How can I determine the stability of equilibrium points in dynamics?

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In summary: You might also see if you can find a worked example in your text or on wikipedia,com or some other source, of how to get the energy equation given a potential function.
  • #1
Gwilim
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I'm going to post everything I can't solve from a past exam paper here, with the original numbering, one post per question. Some I will attempt solutions for, others have me stumped completely. Any and all contributions will be appreciated.
 
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  • #2
2.

Let P be a particle with unit mass and position vector ->OP=r(t) at time t in some inertial frame. Let

r(t) = t^(-1/2)(cos (t^2 /2)i + sin (t^2 /2)j)

with usual conventions, for t > 0. Prove that r(t)=t^(-1/2) and find r'(t), and |r'|(t). Deduce that h(t)=r(t)^r'(t) is a constant vector.

(note: the final ^ character is not an attempt at rendering indices through text.. it is the actual character used on the paper and I have no idea what it means)
 
  • #3
The first part is fairly easy. I think the scalar quantity, indicated by the lack of boldface, must be determined using Pythagoras' theorem. Then the i component and the j component squared and added reduce to 1 by trigonometric identity and we are left with t^(-1/2).

I don't know how to find the derivative of a vector quantity.. I was thinking of finding the derivative of the i component and the j component then adding them. Is this correct?

I simply do not know what r(t)^r'(t) means so I cannot discern what I am being asked to show.
 
  • #4
3.

Briefly describe a sidereal frame of reference, and then state Newton's laws of motion

State the properties satisfied by a conservative field of force, and then derive the energy equation for a particle of constant mass m moving under the influence of such a force.
 
  • #5
I think a sidereal frame of reference means "using the fixed stars as a reference point to describe the motion of a particle". I'm not sure how I would word that in an exam. Newtons laws are of course a simple matter of commiting to memory but I get the feeling the formulation required here includes mention of frames of reference.

I don't know what a conservative field of force is. The energy equation referred to here is a believe E= 1/2mv^2 + V(t) where 1/2mv^2 is kinetic energy and V(t) is potential energy
 
  • #6
4.

A particle of constant mass m moves in one dimension only under the influence of a conservative field of force. If the position at time t is given by x(t), write down the energy equation. Hence give definitions of turning points and equilibrium points.

For a potential function V(x) = -x/(1+x^2), determine the equilibrium points, and comment on their stability. What range of values for x' at the stable equilibrium point gives rise to a bound motion?
 
  • #7
This one screams calculus at me. Again there's the problem of not knowing what is meant by 'conservative field of force'. Turning points I would guess are points at which the derivative is 0. Equilibrium points are unkown to me, and consequently their stability also.

More of these later.
 
  • #8
You really should have put each problem in a separate thread, not put all of the problems as separate posts in a single thread.
Gwilim said:
2.

Let P be a particle with unit mass and position vector ->OP=r(t) at time t in some inertial frame. Let

r(t) = t^(-1/2)(cos (t^2 /2)i + sin (t^2 /2)j)

with usual conventions, for t > 0. Prove that r(t)=t^(-1/2)
I think that r(t) here means |r(t)|; i.e., the magnitude of the vector r(t). This is pretty easy once you realize that the part inside the parentheses is a unit vector.
Gwilim said:
and find r'(t), and |r'|(t). Deduce that h(t)=r(t)^r'(t) is a constant vector.

(note: the final ^ character is not an attempt at rendering indices through text.. it is the actual character used on the paper and I have no idea what it means)
You're going to have to find out what that ^ symbol means. The most common multiplication operations with two vectors are the dot product and the cross product (which is defined on pairs of 3-d vectors). I saw the caret symbol used for an outer product once, but that's not something I'm very familiar with. Also, what's the significance of h(t)? Could that be the tangent vector?
 
  • #9
Mark44 said:
You really should have put each problem in a separate thread, not put all of the problems as separate posts in a single thread.

Okay I'll post the rest in different threads. If a moderator happens across this please can you separate this one out?

I think that r(t) here means |r(t)|; i.e., the magnitude of the vector r(t). This is pretty easy once you realize that the part inside the parentheses is a unit vector.

Yes

You're going to have to find out what that ^ symbol means. The most common multiplication operations with two vectors are the dot product and the cross product (which is defined on pairs of 3-d vectors). I saw the caret symbol used for an outer product once, but that's not something I'm very familiar with. Also, what's the significance of h(t)? Could that be the tangent vector?

