Question

On an ice rink, two skaters of equal mass grab hands and spin in a mutual circle once every 2.5 seconds.

If we assume their arms are each 0.80m long and their individual masses are 55.0kg , how hard are they pulling on one another?

Answer 1

The basic equation is:

F = mv2/r where F is the force, m is the mass v is the tangential velocity and r is the radius

But we need to find the velocity first so we need to work out the circumference of the circle which is pi*diameter (diameter is twice the length of the radius) of the circle

Therefore the circumference is 1.6*pi

We can now work out the velocity by dividing the circumference by the time it takes to complete one rotation:

v= (1.6*pi)/2.5 => v=0.5517*pi

now we can apply this to the original equation:

F=55*((0.5517pi)^2)/0.8
F= 214.038 Newtons (kgm/s^2)

where pi = 3.1416