A particle moves on a frictionless horizontal (r,) plane under F=-k r/(2r3). For conjugate momenta p r,p , what is dp r/dt?

GATE 2023 PH – Question 42

Classical Mechanics · Hamiltonian and Hamilton's equations of motion · 2 marks · Multiple choice

A particle moves on a frictionless horizontal $(r,\theta)$ plane under $\mathbf F=-k\hat r/(2r^3)$. For conjugate momenta $p_r,p_\theta$, what is $dp_r/dt$?

$(p_\theta^2-2mk)/(2mr^3)$
$(2p_\theta^2-mk)/(mr^3)$
$(p_\theta^2-2mk)/(mr^3)$
$(2p_\theta^2-mk)/(2mr^3)$

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Correct answer: (D) $(2p_\theta^2-mk)/(2mr^3)$

Explanation

**Find the potential.** The force $\mathbf F=-\dfrac{k}{2r^3}\hat r$ is attractive and radial. It comes from a potential $U(r)$ with $F=-\dfrac{dU}{dr}$:
$$\frac{dU}{dr}=\frac{k}{2r^3}\;\Rightarrow\;U=-\frac{k}{4r^2}.$$

**Hamiltonian** in polar coordinates (mass $m$):
$$H=\frac{p_r^2}{2m}+\frac{p_\theta^2}{2mr^2}-\frac{k}{4r^2}.$$

**Hamilton's equation for $p_r$:**
$$\dot p_r=-\frac{\partial H}{\partial r}=\frac{p_\theta^2}{mr^3}-\frac{k}{2r^3}=\frac{2p_\theta^2-mk}{2mr^3}\quad(\text{option D}).$$