I think the h(t) was being defined by the r(t)^r'(t)
 
  • #10
Gwilim said:
I think the h(t) was being defined by the r(t)^r'(t)
Yes, it was, but as this is a problem in dynamics, it's likely to have some physical meaning. In any case, I'm blocked by not knowing what operation ^ represents.
 
  • #11
DO you have any insight into the other questions?
 
  • #12
Some.
The derivative of a vector function is the vector made up of the derivatives. For example, if r(t) = (x(t), y(t)) = x(t)i + y(t)j, then r'(t) = (x'(t), y'(t)) = x'(t)i + y'(t)j. This should help you get the other parts of problem 2.

For the terms "sidereal frame of reference" and "conservative field" I would look them up on wikipedia.org to see if what I thought they were was reasonable. Same advice for Newton's laws of motion, of which there are three.

For #4, the best I can do is advise you to check your text for similar worked problems, as well as definitions of "equilibrium point" and "turning point."
 
  • #13
Mark44 said:
For #4, the best I can do is advise you to check your text for similar worked problems, as well as definitions of "equilibrium point" and "turning point."

I don't have any, I'm in kind of a crisis situation.

Guess I'll just have to think really really hard.

Wikipedia doesn't have "sidereal frame" just "sidereal time" but I think I can deduce the meaning from there.

Thanks for confirming the method for differentiating vectors though.
 
  • #14
Gwilim said:
I don't know how to find the derivative of a vector quantity.. I was thinking of finding the derivative of the i component and the j component then adding them. Is this correct?

Let [itex]\textbf{r}(t)=x(t)\textbf{i}+y(t)\textbf{j}+z(t)\textbf{k}[/itex], then

[tex]\textbf{r}'(t)=\frac{d \textbf{r}}{dt}=\frac{d}{dt}\left(x(t)\textbf{i}+y(t)\textbf{j}+z(t)\textbf{k}\right)=\frac{d}{dt}\left(x(t)\textbf{i}\right) +\frac{d}{dt}\left(y(t)\textbf{j}\right)+\frac{d}{dt}\left(z(t)\textbf{k}\right)[/tex]

Now, the Cartesian unit vectors do not change in space or time (the same is not true of curvilinear unit vectors), so they can be treated as constants for these derivatives and hence [itex]\textbf{r}(t)=x'(t)\textbf{i}+y'(t)\textbf{j}+z'(t)\textbf{k}[/itex] ...make sense?
I simply do not know what r(t)^r'(t) means so I cannot discern what I am being asked to show.

I think the [itex]\wedge[/itex] symbol represents a cross product, and [itex]\textbf{h}(t)[/itex] is the angular momentum of the particle (position crossed with momentum)
 
  • #15
Gwilim said:
4.

A particle of constant mass m moves in one dimension only under the influence of a conservative field of force. If the position at time t is given by x(t), write down the energy equation. Hence give definitions of turning points and equilibrium points.

For a potential function V(x) = -x/(1+x^2), determine the equilibrium points, and comment on their stability. What range of values for x' at the stable equilibrium point gives rise to a bound motion?

A conservative field has zero curl, and can therefor be written as the gradient of a scalar function. By convention, one usually takes [itex]\textbf{F}=-\mathbf{\nabla}V[/itex]...if the particle moves only in one direction, the gradient becomes just a normal derivative and so [itex]\textbf{F}=-\frac{dV}{dx}\textbf{i}[/itex]

For the energy equation, I think they want you to write the potential as V(x) (to make it clear that it depends only on 'x') and to write the kinetic energy in terms of the speed x'(t).

Equilibrium points are points where the force on the particle is zero, so that if you place a particle at an equilibrium point and give it no initial velocity, it will simply stay there forever. A stable equilibrium point is one where if you move the particle a very small distance away from the equilibrium in any direction, the force will point it back towards the equilibrium point.
 
