Question 15: Rotating Rope Dynamics
Solution Analysis
We analyze the system in the rotating frame of reference. We focus on two conditions: the force equilibrium of the particle and the torque equilibrium of a rope quadrant.
1. Force Equilibrium of the Particle:
The particle of mass $m$ moves in a circle of radius $l/2$. It is pulled by two ropes (top and bottom). Let $T_1$ be the tension in the rope at the particle end. The angle between the two ropes is $\theta$, so each rope makes an angle $\theta/2$ with the horizontal (radial) line.
The horizontal components of the tension provide the centripetal force:
$$ 2 T_1 \cos(\theta/2) = m \omega^2 \left(\frac{l}{2}\right) $$
$$ T_1 = \frac{m \omega^2 l}{4 \cos(\theta/2)} \quad \dots(1) $$
2. Rotational Equilibrium of Rope Segment:
Consider the upper-left quadrant of the rope system. In the rotating frame, the rope is in equilibrium. We take torques about the center of the structure (intersection of the rotation axis and the line connecting particles).
- Torque from Midpoint Tension $T$: Acts horizontally at a vertical distance $d/2$. Torque magnitude: $T (d/2)$.
- Torque from Particle Tension $T_1$: The vertical component of $T_1$ is $T_1 \sin(\theta/2)$. It acts at a horizontal distance $l/2$. Torque magnitude: $(T_1 \sin(\theta/2)) (l/2)$.
3. Solving for Length $l$:
Substitute $T_1$ from equation (1) into equation (2):
$$ T d = \left[ \frac{m \omega^2 l}{4 \cos(\theta/2)} \right] l \sin(\frac{\theta}{2}) $$
$$ T d = \frac{m \omega^2 l^2}{4} \tan(\frac{\theta}{2}) $$
Rearranging for $l$:
$$ l^2 = \frac{4 T d}{m \omega^2 \tan(\theta/2)} $$
$$ l = \frac{2}{\omega} \sqrt{\frac{T d}{m \tan(\theta/2)}} $$