Solution: Dynamics of Bead-Rod System
1. Geometric Proof: Perpendicularity of Cords
Before analyzing the dynamics, we must establish the trajectory of the intersection point P. We prove that the angle between the cords is always $90^\circ$.
Proposition: The angle $\beta$ between the cords AC and BD is constant at $90^\circ$.
From the diagram, consider the chords $AB$ and $CD$ which both have length $r\sqrt{2}$. This length implies they subtend an angle of $90^\circ$ at the center, and consequently $45^\circ$ at the circumference.
Using the property of cyclic quadrilaterals and the angles subtended by these arcs, we have the relation:
$$\alpha – 45^\circ = 45^\circ – \theta$$Rearranging this yields:
$$\alpha + \theta = 90^\circ$$In the triangle formed by the intersection P and the base points, the intersection angle $\beta$ is the exterior angle sum (or simply using the sum of triangle angles):
$$\beta = 180^\circ – (\alpha + \theta)$$ $$\beta = 180^\circ – 90^\circ = 90^\circ$$Implication: The intersection point P always subtends $90^\circ$ to the fixed chord $CD$. Therefore, the locus of P is a circle with diameter $CD$.
2. Energy Analysis
Initial State: The cords are initially diameters of length $2r$. The relaxed length is $L_0 = r\sqrt{2}$. The extension is $x_0 = r(2-\sqrt{2})$.
Stored Energy:
$$U_{initial} = 2 \times \frac{1}{2} k x_0^2 = k r^2 (2-\sqrt{2})^2$$Stability: Any displacement shortens the cords towards their natural length, reducing potential energy. The equilibrium is unstable. The system accelerates until tension vanishes (cords relax).
Conservation of Energy: At the moment of relaxation ($x=0, U_f=0$):
$$K_{final} = U_{initial}$$ $$\frac{1}{2} I \omega^2 = k r^2 (2-\sqrt{2})^2$$With $I = 2mr^2$ (moment of inertia of two beads), we get:
$$m r^2 \omega^2 = k r^2 (2-\sqrt{2})^2 \implies \omega^2 = \frac{k}{m}(2-\sqrt{2})^2$$3. Calculation of Acceleration
At the instant the cords relax, the tension becomes zero. Therefore, the restoring torque is zero, and the angular acceleration $\alpha = 0$.
The acceleration of point P is purely centripetal (normal) towards the center of its locus circle (midpoint of CD).
- Radius of Locus ($R_L$): $CD/2 = r\sqrt{2}/2 = r/\sqrt{2}$.
- Velocity of P ($v_P$): $v_P = R_L \omega$.
Acceleration magnitude:
$$a_P = \frac{v_P^2}{R_L} = R_L \omega^2$$Substituting $R_L$ and $\omega^2$:
$$a_P = \left( \frac{r}{\sqrt{2}} \right) \cdot \left[ \frac{k}{m} (2-\sqrt{2})^2 \right]$$ $$a_P = \frac{k r (2-\sqrt{2})^2}{m\sqrt{2}}$$Final Answer
The acceleration of the intersection point P is:
$$- \hat{i} \left\{ \frac{k(2-\sqrt{2})^2 r}{m\sqrt{2}} \right\}$$