Discussion
A particle falls vertically under the action of gravity and a frictional force obtainable from dissipation function \(G(v) = \frac{1}{2}kv^{2}\), k is constant. The Lagrange's equation of motion is :
Answer: A
The equation is \(m\ddot{y} + k\dot{y} + mg = 0\), with y measured upward.
With a dissipation function G, Lagrange's equation becomes
\[ \begin{aligned} &\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} \\ &\quad - \frac{\partial L}{\partial y} + \frac{\partial G}{\partial \dot{y}} = 0 \end{aligned} \]
Taking \(L = \frac{1}{2}m\dot{y}^{2} - mgy\):
\[ \begin{aligned} \frac{d}{dt}(m\dot{y}) + mg + k\dot{y} &= 0 \\ m\ddot{y} + k\dot{y} + mg &= 0 \end{aligned} \]
The friction term \(k\dot{y}\) always opposes the velocity, so it enters with the same sign as \(m\ddot{y}\). Option D omits friction altogether, and options B and C give the friction term the wrong sign, which would make the resisting force speed the particle up.
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