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Homework Help: The limit energy system

  1. Jul 28, 2017 #1
    Hi everyone. I'll be grateful if someone can help me with this problem.

    1. The problem statement, all variables and given/known data

    I have a closed system composed of one particle. The maximal velocity that this particle can have is equal to Vc.

    Here we consider only 2D space: X and Y direction. The particle velocity is V (which is limited to Vc) and have two velocity components Vx and Vy, each cab vary from 0 to Vc.

    The existence of Vc creates the relation between Vx and Vy. For example if Vx = Vc then Vy = 0.
    We can write this relation as follows: Vx2 + Vy2 ≤ Vc2.

    I want to express the component Vy as a function of Vx.

    2. The attempt at a solution

    In order to resolve this, first I studied the case when the overall velocity V is equal to Vc. In this case:

    Vx2 + Vy2 = Vc2.


    Vy = sqrt( Vc2 - Vx2) . (1)

    Can I conclude that this last relation (1) is correct for all cases (and not only when V = Vc)? My problem is that I don't know if it is mathematically justified to take a special case: V = Vc and then conclude that the final relation (1) between Vy and Vx works for general case for any value of V.
  2. jcsd
  3. Jul 28, 2017 #2

    Ray Vickson

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    Homework Helper

    No, it is not correct. Your ##v_x,v_y## satisfy ##v_x^2+v_y^2 \leq v_c^2##, but are not otherwise restricted. That means that ##(v_x,v_y)## lies inside on on the edge of a circular disc of radius ##v_c##. You have ##v_y = \sqrt{v_c^2 - v_x^2}## or (a case you forgot) ##v_y = -\sqrt{v_c^2 - v_x^2}## only for points ##(v_x,v_y)## that lie on the boundary of the disc. For example, ##v_x = v_y = 0## satisfies the restriction, as does ##v_x = 0, v_y = \frac{1}{2} v_v## or ##v_x = \frac{1}{2} v_c, v_y = \frac{3}{4} v_c##, etc. There are infinitely many possible solutions!
  4. Jul 28, 2017 #3
    Hi. thank you for the response.

    Actually I made a mistake, the problem is somehow more complicated. I'll do my best to describe it in detail.

    The condition with the limit velocity is the same. But, in order to accelerate our particle to the speed Vc we need to apply energy equal to some precise value, let's call it Ec. Even If we give more energy to the particle will never go beyond the limit velocity Vc.
    (This is somehow similar to the special relativity. Example: the limit speed for the electron is the speed of light, even if you give it an infinetly high amount of energy its velocity will never go beyond the speed of light and will only converge to it).

    Lets return to our problem. If we take for example the case when the particle has already an initial velocity in the X direction Vx = 0.9*Vc. Now, if we will try to accelerate this moving particle in the Y direction using the energy = Ec, the overall velocity will be still equal to Vc (V = Vc). We can calculate Vy' = sqrt(Vc2 - Vx2); since we know that the component Vx = 0.9*Vc;

    IMPORTANT: we need to make the difference between Vy and Vy'

    We call Vy the velocity that our particle should have when initially Vx is equal to 0. And Vy' is the reduced velocity due to the initial non-zero Vx component.

    we can write: Vy' = Vy*sqrt(Vc2 - Vx2) (1)

    This surely works when we use the energy Ec in order to accelerate the moving particle, since the overall velocity V will always be equal to Vc (as you said we are on the edge of a circular disc). BUT, can we use the equation (1) in order to calculate Vy' even when the overall velocity V is inferior to Vc (when we are inside of a disc)?

    If we finally write V2 = Vy'2 + Vx2 (2)

    with Vy' (see equation (1))

    We can see through the equation (2) that we will never obtain a velocity higher than Vc, so this equation suits the answer very well. But how to justify that the relation (1) can be used not only on the edge of a circle but also inside of it.

    Can we use this argument: in order to find the general relation between Vy and Vy' we resolve the equation for the special case Vy = Vc?

    i'm really sorry for my English.
    Last edited: Jul 28, 2017
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