Question

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 1

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
`F  = 316.8 \  N`

Step-by-step 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
`m =  65.0 \  kg`

Generally their angular velocity is mathematically represented as

`w = (2 \pi)/(T)`

=>
`w = (2 *  3.142 )/(2.7)`

=>
`w =2.327 \ rad/ s`

Generally the pulling force is mathematically represented as

`F  = m *  w ^2 *  r`

=>
`F  = 65 *  2.327^2 *  0.90`

=>
`F  = 316.8 \  N`