Question

in short-track speed skating, the track has straight sections and semicircles 16 m in diameter. assume that a 65 kg skater goes around the turn at a constant 13 m/s .

Answer 1

The track comprises straight stretches and semicircles with a diameter of 16 meters for short-track speed skating. Consider a 65 kg skater traveling at a constant 13 m/s with a force of 1373.12 N.

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 = (65 kg)*(21.125 m/s²) = 1373.12 N

In mechanics, a force is any action that has the potential to change, maintain, or deform a body's motion. The three principles of motion outlined by Isaac Newton in his Principia Mathematica are frequently used to illustrate the concept of force (1687). Newton's first law states that a body at rest or moving uniformly in a straight line will stay in that state until a force is applied to it. According to the second law, a body will accelerate (change in velocity) in the direction of any external force acting on it. The strength of the external force directly correlates with the strength of the acceleration.

Learn more about  force here:

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