An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 11 % . By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)

  K_{f} / K₀ =1.12

Explanation:

This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.

Initial moment. With arms outstretched

         L₀ = I₀ w₀

the wo value is 5.0 rad / s

final moment. After he shrugs his arms

          = I_{f}  w_{f}

indicate that the moment of inertia decreases by 11%

        I_{f} = I₀ - 0.11 I₀ = 0.89 I₀

        L_{f} = L₀

        I_{f} w_{f}  = I₀ w₀

        w_{f} = I₀ /I_{f}    w₀

let's calculate

        w_{f} = I₀ / 0.89 I₀   5.0

        w_{f} = 5.62 rad / s

Having these values we can calculate the change in kinetic energy

          / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)

         K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²

         K_{f} / K₀ =1.12

wavelength variability is the right one

prolly a beacuse the momentuem the car has increases while going down hill