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

w = 7.89 10⁻² rad/s

Step-by-step 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