Question41. (II) On an ice rink two skaters of equal mass grab hands and spin in a mutual circle once every 2.5 s . If we assume their arms are each 0.80 m long and their individual masses are 55.0 kg , how hard are they pulling on one another?

Studdy Solution
Calculate the force exerted by each skater. Simplify and compute the expression for $F_c$:
$$F_c = 55.0 \times 0.80 \times \left(\frac{2\pi}{2.5}\right)^2$$
$$F_c \approx 55.0 \times 0.80 \times 6.32$$
$$F_c \approx 278.3 \, \text{N}$$
Thus, each skater is pulling on the other with a force of approximately:
$$\boxed{278.3 \, \text{N}}$$