore it is not conserved in general, although it is in special configurations or for a constant potential.

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