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

Marquis Hatfield

18 June, 14:55

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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Collin Estes · 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