Question

Interactive Solution 9.63 illustrates one way of solving a problem similar to this one. A thin rod has a length of 0.620 m and rotates in a circle on a frictionless tabletop. The axis is perpendicular to the length of the rod at one of its ends. The rod has an angular velocity of 0.185 rad/s and a moment of inertia of 1.43 x 10-3 kg·m2. A bug standing on the axis decides to crawl out to the other end of the rod. When the bug (whose mass is 5 x 10-3 kg) gets where it's going, what is the change in the angular velocity of the rod?

Answer 1

Answer:

w = 7.89 10⁻² rad/s

Explanation:

We will solve this exercise with the conservation of the annular moment, let's write it in two moments

Initial. With the insect in the center

      L₀ = I w₀

End with the bug on the edge

     $L_{f}$= I w +  $I_{bug}$ w

The moments of inertia are

For a rod

       I = 1/3 M L²

For the insect, taken as a particle

       I = m L²

The system is formed by the rod and the insect, this is isolated, therefore the external torque is zero and the angular momentum is conserved

      L₀ =  $L_{f}$

      I w₀ = I w + $I_{bug}$ w

      w = I / (I +  $I_{bug}$) w₀

 

      w = I / (I + m L²) w₀

Let's calculate

      w = 1.43 10⁻³ / (1.43 10⁻³ + 5 10⁻³ 0.620²)²   0.185

      w = 1.43 10⁻³ / 3.352 10³ 0.185

      w = 7.89 10⁻² rad/s