A person stands, hands at his side, on a platform that is rotating at a rate of 0.80 rev/s. If he raises his arms to a horizontal position, Fig. 11–31, the speed of rotation decreases to 0.60 rev/s. (a) Why? (b) By what factor has his moment of inertia changed? <IMAGE>

You stand on a stool that is free to rotate about an axis perpendicular to itself and through its center. The stool's moment of inertia around its central axis is 1.50 kg m² . Suppose you can model your body as a vertical solid cylinder (height = 1.80 m, radius = 20 cm, mass = 80 kg) with two horizontal thin rods as your arms (each:length = 80 cm, mass = 3 kg) that rotate at their ends, about the same axis, as shown. Suppose that your arms' contribution to the total moment of inertia is negligible if you have them pressed against your body, but significant if you have them wide open. If you initially spin at 5 rad/s with your arms against your body, how fast will you spin once you stretch them wide open? (Note:The system has 4 objects (stool + body + 2 arms), but initially only stool + body contribute to its moment of inertia)

Model a figure skater’s body (mass M) as a solid cylinder ( ≈ 0.75 M including head and thighs) and her arms as thin rods ( 0.13 M for both), making reasonable estimates for the dimensions. Then calculate the ratio of the angular speeds for a spinning skater with outstretched arms, and with arms held tightly against her body. (Ignore contributions from lower legs and feet.)

Competitive ice skaters commonly perform single, double, and triple axel jumps in which they rotate $1\frac{1}{2}$, $2\frac{1}{2}$ and $3\frac{1}{2}$ revolutions, respectively, about a vertical axis while airborne. For each of these jumps, a typical skater remains airborne for about 0.70 s. Suppose a skater leaves the ground in an “open” position (e.g., arms outstretched) with moment of inertia I₀ and rotational frequency ƒ₀ = 1.2 rev/s , maintaining this position for 0.10 s. The skater then assumes a “closed” position (arms brought closer) with moment of inertia I, acquiring a rotational frequency f, which is maintained for 0.50 s. Finally, the skater immediately returns to the “open” position for 0.10 s until landing (see Fig. 11–51). Why is angular momentum conserved during the skater’s jump? Neglect air resistance.

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Competitive ice skaters commonly perform single, double, and triple axel jumps in which they rotate $1\(\frac\)12$, $2\(\frac\)12$ and $3\(\frac\)12$ revolutions, respectively, about a vertical axis while airborne. For each of these jumps, a typical skater remains airborne for about 0.70 s. Suppose a skater leaves the ground in an “open” position (e.g., arms outstretched) with moment of inertia I₀ and rotational frequency ƒ₀ = 1.2 rev/s , maintaining this position for 0.10 s. The skater then assumes a “closed” position (arms brought closer) with moment of inertia I, acquiring a rotational frequency f, which is maintained for 0.50 s. Finally, the skater immediately returns to the “open” position for 0.10 s until landing (see Fig. 11–51). Determine the minimum rotational frequency f during the flight’s middle section for the skater to successfully complete a single and a triple axel.

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A professor sits on a rotating stool that is spinning at an initial angular velocity of ${\omega }_i$. If the professor pulls her arms inward, reducing her moment of inertia from $I_i$ to $I_f$, what happens to her angular velocity $\omega$?