Solving the Flocking Problem: Finding A,B,C Meeting Point

[Tanya Sharma]

Maybe, look at it this way:
$$\tan\frac{\pi}{6}=\frac{r\dot{\theta}}{-\dot{r}}$$

Sorry...I still do not understand why this relation holds .I understand the angle you are talking about but why the tangent of that angle is equal to the ratio of tangential and radial components of velocity.

The one you want. From two equations, you can find a differential equation in either $r$ and $t$ or $\theta$ and $t$.

I think that gives a DE in r and θ ? How do you get DE in r and t ?

 

[Saitama]

Sorry...I still do not understand why this relation holds .I understand the angle you are talking about but why the tangent of that angle is equal to the ratio of tangential and radial components of velocity.

I have attached one more picture, I wonder if it is going to help.

You can do it another way too. Assume that the first mass is at $\vec{r_1}$, the second one at $\vec{r_2}$ and the third one at $\vec{r_3}$. You can write these three vectors in the following way:
$$\vec{r_1}=r\cos\theta \hat{i}+r\sin\theta \hat{j}$$
$$\vec{r_2}=r\cos\left(\theta+\frac{2\pi}{3}\right)\hat{i}+r\sin\left( \theta+\frac{2\pi}{3}\right)\hat{j}$$
$$\vec{r_3}=r\cos\left(\theta+\frac{4\pi}{3}\right)\hat{i}+r\sin\left( \theta+\frac{4\pi}{3}\right)\hat{j}$$
Do you see that the following relation holds?
$$\left(\vec{r_2}-\vec{r_1}\right)\times \frac{d\vec{r_1}}{dt}=0$$
If so, can you use the above to find some useful info?

Remember, we are working in a coordinate system where the centroid of the triangle formed by the initial position of masses is at origin.

I think that gives a DE in r and θ ? How do you get DE in r and t ?

From the relation above, you get
$$r\dot{\theta}=-\frac{\dot{r}}{\sqrt{3}}$$
Plug this in
$$u^2=\dot{r}^2+(r\dot{\theta})^2 \Rightarrow 3u^2=4\dot{r}^2$$

 

[Chestermiller]

Hi guys,

This problem is much simpler than it seems. Take the center of the inscribed/circumscribed circle as the origin. By symmetry, the three masses are always going to be separated from one another by angles of 120 degrees. The component of the velocity in the radial direction will therefore always be -ucos(π/6) = -√3u/2. So,

\frac{dr}{dt}=-\frac{\sqrt{3}}{2}u

The initial r is the radius of the circumscribed circle for an equilateral triangle of side a.

There is nothing wrong with Tanya's original formulation. It's just too hard to solve that way.

Chet

 

[Tanya Sharma]

This problem is much simpler than it seems. Take the center of the inscribed/circumscribed circle as the origin. By symmetry, the three masses are always going to be separated from one another by angles of 120 degrees. The component of the velocity in the radial direction will therefore always be -ucos(π/6) = -√3u/2. So,

\frac{dr}{dt}=-\frac{\sqrt{3}}{2}u

The initial r is the radius of the circumscribed circle for an equilateral triangle of side a.

Nicely done :)

Is it possible to find the distance traveled by each mass ?

 

[TSny]

Is it possible to find the distance traveled by each mass ?

Yes. The particles travel at constant speed for how much time?

 

[Tanya Sharma]

Thanks