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?

Question

Dragonfly

Physics

14 May, 05:01

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?

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Answers (1)

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Mireya · 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