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How to graph acceleration vectors on position axis?

  1. Feb 23, 2012 #1
    suppose I have two position functions in terms of time. One position function for the x component and the other for the y component. The y component is of constant acceleration. The x component is of constant velocity. How do I represent acceleration on a graph of this type?

    Here is an illustration of my attempt. Will acceleration always be vertical if the y component has constant acceleration and the x component has zero acceleration?


    graph2.png
     
  2. jcsd
  3. Feb 23, 2012 #2

    jedishrfu

    Staff: Mentor

  4. Feb 23, 2012 #3
    Well, that was easy, thanks!!

    Okay, so kinda zooming into one of the velocity vectors on the graph of the post above...

    following up on this question, can you tell me if these statements i'm making are accurate:


    If the acceleration vectors are to point vertically at 90 degrees, that would mean there is no x component of acceleration, right or wrong?

    However, the path of the object described by the position functions in the first post's graph is in a curved trajectory. Thus it must have a centripetal acceleration, yes or no? And that centripital acceleration is ALWAYS supposed to be exactly perpendicular to the velocity vector, correct?

    And centripetal acceleration is defined as a component of total acceleration. And if the vertical acceleration in the above graph is defined as total acceleration, then it does in fact have a component called centripital acceleration.

    And the other component of total acceleration would be parallel to the velocity vector.

    So while total acceleration is completely vertical, thus not having any components on the x vs y axis, it does have components on a different set of axis that run parallel and perpendicular to the velocity vector. So there are two sets of axis to consider when evaluating/examining velocity vectors, the x y cooridinates and the x' y' coordinates.. Would this be right?

    graph3-1.png

    graph4.png
     
    Last edited: Feb 24, 2012
  5. Feb 24, 2012 #4
    Just in case it's not so clear, the two graphs in the above post (post # 3) are supposed to represent zooming into one of the velocity vectors on the graph in the original post (post # 1)...and with that in mind, hopefully I can find out if any of the statements I made are right or wrong, and thus being able to pin point any portion of acceleration I may still be confused about.

    So if anyone can tell me if i'm confusing any details here, i'd really appreciate it.
     
  6. Feb 24, 2012 #5

    gneill

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    Yes, it means that the x-component of the acceleration is zero at least for that moment in time.
    Yes, if by that you mean an instantaneous centripetal acceleration for the given instant of time, and it is with respect to the instantaneous radius of curvature for the curve at that point and instant of time.
    That is true, yes. Note that there may be other acceleration, too. In keeping with the notion of perpendicular coordinate systems and therefore perpendicular vector components, there could also be a "linear" acceleration tangent to the curve at that that same instant. If centripetal acceleration is constant and non-zero for all time, and linear acceleration is constant and zero for all time, then you have true circular motion.
    Again, centripetal acceleration is a component the acceleration as judged from a different coordinate system. One doesn't usually want to mix them all together in the same bag without careful consideration of how the various coordinate systems are related. The instantaneous total acceleration can always be had by adding together the component vectors in a given coordinate system. But that is the total acceleration as expressed in that coordinate system. You would have to know how to convert from one coordinate system to another to express the same vector in another coordinate system.
     
  7. Feb 24, 2012 #6
    I had never heard of instantaneous radius of curvature before you mentioned it in post #16 of the other thread and also here. I looked it up in my textbook to read up on it, and while it exists in the index, it begins on page 1116 and 1133, and my text book only goes up to page 685. I have volume one of "University Physics" by Young and Freedman. Acceleration vectors are covered in chapter 2 and 3 of my book. Why would something essential to understanding acceleration vectors be non existent in the fundamental chapters that cover acceleration vectors along curves?


    Is it a goal of the book to not explain acceleration in its entirety?
     
  8. Feb 24, 2012 #7

    gneill

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    No, that is not the goal of the book. The book presents concepts in sequence as required. It's not likely that you'll be needing to apply the concept of instantaneous radius of curvature until much later. I mentioned it because it was applicable to your query on types of acceleration where a particle is following some general trajectory. In your present studies the radius of curvature that you'll run into will almost invariably be associated with circular motion wherein the curvature remains constant and the concepts of that form of motion are presented fairly early on.
     
