As an ice skater begins a spin, his angular speed is 3.04 rad/s. After pulling in his arms, his angular speed increases to 5.24 $\mathrm{rad} / \mathrm{s}$ . Find the ratio of the skater's final moment of inertia to his initial moment of inertia.
Added by Philip H.
Step 1
The angular momentum before the spin (when the skater's arms are extended) is equal to the angular momentum after the spin (when the skater's arms are pulled in). The formula for angular momentum is L = Iω, where L is the angular momentum, I is the moment of Show more…
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$\cdot$ The spinning figure skater. The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center. (See Figure $10.53 .$ ) When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled hollow cylinder. His hands and arms have a combined mass of 8.0 $\mathrm{kg}$ . When outstretched, they span 1.8 $\mathrm{m}$ ; when wrapped, they form a cylinder of radius 25 $\mathrm{cm} .$ The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to 0.40 $\mathrm{kg} \cdot \mathrm{m}^{2} .$ If the skater's original angular speed is 0.40 $\mathrm{rev} / \mathrm{s}$ what is his final angular speed?
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