Question:

A student conducts an experiment to demonstrate the conservation of angular momentum. The student starts by sitting on a freely rotating chair with no external torques acting on the system. The initial moment of inertia of the system is 2 kg·m² and the student is spinning at an initial angular velocity of 3 rad/s. When the student extends their arms, the moment of inertia of the system changes to 3 kg·m². Determine the final angular velocity of the student after extending their arms. Assume no external torques act throughout the experiment.

Answer:

Given: Initial moment of inertia, I₁ = 2 kg·m² Initial angular velocity, ω₁ = 3 rad/s Final moment of inertia, I₂ = 3 kg·m²

According to the conservation of angular momentum, the initial angular momentum (L₁) is equal to the final angular momentum (L₂) when there are no external torques acting on the system.

The initial angular momentum (L₁) can be calculated as: L₁ = I₁ω₁

Let's substitute the known values: L₁ = (2 kg·m²) * (3 rad/s) L₁ = 6 kg·m²/s

The final angular momentum (L₂) can be calculated as: L₂ = I₂ω₂

We need to find the final angular velocity (ω₂). To do this, we rearrange the equation as follows: ω₂ = L₂ / I₂

Since L₁ = L₂, we can substitute L₁ into the equation: ω₂ = L₁ / I₂

Substituting the values we know: ω₂ = (6 kg·m²/s) / (3 kg·m²)

Simplifying: ω₂ = 2 rad/s

Therefore, the final angular velocity of the student after extending their arms is 2 rad/s.