  • #16
gabbagabbahey said:
Let [itex]\textbf{r}(t)=x(t)\textbf{i}+y(t)\textbf{j}+z(t)\textbf{k}[/itex], then

[tex]\textbf{r}'(t)=\frac{d \textbf{r}}{dt}=\frac{d}{dt}\left(x(t)\textbf{i}+y(t)\textbf{j}+z(t)\textbf{k}\right)=\frac{d}{dt}\left(x(t)\textbf{i}\right) +\frac{d}{dt}\left(y(t)\textbf{j}\right)+\frac{d}{dt}\left(z(t)\textbf{k}\right)[/tex]

Now, the Cartesian unit vectors do not change in space or time (the same is not true of curvilinear unit vectors), so they can be treated as constants for these derivatives and hence [itex]\textbf{r}(t)=x'(t)\textbf{i}+y'(t)\textbf{j}+z'(t)\textbf{k}[/itex] ...make sense?

yes. I've got r'(t) in the question. It's:

(-t^(1/2)sin((t^2)/2) - 1/2t^(-3/2)cos((t^2)/2)i + (t^(1/2)cos((t^2)/2) - 1/2t^(-3/2)sin((t^2)/2)j

The expression for |r'|(t) looks truly horrible though, I think I'm missing a trick somewhere.
 
  • #17
|r'(t)| shouldn't be too bad, just use [itex]\sin^2 u + \cos^2 u=1[/itex] and [itex]2\sin u \cos u=\sin(2u)[/itex]...what do you get?
 
  • #18
gabbagabbahey said:
|r'(t)| shouldn't be too bad, just use [itex]\sin^2 u + \cos^2 u=1[/itex] and [itex]2\sin u \cos u=\sin(2u)[/itex]...what do you get?

You're right it just looked like it should be ugly..

Came out as t+(1/4)t^-3

edit: uhh I mean (t+(1/4)t^-3)^1/2
 
  • #19
Right, now try to show h(t) is a constant vector...
 
  • #20
^ I'll get back to that one later.

Gwilim said:
For a potential function V(x) = -x/(1+x^2), determine the equilibrium points, and comment on their stability. What range of values for x' at the stable equilibrium point gives rise to a bound motion?

I've started out by trying to differentiate V(x)?

I've got V'(x) = (1/1+x^2)((2x^2/1+x^2) - 1)

Ignoring non-real solutions this leaves x = 1.

Where do I go from there? Do I differentiate again?
 
  • #21
Gwilim said:
I've started out by trying to differentiate V(x)?

Sounds good to me. You know that equilibrium points have zero net force acting on the particle, so you want to find where [itex]F=-\frac{dV}{dx}=0[/itex]

I've got V'(x) = (1/1+x^2)((2x^2/1+x^2) - 1)

Good, but it is probably easiest to see where the roots are if you put everything over a common denominator:

[tex]F=-\frac{dV}{dx}=-\frac{2x^2-(1+x^2)}{(1+x^2)^2}=\frac{1-x^2}{(1+x^2)^2}[/tex]

Ignoring non-real solutions this leaves x = 1.

Really? Aren't there two real roots to [itex]1-x^2=0[/itex]? :wink:

Where do I go from there? Do I differentiate again?

Now there are a few different ways to test the stability of your equilibrium points:

(1) You know that at a stable equilibrium [itex]x_0[/itex], a small displacement in either direction [itex]\pm\delta[/itex] will result in a force that tends to push the particle back towards [itex]x_0[/itex]...so, what is [itex]F(x_0\pm\delta)[/itex] for your two equilibrium points? Does this force push the particle towards or away from [itex]x_0[/itex]?

(2) You should also have learned that a conservative force always pushes a particle towards a potential minimum, so just test whether your equilibrium points are local minima or local maxima via the 2nd derivative test.
 

What is dynamics?

Dynamics is the branch of physics that deals with the motion of objects and the forces that cause the motion.

What are the different types of dynamics?

The three main types of dynamics are kinematics, kinetics, and statics. Kinematics deals with the motion of objects without considering the forces that cause the motion. Kinetics deals with the forces that cause motion. Statics deals with objects that are not moving or are in equilibrium.

What is Newton's first law of motion?

Newton's first law of motion, also known as the law of inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

What is the difference between linear and angular dynamics?

Linear dynamics deals with the motion of objects in a straight line, while angular dynamics deals with the motion of objects in a circular path.

Can dynamics be applied to everyday life?

Yes, dynamics can be applied to everyday life, from the motion of vehicles to the forces acting on the human body during physical activities. Understanding dynamics can also help in designing and improving everyday objects and technologies.

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