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Centripetal acceleration vector

  1. Oct 4, 2013 #1
    Purely mathematically, how does the expression for the centripetal acceleration a=v^2/R give information about the direction of the acceleration vector? I don't seem to notice any indication that the acceleration vector should be perpendicular to the velocity, or pointing towards the center.

    It's more of a conceptual question, really: I understand that we get the result with the ASSUMPTION that the acceleration has such direction, but I am only wondering whether it should be explicit in the result.

    Thank you.
     
  2. jcsd
  3. Oct 4, 2013 #2

    jhae2.718

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    [itex]\renewcommand{\vec}[1]{\boldsymbol #1}[/itex]
    The acceleration vector comes from performing vector kinematics. Let's consider a case of motion of a point mass in the plane.

    Consider a rotating reference frame [itex]b^+[/itex] relative to an inertial frame [itex]n^+[/itex], where the ##\hat{\vec{b}}_1## unit vector always points toward the point mass.
    RBwuEmC.png
    We can write a position vector of a point mass in the ##b^+## frame as:[tex]
    \vec{p}=r\hat{\vec{b}}_1
    [/tex]where the ##\hat{\vec{b}}_i\text{'s}## are the unit vectors defining ##b^+##. In the plane, we can write the angular velocity of the frame ##b^+## relative to the ##n^+## frame as [itex]\vec{\omega}_{b/n} = \dot{\theta}\hat{\vec{b}}_3[/itex].

    There is a neat result that we call the kinematic transport theorem. Simply put, it says that the derivative of a vector in one frame ##b^+## as viewed by an observer in another frame, say ##a^+##, is made up of two parts: 1) the change of the vector in the ##b^+## frame as viewed by an observer in the ##b^+## frame, and 2) the angular velocity of ##b^+## relative to ##a^+##. Mathematically,[tex]\frac{{}^a\text{d}\vec{r}}{\text{d}{t}} = \frac{{}^b\text{d}\vec{r}}{\text{d}{t}} + \vec{\omega}_{b/a} \times \vec{r}[/tex]
    We can apply this to our position vector to get an inertial velocity vector:[tex]
    \begin{align*}
    \frac{{}^n\text{d}\vec{p}}{\text{d}{t}} &= \frac{{}^b\text{d}\vec{p}}{\text{d}{t}} + \vec{\omega}_{b/n}\times \vec{p}\\
    &= \dot{r}\hat{\vec{b}}_1 + \dot{\theta}\hat{\vec{b}}_3\times r\hat{\vec{b}}_1\\ \vec{v} = \frac{{}^n\text{d}\vec{p}}{\text{d}{t}} &= \dot{r}\hat{\vec{b}}_1 + r\dot{\theta}\hat{\vec{b}}_2\end{align*}[/tex]
    Apply the KTT once more to obtain the inertial acceleration:[tex]
    \begin{align*}
    \vec{a} = \frac{{}^n\text{d}\vec{v}}{\text{d}{t}} &= \frac{{}^b\text{d}\vec{v}}{\text{d}{t}} + \vec{\omega}_{b/n}\times \vec{v}\\
    &= \ddot{r}\hat{\vec{b}}_1 + \left(\dot{r}\dot{\theta} + r\ddot{\theta}\right)\hat{\vec{b}}_2 + \dot{\theta}\hat{\vec{b}}_3 \times \left(\dot{r}\hat{\vec{b}}_1 + r\dot{\theta}\hat{\vec{b}}_2\right)\\
    &= \ddot{r}\hat{\vec{b}}_1 + \left(\dot{r}\dot{\theta} + r\ddot{\theta}\right)\hat{\vec{b}}_2 + \dot{r}\dot{\theta}\hat{\vec{b}}_2 - r\dot{\theta}^2\hat{\vec{b}}_1 \\
    \vec{a} &= \left(\ddot{r} - r\dot{\theta}^2\right)\hat{\vec{b}}_1 + \left(2\dot{r}\dot{\theta} + r\ddot{\theta}\right)\hat{\vec{b}}_2
    \end{align*}[/tex]
    From this last result, we have the centripetal acceleration given by ##-r\dot{\theta}^2\hat{\vec{b}}_1## (this result may look more familiar if we define ##\omega \equiv \dot{\theta}## and write ##-r\dot{\theta}^2\hat{\vec{b}}_1 = - r\omega^2\hat{\vec{b}}_1 = -v^2/r\hat{\vec{b}}_1##). But we know that ##\hat{\vec{b}}_1## points outwards, so the centripetal acceleration must point toward the center.

    You can easily extend this result to 3-D.
     
