1. Not finding help here? Sign up for a free 30min 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!

Radius of an Arc Inside an Arc

  1. Jul 3, 2010 #1

    IBY

    User Avatar

    1. The problem statement, all variables and given/known data
    Say there are two arcs which are part of the circumference of a circle within a circle with the same angles, but since one arc belongs to the circle outside, that arc is longer than the arc inside it. If the first arc has lenght s, then the second arc has length [tex] s + \Delta s [/tex]. The distance between the two arcs are [tex]\Delta r[/tex]. Supposing that we don't know what the lenght of the inner circle is, and the information we know is the angle of the arcs, the difference of lenght between the two arcs, and the distance between them, what is the radius of the inner circle, supposing that the outer circle is [tex]r+\Delta r[/tex]?

    This is a problem I made up, so there is no hurry in answering it. Anyways, I want someone to check the logic of my procedure because when I plugged in some numbers, they did not compute.
    2. Relevant equations

    [tex]\theta=\frac{s}{r}=\frac{s+ \Delta s}{r+ \Delta r}[/tex]

    [tex]\Delta r=\frac{s+\Delta s}{\theta}-\frac{s}{\theta}[/tex]
    (the second one is the one where I start from, basically the distance between the two arcs is the radius of the outside minus the radius of the inside)

    3. The attempt at a solution

    [tex]\Delta r=\frac{s+\Delta s}{\theta}-\frac{s}{\theta}[/tex]

    Since s over theta is the radius:

    [tex]\Delta r=\frac{s+\Delta s}{\theta}-r[/tex]

    And since theta is s over r:

    [tex]\Delta r=(s+\Delta s)\frac{r}{s}-r[/tex]

    [tex]\Delta r=(1+\frac{\Delta s}{s})*r-r[/tex]

    Factoring out the r:

    [tex]\Delta r=r*((1+\frac{\Delta s}{s})-1)[/tex]

    [tex]\Delta r=r*\frac{\Delta s}{s}[/tex]

    Looking for r, I get:

    [tex]r=\Delta r \frac{s}{\Delta s}[/tex] ***

    Now, using equation 1 of the relevant equations:

    [tex]\theta=\frac{s+ \Delta s}{r+ \Delta r}[/tex]

    I solve for s, which is:

    [tex]\theta*r+\theta*\Delta r-\Delta s[/tex]

    Now, using the equation I asterisked, I substitute for s:

    [tex]r=\Delta r*\frac{\theta*r+\theta*\Delta r-\Delta s}{\Delta s}[/tex]

    I simplify:

    [tex]r=\frac{\theta*r*\Delta r}{\Delta s}+\frac{\theta*\Delta r^2}{\Delta s}-\Delta r[/tex]

    Substracting both sides with the term at the right with the r in it, I get:

    [tex]r-\frac{\theta*r*\Delta r}{\Delta s}=\frac{\theta*\Delta r^2}{\Delta s}-\Delta r[/tex]

    Factoring out r, I get:

    [tex]r(1-\frac{\theta*\Delta r}{\Delta s})=\frac{\theta*\Delta r^2}{\Delta s}-\Delta r[/tex]

    Solving for r:

    [tex]r=\frac{\frac{\theta*\Delta r^2}{\Delta s}-\Delta r}{1-\frac{\theta*\Delta r}{\Delta s}}[/tex]
     
  2. jcsd
  3. Jul 3, 2010 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member



    [tex]\Delta r=\frac{s}{\theta}+\frac{\Delta s}{\theta}-\frac{s}{\theta}=\frac{\Delta s}{\theta}[/tex]

    Δs/Δr= θ, they are not independent quantities.


    ehild
     
  4. Jul 3, 2010 #3

    IBY

    User Avatar

    I am sorry to tell you that I don't know what you mean by that. :confused: Perhaps you could clarify that part to me a bit more.
     
  5. Jul 3, 2010 #4

    ehild

    User Avatar
    Homework Helper
    Gold Member

    I mean that you can not choose Δs and Δr and θ arbitrarily, and you can not find r from them. Your resulting formula for r is equivalent to r=0/0.
     
  6. Jul 3, 2010 #5

    IBY

    User Avatar

    Does this mean the problem is unsolveable?
     
  7. Jul 3, 2010 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes.

    ehild
     
  8. Jul 3, 2010 #7

    IBY

    User Avatar

    Ok, thanks!
     
  9. Jul 3, 2010 #8

    Char. Limit

    User Avatar
    Gold Member

    Wait, but it isn't. You're thinking of differentials. Since he has no limits in his equation, he's not dealing with differentials. So, there's no 0/0 involved.
     
  10. Jul 4, 2010 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Sorry, I do not understand. What do you speak about?

    ehild
     
  11. Jul 4, 2010 #10
    A diagram explaining your problem would really help. From your explanation, this is what I'm imagining:

    sqfgZ.png

    So, to define our variables:
    • θ is the angle AOB
    • r = AO = BO
    • Δr = AC = BD
    • s = lesser arc AB
    • s + Δs = lesser arc CD
    Is that what you had in mind?

    I've also attached file, in case the image above goes down in the future.
     

    Attached Files:

    • IBY.bmp
      IBY.bmp
      File size:
      107.6 KB
      Views:
      84
    Last edited: Jul 4, 2010
  12. Jul 4, 2010 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, this is correct.

    Why did you go back to r? Just do the subtraction:
    [tex]\Delta r= \frac{s}{\theta}+ \frac{\Delta s}{\theta}- \frac{s}{\theta}= \frac{\Delta s}{\theta}[/itex]

    Isn't that what you want?

     
  13. Jul 5, 2010 #12

    Mentallic

    User Avatar
    Homework Helper

    It is unsolvable because the values of [tex]\Delta r, \Delta s, \theta[/tex] don't depend on the inner circle's radius. The only relationship you need (and can deduce for that fact) is [tex]\Delta s=\frac{2\pi \Delta r}{\theta}[/tex] which can be easily shown.
     
  14. Jul 5, 2010 #13

    ehild

    User Avatar
    Homework Helper
    Gold Member

    This problem was solved long ago in this thread, and it is really very simple, I do not understand why was it started again.
    The length of arc is angle times radius, just like mass is density times volume. If you have two pieces of gold, and you know the density and you know that one piece has 2 g more mass than the other, you can find the difference of the volumes, but there is no way to find out the individual volumes or masses.

    ehild
     
    Last edited: Jul 5, 2010
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Radius of an Arc Inside an Arc
  1. Arc Tangent Squared (Replies: 3)

  2. Arc lenght (Replies: 11)

Loading...