Question

A diver in midair has an angular velocity of 6.0 rad/s and a moment of inertia of 1.2 kg·m2. He then pulls is arms and legs into a tuck position and his angular velocity increases to 12 rad/s. The net external torque acting on the diver is zero. What is his moment of inertia in the tuck position?

Answer 1

The diver's moment of inertia in the tuck position is found to be 0.6 kg·m^2 using the conservation of angular momentum, given that no external torque acts on him.

Step-by-step explanation:

The question relates to the concept of conservation of angular momentum, which is a principle in physics stating that if no external torque acts on a system, the total angular momentum of the system remains constant. In this problem, a diver's angular velocity increases as he changes from a relaxed position to a tucked position, indicating that his moment of inertia must decrease to conserve angular momentum because external torque is zero.

To solve for the diver's moment of inertia in the tucked position, we use the formula for conservation of angular momentum:

1. L_initial = L_final
2. I_initial * ω_initial = I_final * ω_final
3. (1.2 kg·m2) * (6.0 rad/s) = I_final * (12 rad/s)
4. I_final = (1.2 kg·m2 * 6.0 rad/s) / 12 rad/s
5. I_final = 0.6 kg·m2

Therefore, the moment of inertia of the diver in the tuck position is 0.6 kg·m2.

Answer 2

0.6 kg m²

Step-by-step explanation:

Angular momentum is conserved.

Iω = Iω

(1.2 kg m²) (6.0 rad/s) = I (12 rad/s)

I = 0.6 kg m²