    Last edited: Oct 4, 2013
  4. Oct 4, 2013 #3
    If you differentiate the velocity vector of a particle undergoing uniform circular motion you should get a vector perpendicular to it pointing towards the centre.
     
  5. Oct 4, 2013 #4

    A.T.

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    You mean the net acceleration vector in uniform circular motion? If it wasn't perpendicular to velocity, the speed would change, so it would not be uniform circular motion. In this later case the velocity-perpendicular component of an arbitrary acceleration is often called "centripetal" (the other component being "tangential"). That makes "centripetal acceleration" perpendicular to velocity per definition.
     
  6. Oct 4, 2013 #5

    Andrew Mason

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    Somewhere between Newton and the 20th century the convention developed to call the component of acceleration that is perpendicular to the velocity of the body "centripetal". I am not sure why that occurred. Newton appears to have coined the term in his 1684 treatise "De motu corporum in gyrum" (On the motion of orbiting bodies). It is a confusing term if it is applied to a body in something other than circular orbit.

    AM
     
  7. Oct 5, 2013 #6
    jhae, that was the precise answer I was looking for.

    Andrew Mason, thank you, but I don't think the term is confusing at all. Mine was a question regarding the explicitness of the vectorial nature of the result.
     
  8. Oct 5, 2013 #7

    Philip Wood

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    Think this is quick and straightforward...

    Let [itex]\widehat{\textbf{n}}[/itex] be the unit vector normal to the plane containing the circle.
    Then the velocity vector will be at right angles both to [itex]\widehat{\textbf{n}}[/itex] and to the radius vector, so (for anticlockwise rotation in a circle drawn on the page, with [itex]\widehat{\textbf{n}}[/itex] pointing out of the page)

    [tex]\widehat{\textbf{v}} = \widehat{\textbf{n}} \times \widehat{\textbf{r}}[/tex]
    that is [itex]\frac{\textbf{v}}{v} = \widehat{\textbf{n}} \times \frac{\textbf{r}}{r}[/itex], that is [itex]\textbf{v} = \frac{v}{r}\widehat{\textbf{n}} \times \textbf{r}[/itex].

    Remembering that v, r and [itex]\widehat{\textbf{n}}[/itex] are all constants:

    [itex]\frac{d\textbf{v}}{dt} = \frac{v}{r} \widehat{\textbf{n}} \times \frac{d\textbf{r}}{dt}[/itex] that is [itex]\frac{d\textbf{v}}{dt} = \frac{v}{r} \widehat{\textbf{n}} \times \textbf{v}[/itex] that is [itex]\frac{d\textbf{v}}{dt} = \frac{v^2}{r} \widehat{\textbf{n}} \times \widehat{\textbf{v}}[/itex] so [itex]\frac{d\textbf{v}}{dt} = \frac{v^2}{r} (-\widehat{\textbf{r}})[/itex].
     
    Last edited: Oct 6, 2013
  9. Oct 6, 2013 #8

    Andrew Mason

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    My point is that the direction of the centripetal acceleration vector, [itex]\frac{v^2}{r}\hat{n} \text{ where } \hat{n} \text{ is the unit vector normal to the velocity }[/itex] is not necessarily the direction of acceleration. It is only the direction of acceleration for uniform circular motion. It is confusing because the term "centripetal" means "centre seeking" and [itex]\hat{n}[/itex] is not toward the centre, except for uniform circular motion.

    AM
     
  10. Oct 6, 2013 #9

    A.T.

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    It is towards the instantaneous centre of path curvature.
     
  11. Oct 6, 2013 #10

    Andrew Mason

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    I am not sure what that means. How is that point determined? For example, how would you determine the centripetal acceleration of a projectile in flight? If the point keeps changing it is confusing to call it centripetal. That's all I am saying.

    AM
     
  12. Oct 7, 2013 #11

    A.T.

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  13. Oct 7, 2013 #12
    If the centripetal acceleration is a=v^2/R you are dealing with a Circular orbit.

    For a non-Circular orbit: [tex]a=\frac{|v|^2}{\rho} * u_n[/tex]where [tex]\rho = \frac{1}{\kappa}[/tex] and ##\kappa## is the curvature. and ##u_n## is the normal.
     
  14. Oct 9, 2013 #13

    Andrew Mason

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    I understand how to determine the radius of curvature. The problem is that the radius keeps changing if the motion is not circular.

    The issue is very simple: an object prescribing non-circular curved motion about a central point (zero torque) is always accelerating always toward that central point. It is incorrect to speak about acceleration toward an arbitrary point that it is not accelerating toward (let alone one that keeps changing). If the body is not experiencing a change in tangential speed, then the motion will be circular and the acceleration is indeed centripetal acceleration in the direction of the radius of curvature (which is constant). But if it is experiencing a change in tangential speed, the body's acceleration is not along the radius of path curvature. Its acceleration is always toward the central point.

