Irodov Equilateral Triangle Problem

In summary, a freshman physics major is trying to improve their problem solving skills by going through Irodov's problems. They have a question about two different solutions to a problem involving three points moving simultaneously towards each other in an equilateral triangle. The first solution considers the area of the triangle and calculates the time needed for the points to converge as a/2v, while the second solution takes into account the changing rate of change of area and considers the velocity of each point towards the circumcentre. The second solution calculates the time needed for the points to converge as 2a/3v. However, the first solution contains errors in the calculation of the rate of change of area and the velocity of the particles. The correct solution takes
  • #1
hello95
33
0

Homework Statement



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.

Homework Statement



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?

Homework Equations





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.
 
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  • #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.
 
  • #3
Wow, I can't believe I missed that. Thank you.
 
  • #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.
 
  • #5
I forgot to point out another mistake in your answer..

v = da/dt

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.
 
  • #6
Thanks!
 
  • #7
What will be the equation of the trajectory of all the three particles
 

1. What is the "Irodov Equilateral Triangle Problem"?

The "Irodov Equilateral Triangle Problem" is a mathematical problem proposed by the Russian physicist and mathematician, Igor Irodov, in his book "Problems in General Physics". It involves finding the ratio of the radii of three small circles inscribed within an equilateral triangle.

2. How do you solve the "Irodov Equilateral Triangle Problem"?

The solution to the "Irodov Equilateral Triangle Problem" involves using geometric principles, such as the properties of an equilateral triangle, the Pythagorean theorem, and the law of sines. It also requires some algebraic manipulation and trigonometric calculations.

3. Why is the "Irodov Equilateral Triangle Problem" considered challenging?

The "Irodov Equilateral Triangle Problem" is considered challenging because it requires a solid understanding of geometry, algebra, and trigonometry. It also involves multiple steps and concepts, making it a complex problem to solve.

4. What is the significance of the "Irodov Equilateral Triangle Problem"?

The "Irodov Equilateral Triangle Problem" is a popular problem in physics and mathematics, and it is often used to test students' problem-solving abilities and understanding of mathematical concepts. It also has practical applications in fields such as engineering and astronomy.

5. Are there any variations of the "Irodov Equilateral Triangle Problem"?

Yes, there are variations of the "Irodov Equilateral Triangle Problem" that involve finding the ratio of the radii of circles inscribed within other shapes, such as squares or regular polygons. These variations can also be solved using similar geometric and trigonometric principles.

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