Question

Warsaw, Poland, Olivia Oliver of Nova Scotia broke the world record for being the fastest spinner on ice skates. She achieved a record 342 rev/min, beating the existing Guinness World Record by 34 rotations. If an ice skater extends her arms at that rotation rate, what would be her new rotation rate? Assume she can be approximated by a 45-kg rod that is 1.7 m tall with a radius of 15 cm in the record spin. With her arms stretched take the approximation of a rod of length 130 cm with 10% of her body mass aligned perpendicular to the spin axis. Neglect frictional forces.

Answer 1

Answer:

155.5 rev/min

Explanation:

First, we will calculate the initial moment of inertia I_o. We will consider the ice skater as a rod rotating around its axis. Then, we calculate the final moment of inertia I_f. In this occasion we consider the arms as a rod of length L that is horizontally positioned, L so that the length of an arm is L/2. We will call M_1 the mass that remains close to the rotation axis (90 percent) and M_2 the mass located at the arms (10 percent). Finally, we write the equation for the conservation of angular momentum and we solve for  ω_f.

I_o=MR^2/2

     =(45)(0.15)^2/2

     =0.5 kgm^2

M_1=(0.9)(45)

      =40.5 kg

M_2=(0.1)(45)

      =4.5 kg

  I_f=M_1*R^2/2+M_2*L^2/12

      =1.1 kg m^2

I_f*ω_f=I_o*ω_o

     ω_f = I_o*ω_o/ I_f

            =155.5 rev/min