We have three particles at the vertices of equilateral triangle of side $d$. At $t=0$ they start moving in such away that at all instant of time each of them has a speed $v$ towards adjacent one. We have to find the time after which they will collide. I know that you can solve this by calculating from reference frame of any one particle, but I want a way to solve this by calculating the total distance traveled first in ground frame and then dividing by speed.


1 Answer 1


Let me first describe the usual way one solves the problem as mentioned by the OP. I will then work the problem out in the ground frame.

Nice Answer: Let us work in the frame of particle 2. In this frame, particle 3 is spiralling in towards 2, while particle 1 has a complicated motion that is always point towards 3. It is easy to show that the component of the velocity of 3 along the line joining 2 and 3 is $\frac{3v}{2}$. The net time taken is $t_0 = \frac{2d}{3v}$.

Nice Answer in ground frame: In the ground frame, there is a clear symmetry. The particles slowly move in towards the center. Due to the symmetry, the component of the velocity of each particle towards the center of the triangle is always equal and is a constant. At $t=0$, the component in this direction is $v \cos \frac{\pi}{6} = \frac{\sqrt{3}v}{2}$. The total distance every particle travels in that direction is $\frac{d}{\sqrt{3}}$. The total time is then $\frac{2d}{3v}$.

Full answer in ground frame: Choose the origin to be the center of the equilateral triangle. At any time $t$, let the polar coordinates of the $i$th particle be $(r_i(t),\theta_i(t))$ with $0 < \theta_i(t) \leq 2\pi$ and $i=1,2,3$. The position vector is then given by $$ \vec{r}_i = r_i \left( \cos\theta_i {\hat i} + \sin\theta_i {\hat j} \right) $$ Now, before we proceed, we will use symmetry. Note that, no matter how the particles travel they will always form an equilateral triangle. Mathematically this implies the following equations $$ r_1(t) = r_2(t) = r_3(t) = r(t) $$ $$ \theta_1(t) - \theta_2(t) = \theta_2(t) - \theta_3(t) = \frac{2\pi}{3} $$ $$ \implies \theta_1(t) = \theta(t) + \frac{2\pi}{3} ,~\theta_2(t) = \theta(t),~\theta_3(t) = \theta(t) - \frac{2\pi}{3} $$ Thus, the original 6 variables have been reduced to two variables $r(t)$ and $\theta(t)$. Let us now write down the equations of motion. These are $$ \frac{d}{dt}\vec{r}_2(t) = \frac{v \left( \vec{r}_1 - \vec{r}_2\right) }{|\vec{r}_1 - \vec{r}_2|} $$ and similarly for $ \vec{r}_1(t) $ and $ \vec{r}_3(t) $, but given symmetry, these equations can be derived from the equation above. Let us now explicitly write out the equation above. $$ \vec{r}_1 - \vec{r}_2 = r(t) \left[ \left( \cos\theta_1(t) - \cos\theta_2(t) \right) {\hat i} + \left( \sin\theta_1(t) - \sin\theta_2(t) \right) {\hat j} \right] $$ $$ = \frac{\sqrt{3}}{2} r(t) \left[ \left( - \sqrt{3} \cos\theta - \sin\theta \right) {\hat i} + \left( \cos\theta - \sqrt{3} \sin\theta \right) {\hat j} \right] $$ Also $$ |\vec{r}_1 - \vec{r}_2 | = \sqrt{3} r(t) $$ The full equation is then $$ \frac{d}{dt} \left[ r(t) \left( \cos\theta {\hat i} + \sin\theta{\hat j} \right) \right]= \frac{v}{2} \left[ \left( - \sqrt{3} \cos\theta - \sin\theta \right) {\hat i} + \left( \cos\theta - \sqrt{3} \sin\theta \right) {\hat j} \right] $$ This splits into two equations $$ {\dot r} \cos\theta - r {\dot \theta} \sin\theta = - \frac{v}{2} \left( \sqrt{3} \cos\theta + \sin\theta \right) $$ $$ {\dot r} \sin\theta + r {\dot \theta} \cos\theta = \frac{v}{2} \left( \cos\theta - \sqrt{3} \sin\theta \right) $$ The boundary conditions are $r(0) = \frac{d}{\sqrt{3} },~\theta(0) = \frac{7\pi}{6} $. We now wish to decouple the two first-order differential equations into two decoupled second order differential equations. This is quite easy to do. Multiply the first equation by $\cos\theta$, the second by $\sin\theta$ and add the two. We get $$ {\dot r} = - \frac{\sqrt{3}}{2} v $$ We must now integrate this. At time $t=0$, $r(0) = \frac{d}{\sqrt{3}}$. The radial coordinate at arbitrary time $t$ is then $$ r(t) = - \frac{\sqrt{3}}{2} v t + \frac{d}{\sqrt{3} } $$ Though we are done here, for completeness, we can also solve for $\theta(t)$. The solution is $$ \theta(t) = \frac{7}{6} \pi - \frac{1}{\sqrt{3} } \log \left( 1 - \frac{3 v t }{2 d } \right) $$ The total time of travel is the time when $r(t_0) = 0 \implies t_0 = \frac{2d}{3v}$. Let us also make sure that we understand what the total distance is. Squaring and adding the two equations of motion we wrote above, we find $$ {\dot r}^2 + r^2 {\dot \theta}^2 = v^2 $$ The total distance travelled is then $$ s = \int_0^{t_0} \left[ {\dot r}^2 + r^2 {\dot \theta}^2 \right]^{1/2} = v t_0 = \frac{2d}{3} $$ Note that as $t \to \frac{2d}{3v}$, $\theta(t) \to \infty$, i.e. the particles spiral an infinite number of times before reaching the center of the equilateral triangle.


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