## Physics: Principles with Applications (7th Edition)

$1.2 kg \cdot m^2$ By bringing her arms close to her body.
There is no net external torque on the skater, so her net angular momentum is conserved. $$L_{i}=L_{i}$$ $$I_i \omega_i = I_f \omega_f$$ $$I_f = \frac{\omega_i }{\omega_f}I_i$$ $$I_f = \frac{1.0rev/1.5s }{2.5rev/s}(4.6 kg \cdot m^2)=1.2 kg \cdot m^2$$ Physically, she pulls her arms closer to her body (the rotation axis) to reduce her moment of inertia.