Discussion
A particle of mass m is constrained to move on the arc of a parabola \(x^{2} = 4ay\), where y is vertical axis, under gravity. Then the canonical momentum is given by :
Answer: D
The canonical momentum is \(m\dot{x}\left(1 + \frac{x^{2}}{4a^{2}}\right)\).
On \(x^{2} = 4ay\), \(y = \frac{x^{2}}{4a}\) and \(\dot{y} = \frac{x\dot{x}}{2a}\). With x as the generalised coordinate,
\[ \begin{aligned} L &= \tfrac{1}{2}m\left(\dot{x}^{2} + \tfrac{x^{2}\dot{x}^{2}}{4a^{2}}\right) - mg\tfrac{x^{2}}{4a} \\ p_{x} &= \frac{\partial L}{\partial \dot{x}} = m\dot{x}\left(1 + \frac{x^{2}}{4a^{2}}\right) \end{aligned} \]
\(m\dot{x}\) ignores the vertical motion forced by the constraint. Option B keeps \(\dot{y}\) as if it were independent, and option C is not even linear in \(\dot{x}\).
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