Force on a skater’s wrist.?

Question

A 48.0 kg ice skater spins about a vertical axis through her body with her arms horizontally outstretched, making 2.50 turns each second. The distance from one hand to the other is 1.5 m. Biometric measurements indicate that each hand typically makes up about 1.25 % of body weight.

A) What horizontal force must her wrist exert on her hand?

B) Express the force in part (a) as a multiple of the weight of her hand.

Answer 1

m=48*0.0125
omega=2*pi*2.5
r=1.5/2
A) centripetal force f = m*omega^2*r
B) centripetal acceleration f/m = omega^2*r