Final answer:
- Mass of hands and arms: 8.0 kg
- Span of outstretched arms: 1.6 m
- Radius of the thin-walled hollow cylinder (wrapped arms): 0.24 m
- Constant moment of inertia of the rest of the body: 0.4 kg·m2
- Initial angular speed: 0.50 rev/s
Step-by-step explanation:
To calculate the figure skater's final angular speed, we need to use conservation of angular momentum, since no external torques are acting on the system. The initial moment of inertia when the skater's arms are outstretched is that of a slender rod pivoting about its center: Iinitial = (1/12) * m * L2, where m is the mass of the arms and L is the span of the arms. When the arms are wrapped around the body, forming a thin-walled hollow cylinder, the moment of inertia is Iwrapped = m * r2, with m being the mass and r the radius of the cylinder formed by the skater's arms.
Using the data provided:
- Mass of hands and arms: 8.0 kg
- Span of outstretched arms: 1.6 m
- Radius of the thin-walled hollow cylinder (wrapped arms): 0.24 m
- Constant moment of inertia of the rest of the body: 0.4 kg·m2
- Initial angular speed: 0.50 rev/s
Applying conservation of angular momentum Linitial = Lfinal, we get:
Iinitial * ωinitial = (Iwrapped + Ibody) * ωfinal
Where ω represents the angular speed. Plugging in the values and solving for the final angular speed, we can determine how much faster the skater will spin after bringing in his arms.