Physics help: On an ice rink, two skaters of equal mass grab hands and spin in a mutual circle once every 2.9s?

If we assume their arms are each 0.80m long and their individual masses are 57.0kg , how hard are they pulling on one another?

2 Answers

  • 1 decade ago
    Favorite Answer

    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)/2.9 => v=0.5517*pi

    now we can apply this to the original equation:


    F= 214.038 Newtons (kgm/s^2)

    where pi = 3.1416

    This is the force that one person is exerting on the other.

  • Anonymous
    5 years ago

    common velocity V = m1 u1 / m1+m2 = m x4 / 2 m = 2 m/sec so, the answer is (3).

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