In a skating stunt, a number of skaters hold hands and form a straight line. The
ID: 2185042 • Letter: I
Question
In a skating stunt, a number of skaters hold hands and form a straight line. They try to skate so that the line rotates about the skater at one end, who acts as the pivot. The skater farthest out has a mass of 80.0 kg and is 6.10 m from the pivot. He is skating at a speed of 6.80 m/s. Determine the magnitude of the centripetal force that acts on him?
please explain
Explanation / Answer
Answer --> 485.9 N Equation F = [m * v^2] / r Where m = Mass of outer guy v = The guy's velocity r = Distance to the guy Values: m = 80.0 kg v = 6.45 m/s r = 6.85 m Solve F = [ (80.0 kg) * (6.45 m/s)^2 ] / (6.85 m) F = [ (80.0 kg) * (41.6025 m^2/s^2) ] / (6.85 m) F = [ 3,328.2 kg-m^2/s^2 ] / (6.85 m) F = 485.9 N