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 hand 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 cilinder. His hands and arms have a combine and arms have a combined mass of 8kg. When outstreched, they span 1.8m; when wrapped, they form a cylinder of radius 25cm. The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to .40kg*m^2. If the skater's original angular speed is .40rev/s, what is his final angular speed?
moment of inertia for slender rod, axis through center I= 1/12ML^2
moment of inertia for hollow cylinder 1/2M(R_1^2 + R_2^2)
conservation of angular momentum I_i*ω_i = I_f*ω_f
The Attempt at a Solution
.4rev/s = 2.51 rad/s
I_i = 1/2*8*1.8^2 = 12.96
ω_i = 2.51
I_f = 1/2*8*(.9^2 + .125^2) = 3.3 +.4 = 3.7
putting those numbers in the conservation of energy forumla and i get ω_f = 8.7 rad/s. the back of my book says the answer is 7.16 rad/s.