Question

In short-track speed skating, the track has straight sections and semicircles 16 m in diameter. assume that a 64 kg skater goes around the turn at a constant 13 m/s. what is the horizontal force on the skater?

Answer 1

The diameter of the semicircular portion of the track is 16 m.
Therefore its radius is
r = 8 m.

The tangential velocity of the skater is 13 m/s.
Therefore the angular speed is
ω = v/r = (13 m/s)/(8 m) = 1.625 rad/s

The horizontal force on the skater is due to centripetal acceleration of
a = r*ω² = (8 m)*(1.625 rad/s)² = 21.125 m/s².

The force acting on the 64-kg skater is
F = m*a = (64 kg)*(21.125 m/s²) = 1352 N

Answer: 1352 N