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I Trajectory equations

  1. Nov 3, 2016 #1

    Just starting to learn kinematics and the book says that from the kinematic equations of motion:

    x=ƒ1(t) ; y=ƒ2(t) ; z=ƒ3(t)

    we can "eliminate de time" from the above and get:

    F1(x,y,z)=0 ; F2(x,y,z)=0 -> aka. trajectory eq.

    their "ensemble" being 2 surfaces giving the trajectory

    So my questions would be:

    - how does one get those 2 functions out of the starting 3 ?
    - how does one "eliminate" time and why does it makes sense, since without time there is no motion ?
    - does "ensemble" mean intersection here, and are these curved or plane surfaces ?

    Thanks, and sorry if stupid questions :)
  2. jcsd
  3. Nov 3, 2016 #2


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    Hello Adrian,

    The description ##\ (x,y,z) = \left ( x(t), y(t), z(t) \right ) \ ## is general, so there's no telling whether these are planar, linear or whatever. Nor 'how to get those 2 functions' -- usually some math is required :smile:

    You could try to check with some simple examples (circular motion in a plane, a helix, a cycloid, etc.)

    Not stupid questions at all.
  4. Nov 3, 2016 #3


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    Conceptually at least, you solve one of the equations for t and substitute into the other two to get two equations that do not depend on t. That is, you have eliminated t from the equations.

    The result is something that describe the path of the object rather than its position as a function of time. As BvU says, your question is rather general. Consider ballistic motion. You can describe the x and y coordinates in terms of initial velocity, gravity, and time. But it's easy to eliminate the time and get the y position as a function of x, which is useful for answering questions like "will this shot go over that wall".
  5. Nov 4, 2016 #4

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    Suppose f1 is invertible on at least part of the domain.
    Then we have ##y=f_2(f_1^{-1}(x))## or ##F_1(x,y,z)=y-f_2(f_1^{-1}(x))=0##.
  6. Nov 4, 2016 #5
    I guess I'm just trying to visualize how this all works and linked because I can't seem to solve one problem right. This transition through position vectors, trajectory, the "law of space", tangents/differentials, angles, trigonometric functions and rotations is all very confusing...

    Btw. is it normal not being able to make one problem correctly in 5 weeks of classical mechanics undergrad course ? I know how to solve equations, dot, cross product (which btw nobody can explain intuitively, and where they came from, and why they are thus, without involving tons of algebra and tongue twisting mathematical formalism terminology), take derivatives, integrals but unfortunately they're useless without knowing what happens in the theory and how to apply them..

    Sorry for the rant, but it's been 3 weekends and still on 1st chapter kinematics. Mid-term is coming :)
  7. Nov 4, 2016 #6

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    No worries. I don't consider this particular problem statement very important.
    I think it's only intended as a trigger to get people thinking - not as a requirement to understand the theory.
  8. Nov 4, 2016 #7


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    I shouldn't think it is. At the very least it's not motivating at all. What do your fellow students have to say about this ?

    Some 45 years ago we would have about 4 or 6 hours of lectures a week and two afternoons (so 6 to 7 hours) of doing exercises with assistance from staff and senior students. Exercises were rather tough, but the satisfaction of getting through (with some help) was rewarding and motivating. Classical mechanics was the first subject at university (in parallel with tons of math) and considered very important. I must say: rightly so.

    In your case I would suggest finding a textbook you like with lots of exercises. Many books have re-hashing/digesting exercises, followed by a few more challenging ones.
    Does your institution offer problem solving courses ?
  9. Nov 5, 2016 #8


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    What book are you using? Is it about "naive" or "analytical" mechanics. The irony is that the "naive" mechanics is usually more difficult than "analytical" mechanics, because you deal with forces rather than Lagrangians and Hamiltonians.

    Nevertheless the fundamental concept is of course the same. What you like to achieve with mechanics is to describe your system of point particles through the dynamics determined by the forces acting between these particles. The most simple case is that you have one particle which moves under the influence of some external force which is due to other matter around it but not described in all microscopic detail, which is often not possible and also not necessary. The most simple example is the motion of some particle with mass ##m## under the influence of the gravity due to the presence of the Earth. If you stay very close to Earth you can neglect the variation of the gravitational force with position of the particle and then the force is very simple, i.e., constant
    $$\vec{F}=-m g \vec{e}_z,$$
    where I use a Cartesian coordinate system with the ##z## axis pointing upwards, i.e., against the direction of the gravitational force, which is here approximated as constant with the gravitational acceleration ##g \simeq 9.81 \text{m}/\text{s}^2##, and ##m## is the mass of the particle (it's the very fundamental feature of Newtonian gravity that it is proportional to the mass of the particle it's acting on).

    Now to describe the motion of the particle, you have to tell, where it is at any given time ##t##. That's described by a function ##\vec{x}(t)##, which gives the positive vector (pointing from the origin of the coordinate system to the point where the particle is at time ##t##, and this function obeys Newton's equation of motion,
    $$m \frac{\mathrm{d}^2}{\mathrm{d} t^2} \vec{x}=\vec{F}=-m \vec{g} \vec{e}_z.$$
    Now this is really very simple, but it gets the concept pretty clear. You can now solve this equation of motion by just integrating with respect to time. That's because the force is very simple in this case, i.e., it doesn't depend on the position or velocity of the particle as usually forces do; than you have more complicated differential equations to solve, but the fundamental concept is always the same: You figure out the force acting on the particle and solve for the differential equations to get the particle trajectory.

