Problem 6.03 Force on a skater\'s wrist What horizontal force must her wrist exe

Question

Problem 6.03 Force on a skater's wrist What horizontal force must her wrist exert on her hand? A 50.0 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. F 41.6 Submit Previous Answers Request Answer X Incorrect; Try Again; 7 attempts remaining Part B Express the force in part (a) as a multiple of the weight of her hand. F 0.0849 Submit Previous Answers Request Answer 15. Problem 6.26 Practice1 pt 86%

Explanation / Answer

Part A:

Centripetal acceleration is given by:

ac = V^2/R

V = w*R

ac = w^2*R

w = 2*pi/T

ac = 4*pi^2*R/T^2

Force on her hand will be

F = m*ac

F = 4*pi^2*m*R/T^2

r = d/2 = 1.5/2 = 0.75 m

m = 1.25% of total weight = 1.25% of 50 = 0.625 kg

T = 2.5 turns per second = 1/2.5 = 0.4 sec

Using known values:

F = 4*pi^2*0.625*0.75/0.4^2

F = 115.66 N

Part B

weight of each hand will be

W = Mh*g

Mh = mass of hand = 0.625 kg

W = 0.625*9.81 = 6.131 N

F in part A, in terms of W will be

F = 115.66 N = (115.66/6.131)*W

F = 18.86*W

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