Question

A 45 kg figure skater is spinning on the toes of her skates at 0.60 rev/s. Her arms are outstretched as far as they will go. In this orientation, the skater can be modeled as a cylindrical torso (40 kg, 20 cm average diameter, 160 cm tall) plus two rod-like arms (2.5 kg each, 65 cm long) attached to the outside of the torso. The skater then raises her arms straight above her head, where she appears to be a 45 kg, 20-cm-diameter, 200-cm-tall cylinder.

Part A

 What is her new rotation frequency, in revolutions per second? 

Express your answer to two significant figures and include the appropriate units Value Units.

Answer 1

Initial moment of inertia of the system = Moment of inertia of person (I_C)+ Moment of inertia of his arms (I_A)
Moment of inertia of personas cylinder (I_C) = (1/2)MR² = (1/2)*40*(0.1)² = 0.2 kg-m²
where M is mass of person as cylinder = 40 kg
R is radius of person as cylinder = 10 cm = 0.1 m
Now moment of inertia of arms
I_A= I_Center + m*x²
Where m is the mass of the arms = 2.5*2 = 5 kg
where x is the distance between the center of arms and the center of cylinder (x) = (L/2)+R = 32.5+10 = 42.5 cm
where L is length of the arms = 65 cm
= mL²/12) +(mx²) = (5*0.65²/12) +(5*0.425²) = 1.079 kg-m²
Now the total moment of inertia of system
I_initial = I_C + I_A = 1.279 kg-m²
Now the final moment of inertia of the system
I_final = (1/2)(m+M)R² = 45*(0.1)² = 0.225 kg-m²
Now using the angular Conservation of momentum
I_initial*W_intial   = I_Final*W_final
1.279*0.6 = 0.225*W_final
W_final = 3.41 rad/s