1. **Problem statement:** A rotating ring (portal) has an initial moment of inertia $I_1=6.0$ kgm$^2$ and initial angular velocity $\omega_1=1.5$ rad/s. It shrinks, reducing its moment of inertia to $I_2=2.0$ kgm$^2$. We need to find: a) The new angular velocity $\omega_2$. b) The change in rotational kinetic energy $\Delta K$. 2. **Relevant formulas and principles:** - Conservation of angular momentum applies since no external torque acts: $$L = I_1 \omega_1 = I_2 \omega_2$$ - Rotational kinetic energy is: $$K = \frac{1}{2} I \omega^2$$ 3. **Find the new angular velocity $\omega_2$: ** From conservation of angular momentum: $$I_1 \omega_1 = I_2 \omega_2$$ Solve for $\omega_2$: $$\omega_2 = \frac{I_1}{I_2} \omega_1$$ Substitute values: $$\omega_2 = \frac{6.0}{2.0} \times 1.5 = 3 \times 1.5 = 4.5 \text{ rad/s}$$ 4. **Calculate the change in rotational kinetic energy $\Delta K$: ** Initial kinetic energy: $$K_1 = \frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} \times 6.0 \times (1.5)^2 = 3.0 \times 2.25 = 6.75 \text{ J}$$ Final kinetic energy: $$K_2 = \frac{1}{2} I_2 \omega_2^2 = \frac{1}{2} \times 2.0 \times (4.5)^2 = 1.0 \times 20.25 = 20.25 \text{ J}$$ Change in kinetic energy: $$\Delta K = K_2 - K_1 = 20.25 - 6.75 = 13.5 \text{ J}$$ **Answer:** - a) The new angular velocity is $\boxed{4.5}$ rad/s. - b) The rotational kinetic energy increases by $\boxed{13.5}$ joules.