Question

Three particles A, B and C are situated at the vertices of an equilateral triangle. They start moving with equal speeds (remains constant during the whole motion) such that A always heads towards B, B towards C & C towards A. Finally they meet at a point. Find out the angle by which the line connecting A and B rotates with respect to initial orientation in space by the time distance between A and B reduces to half of the initial distance between A and B

Answer 1

\frac{\sqrt{3}}{3} \ln 2

Answer 2

Let the initial distance between the particles A, B, and C be $L_0$. They form an equilateral triangle. Let the speed of each particle be $v$. Particle A moves towards B, B towards C, and C towards A. Due to symmetry, the particles always form an equilateral triangle, and the centroid of the triangle remains stationary. The distance between any two particles decreases at a constant rate.

Consider the relative velocity of particle A with respect to particle B, $\vec{v}_{A/B} = \vec{v}_A - \vec{v}_B$. Let $L(t)$ be the distance between A and B at time $t$. The rate of change of this distance is the component of $\vec{v}_{A/B}$ along the line AB. Let $\hat{r}_{AB}$ be the unit vector from A to B. The velocity of A is $\vec{v}_A = v \hat{r}_{AB}$. The velocity of B is $\vec{v}_B$, which is directed towards C. The angle between $\vec{v}_A$ (along AB) and $\vec{v}_B$ (along BC) is $120^\circ$. Thus, the angle between $\hat{r}_{AB}$ and $\vec{v}_B$ is $120^\circ$. The rate of change of the distance $L$ is $dL/dt = \frac{d}{dt}|\vec{B}-\vec{A}|$. This is the component of $\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A$ along $\hat{r}_{AB}$. $dL/dt = (\vec{v}_B - \vec{v}_A) \cdot \hat{r}_{AB} = \vec{v}_B \cdot \hat{r}_{AB} - \vec{v}_A \cdot \hat{r}_{AB}$. $\vec{v}_A \cdot \hat{r}_{AB} = v \hat{r}_{AB} \cdot \hat{r}_{AB} = v$. $\vec{v}_B \cdot \hat{r}_{AB} = v \cos(120^\circ) = -v/2$. So, $dL/dt = -v/2 - v = -3v/2$. The distance between A and B decreases linearly with time.

Now consider the rotation of the line AB. The rotation is caused by the component of the relative velocity $\vec{v}_{B/A} = \vec{v}_B - \vec{v}_A$ perpendicular to the line AB. The component of $\vec{v}_A$ perpendicular to AB is 0. The component of $\vec{v}_B$ perpendicular to AB is $v \sin(120^\circ) = v\sqrt{3}/2$. This component is directed such that it causes the line AB to rotate. The angular velocity $\omega$ of the line AB is given by the perpendicular component of $\vec{v}_{B/A}$ divided by the distance $L$. The perpendicular component of $\vec{v}_{B/A}$ is $(v_B)_\perp - (v_A)_\perp = v\sqrt{3}/2 - 0 = v\sqrt{3}/2$. The angular velocity is $\omega = \frac{v\sqrt{3}/2}{L}$. This angular velocity is positive, indicating a counter-clockwise rotation (if we assume the standard orientation of the triangle and the direction of motion).

We want to find the total angle of rotation $\Delta\phi$ when the distance $L$ reduces from $L_0$ to $L_0/2$. The angular velocity is $\omega = d\phi/dt = \frac{v\sqrt{3}}{2L}$. We know $dL/dt = -3v/2$, so $dt = -\frac{2}{3v} dL$. Substitute $dt$ into the equation for $d\phi$: $d\phi = \omega dt = \frac{v\sqrt{3}}{2L} \left(-\frac{2}{3v} dL\right) = -\frac{\sqrt{3}}{3L} dL$.

To find the total angle of rotation, integrate $d\phi$ as $L$ goes from $L_0$ to $L_0/2$: $\Delta\phi = \int_{L_0}^{L_0/2} -\frac{\sqrt{3}}{3L} dL = -\frac{\sqrt{3}}{3} [\ln|L|]_{L_0}^{L_0/2}$ $\Delta\phi = -\frac{\sqrt{3}}{3} (\ln(L_0/2) - \ln(L_0)) = -\frac{\sqrt{3}}{3} \ln\left(\frac{L_0/2}{L_0}\right) = -\frac{\sqrt{3}}{3} \ln(1/2)$ $\Delta\phi = -\frac{\sqrt{3}}{3} (-\ln 2) = \frac{\sqrt{3}}{3} \ln 2$.

The angle by which the line connecting A and B rotates is $\frac{\sqrt{3}}{3} \ln 2$ radians.