A Dipole in an Electric Field — Question 3

Question 3

Free rotation converts dipole energy into angular kinetic energy

A rigid permanent dipole with moment magnitude p and moment of inertia I is pivoted at its center in a uniform field E. It is released from rest at $\theta_0=60^\circ$. Neglect friction; the pivot does no work.

1. Find its angular speed on first reaching alignment with the field.

2. Find the subsequent turning angle and determine whether it can reach the anti-aligned orientation without extra energy.

Question 3 - Solution

Free rotation converts dipole energy into angular kinetic energy

(a) Method and calculation

$$\tfrac12I\dot\theta^2-pE\cos\theta=-pE\cos\theta_0.$$$$|\dot\theta|_{\theta=0}=\sqrt{\frac{2pE(1-\cos60^\circ)}I}=\sqrt{\frac{pE}I}.$$

(b) Interpretation and further calculation

$$\dot\theta=0\Rightarrow\cos\theta=\cos60^\circ;\quad\theta_{\mathrm{next}}=-60^\circ.$$$$\mathcal E=-pE/2<U(\pi)=+pE,\quad\Delta\mathcal E_{\mathrm{barrier}}=3pE/2.$$It oscillates through alignment instead of settling there. Reaching the barrier with zero speed is a limiting separatrix motion; crossing it requires energy above the barrier.

Physical check

Without dissipation, the conserved mechanical energy fixes both turning angles and the accessible orientations.