Question

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?

Answer 1

Answer:

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

Answer 2

The force exerted by the skater is 575.2 N.

Force exerted by the skater

The force exerted by the skater is calculated by taking net force as follows;

$Fcos(\theta) = F_c$

where;

• Fc is the centripetal force on the circular path

$Fcos(\theta) = m\omega ^2 r$

The angular speed for each complete turn in 2 seconds is calculated as follows;

$\omega = \frac{2\pi \ rad}{2 \ s} = \pi \ rad/s$

The force exerted by the skater is calculated as follows;

$F = \frac{m \omega ^2 r cos(\theta)}{cos(\theta)} \\\\F = \frac{47 \times (\pi)^2 \times 1.24cos(45)}{cos(45)} \\\\F = 47 \times (\pi)^2 \times 1.24\\\\F = 575.2 \ N$

Learn more about centripetal force here: https://brainly.com/question/20905151