**Problem 4: Rotational Dynamics and Angular Velocity****Problem Statement:**Two identical, 1.50-kg point masses are supported on a rod with negligible mass. The rod and masses are rotating about its center, supported in a frictionless bearing. Initially, the masses are located 0.25 meters from the pivot and held in position by small clips. The original angular velocity, \(\omega_o\), is 50 rad/s. Calculate the angular velocity, \(\omega_f\), when the masses are at the ends of the rod, spaced 1.25 meters from the pivot.**Illustrations:**1. **Initial Configuration:** - Two masses of 1.50 kg each are shown attached to a rod. - Each mass is positioned 0.25 meters from the pivot point at the center. - The rod rotates with an angular velocity, \(\omega_o\), of 50 rad/s. ![Initial Setup](diagram_initial.png) ``` 0.25 m 0.25 m O---●---O---●---O Rotation: ωo ```2. **Final Configuration:** - The same masses are moved to the ends of the rod. - Each mass is now 1.25 meters from the pivot point. - The rod now rotates with a new angular velocity, \(\omega_f\). ![Final Setup](diagram_final.png) ``` 1.25 m 1.25 m O-------------●-------------O Rotation: ωf ```**Solution Outline:**To find the final angular velocity \(\omega_f\), conservation of angular momentum is used, given that no external torques are acting on the system. This implies:\[ L_i = L_f \]Where \(L_i\) is the initial angular momentum and \(L_f\) is the final angular momentum.The angular momentum \(L\) of a rotating system is given by:\[ L = I \cdot \omega \]Here, \(I\) is the moment of inertia, and \(\omega\) is the angular velocity.1. **Initial Moment of Inertia (\(I_i\))**: \[ I_i = 2 \cdot m \cdot r_i^2 \] \[

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