  9. Feb 24, 2012 #8
    So, I suppose you're also saying that, in this case, it may not be absolutely essential or necessary to have to read the 1000 pages of physics text in order to understand instantenous radius of curvature as it relates to the types of acceleration components that i'm asking about. Its just, as you say, a sequence of concepts the book requires. If understanding instantaneous radius of curvature can bring more clarity to acceleration vectors along a curve, then i'd gladly take the extra time to examine those concepts if they are within my ability to understand it.

    I'm going to begin searching for info on instantaneous radius of curvature. In the meantime, I do have a few more questions about acceleration along curves.

    Okay, so, we have established that the x and y axis position equations as a function of time are the equations that describe motion along a trajectory. And if one or both of those position equations are to the power of 2, then there will be some constant acceleration in some direction. So, in regard to:

    Then it should be able to be said that since the motion of an object follows the position equations, if the position equations represented the path of a circle, then an object would have a circular motion.

    However, coming up with an equation that describes the path of a circle as a function of time is proving difficult. I know the equation of a circle is r = [itex]\sqrt{(x-h)^2 + (y-k)^2}[/itex]. But that is not a function. It can be made into a piecewise function where f(x) = ±√(1-x^2), but I don't think that describes position in terms of x and y components on a graph, nor does it seem to describe position as a function of time.

    If I could find an equation for both x and y axes that describes motion of a circle, it would seem I could then take the second derivative of those functions and calculate circular acceleration as a function of time.

    So the problem here is: how could I find an x and y components of a position equation as a function of time that describes a circlular motion?
     
  10. Feb 24, 2012 #9

    gneill

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    Extra investigation outside basic curriculum can be excellent, particularly if it broadens understanding or at least sheds light on why the current topics of study will form an important basis for future methods. The concepts presented at the current level will be elaborated upon with 'heavier machinery' (mathematical methods, concepts, techniques) in the future.
    I will mention that there are other ways that acceleration can change besides a dependance on time. For example, the forces experienced by an object can also be a function of position. Such is the case, for example, in a gravitational field when we're not restricted to a small region close to the surface of the Earth. We approximate the gravitational acceleration with g for such cases, but the reality is the acceleration changes with radial distance from the center of the Earth.
    The 'failure' you describe above is a failure of the chosen coordinate system for describing the problem. Quite simply, a circle is not a function. You can get around the problem through the use of piecewise continuous functions to describe the motion (as in the two halves of the circle you described above), or more elegantly, change the coordinate system to something more conducive to the problem at hand. In this case a polar coordinate system might be appropriate. It describes position as a function of radius and angle. Integrations and derivatives can be done in such a coordinate system too, with the related calculus techniques for such coordinates. You'll probably run into this subject when you cover planetary motion (astrodynamics).
     
  11. Feb 24, 2012 #10


    I am eating my words from post #17 of the other thread https://www.physicsforums.com/showthread.php?t=580725&page=2

    It would seem that simply being familiar with the unit circle would not qualify one as being proficient with the polar coordinate system . I just looked at my trig book and can confirm that most of chapter nine, which contains polar coordinates, was skipped in my trig course.

    So before moving on, I think it would be imparative to do some reading. Thanks.
     
  12. Feb 25, 2012 #11
    Okay, here is what I have so far:


    The equation of a line is [itex]x^{2}+y^{2}[/itex] = r[itex]^{2}[/itex]

    And the trig identity for a circle is cosθ[itex]^{2}+sinθ^{2} = 1 [/itex]

    so x = cosθ and y = sinθ

    or as a function of time, x(t) = cost and y(t) = sint

    However, I'm not sure whether those would be the position functions or if

    x(t) = cos[itex]^{3}[/itex]t and y(t) = [itex]sin^{2}[/itex]t

    I'm leaning toward the latter since it more resembles the circle equation.

    Or... i could be going in a completely wrong direction here.

    Assuming i'm heading in the right direction, I still can't figure out how I would use those x and y component functions to actually graph a circle.
     
  13. Feb 25, 2012 #12

    gneill

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    This is still whacking away at the problem with Cartesian coordinates (note that you're writing expressions like x(t) = .... ).

    In polar coordinates we do away with x and y coordinate pairs (x,y) and replace them with two new coordinates: (r,θ). That is, radius and angle, the radius being measured from a defined origin and the angle being measured from a defined reference direction. Both r and θ can be made to depend upon functions of time just as x and y were.

