Complete Question
On an ice rink two skaters of equal mass grab hands and spin in a mutual circle once every 2.7 s .
If we assume their arms are each 0.90 m long and their individual masses are 65.0 kg , how hard are they pulling on one another?
Answer:
The force is [tex]F = 316.8 \ N[/tex]
Explanation:
From the question we are told that
The period is T = 2.7 s
The radius of the circle formed by their arms is r = 0.90 m
Their individual mass is [tex]m = 65.0 \ kg[/tex]
Generally their angular velocity is mathematically represented as
[tex]w = \frac{2 \pi}{T}[/tex]
=> [tex]w = \frac{2 * 3.142 }{2.7}[/tex]
=> [tex]w =2.327 \ rad/ s [/tex]
Generally the pulling force is mathematically represented as
[tex]F = m * w ^2 * r[/tex]
=> [tex]F = 65 * 2.327^2 * 0.90[/tex]
=> [tex]F = 316.8 \ N[/tex]
