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
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
Related questions
Question
```
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, \(\omega_f\).
```
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 (\(I_i\))**:
\[
I_i = 2 \cdot m \cdot r_i^2
\]
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1829f362-1a16-45e8-8282-990908270dc7%2F16e9b2bd-b2fe-406e-8ed8-a72d8d933a69%2Fwtekkqu_processed.png&w=3840&q=75)
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.
```
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, \(\omega_f\).
```
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 (\(I_i\))**:
\[
I_i = 2 \cdot m \cdot r_i^2
\]
\[
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 4 images
