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

Andrew Fletcher

Physics

17 October, 07:33

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?

+4

Answers (1)

Know the Answer?

Lorenzo Arroyo · 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