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In this question, how can we comment on the position of stable/unstable equilibrium if we don't know the total energy of the system? We know force is zero at points 'a', 'b', 'c' and 'd' but there might be some kinetic energy at those points. Moreover, the answer given in the book is (d). But how can (c) be True?

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    $\begingroup$ Equilibrium points are defined by the potential. Whether or not a particle is at that point and what it's total energy in is immaterial to the definition. You are correct: (b) is an equilibrium point. $\endgroup$ – garyp Feb 4 '18 at 18:56
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Static equilibrium means that if you place an object at rest in that position then it will stay there. The kinetic energy of the particle is zero because it is at rest.

Option (c) is false. Points b and d are positions of unstable equilibrium. The book answer is incorrect.

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In natural systems, the kinetic energy will be of the form $$ T=\sum_i \frac{p_i^2}{2m_i} $$ and the Hamiltonian is thus $$ H=T+U\, . $$ The fixed points of the system are determined by $$ \frac{\partial H}{\partial p_i}=0\, ,\qquad \frac{\partial H}{\partial q_i}=0 $$ and, given the form of $T$, we have $p_i=0$ together with $$ \frac{\partial H}{\partial q_i}=\frac{\partial U}{\partial q_i}=0\, . $$ One easily shows by expanding $H$ about the fixed points that, if $U$ is a minimum, the linearized equations of motions are those of a harmonic oscillator, so that the solution stay close to the fixed point for small perturbations away from that point: this is stable equilibrium. On the other hand, if $U$ is a max (or a saddle in more than 1d), then there is a direction in which the motion will be unbounded: this is unstable equilibrium.

The same will work when $H$ is not the total energy but is nevertheless conserved. Examples include beads on rotating hoops of various shapes. In these cases, $\sum_i \dot q_i p_i \ne 2T$ typically, and it is difficult to isolate a "potential", but one can sometimes identify an "effective" potential and proceed as before.

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I'll try to focus on your concrete questions.

  1. Kinetic energy is made of a squared quantity, so it can be possitive or zero, but never negative.

  2. If energy is conserved, and you have $E=T+V$ (or as I like to write it, $E=E_k+E_p$, which I find infinitely better), then there is an inmediate consequence: the total energy will always be $\mathbf{\geq E_{p_{min}}}$.

That's the only thing you need to know. Whatever the Energy is, it will ALWAYS be above the minimum of the curve you show. Does it matter the concrete value? Nothing changes if it is 2 or 2.1. This is a qualitative analysis.

So you can just think of Total Energy as an horizontal straight line (because it is the same along all positions), a horizontal line which lies above the curve's absolute minimum.

You only have yo suppose that it is "above enough", and that's all. The particle will be preferably moving around minima. If the energy goes down (due to friction or whatever), the particle will get confined around the minimum in which it was in the moment of the energy lowering. The movement will then be restricted to that area.

However, the other minima are still equilibrium points, they're just forbidden, but they will become allowed agian as soon as the energy rises again due to an external action.

And yes, point b is an equilibrium point too, it's just unstable.

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Whether an equilibrium is stable or not has got nothing to do with kinetic energy. It has got nothing to do with motion. Yes, if there was enough kinetic energy in a low valley, then an object would be able to roll out of it again.

But everything can escape anything with the right kinetic energy - we call something stable, not because it can't escape at any value of kinetic energy, but because it can't escape at some.

Place a marple on a hill top and another in a valley crest. Which of these positions is stable? The marples can lie still at both, so they are at equilibrium at both positions. But which would we call a "stable" equilibrium? The valley, of course. Because at the slightest offsetting from the exact top-point it will roll away - so we call it unstable - while at the slightest offsetting from the exact bottom-point it will roll back - so we call it stable.

In general, objects want to move the way that decreases the potential energy (that's why objects fall down, if they have the chance, and not up).

  • At a hill top, the potential energy is lower at all other nearby positions, so the object will roll down.

  • While at a valley bottom, the potential energy is higher at all nearby positions, so it will not want to go there but will rather stay down at that bottom.

Therefore, option c) is false. Point b is definitely a point of equilibrium, just an unstable equilibrium. If the text book says that c) is a right answer, then that must be a mistake.

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protected by Qmechanic Feb 5 '18 at 0:00

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