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 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 8.00kg . When outstretched, they span 1.80m ; when wrapped, they form a cylinder of radius 25.0 cm. The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to 0.350kg*m2. If the skater's original angular speed is 0.450rev/s , what is his final angular speed?

Answer 1

I_i=I_body+I_arms =0.35+M_armsR_arms²/12

I_i =0.35+(8*1.8²/12)=2.51Kg-m²

I_f =0.35+M_armR_cylinder²

I_f =0.35+8*0.25²=0.85kg-m²

I_fω_f =I_iω_i

ω_f =I_iω_i/I_f =2.51*0.45/0.85

ω_f=1.297 rev/s ≈1.3rev/s

Answer 2

I_i=I_body+I_arms =0.35+M_armsR_arms²/12
I_i =0.35+(8*1.8²/12)=2.51Kg-m²
I_f =0.35+M_armR_cylinder²
I_f =0.35+8*0.25²=0.85kg-m²
I_fω_f =I_iω_i
ω_f =I_iω_i/I_f =2.51*0.45/0.85
ω_f=1.297 rev/s ≈1.3rev/s

Answer 3

From conservation of angular momentum (I*omega)i = (I*omega)f

I (initial) = Ibody + I arm = 0.350 kg-m^2 + 1/12*m*l^2 = 0.350 kg-m^2 + 1/12*8.00kg*(1.80m)^2 =in kg-m^2

I (final) = 0.350kg-m^2 + m*r^2 = 0.350 + 8.00*0.25^2 = 0.85 kg-m^2

Now I (initial) *0.450 = 0.85*omega(f) So omega(f) = in rev/s( calculate)