1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Describing the radius with density

  1. Jul 13, 2013 #1
    How can I describe mathematically that if you take some mass [itex]m[/itex] with some density [itex]d[/itex] as the radius tends to [itex]0[/itex] the density tends to [itex]\infty[/itex]

    I'm not exactly sure what it is I'm trying to describe I'm just looking for something that describes when the radius of an object decreases, the density increases.

    How can I write this using mathematical notation?
     
  2. jcsd
  3. Jul 13, 2013 #2
    You could say [itex]d \propto r^{-1}[/itex], or more accurately [itex]d \propto v^{-1}[/itex].
     
  4. Jul 13, 2013 #3
    Ok so how would I read that?
     
  5. Jul 13, 2013 #4
    Density is inversely proportional to volume, given constant mass.
     
  6. Jul 15, 2013 #5
    OK I think I understand. so let's just use some random figures. If I had a steel ball bearing which had a radius of 8cm, a mass of 1lb, volume of 2144.66 and a density of 2.76 g/cm3. If I decreased the radius by half, what factor does the density increase by?
     
  7. Jul 25, 2013 #6
    This will fully explain.

    Remember,
    [itex]D=\frac{m}{V}[/itex]
    and, for a sphere,
    [itex]V=\frac{4}{3}[/itex][itex]\pi[/itex]r2
    so if
    [itex]V[/itex]1=[itex]\frac{4}{3}[/itex][itex]\pi[/itex]r2
    with full radius r, then
    [itex]V[/itex]2=[itex]\frac{4}{3}[/itex][itex]\pi[/itex][itex](\frac{r}{2})[/itex]2
    with half the radius or
    [itex]V[/itex]2=[itex]\frac{1}{4}[/itex][itex](\frac{4}{3}[/itex][itex]\pi[/itex]r2)
    which yields the proportion
    [itex]V[/itex]2=[itex]\frac{1}{4}[/itex][itex]V[/itex]1

    So, since
    [itex]D[/itex]1=[itex]\frac{m}{V_{1}}[/itex]
    and
    [itex]D[/itex]2=[itex]\frac{m}{V_{2}}[/itex]
    then
    [itex]D[/itex]2=[itex]\frac{4(m)}{V_{1}}[/itex]
    and
    [itex]D[/itex]2=[itex]4D[/itex]1

    Notice how they're inversely proportional by a square of the radius? (I mostly wrote this up to practice Latex, but I hope this helped)
     
  8. Jul 25, 2013 #7

    chiro

    User Avatar
    Science Advisor

    Hey MathJakob.

    Take what the above posters have said and pose your problem in terms of mathematical limits.
     
  9. Jul 28, 2013 #8
    Well mathematically all I can think of is [tex]\lim_{r \to 0} d \to \infty[/tex] As the limit of radius goes to 0, the limit of density goes to infinity
     
  10. Jul 28, 2013 #9
    Syntax wise, that's not quite right.

    The limit of density as radius tends to 0 is infinity (a lot of people say "grows without bound.")

    But, by syntax wise, I mean that you spoke of two limits, but this is one limit. Furthermore, you would write:

    [tex]\lim_{r \to 0} d = \infty[/tex]

    (An equals sign as opposed to a second arrow)

    But this is just my personal experience, it may be the way that you wrote it is acceptable, it just looks and sounds strange to me.
     
  11. Jul 28, 2013 #10

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This just does not work. You have SPECIFIED a specific and fixed density. If you compute the the mass of your sphere given the above data either the density is incorrect or the mass is you cannot have it both ways. Given your density and radius the mass must be near 6kg or 13lbs.

    Why are you mixing systems? Specify your quantities in either metric or American, don't mix.

    I think what you want to do is hold mass constant then shrink the radius. Now as the radius tends to zero the density will grow without bound.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Describing the radius with density
  1. Radius of an Ellipse (Replies: 6)

  2. Radius of Ellipsoid (Replies: 13)

  3. Triangulated radius (Replies: 8)

Loading...