The Spinning Figure Skater. The outstretched hands and arms of a figure skater p

Question

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. 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 7.0 kg. When outstretched, they span 1.9 m; when wrapped, they form a cylinder of radius 35 cm. The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.50 kg·m2. If his original angular speed is 0.40 rev/s, what is his final angular speed?
rev/s

Explanation / Answer

using the principle of conservation of angular momentum -

initial angular momentum=final angular momentum

angular momentum = I(MOI)*(ang vel)

(mL2/12 + I)=(mr2 + I)

r = 0.35 m

L = 1.9 m

I = 0.50 kg·m2

m= 7 kg

= 0.4 rps

= final angular speed

by calculation we get

0.4 (7*1.92 /12 + 0.5) = (7*0.352 + 0.5)

= 0.76 rev per second

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.