    Let's follow our simple example of free fall to the end: We simply can integrate with respect to time once to get
    $$\dot{\vec{x}}=-\vec{g} t \vec{e}_z+\vec{v}_0,$$
    where ##\vec{v}_0## is a constant vector. It's the velocity at time ##t=0##. It is clear that you need this information to integrate the equation of motion. You can just think about a ball you through with some initial velocity, and you can choose this velocity more or less as you like. So you need to give the initial velocity in order to uniquely solve for the velocity of the particle at any later time.

    Now you can integrate once more with respect to ##t## to get the trajectory,
    $$\vec{x}(t)=-\frac{g}{2} t^2 \vec{e}_z+\vec{v}_0 t + \vec{x}_0.$$
    Again the same as we've seen concerning velocity also happens here: You can not only choose the velocity you through your ball at time ##t## but also the initial position ##\vec{x}_0## at ##t=0##.

    So the usual task to figure out the trajectory of particle is: Get the force acting on it and specify the initial conditions ##\vec{x}(t=0)=\vec{x}_0## and ##\dot{\vec{x}}(t=0)=\vec{v}_0##, where the dot means the time derivative.

    Perhaps we can help you better when you tell us, where you have problems with these concepts.
  10. Nov 5, 2016 #9
    They also have difficulties. We sometimes meet and brainstorm but progress is slow.

    We didn't have seminaries until this week, only laboratory doing experiments. But not too useful since the same course teacher (no assistants) explains exercises at the whiteboard and keeps saying "obviously" to things only 2 olympians in front row understand.

    I'm not sure but it uses only differentials, integrals and vector operations until now. It starts with theory of kinematics (position, velocity, tangential/normal acceleration, curvature/radius of curvature) then it particularizes, for movement in gravitational field (parabola movement, circular motion, harmonic motion) and then is followed by 24 exercises which I guess should be solved by just correctly understanding the theory before.

    My interest is to understand logically and visually how the theory was build together from scratch (which btw. I couldn't find one YouTube video explaining kinematics thoroughly starting from defining the position vector all the way to Frenet-Serret equations and circular motion). I feel like this theory is at the basis of infinitesimal calculus and how, I think Newton for the first time (although I looked through his Principia and couldn't find it there :) ) used it so beautifully to describe motion. I feel like it is fundamental to understand this so that I can truly appreciate calculus.

    On a more practical note, an exercise that I think is meant to teach how to manipulate equations by understanding the theory states that:

    The kinematic equations of a material point (assume no mass - btw how is that called in English, here is coined "a mobile") are

    x=A⋅cosωt ; y=B⋅sinωt ; A,B,ω - positive constants.

    Find: a) implicit equation F(x,y)=0 ; b) velocity v(x,y) and it's hodograph c) total,tangential,normal accelerations d) radius of curvature R(x,y) of the trajectory

    So how I imagine this problem according to the theory, I have the movement of the material point (mobile) that is drawn by the position vector r(t) such that the components of i, j unit vectors vary as the functions x(t) and y(t) given. If I understood correctly, F(x,y) is obtained by taking t from say x(t) and substitute it in y(t) therefore getting the actual path which is the function F(x,y). Unfortunately I don't know how to plot this trajectory especially since t=arccos(x/A)/ω. But assuming I can plot this function of 2 variables, then it should follow that from this known trajectory to find the velocity vector defined as the tangent on this curve/trajectory, that is actually another "position-like" vector of the form v(t) with components say vx(t), vy(t) that form another V(x,y) function that describes I guess a velocity curve/trajectory ? By the time this is similarly repeated for the acceleration I'm totally lost because of the tangential/normal/bi-normal acceleration namings, that doesn't tell me anything after so many tangents and curves drawn and how it's related to the original trajectory intuitively. Also further on, where does the circular motion fit in all of this ?

    Anyway I hope is not total gibberish what I'm explaining here but that's how I understand the theory so far...

    Also thanks a lot for all your efforts to write back so much!
  11. Nov 6, 2016 #10


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    How good of you to (almost) take the first simple example I mentioned :wink:. To get a feeling what it represents, take A=B=1 and ##\omega = 2\pi ##; draw a few points for t=0 to t=1.

    On a serious note: The olympians in the front row block teacher's view on what's really going on in his classroom. I'm not an expert, but it has the ring of a beginner's mistake. If your fellow students feel the same way you do, perhaps you could form a delegation and discuss this with teacher. And if that doesn't help there might be a dean or some other mediation possibility ?
  12. Nov 7, 2016 #11


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    Well, it's not too helpful to use constants without the correct dimension, particularly at this state of understanding (or unfortunately confusion). Just take the given equation and write it in the concise form of Cartesian vector components:
    $$\vec{x}(t)=\begin{pmatrix} A \cos (\omega t) \\ B \cos(\omega t) \\ 0 \end{pmatrix}.$$
    Now you can calculate the velocity as well as the acceleration by remembering how these vector quantities are defined (do it!). For the implicit form (what this is good for, is the secret of the textbook author/professor ;-)) remember that for any angle ##\alpha## you have ##\cos^2 \alpha+\sin^2 \alpha=1##. For the decomposition into tangential and normal components of the acceleration as well as the curvature, look for "Frenet formulae", e.g., here

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