1

Question.
My confusion comes from a problem and its solution (both partially).

Here is the problem. Problem

Here is the solution.Solution

Now, what baffles me is the highlighted part in the solution; I do not understand it.

My Attempt. I tried to understand the derivative, by definition, as the rate of change in potential energy following some change in the angle $\theta$. But why its being zero represents equilibrium? Does equilibrium occur at some turning point of potential energy? I am puzzled.

Comment. Any kind of help would be appreciated. Personally I prefer some short but helpful hint. Thank you in advance!

IncredibleSimon
  • 933
  • 2
    Energy of a system is lowest when it's in equilibrium so total energy is $T=U+K$ where $U$ is potential energy and since particles are just lying over smooth surface, kinetic energy K is $0$. When is $T$ minimum? Use maxima-minima theorem. – Koro Apr 10 '21 at 16:10
  • Thanks for the comment Koro. But I still don't get it: why a system's energy is lowest in equilibrium? Say no external forces act on an object, then its mechanical energy which is potential plus kinetic is conserved, then no matter it is accelerating (i.e. not in equilibrium; potential transforms into kinetic in progress) or stays still (i.e. in equilibrium) its total energy keeps the same, doesn't it? Thanks. @Koro – IncredibleSimon Apr 10 '21 at 16:24
  • Hmm, but conservation of mechanical energy excludes gravitational pull as an external force I think? Well, at least from my textbook. @Koro – IncredibleSimon Apr 10 '21 at 16:31
  • 1
    Change in energy of system = work done on the system by external force. In this case (with appropriate sign convention), $dU=Fa d\theta\implies dU/d\theta =F=0$ (Net external force F=0 since the system is in equilibrium.) I don't currently remember more intricacies of this. :'( – Koro Apr 10 '21 at 16:40
  • No worries Koro. Though I still don't understand, I am giving you an up vote for the concept of lowest total energy. Thank you sincerely. :) @Koro – IncredibleSimon Apr 10 '21 at 16:45
  • Your question isn't about mathematics, but physics, and would be better asked in the physics forum. – Paul Sinclair Apr 11 '21 at 02:37
  • Well, I cannot say what you said is totally wrong, but mechanics is a legitimate topic of maths, especially in UK. Otherwise, the classical-mechanics tag would be non-existent; by the way, there is even quantum-mechanics tag which sounds very physics. Indeed, as long as a model could be set up, there is no clear boundary between maths and physics; physics is to some extent just maths with some formulas. Anyway, I am not arguing further. Thanks for the advice. @PaulSinclair – IncredibleSimon Apr 11 '21 at 03:42
  • 1
    When you were discussing the mathematical model, yes, it is a mathematical question, but in your discussion with Koro, the mathematical aspect was answered, and the question evolved into a discussion of the physics leading to the model. At that point Koro couldn't help anymore. Why? Because their expertise is in mathematics, not the physical underpinnings. At that point, you should consider where to find the better experts in those questions, and that is the physics forum. There are plenty of overlap, but the physics forum is where you will find the greater expertise on this question. – Paul Sinclair Apr 11 '21 at 13:35

0 Answers0