# Particles moving on a regular hexagon

• Monsterboy
In summary: I am trying to understand this statement. I am still not convinced that the particles will meet at the center.In summary, the question discusses a scenario where particles are traveling in a hexagonal pattern, with each particle maintaining a direction towards the next particle. The question asks to calculate the time it takes for the particles to meet each other. After some discussion and calculations, it is determined that the particles will eventually meet at the center of the hexagon, as the hexagon rotates and shrinks, causing the particles to accelerate towards each other. The time it takes for them to meet is calculated to be ##\frac{2a}{v}##. There is also a discussion about the infinite number of rotations that occur as the particles approach
Monsterboy
Homework Statement
Six particles situated at the corners of a regular hexagon of side a move at ##a## constant speed ##v##. Each particle maintains a direction towards the particle at the next corner. Calculate the time the particles will take to meet each other.
Relevant Equations
Speed = distance/time
When I first read the question, it didn't occur to me that these particles would ever meet or catch up with their neighbors. They are all traveling from one vertex to another with a velocity ##v## and a distance ##a##, all either clockwise or anticlockwise right ?

The question says "Each particle maintains a direction towards the particle at the next corner." it's not clear if all the particles are moving around the center of the hexagon in the same direction (clockwise or anticlockwise) or whether some particles are moving towards each other.

If we consider two particles moving towards each other, we get ##time = \frac{distance}{speed}## so ##t = \frac{\frac{a}{2}}{v}##

But the answer is ## \frac{2a}{v} ##

The question also says "Calculate the time the particles will take to meet each other." does this mean the all particles are going to meet each other ?
.

I saw one answer online which said for the particle at point C.
##velocity = v cos \theta = v cos 60 = \frac{v}{2}##
## t = \frac{a}{v/2} = \frac{2a}{v} ## but why would the particle at B stay there ? It would have moved to point A.

The hexagon must simultaneously be shrinking and rotating.

hutchphd and Monsterboy
This is a fun problem once you "see" it. Each particle chases the one in front, at speed v and always directed at the (moving) particle ahead all in the same direction. The key to the solution is to see that the particles continue to maintain a perfect hexagonal arrangement. Then you figure their relative speed using vectors. The result then follows as you describe.

Hamiltonian and Monsterboy
@Lnewqban and @hutchphd beat me to it. But since I've already drafted it...

- the particles are all moving in the same sense (all clockwise or all anticlockwise).;

- each particle follows an inwards spiral path;

- the particles always form a hexagon, but the hexagon is rotating and shrinking;

- eventually all the particles reach the centre – that’s the moment they ‘meet each other’.

Lnewqban
Alright, so the hexagon is shrinking and rotating, but then all the particles are traveling at the same speed towards the next vertex.

When the particle at C has a velocity ##v## at a direction 600 to the horizontal.
Its horizontal component will be ## v\over 2 ##, this is less than the magnitude of ##v## that the next particle is traveling at, in the horizontal direction from point B to A.

So, the relative speed will be ##v - \frac{v}{2} = \frac{v}{2}##.

As ##a## decreases, the relative speed remains the same ? And then at some point ##a## becomes zero that's when they meet right ?

For every 600 rotation, the particles manage to move to the next vertex ?

I don't know how the answer can as simple as ##\frac{2a}{v}##. For me looks like some kind of a summation, something where ##a## decreases from infinity(or any number) to zero and the magnitude of ##v## remains the same.

Can we calculate how many rotations it will take for the particles to meet each other in terms of ##a## and ##v##?

hutchphd
Monsterboy said:
... I don't know how the answer can as simple as ##\frac{2a}{v}##. For me looks like some kind of a summation, something where ##a## decreases from infinity(or any number) to zero and the magnitude of ##v## remains the same.

Can we calculate how many rotations it will take for the particles to meet each other in terms of ##a## and ##v##?
Instead, place yourself on the center point of the rotating hexagon and focus your sight on one single particle.
Do you see the tangential speed of that particle?
If not, what type of movement, in what direction, and at what speed do you perceive?

Hint: Observing the geometry of one of the six triangles of the regular hexagon may reveal to you the reason for that number 2 in the correct answer.

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hutchphd and jbriggs444
Its simpler than that. You start a distance "a" from the particle ahead. You travel directly at it with relative speed v/2 as you have shown. How long to meet? Not a difficult calculation (in the appropriate reference frame!)

.

Steve4Physics
Monsterboy said:
Can we calculate how many rotations it will take for the particles to meet each other in terms of ##a## and ##v##?
That's an interesting question and presumably not part of the original homwork.

There is a constant angle (60º) between each particle's velocity and the particle's instantaneous radius of rotation. That means each particle moves along a logarithmic (equiangular) spiral.

The speed of rotation increases as the hexagon shrinks. In fact the angular speed tends to infinity as the 6 particles approach each other at the centre. So (I think) there will be an infinite number of rotations.

Monsterboy, Lnewqban and hutchphd
We have argued that the shape doesn't change, and the hexagon just gets smaller. Hence (as noted by @Steve4Physics this is a https://en.wikipedia.org/wiki/Logarithmic_spiralic_spiral and so scale change is equivalent to a rotation: "its spirals all the way down.."

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Monsterboy
Lnewqban said:
Hint: Observing the geometry of one of the six triangles of the regular hexagon may reveal to you the reason for that number 2 in the correct answer.
Well, its an equilateral triangle with sides of length ##a##, that is rotating and shrinking. I don't know what else to make of it.

hutchphd said:
Its simpler than that. You start a distance "a" from the particle ahead. You travel directly at it with relative speed v/2 as you have shown. How long to meet? Not a difficult calculation (in the appropriate reference frame!)

.
If you are saying ## time = \frac{distance}{speed} ## so, ## t= \frac{a}{v/2} = \frac{2a}{v}##, I get that, but I can't visualize it. The particle ahead is not slowing down, nor is the particle chasing it speeding up, how are they ever going to meet if the particles are dimensionless and the hexagon shrinks forever ?

Monsterboy said:
the hexagon shrinks forever ?
No. Because the rotation accelerates they meet in a finite time.

haruspex said:
No. Because the rotation accelerates they meet in a finite time.
Okay but, Steve4Physics said

Steve4Physics said:
The speed of rotation increases as the hexagon shrinks. In fact the angular speed tends to infinity as the 6 particles approach each other at the centre. So (I think) there will be an infinite number of rotations.
If the particles meet in finite time, how can there be infinite rotations before they meet ?

Monsterboy said:
Okay but, Steve4Physics saidIf the particles meet in finite time, how can there be infinite rotations before they meet ?
Because the rotation rate increases geometrically. It's like 1+1/2+1/4+..., an infinite numbers of terms converging to a finite value.

hutchphd, Monsterboy and Steve4Physics

## 1. How do particles move on a regular hexagon?

Particles on a regular hexagon move in a zigzag pattern, following the edges of the hexagon. They move in a straight line until they reach a vertex, then turn and continue along the next edge.

## 2. What determines the direction of particle movement on a regular hexagon?

The direction of particle movement on a regular hexagon is determined by the angle of the edges. The angle between any two adjacent edges of a regular hexagon is 120 degrees, which causes the particles to move in a zigzag pattern.

## 3. Can particles move in a circular pattern on a regular hexagon?

No, particles cannot move in a circular pattern on a regular hexagon. The angles between the edges of a regular hexagon do not allow for circular movement, only a zigzag pattern.

## 4. How does the speed of particle movement on a regular hexagon compare to other shapes?

The speed of particle movement on a regular hexagon is relatively fast compared to other shapes. This is because the zigzag pattern allows for efficient movement along the edges of the hexagon.

## 5. Can particles move in a straight line on a regular hexagon?

Yes, particles can move in a straight line on a regular hexagon, but only along the edges. Once they reach a vertex, they must turn and continue along the next edge in a zigzag pattern.

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