The outstretched hands and arms of a figure skater preparing for a spin can be c

Question

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 his 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 . When outstretched, they span 1.8 ; when wrapped, they form a cylinder of radius 25 . The moment of inertia about the rotation axis of the remainder of his body is constant and equal to .
f his original angular speed is 0.30 , what is his final angular speed?

Explanation / Answer

From conservation of angular momentum (I*omega)i = (I*omega)f I (initial) = Ibody + I arm = 0.450 kg-m^2 + 1/12*m*l^2 = 0.450 kg-m^2 + 1/12*8.00kg*(1.80m)^2 = 2.61 kg-m^2 I (final) = 0.450kg-m^2 + m*r^2 = 0.450 + 8.00*0.25^2 = 0.95 kg-m^2 Now 2.61*0.400 = 0.95*omega(f) So omega(f) = 2.61*0.400/0.95 = 1.10 rev/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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