an ice skater starts her spin with her arms stretched straight out. She then pulls her armsin close to her body. Assume her moment of inertia when her arms are stretched out is 3.0 kg-m sqaured and it is 2.0 kg-m squared when she pulls them in tight. If she starts her spin at 5.0 rad/s, what is her final angular velocity?

To solve this problem, we can use the law of conservation of angular momentum, which states that the initial angular momentum is equal to the final angular momentum.

The formula for angular momentum is given by:

L = Iω

Where:
L = Angular momentum
I = Moment of inertia
ω = Angular velocity

We are given:
Initial moment of inertia, I₁ = 3.0 kg-m²
Final moment of inertia, I₂ = 2.0 kg-m²
Initial angular velocity, ω₁ = 5.0 rad/s

Let's calculate the final angular velocity, ω₂:

Using the law of conservation of angular momentum:
L₁ = L₂

I₁ ω₁ = I₂ ω₂

Rearranging the equation, we get:
ω₂ = (I₁ ω₁) / I₂

Substituting the given values, we have:
ω₂ = (3.0 kg-m² × 5.0 rad/s) / 2.0 kg-m²

Calculating this expression gives us:
ω₂ = 7.5 rad/s

Therefore, the final angular velocity of the ice skater will be 7.5 rad/s.

To find the final angular velocity of the ice skater, we can use the principle of conservation of angular momentum. According to this principle, the initial angular momentum of the skater must be equal to the final angular momentum.

The formula for angular momentum is given by:

Angular Momentum = Moment of Inertia × Angular Velocity

Initially, when the skater's arms are stretched out, the moment of inertia is 3.0 kg-m^2, and the initial angular velocity is 5.0 rad/s. Therefore, the initial angular momentum can be calculated as:

Initial Angular Momentum = 3.0 kg-m^2 × 5.0 rad/s
= 15.0 kg-m^2/s

When the skater pulls her arms in close to her body, the moment of inertia decreases to 2.0 kg-m^2. Let's assume her final angular velocity as "ω" (omega). Using the conservation of angular momentum, we can write:

Initial Angular Momentum = Final Angular Momentum

(3.0 kg-m^2) × (5.0 rad/s) = (2.0 kg-m^2) × (ω)

Now, we can solve for ω:

15.0 kg-m^2/s = 2.0 kg-m^2 × ω
ω = 15.0 kg-m^2/s / 2.0 kg-m^2
ω = 7.5 rad/s

Therefore, the final angular velocity of the ice skater is 7.5 rad/s.