Math Problem Statement

2

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.

Solution

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

Problem Statement

Two masses, mm and 2m2m, 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 α\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 (τ\tau) generated by each mass is given by τ=rF\tau = r \cdot F, where rr is the distance from the pivot point and FF is the force exerted by each mass due to gravity (F=mgF = mg).

    • Given:

      • Mass mm is located at a distance LL from the pivot.
      • Mass 2m2m is located at a distance 2L2L from the pivot.
    • Calculate the torques:

      • Torque due to mass mm: τm=mgL\tau_m = m \cdot g \cdot L.
      • Torque due to mass 2m2m: τ2m=2mg2L=4mgL\tau_{2m} = 2m \cdot g \cdot 2L = 4m \cdot g \cdot L.
    • Since τ2m\tau_{2m} is greater than τm\tau_m, the torques are not balanced, and thus the system is not in equilibrium.

  2. Finding Angular Acceleration α\alpha:

    • Since the system is not in equilibrium, we apply Newton's second law for rotation: τnet=Iα\tau_{\text{net}} = I \cdot \alpha where II is the moment of inertia of the system about the pivot point, and τnet\tau_{\text{net}} is the net torque.

    • Calculating the Net Torque: τnet=τ2mτm=(4mgL)(mgL)=3mgL\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 II: I=mL2+2m(2L)2=mL2+2m4L2=9mL2I = m \cdot L^2 + 2m \cdot (2L)^2 = m \cdot L^2 + 2m \cdot 4L^2 = 9m \cdot L^2

    • Calculating α\alpha: α=τnetI=3mgL9mL2=3g9L=g3L\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 α\alpha is given by: α=g3L\alpha = \frac{g}{3L}

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


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 LL 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.

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Math Problem Analysis

Mathematical Concepts

Rotational Dynamics
Torque
Angular Acceleration
Moment of Inertia

Formulas

Torque (τ) = r × F
Net Torque (τ_net) = I × α
Moment of Inertia (I) = Σ m × r^2

Theorems

Newton's Second Law for Rotation

Suitable Grade Level

Undergraduate Level - Physics or Engineering