A baggage carousel at an airport is rotating with an angular speed of 0.397 rad/s when the baggage begins to be loaded onto it. The moment of inertia of the carousel is 2440 kg·m2. Ten pieces of baggage with an average mass of 16.5 kg each are dropped vertically onto the carousel and come to rest at a perpendicular distance of 2.95 m from the axis of rotation. Assuming that no net external torque acts on the system of carousel and baggage, find the final angular speed.

YOu know the initial angular moentum, the final moment of inertia= initial + 10mass*r^2

solve the conservation of angular momentum problem.

To find the final angular speed of the carousel, we can apply the law of conservation of angular momentum. The angular momentum of the system before the bags are dropped is equal to the angular momentum of the system after the bags have come to rest.

The formula for angular momentum is given by: L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.

Before the bags are dropped, the angular momentum of the carousel alone is given by: L1 = I1ω1, where I1 is the moment of inertia of the carousel and ω1 is the initial angular speed.

After the bags have come to rest, the angular momentum of the system is given by: L2 = (I1 + Σ(mr^2))ω2, where m is the mass of each bag, r is the perpendicular distance of the bags from the axis of rotation, and ω2 is the final angular speed.

Since no net external torque acts on the system, the angular momentum is conserved, so L1 = L2.

Thus, we have I1ω1 = (I1 + Σ(mr^2))ω2.

Substituting the given values, we have:
2440 kg·m^2 * 0.397 rad/s = (2440 kg·m^2 + 10 * 16.5 kg * (2.95 m)^2) * ω2.

Simplifying this equation, we can solve for ω2:
omega2 = (2440 * 0.397) / (2440 + 10 * 16.5 * (2.95)^2).

Calculating this expression, we find that the final angular speed of the carousel is approximately 0.275 rad/s.