Discussion
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 \(\vec{F} = -\frac{k}{2r^{3}}\hat{r}\) with \(k\) being a positive constant. If \(p_{r}\) and \(p_{\theta}\) are the generalised momenta corresponding to \(r\) and \(\theta\) respectively, then what is the value of \(\frac{dp_{r}}{dt}\)?
Answer: D
The result is \(\frac{dp_{r}}{dt} = \frac{2p_{\theta}^{2} - mk}{2mr^{3}}\).
The force \(-\frac{k}{2r^{3}}\hat{r}\) comes from the potential \(V = -\frac{k}{4r^{2}}\). The Lagrangian in polar coordinates is \(\mathcal{L} = \frac{m}{2}(\dot{r}^{2} + r^{2}\dot{\theta}^{2}) - V\), with \(p_{\theta} = mr^{2}\dot{\theta}\). Lagrange's equation for \(r\) gives
\[ \begin{aligned} \frac{dp_{r}}{dt} &= \frac{\partial \mathcal{L}}{\partial r} = mr\dot{\theta}^{2} - \frac{k}{2r^{3}} \\ &= \frac{p_{\theta}^{2}}{mr^{3}} - \frac{k}{2r^{3}} = \frac{2p_{\theta}^{2} - mk}{2mr^{3}} \end{aligned} \]
The first term is the centrifugal term and the second is the applied force. The options with \(p_{\theta}^{2} - 2mk\) double the force term, and \(\frac{2p_{\theta}^{2} - mk}{mr^{3}}\) doubles both terms.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development