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!

Irodov Equilateral Triangle Problem

  1. Jun 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Hello, so I'm going to be a freshman physics major next year, and over the summer I'm trying to improve my problem solving skills. I've started going through Irodov's problems in general physics, and I have a question about two different solutions to one of the problems.

    1. The problem statement, all variables and given/known data

    Here's the problem:

    Three points are located on the vertices of an equilateral triangle whose side equals a. They all start moving simultaneously with velocity v constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?

    2. Relevant equations



    3. The attempt at a solution

    And here's my first solution:

    Each point then moves towards the one in front of it with velocity v. Since all of the particles are moving with equal velocities, all of the sides remain equal in length. Once the particles begin moving, the triangle will then begin to simultaneously shrink and rotate through a spiral. We approach the problem by looking at the area of this equilateral triangle.

    The area of the triangle will be given by the following equation:

    A = a^2(sqrt 3)/ 4

    We take the derivative of this area with respect to the variable a (length of each side of the triangle):

    a(sqrt 3)/2

    Next, we examine the velocity of the particles, which will remain constant throughout the shrinkage of the triangle:

    v = da/dt

    We then multiply the derivative of the area with respect to the side length (dA/da) by the velocity (da/dt) to get the derivative of the area of the triangle with respect to time:


    dA/dt = (v)a(sqrt 3)/2

    Now, we have the rate of change of the area of the triangle with respect to time. Next, we find the net change in area of the triangle and divide it by the rate of change of the area with respect to time to find the time necessary to make that change occur.

    This net change will equal the original area of the triangle, since we are attempting to find the time needed for the particles to converge (I.e: for the area of the triangle between them to equal zero). Thus, we take the area of the triangle at t = 0 and divide it by dA/dt:

    a^2(sqrt 3)/4 over va(sqrt 3)/2 = a /2v

    Thus, the time needed for the points to converge is a/2v.

    Now, I thought my math was foolproof, however when I searched for alternative solutions to the problem, I got the following:

    "Now this is a tricky question and one has to understand the kinematics behind it carefully. Few points one should remember are:
    1. Since the triangle is equilateral and all the points are moving a same speed and uniform acceleration (change in direction is also change in velocity) the points would converge at a point equidistant from all the vertices of the triangle which is or the circumcentre of the triangle. In this case circumcentre is also the ortho-centre.
    2. This circumcentre would remain the circumcentre of all the subsequent triangles the points would make in course of time as they move at equal speed towards each other.
    3. The velocity component of each point towards the circumcentre at any point in time would remain vCos(30). (Draw a triangle and check this)

    Hence the initial distance of the points from the circumcentre = a/√3
    Relative velocity of each point towards circumcentre = vCos(30)
    Hence the time elapsed to converge = [a/√3]/[vCos(30)]"

    They get a time of 2a/3v, which appears to be equally valid according to their reasoning.

    Is there some kind of conceptual error I made in this problem? I can't seem to find any errors in my reasoning.

    Thanks for your help.
     
  2. jcsd
  3. Jun 6, 2012 #2
    Hello, hello95!! :tongue2:!

    Welcome to PF! :smile:

    The problem with your method is that the rate of change of area itself changes with time :wink: So dividing total area by the initial rate of change of area will not give you the correct answer.
     
  4. Jun 6, 2012 #3
    Wow, I can't believe I missed that. Thank you.
     
  5. Jun 6, 2012 #4
    Would it work to take the average rate of change of area and divide the initial area by that? The average rate would simply be the rate I got divided by two, since the second derivative is constant, correct? Then I would simply multiply the time by two to get a/v, which is still different from the alternative solution.
     
  6. Jun 6, 2012 #5
    I forgot to point out another mistake in your answer..

    This is not the velocity of the particle. It is the rate of change of separation between particles. So, velocity relation in the beginning of motion would be,

    [tex]v(1+cos\frac{\pi}{3}) = \frac{da}{dt}[/tex]

    Now apply the average rate change concept.
     
  7. Jun 7, 2012 #6
    Thanks!
     
  8. May 30, 2013 #7
    What will be the equation of the trajectory of all the three particles
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Irodov Equilateral Triangle Problem
  1. Irodov Problem (Replies: 1)

  2. Irodov Problem (Replies: 5)

Loading...