Question

in a regular hexagon each particle is moving with a speed v and always point towards the adjacent particle. Find accleration of the particel as a function of time t and also radius of curvature as a function of time t

Answer 1

Acceleration of the particle as a function of time $t$:

$a(t) = \frac{v^2\sqrt{3}}{2\left(L_0 - \frac{3v}{2}t\right)}$

Radius of curvature as a function of time $t$:

$\rho(t) = \frac{2}{\sqrt{3}}\left(L_0 - \frac{3v}{2}t\right)$

where $L_0$ is the initial side length of the hexagon.

Answer 2

1. Determine the rate of change of side length: By analyzing the relative velocity component along the line joining adjacent particles, the side length $L(t)$ is found to decrease at a constant rate $dL/dt = -3v/2$. Thus, $L(t) = L_0 - (3v/2)t$, where $L_0$ is the initial side length. Since it's a regular hexagon, the distance from the center to any vertex $R(t)$ is equal to $L(t)$.

2. Calculate the acceleration: Each particle moves with constant speed $v$. The acceleration is purely centripetal, perpendicular to the velocity vector. The magnitude of acceleration is $a = v\omega$, where $\omega$ is the angular velocity of the particle around the center of the hexagon. The tangential component of the particle's velocity with respect to the center is $v_t = v \sin(60^\circ) = v\sqrt{3}/2$. The angular velocity $\omega = v_t/R(t) = (v\sqrt{3}/2)/R(t)$. Substituting this into the acceleration formula gives $a(t) = v (v\sqrt{3}/(2R(t))) = v^2\sqrt{3}/(2R(t))$.

3. Calculate the radius of curvature: For motion with constant speed, the acceleration is entirely normal (centripetal), given by $a = v^2/\rho$, where $\rho$ is the radius of curvature. Therefore, $\rho(t) = v^2/a(t)$. Substituting the expression for $a(t)$ yields $\rho(t) = v^2 / (v^2\sqrt{3}/(2R(t))) = 2R(t)/\sqrt{3}$.

4. Substitute time dependence: Replace $R(t)$ with $L_0 - (3v/2)t$ in both expressions to get the acceleration and radius of curvature as functions of time.