Question

At the end of cylindrical rod of length l = 1 m and mass M = 1 kg rotating horizontaly along the vertical axis in its center with an angular speed  a small mouse of mass m = 0.02 kg is located Determine the angular speed 
of this double system when a mouse will move to the center of rod close to the rotating axis, taking into
account that the mouse can be treated as the material point.

Answer 1

Answer:

w = 0.943 rad / s

Explanation:

For this problem we can use the law of conservation of angular momentum

       

Starting point. With the mouse in the center

            L₀ = I w₀

Where The moment of inertia (I) of a rod that rotates at one end is

         I = 1/3 M L²

Final point. When the mouse is at the end of the rod

          $L_{f}$ = I w + m L² w

As the system is formed by the rod and the mouse, the forces during the movement are internal, therefore the angular momentum is conserved

        L₀ = L_{f}

        I w₀ = (I + m L²) w

        w = I / I + m L²) w₀

We substitute the moment of inertia

        w  = 1/3 M L² / (1/3 M + m) L²    w₀

        w = 1 / 3M / (M / 3 + m) w₀

We substitute the values

      w = 1/3 / (1/3 + 0.02) w₀

      w = 0.943 w₀

To finish the calculation the initial angular velocity value is needed, if we assume that this value is w₀ = 1 rad / s

        w = 0.943 rad / s