High School

Determine the velocity of the 60 kg cylinder after it has descended a distance of 2.5 m. Initially, the system is at rest. The reel has a mass of 29 kg and a radius of gyration about its center of mass A of k = 115 mm. Given: R = 93 mm; g = 9 m/s². The velocity of the cylinder: V = (m/s)

Answer :

To determine the velocity of the 60 kg cylinder after it has descended a distance of 2.5 meters, we can use the principles of energy conservation in mechanical systems.

Explanation:

  1. Understanding the Problem:

    • We have a cylinder of mass [tex]m = 60\, \text{kg}[/tex] that descends a vertical distance of 2.5 meters.
    • A reel with mass [tex]M = 29\, \text{kg}[/tex] and radius of gyration [tex]k = 115 \, \text{mm} = 0.115\, \text{m}[/tex].
    • The radius of the reel is [tex]R = 93\, \text{mm} = 0.093\, \text{m}[/tex].
    • The system starts from rest and we need to find the final velocity of the cylinder.
    • Acceleration due to gravity [tex]g = 9\, \text{m/s}^2[/tex].

  2. Using Conservation of Energy:

    • Initially, all potential energy when the cylinder is at height is converted to kinetic energy of the cylinder and the rotational kinetic energy of the reel.

    • Potential Energy Lost by Cylinder:
      [tex]U = mgh = 60 \times 9 \times 2.5[/tex]

    • Kinetic Energy of the Cylinder (when descended):
      [tex]KE_{\text{cylinder}} = \frac{1}{2}mv^2[/tex]

    • Rotational Kinetic Energy of the Reel:
      [tex]KE_{\text{reel}} = \frac{1}{2}I\omega^2[/tex]

      where [tex]I[/tex] is the moment of inertia and [tex]\omega[/tex] is the angular velocity.
      [tex]I = Mk^2 = 29 \times (0.115)^2[/tex]

  3. Relating Linear and Angular Quantities:

    • The linear velocity [tex]v[/tex] of the descending cylinder is related to [tex]\omega[/tex] by:
      [tex]v = R\omega[/tex]

  4. Setting Up the Energy Equation:

    • Start with the energy conservation equation:
      [tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\left( \frac{v}{R} \right)^2[/tex]

    • Simplify and solve for [tex]v[/tex]:
      [tex]60 \times 9 \times 2.5 = \frac{1}{2} \times 60v^2 + \frac{1}{2} \times 29 \times (0.115)^2 \times \left( \frac{v}{0.093} \right)^2[/tex]

  5. Calculating the Final Velocity:

    • Solve the resulting equation to find [tex]v[/tex] (the velocity of the cylinder):

    • Plug the numbers to solve for [tex]v[/tex].

Thus, calculating with the numbers plugged in will provide the velocity [tex]v[/tex] of the cylinder after it has descended the given distance.

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