On an ice rink, two skaters of equal mass grab hands and spin in a mutual circle once every 2.3 s. if we assume their arms are each 0.85 m long and their individual masses are 65.0 kg, how hard are they pulling on one another

We have to

m = 65 kg

r = 0.85 m

t = 2.3 s

The perimeter of the circle is given by:

P = 2 * pi * r

P = 2 * pi * (0.85 m)

P = 5.34 m

Then, by definition, the distance equals the speed by time:

d = v * t

v = d / t

v = (5.34 m) / (2.3 s)

v = 2.32 m / s

Then, to find the radial acceleration, we must use the speed found and the radius of the circle:

a = v ^ 2 / r

a = (2.32 m / s) ^ 2 / (0.85 m)

a = (5.38 m ^ 2 / s ^ 2) / (0.85 m)

a = 6.32 m / s ^ 2

Finally, we have that by definition the force is equal to the mass by acceleration:

F = m * a

F = (65 kg) * (6.32 m / s ^ 2)

F = 410.8 N

answer

F = 410.8 N