Modeling the female skater, of mass 47.0 kg, as a particle, and assuming that the combined length of the two outstretched arms is 124 cm and that arms make an angle of 45.0° with the horizontal, what is the magnitude of the force (in N) exerted by the male skater's wrist if each turn is completed in 2.00 s?

Question

Physics

Dorsey

3 May, 08:14

Modeling the female skater, of mass 47.0 kg, as a particle, and assuming that the combined length of the two outstretched arms is 124 cm and that arms make an angle of 45.0° with the horizontal, what is the magnitude of the force (in N) exerted by the male skater's wrist if each turn is completed in 2.00 s?

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Answers (1)

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Jordin Camacho · Answer 1

The magnitude of the force is 575.2 N

Explanation:

the force exerted by the male skater must be equal to the centripetal force:

F*cos45° = m*r*w^2, where

m = 47 kg

r = 1.24*cos45° m

w = (2*π) / 2 = π radians/s

Replacing values:

F*cos45° = 47 * (1.21*cos45°) * π^2

F = 47*1.24*π^2 = 575.2 N