Question

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

Answer 1

Time taken to complete on complete circle = 2.3 seconds

Radius of circle (r) = 0.85 (length of each arm)

Speed of skaters = $\frac{Distance}{time}$

$Speed = \frac{2\pi r}{t}$

$Speed = \frac{2 \times 3.14 \times 0.85}{2.3}$

Speed (v) = 2.32 m/s

Let the force applied by one skater on the other be F

F(net) = centripetal force

$F = \frac{mv^2}{r}$

$F = \frac{65 \times 2.32^2}{0.85}$

F = 411.59 Newtons

Hence, the force applied by each skater on the other is: F = 411.59 Newtons

Answer 2

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