Question

A particle of mass $m$ is free to move on a frictionless horizontal two-dimensional $(r, \theta)$ plane, and is acted upon by a force $\vec{F} = -\dfrac{k}{r^3}\hat{r}$ with $k$ being a positive constant. If $p_r$ and $p_\theta$ are the generalized momenta corresponding to $r$ and $\theta$ respectively, then what is the value of $\dfrac{dp_r}{dt}$?

• A. $\dfrac{p_\theta^2 - 2mk}{2mr^3}$
• B. $\dfrac{2p_\theta^2 - mk}{mr^3}$
• C. $\dfrac{p_\theta^2 - 2mk}{mr^3}$
• D. $\dfrac{2p_\theta^2 - mk}{2mr^3}$

Hint:

For a particle moving under a central force in polar coordinates, the rate of change of the radial momentum is related to the central force and angular momentum.

Answer 1

The Correct Option is D

- The generalized momentum for $r$ and $\theta$ are given as $p_r = m\dot{r}$ and $p_\theta = mr^2\dot{\theta}$.
- Using Lagrangian mechanics, the equation of motion for $r$ is given by \[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{r}} \right) - \frac{\partial L}{\partial r} = 0, \] where $L$ is the Lagrangian of the system.
- In this case, the force $\vec{F}$ is derived from the potential energy associated with the inverse cubic law. The solution for $\dfrac{dp_r}{dt}$ leads to the equation: \[ \dfrac{dp_r}{dt} = \dfrac{2p_\theta^2 - mk}{2mr^3}. \] Thus, the correct answer is (D).