https://file.aiquickdraw.com/mathbot/file/1730386841966jkkccef3.jpg

Let's analyze and solve the problem in item 2 from the image.

### Problem Statement
Two masses, \( m \) and \( 2m \), are attached to a massless rod that can rotate around a pivot point, as shown in the image. The rod is free to rotate about this pivot. The question asks:
1. To determine if the system is in equilibrium.
2. If it is not in equilibrium, to find the expression for the angular acceleration \( \alpha \).

### Solution Approach

1. **Checking for Equilibrium:**
   - For the system to be in equilibrium, the net torque around the pivot point must be zero.
   - The torque (\( \tau \)) generated by each mass is given by \( \tau = r \cdot F \), where \( r \) is the distance from the pivot point and \( F \) is the force exerted by each mass due to gravity (\( F = mg \)).
   - Given:
     - Mass \( m \) is located at a distance \( L \) from the pivot.
     - Mass \( 2m \) is located at a distance \( 2L \) from the pivot.
   
   - Calculate the torques:
     - Torque due to mass \( m \): \( \tau_m = m \cdot g \cdot L \).
     - Torque due to mass \( 2m \): \( \tau_{2m} = 2m \cdot g \cdot 2L = 4m \cdot g \cdot L \).
   
   - Since \( \tau_{2m} \) is greater than \( \tau_m \), the torques are not balanced, and thus the system is **not in equilibrium**.

2. **Finding Angular Acceleration \( \alpha \):**
   - Since the system is not in equilibrium, we apply Newton's second law for rotation:
     \[
     \tau_{\text{net}} = I \cdot \alpha
     \]
     where \( I \) is the moment of inertia of the system about the pivot point, and \( \tau_{\text{net}} \) is the net torque.

   - **Calculating the Net Torque:**
     \[
     \tau_{\text{net}} = \tau_{2m} - \tau_m = (4m \cdot g \cdot L) - (m \cdot g \cdot L) = 3m \cdot g \cdot L
     \]

   - **Calculating the Moment of Inertia \( I \):**
     \[
     I = m \cdot L^2 + 2m \cdot (2L)^2 = m \cdot L^2 + 2m \cdot 4L^2 = 9m \cdot L^2
     \]

   - **Calculating \( \alpha \):**
     \[
     \alpha = \frac{\tau_{\text{net}}}{I} = \frac{3m \cdot g \cdot L}{9m \cdot L^2} = \frac{3g}{9L} = \frac{g}{3L}
     \]

### Answer
1. The system is **not in equilibrium**.
2. The angular acceleration \( \alpha \) is given by:
   \[
   \alpha = \frac{g}{3L}
   \]

Would you like more details or have any questions on this solution?

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### Related Questions
1. What is the role of torque in determining rotational equilibrium?
2. How would the solution change if both masses were located at equal distances from the pivot point?
3. What happens to angular acceleration if we increase the distance \( L \) between the pivot and the masses?
4. How does the moment of inertia affect rotational acceleration?
5. If the pivot point had friction, how would that influence the equilibrium condition?

#### Tip:
When analyzing rotational systems, always check both the torques and the moment of inertia to understand the system's dynamics accurately.

Dos masas, una de valor m y la otra de valor 2m, están firmemente sujetas a una varilla de masa despreciable. La varilla es libre de rotar respecto del punto marcado como pivot point en la figura. Si se libera el sistema, determine si está en equilibrio y, si no lo está, encuentre la expresión de la aceleración angular α.

Solve a rotational dynamics problem involving two masses m and 2m attached to a pivoted rod. Determine if the system is in equilibrium, and if not, find the expression for angular acceleration. This solution covers key physics concepts, including torque, moment of inertia, and Newton's Second Law for Rotation.

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