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Homework Help: Many Questions on Dynamics (please help)

  1. Aug 13, 2009 #1
    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.
    Last edited: Aug 13, 2009
  2. jcsd
  3. Aug 13, 2009 #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)
  4. Aug 13, 2009 #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.
  5. Aug 13, 2009 #4

    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.
  6. Aug 13, 2009 #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
  7. Aug 13, 2009 #6

    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?
  8. Aug 13, 2009 #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.
  9. Aug 13, 2009 #8


    Staff: Mentor

    You really should have put each problem in a separate thread, not put all of the problems as separate posts in a single thread.
    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.
    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?
  10. Aug 13, 2009 #9
    Okay I'll post the rest in different threads. If a moderator happens across this please can you seperate this one out?


    I think the h(t) was being defined by the r(t)^r'(t)
  11. Aug 13, 2009 #10


    Staff: Mentor

    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.
  12. Aug 13, 2009 #11
    DO you have any insight into the other questions?
  13. Aug 13, 2009 #12


    Staff: Mentor

    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."
  14. Aug 13, 2009 #13
    I don't have any, I'm in kind of a crisis situation.

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

    Wikipedia doesnt 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.
  15. Aug 13, 2009 #14


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    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 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)
  16. Aug 13, 2009 #15


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    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.
  17. Aug 14, 2009 #16
    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.
  18. Aug 14, 2009 #17


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    |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?
  19. Aug 14, 2009 #18
    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
  20. Aug 14, 2009 #19


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    Right, now try to show h(t) is a constant vector...
  21. Aug 15, 2009 #20
    ^ I'll get back to that one later.

    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?
  22. Aug 15, 2009 #21


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    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]

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


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

    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.
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