A thin uniform rod is rotating at an angular velocity of 7.70 rad/s about an axis that is perpendicular to the rod at its center. As the figure indicates, the rod is hinged at two places, one-quarter of the length from each end. Without the aid of external torques, the rod suddenly assumes a "u" shape, with the arms of the "u" parallel to the rotation axis. What is the angular velocity of the rotating "u"?

To find the angular velocity of the rotating "u" shape, we need to apply the principle of conservation of angular momentum. Angular momentum is conserved when there are no external torques acting on the system.

Initially, the thin rod is rotating at an angular velocity of 7.70 rad/s. Assuming the moments of inertia about the rotation axis for both halves of the rod are equal, and denoting the total moment of inertia as "I", we can write the initial angular momentum as:

L_initial = 2 * (1/4 * m * (L/4)^2) * 7.70

where m is the mass of the rod, and L is its length.

When the rod assumes the "u" shape, it splits into two halves, each rotating about its respective hinge. The moment of inertia of each half is given by:

I_half = 1/2 * m * (L/4)^2

Now, let's denote the angular velocity of each rotating half as "ω".

The final angular momentum of the system is then given by:

L_final = 2 * I_half * ω

Since angular momentum is conserved, we equate the initial and final angular momentum:

L_initial = L_final

2 * (1/4 * m * (L/4)^2) * 7.70 = 2 * (1/2 * m * (L/4)^2) * ω

Simplifying and solving for ω:

7.70 = ω

Therefore, the angular velocity of the rotating "u" shape is 7.70 rad/s.