12
$\begingroup$

I have come across a question that asked me to find the radius of curvature of a projectile.

As far as I know, the path of a projectile is a parabola and I have found mention of the radius of curvature referring to lenses and mirrors. But in optics, the lens and mirrors were assumed to be part of a circle.

My questions are:

  1. How can a parabola have a center from which a radius is to be measured?

  2. Does the radius of curvature change with the position of body (in projectile motion)?

  3. In mechanics and mathematics, what is the radius of curvature and how does one calculate it (in the case of a parabola)?

$\endgroup$
  • 2
    $\begingroup$ This is osculating circle. $\endgroup$ – user36790 Jun 19 '15 at 14:15
  • 3
    $\begingroup$ We normally require "one question per question" - but I think this is an example of a question where the three sub-questions are so closely connected that breaking it up would make no sense. $\endgroup$ – Floris Jun 19 '15 at 14:26
  • 1
    $\begingroup$ You are using the wrong meaning of that word. Check Radius of curvature (mathematics), not Radius of curvature (optics), and suddenly everything will make more sense. $\endgroup$ – Federico Poloni Jun 20 '15 at 9:40
32
$\begingroup$

So let's start with your last question, informally, the radius of curvature is a measure of how much a certain curve is pointy and has sharp corners. Given a curve $y$, you can calculate its radius of curvature using this formula:

$$\dfrac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^\dfrac{3}{2}}{\left|\dfrac{d^2y}{dx^2}\right|}$$

You might ask what radii of circles have to do with curvature, so it's worthwhile explaining it. This is a parabola:

enter image description here

As you can see, the sides of the parabola are pretty flat, whereas its vertex and the surrounding region (i.e.: at $x=0$) have a relatively sharp corner.

So the question is how to mathematically describe this property?

Well, one way to do it is to use circles. The part of the curve that is pretty flat can be considered to be a section of a really large circle(as shown in the picture), this circle has large radius, and hence we say this part of the curve has large radius of curvature, that is, it's very flat.

On the other hand, the vertex of the parabola and the surrounding region are relatively sharp and pointy, hence you'll notice it takes a circle with small radius to fit it on this edgy section of the parabola, we say this region has small radius of curvature.

You'll also note that, the radius of curvature for a curve, changes from one point on the curve to another, you'll further notice that, when the region is flat, the rate of change of the radius of curvature is small(you can use small number of huge circles to describe a flat region), whereas it takes a lot of circles with small radii to describe a sharp corner, and hence the rate of change of the radius of curvature is great at these regions.

How can a parabola have a center from which radius is to be measured?

No, it does not, but every point on the parabola and and the surrounding region can be regarded as a part of a circle with certain radius.

Does the radius of curvature changes with the position of body(in projectile motion)?

Yes, as stated earlier, the radius of curvature changes from point to point on a curve, since the path of the projectile can be modeled as its position on a parabola, hence the radius of curvature will change with the change of position of the projectile.

$\endgroup$
  • 4
    $\begingroup$ A great answer; +1. $\endgroup$ – user36790 Jun 19 '15 at 14:16
  • 2
    $\begingroup$ Very nicely explained. This answer deserves a lot of upvotes. $\endgroup$ – Floris Jun 19 '15 at 14:24
  • 1
    $\begingroup$ @Kai Yes, this is the equation that calculates your vertical position $y$ as a function of your $x$ coordinate, you should find the first and second derivative of this equation and plug it in the formula I gave you in my answer. $\endgroup$ – Omar Nagib Jun 19 '15 at 15:24
  • 1
    $\begingroup$ this is pure math.... and I like it :) $\endgroup$ – James S. Cook Jun 20 '15 at 3:44
  • 2
    $\begingroup$ This youtube.com/watch?v=bZW6aQmXf_M has a nice derivation complementing the picture, though intuitively you can get everything from the simple relation $s = r \theta$ and noticing $s = r \theta \rightarrow (1/r) = d \theta / ds$ then chain rule. So From $ s = r \theta$ we see curvature comes from asking how much does the angle $\theta$ change per unit of arc $s$, moving one unit of arc length $ds$ on the big circle produces a small change in angle, while the same distance on the small circle produces a bigger change in angle, hence more curved as is visually obvious. $\endgroup$ – bolbteppa Jun 20 '15 at 6:43
1
$\begingroup$

A parabola does not have a radius of curvature. Each point of the parabola does. So, if you are asked to find the radius of curvature on a parabola, you must also be give the point on the parabola.

$\endgroup$
1
$\begingroup$

We can draw a tangent (Call it 't') at any point (call it 'p')on a curve. We can also draw a circle to which 't' is a tangent. The radius of this circle is the radius of curvature to the given curve at the point 'p'.

An analogy from motion of a body along a curved path may help easier understanding.

When a body moves along a curved path, its velocity keeps changing. However, we can talk of the instantaneous velocity of the body at each and every point along the curve.

On similar lines, for a given curve the radius of curvature keeps changing along the curve. However, we can talk of radius of curvature at each and every point along the curve. It refers to the radius of the circle which has a common tangent with the given curve at the point under consideration.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy