Kinematics problem on Tetrahedron

[Satvik Pandey]

Homework Statement

4 ants are arranged in such a way that they make up vertex of a regular tetrahedron, of side length 1m . The ants are named Calvin , Peter , David and Aron. Each ant moves at a speed 1m/s , and moves in such a way that:

Calvin moves toward Peter,
Peter moves toward David,
David moves toward Arron,
Arron moves toward Calvin

If they continue to moves it this direction they will converge somewhere. What time in seconds will it take the ants to meet each other?

Homework Equations

The Attempt at a Solution

The base of tetrahedron will form an equilateral triangle.

Length of side of equilateral triangle(a)=2cos 35.5
It is given Peter moves toward David, David moves toward Arron.
V_{PD}(relative velocity of Peter w.r.t.Arron) =√3V

Projection of V_{PD} on line of motion of Peter=√3Vcos30^{°}=\frac{3V}{2}
now
∫^{0}_{a}dl=∫^{t}_{0}\frac{3V}{2}
t=2a/3V
=4*cos(35.5)/3
But this is not correct.

 

[ehild]

Homework Statement

4 ants are arranged in such a way that they make up vertex of a regular tetrahedron, of side length 1m . The ants are named Calvin , Peter , David and Aron. Each ant moves at a speed 1m/s , and moves in such a way that:

Calvin moves toward Peter,
Peter moves toward David,
David moves toward Arron,
Arron moves toward Calvin

If they continue to moves it this direction they will converge somewhere. What time in seconds will it take the ants to meet each other?

Homework Equations

The Attempt at a Solution

^{0}<span style="font-size: 22px">∫_{a}</span>dl=∫^{t}_{0}\frac{3V}{2}

What do you mean with the equation above?

The problem has a certain symmetry. What do you think where the ants meet?

ehild

 

[Satvik Pandey]

What do you mean with the equation above?

The problem has a certain symmetry. What do you think where the ants meet?

ehild

Sorry ehild I posted incomplete question by mistake
https://d18l82el6cdm1i.cloudfront.net/solvable/1e35c55ad8.df033bb1b8.CVOE2P.png
Here is the figure of the question.
I have edited it.Please take a look.

 

[Satvik Pandey]

What do you mean with the equation above?

The problem has a certain symmetry. What do you think where the ants meet?

ehild

I think they will meet at O.

 

[ehild]

The Attempt at a Solution

The base of tetrahedron will form an equilateral triangle.
View attachment 71859
Length of side of equilateral triangle(a)=2cos 35.5

What do you mean? It is given that the sides of the tetrahedron are of 1 m length.

ehild

 

[Satvik Pandey]

What do you mean? It is given that the sides of the tetrahedron are of 1 m length.

ehild

I thought 1 m length to be a bond length in tetrahedral structure as taken in chemistry.
where as in this question it should be the edge length of the tetrahedron,
According to figure in post#3 it is AC,BD,BC,AB.
Are my further steps correct?

 

[ehild]

Your picture with the ants is magnificent!

I would like to see your real solution. I think the idea might be correct, but I am not sure. I would do it differently. What is the "official" solution?

ehild

 

[Satvik Pandey]

Your picture with the ants is magnificent!

I would like to see your real solution. I think the idea might be correct, but I am not sure. I would do it differently. What is the "official" solution?

ehild

Thank you ehild for the compliment.
https://www.physicsforums.com/attachment.php?attachmentid=71860&d=1407155510
From this figure
If I put the length of side of this triangle=1m
then t=2*1/3
=0.67 sec.
But the answer is 0.75 sec.
This is my attempt at the solution.
What is meant by "official" solution.

 

[ehild]

I can not explain why your solution is incorrect . I think as the followed ant moves also perpendicularly to the edge, the distance to be covered is longer then a.
I would do it differently. The centre of the tetrahedron is stationary and the ants meet there. As the ants move, they always form the vertexes of a tetrahedron which is rotated with respect to the original one and also shrinks. The velocities are along the edges of these tetrahedra which makes an angle with the radius. That is your 35.5° but rather 35.26 °as the tetrahedral angle is 109.5°. The cosine is √(2/3). The displacement is equal to the radius of the tetrahedron, and the radial velocity component is v_r=vcos(35.26)=√(2/3) v. t=radius/v_r
That gives the answer.

ehild

 

[Satvik Pandey]

The displacement is equal to the radius of the tetrahedron, and the radial velocity component is v_r=vcos(35.26)=√(2/3) v. t=radius/v_r
That gives the answer.

ehild

Is this displacement equal to distance between vertex and centroid of Tetrahedral?

 

[Satvik Pandey]

I think that the centroid of the base triangle is just below the centroid of tetrahedral.
Let c be the distance between the vertex and centroid of tetrahedral.

So, by Pythagoras theorem
1/3+b^2=c^2
To find c I need to find another equation in terms of b and c.
Could you please help,ehild?

 

[ehild]

Is this displacement equal to distance between vertex and centroid of Tetrahedral?

Yes.

Spoiler

√(3/8) a

ehild

 

[Satvik Pandey]

I can not explain why your solution is incorrect . I think as the followed ant moves also perpendicularly to the edge, the distance to be covered is longer then a.
ehild

Doesn't the ant move along the edge?
From the figure(in #2) Peter moves along BC and David moves along BD.

 

[Satvik Pandey]

The velocities are along the edges of these tetrahedra which makes an angle with the radius. That is your 35.5° but rather 35.26 °as the tetrahedral angle is 109.5°. The cosine is √(2/3). The displacement is equal to the radius of the tetrahedron, and the radial velocity component is v_r=vcos(35.26)=√(2/3) v. t=radius/v_r
That gives the answer.

ehild

Could you please explain it with figure?

 

[ehild]

They always move along the edges of a tetrahedron, but the tetrahedron changes all time. Rotates and shrinks. The radial component of the velocity counts.

ehild

 

[Satvik Pandey]

I can not explain why your solution is incorrect . I think as the followed ant moves also perpendicularly to the edge, the distance to be covered is longer then a.

ehild

Then,what is meant by that?
I am confused.

 

[ehild]

Never mind. I can not explain it better. Do the other method. Here is a picture.

I have to leave. Hoping you get the desired result.

 

[ehild]

Doesn't the ant move along the edge?
From the figure(in #2) Peter moves along BC and David moves along BD.

They start to move along those edges, but they change position and then it is not BC and BD any more. You get new positions for Peter and David and the others.

ehild

 

[Satvik Pandey]

Never mind. I can not explain it better. Do the other method. Here is a picture.

I have to leave. Hoping you get the desired result.

I got that.
The triangle form by the red lines and black line is an isosceles triangle.
So angle between vertex and centroid is (180-109.5°)/2
V_{r}=Vcos(35.26)
Thank you ehild. Sorry for the inconvenience.

But what is distance between vertex and centroid of Tetrahedron.
I have worked on it in post#11.Please take a look.

 

[ehild]

Thank you ehild. Sorry for the inconvenience.

But what is distance between vertex and centroid of Tetrahedron.
I have worked on it in post#11.Please take a look.

You need to apply Pythagoras a few times. Determine the height h of the tetrahedron from the yellow triangle.
The blue triangle is isosceles. In the green triangle, one leg is h-r, the other is a/√3. r is the hypotenuse.

Check the spoiler in post#12.

ehild

 

[haruspex]

Thank you ehild for the compliment.
https://www.physicsforums.com/attachment.php?attachmentid=71860&d=1407155510
From this figure
If I put the length of side of this triangle=1m
then t=2*1/3
=0.67 sec.
But the answer is 0.75 sec.
This is my attempt at the solution.
What is meant by "official" solution.

I suspect the problem with your first method is that it supposes more symmetry than will be the case.
Do the ants really continue to form the vertices of a regular tetrahedron?
I modeled it in a spreadsheet and found that although the distances between successive ants are always equal, the distances between the first and third and between the second and fourth diverged from the first four distances. By the time they meet in the middle, the ratio is √2:1.

Edit: The same modelling seems to give an answer around 0.8, so I'm not convinced 0.75 is right either.

 

[ehild]

I have doubts, too, if the tetrahedral symmetry is maintained. The given result (t=0.75 s) suggests it was assumed. It would be nice to prove or disprove.

ehild

 

[haruspex]

I have doubts, too, if the tetrahedral symmetry is maintained. The given result (t=0.75 s) suggests it was assumed. It would be nice to prove or disprove.

ehild

It is clear that the four consecutive distances are always equal, and that the two non-consecutive are always equal, but there is no reason to suppose all six are always equal. My spreadsheet model indicates that the shape is asymptotically a flat square.
I obtained the following equation:
$\ddot r r + 2 {\dot r}^2 + 6 \dot r v + 4 v^2 = 0$
where r = r(t) is the distance between two consecutive ants and v is the speed of each ant.
The evolution of this equation matches my earlier model, giving me confidence in both. Don't see how to solve it, though.

 

[Satvik Pandey]

You need to apply Pythagoras a few times. Determine the height h of the tetrahedron from the yellow triangle.
The blue triangle is isosceles. In the green triangle, one leg is h-r, the other is a/√3. r is the hypotenuse.

Check the spoiler in post#12.

ehild

Thank you ehild
I got it √(3/8).
t= √3/√8 *√3/√2=3/4=0.75

 

[Orodruin]

It is clear that the four consecutive distances are always equal, and that the two non-consecutive are always equal, but there is no reason to suppose all six are always equal. My spreadsheet model indicates that the shape is asymptotically a flat square.
I obtained the following equation:
$\ddot r r + 2 {\dot r}^2 + 6 \dot r v + 4 v^2 = 0$
where r = r(t) is the distance between two consecutive ants and v is the speed of each ant.
The evolution of this equation matches my earlier model, giving me confidence in both. Don't see how to solve it, though.

Indeed the sides change at different rates. I solved this in cylinder coordinates with the axis through the center of the sides which are not parallel to the direction of any ant and with z=0 for the center of the tetrahedron. Doing so, I obtained
$$
\frac{dz}{dr} = 2\frac{z}{r}
$$
resulting in
$$
\frac{dz}{z} = 2\frac{dr}{r} \quad \ln(z/z_0) = 2\ln(r/r_0) \quad z = z_0\left(\frac{r^2}{r_0^2}\right).
$$
Thus, the tetrahedron is going to become flat and thus a square. The distance across the square is a factor $\sqrt 2$ larger than the distance between the ants, in accordance with haruspex's numerical findings.

 

[ehild]

Yes, I was wrong when assuming that tetrahedral symmetry is maintained - and I think the problem makers also thought the same. But then how to get the time the ants meet?

ehild

 

[Orodruin]

You can parametrize the curve of any given ant with any of the coordinates in cylinder coordinates (even the angle $\varphi$, although it will wind around several times). You will find that
$$
r\, d\varphi = - dr,
$$
which can be integrated easily (although not really needed for the problem). The curve length is given by
$$
s = \int_{r=0}^{r_0} ds = \int_{r=0}^{r_0} \sqrt{dr^2 + dz^2 + r^2d\varphi^2}
= \int_{r=0}^{r_0} \sqrt{2 + \left(\frac{dz}{dr}\right)^2} dr,
$$
where you can insert the earlier result for $z(r)$ to find the derivative. Divide by the velocity to find the time. (And do some geometry to find the relation between $r_0$ and $z_0$ ...) Substituting some variables you will end up with an integral of the form
$$
\int_0^k \sqrt{1+s^2} ds
$$
multiplied by some constants.

This is all assuming I did not make any trivial errors along the way. ;)

 

[haruspex]

Yes, I was wrong when assuming that tetrahedral symmetry is maintained - and I think the problem makers also thought the same. But then how to get the time the ants meet?

ehild

I eliminated time from my equation, temporarily, to get $r = \frac{\sqrt{\frac v2(2v+y)}}{-v-y}$, where y = dr/dt.
Substituting back in, $\dot y = -2\sqrt{\frac 2v}(v+y)^2(2v+y)^{\frac 12}$.
Looks like this leads to integrating sec-cubed. Haven't gone through the details yet.

 

[Orodruin]

Actually performing the integral in my post and putting the right numbers, I end up with
$$
s = \sqrt{2} r_0 \int_0^1 \sqrt{1+x^2} dx = r_0\left(1 + \frac{\sinh^{-1}(1)}{\sqrt 2}\right) \simeq 1.62 r_0.
$$
Since, in my chosen coordinates, $r_0$ is half the side-length of the original tetrahedron, we have $r_0 = 0.5~{\rm m}$, the end result would be
$$
t = \frac sv \simeq \frac{1.62 \cdot 0.5~{\rm m}}{1~{\rm m/s}} = 0.81~{\rm s}.
$$
Again in accordance with the numerical findings.

 

[haruspex]

Actually performing the integral in my post and putting the right numbers, I end up with
$$
s = \sqrt{2} r_0 \int_0^1 \sqrt{1+x^2} dx = r_0\left(1 + \frac{\sinh^{-1}(1)}{\sqrt 2}\right) \simeq 1.62 r_0.
$$
Since, in my chosen coordinates, $r_0$ is half the side-length of the original tetrahedron, we have $r_0 = 0.5~{\rm m}$, the end result would be
$$
t = \frac sv \simeq \frac{1.62 \cdot 0.5~{\rm m}}{1~{\rm m/s}} = 0.81~{\rm s}.
$$
Again in accordance with the numerical findings.

Well done, and thanks for the confirmation.
Clearly the problem was much harder than intended.