Question

A thin rod has a length of 0.609m 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 0.606rad/s and a moment of inertia of 1.05 x 10^-3 kg•m^2. 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

-0.0069 rad/s

Step-by-step explanation:

The change in angular velocity of the rod can be determined by using the conservation of angular momentum equation: Lf = Li.

The initial angular momentum of the rod is Li = Iwo and the final angular momentum of the rod is Lf = Iwf.

Plugging in the given values, we get:

(1.05 x 10^-3 kg•m^2)(0.606 rad/s) + (5 x 10^-3 kg)(0.609 m)^2(wf) = (1.05 x 10^-3 kg•m^2)(wo).

Solving for wf, we get wf = -0.0069 rad/s.

Therefore, the change in angular velocity of the rod is -0.0069 rad/s.