    In polar coordinates the equation of a circle is simply r = c, where c is a constant, and it is assumed that θ can take on all values -∞ < θ < ∞ . Defining a circle in polar coordinates is as trivial as defining a horizontal line in Cartesian coordinates: y = c.

    That's it. It defines the radius of the function to be some constant value. If the circle is the trajectory that some particle traces out over time then θ becomes a function of time, θ(t). One such function for constant speed around the circle might be θ(t) = ω t.

    In polar coordinates we deal with radial and angular components for quantities. Thus radial speeds and accelerations and angular speeds and angular accelerations. Since these correspond to the similar magnitude and direction characteristics of vectors, the polar coordinates topic forms an important basis before introducing vector calculus... and so it goes with the layering of concepts.
     
  14. Feb 25, 2012 #13

    Ah makes sense. So I could express that in coordinate as (r, -∞<θ<∞). Probably not seeing a specified θ may have thrown me off.


    Okay, I see how this is a function of angle in terms of time. However, I can't relate it to (r,θ) just yet since, while I know θ, I don't yet know r (magnitude), right? So I have to be able to represent r in there somewhere. I believe that Δθ(r)/Δt = velocity. So would I be able to make the function that describes position as a function of time by rearranging that equation to: θ(t) = v(t)/r ? That way there is a definate r and θ for any value of time and it fits into the polar coordinates (r,θ).


    The layering of concepts within the book are becoming more apparent now.
     
  15. Feb 25, 2012 #14

    gneill

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    For circular motion r is a fixed constant, r = c.
    r(t) = c ; θ(t) = ωt

    describes circular motion. ω is called the angular velocity (or angular speed). The instantaneous linear speed of the object would be v = ωr. In polar coordinates the centripetal acceleration would be v2/r = ω2r.
    :smile:
     
  16. Feb 25, 2012 #15
    Okay, so for an object traveling in a circular path as a function of time, it doesn't get assigned seperate equations describing any component of the motion. There's just one equation that describes its motion. And the equation describing the object's position as a function of time would be θ(t) = ωt. No component equations. And that can easily fit into the polar coordinates (r(t), θ(t)) even though r is not specifically expressed in the equation because it is a constant.

    And if I wanted to describe the instantaneous velocity of the object, i'd take the first derivative of θ(t) = ωt, which would be v(t) = ωr = (Δθ(r)/Δt), but that looks tricky because derivatives are defined on the x y axis. So how would you find the derivative of a function on the polar coordinate system in order to go from position to velocity? Or is this not possible because they are apples and oranges. The instanenous velocity being described by the cartesian coordinate system and the θ(t) being described by the polar coordinate system?

    For instance, if i'm describing the motion of an object on a circular path, I would use the polar coordinate system to find the object's position as a function of time using (r,θ). However, if I then wanted to find the angle of the objects instantaneous velocity as it relates to the x y coordinate system, would I then need to incorporate taking the derivative of some position function?
     
  17. Feb 25, 2012 #16

    gneill

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    The components of the motion ARE r(t) and θ(t) in the polar coordinate system. The angular velocity is the rate of change of the angle, thus dθ(t)/dt = ω.
    Derivatives and integrals are not confined to any particular coordinate system. They can be taken in any coordinate system, but you need to know how to identify the differential elements in that system.

    If you want to know the instantaneous speed in the Cartesian frame given that you have the vector components of the position in the polar frame, you take the derivative of the components as usual and then add the resulting vector components in Cartesian style. In the case of circular motion r is a constant so dr/dt is zero, while dθ/dt = ω. The speed is then given by

    ## u = \sqrt{(\omega r)^2 + 0^2} = \omega r ##

    and being a scalar quantity this speed is the same regardless of coordinates (provided that the coordinate systems are mutually at rest), so it is the speed in the Cartesian frame, too.
    The position is described as a vector in either coordinate system. It is the same vector "in reality" despite being described differently (x y coordinates or r θ coordinates). To find the velocity vector you apply the vector calculus applicable to the given coordinate system. It's still derivatives in both cases, but you need to know the applicable calculus rules. You'll cover this when you get to polar coordinates and vector calculus.
    Fortunately, for strictly circular motion the velocity vector happens to always be perpendicular to the position vector. So if you known the θ for the position vector you can add (or subtract as the case may be) 90° to find the direction of the velocity vector.
     
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