4. Two identical, 1.50-kg point masses are supported on a rod with negligible mass. The rod and masses are rotating about its center as shown, supported in a frictionless bearing. In the beginning, the masses are located 0.25 meters from the pivot, and held in position by small clips. The original angular velocity w, is 50 rad/s. What is the angular velocity wf when the masses are at the end of the rods, spaced 1.25 meters from the pivot? 0.25 m 0.25 m Wo 1.25 m 1.25 m Wf

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**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 (\(I_i\))**:
   \[
   I_i = 2 \cdot m \cdot r_i^2
   \]
   \[
Transcribed Image Text:**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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