    For non-circular curved motion, it would be correct to refer to the component of acceleration in the direction of the radius of curvature, which is necessarily the component of acceleration normal to the velocity of the body. Since the body does not accelerate in the direction of the radius of curvature it is a wrong to call this vector the "centripetal acceleration".

    One could call it an instantaneous centripetal component of acceleration . But since Newton called the acceleration toward the central point "centripetal" it is somewhat confusing to call it "centripetal" at all.

    AM
     
  15. Oct 9, 2013 #14

    Philip Wood

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    An interesting case is the acceleration of a planet at the extremes of the major axis of its ellipse. The centres of curvature at these extremes lie on the major axis, as done the Sun, but all in different places on the major axis. So the forces on the planet at either extreme are directed, as we know, towards the Sun, but co-incidentally also towards the centres of curvature! Therefore, I argue, we are allowed to equate the forces to [itex]\frac{mv^2}{\rho}[/itex]. But, appealing to the symmetry of an ellipse, [itex]\rho[/itex] is the same at each end, A and P, of the major axis, so
    [tex]\frac{F_P}{F_A} = \frac{v_P^2}{v_A^2}.[/tex]
    But it's easy to show from Kepler's equal area law that
    [tex]r_Pv_P = r_Av_A.[/tex]
    in which [itex]r_P, r_A[/itex] are distances of the planet from the Sun at P and A.
    So [itex]\frac{F_P}{F_A} = \frac{r_A^2}{r_P^2}.[/itex]
    which is the inverse square law!
     

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    Last edited: Oct 10, 2013
  16. Oct 9, 2013 #15

    Andrew Mason

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    Of course. But at that point there is no tangential acceleration so the acceleration is indeed all toward the central point (which is the focus of the ellipse at which the sun is located), which is also the centre of curvature, and therefore, the acceleration toward the centre of curvature is indeed centripetal.

    AM
     
  17. Oct 9, 2013 #16

    Philip Wood

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    Neither the centre of curvature at P nor the c of c at A is at either focus.
     
  18. Oct 9, 2013 #17

    D H

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    No. You've cherry-picked the two points where this happens to work. What you can say is that ##\frac{v^2}{\rho}## is equal to the normal component of acceleration. But that's true for any space curve.

    You already implicitly assumed an inverse square law force, so it's not that surprising that you derived an inverse square law force.
     
  19. Oct 10, 2013 #18

    Philip Wood

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    DH Don't follow your criticism. (1) Of course I've cherry-picked these two points; they are the only two points relevant to my argument! (2) Where have I implicitly assumed the inverse square law? What I have assumed is Kepler's first Law (elliptical orbit with Sun at one focus) and Kepler's second law (equal areas in equal times). By cherry-picking two special points on the ellipse, I've deduced the inverse square law.

    I've also assumed that the force on the planet is central, directed towards one focus, and is a simple power law. So my argument is in no way a substitute for Newton's proof that an elliptical orbit with equal areas in equal times implies an inverse square law force directed towards a focus.

    Postscript: Post 14 now contains diagram in attachment.
     
    Last edited: Oct 10, 2013
  20. Oct 10, 2013 #19

    D H

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    Exactly. Your argument fails at any other points. That's cherry-picking.

    No, you haven't deduced the inverse square law. First off, you used the only two points where ##v^2/\rho = a## and used these cherry-picked points to make a universal deduction. That's not valid.

    Secondly, you've deduced something that you already assumed. There are only two classes of radial central forces that can result in elliptical motion, force proportional to distance and force inversely proportional to distance squared. You've put the object responsible for that central force at one of the foci of the ellipse, so you're not talking about a harmonic well. You've assumed an inverse square law without knowing it.
     
  21. Oct 10, 2013 #20

    Andrew Mason

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    The argument is more difficult to make at other points. But if you assume that the force is a function of distance from the sun and is radial, all you need are two points (ie. at which you know the speed and the distance from the sun). I don't think Philip has made any implicit assumption about a 1/r^2 force.

    By applying Kepler's second law or the conservation of angular momentum (since there is no torque if the force is central), we can conclude that mvr is constant which implies that v is proportional to 1/r. Since at these two points there is no tangential acceleration (since the force is radial and the velocity is perpendicular to the force, there is no tangential component to the force), the acceleration = v^2/r. This implies that the force is mv^2/r.

    I would agree if you assume that the orbit is an ellipse. But all one has to assume is that it is a closed loop of some kind, that the force is central and is a function of distance. Then I think that all you need are two points where the tangential acceleration is zero.

    AM
     
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