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DECELERATING CARS KangSan Kim 혜혜혜혜
14

Decelerating cars - KYPT

Jun 23, 2015

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Antonio Stark

Experiment presentation for the selection for 혜움나래, KMLA's club for KYPT and IYPT preparation
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Page 1: Decelerating cars - KYPT

DECELERATING CARS

KangSan Kim혜움나래

Page 2: Decelerating cars - KYPT

Problem Statement

1. Simple skid2. reverse

Page 3: Decelerating cars - KYPT

Existing theoretical background𝜇𝑘

velocity

fric

tion

Page 4: Decelerating cars - KYPT

Experiment setup - 1

Page 5: Decelerating cars - KYPT

Experiment setup - 2

Page 6: Decelerating cars - KYPT

Raw dataForce Time (seconds per 10 crank revolutions)

0.4 14.59 12.31 11.89

0.5 6.4 7.07 6.68

0.6 5.29 6.44

Force Time0.7 18.56

0.8 15.44

0.9 10.62

1.1 7.7

Page 7: Decelerating cars - KYPT

Calibrated data

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.604012555561476 x + 0.409807867815422

f(x) = 0.224102928041391 ln(x) + 0.956165805589757f(x) = 0.981458167467079 x^0.341889476070784f(x) = − 0.179906476199951 x² + 0.807388417596375 x + 0.382199699370392

velocity

Fric

tiona

l for

ce

Page 8: Decelerating cars - KYPT

Calibrated data

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.604012555561476 x + 0.409807867815422

f(x) = 0.224102928041391 ln(x) + 0.956165805589757f(x) = 0.981458167467079 x^0.341889476070784

f(x) = − 0.179906476199951 x² + 0.807388417596375 x + 0.382199699370392

velocity

Fric

tiona

l for

ce

Page 9: Decelerating cars - KYPT

Explanation #1

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.604012555561476 x + 0.409807867815422

velocity

Fric

tiona

l for

ce

Page 10: Decelerating cars - KYPT

Explanation #1

time

forc

e

Δ 𝑡 Δ 𝑡 ′

Page 11: Decelerating cars - KYPT

Explanation #2

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.224102928041391 ln(x) + 0.956165805589757f(x) = 0.981458167467079 x^0.341889476070784

velocity

Fric

tiona

l for

ce

Page 12: Decelerating cars - KYPT

Explanation #3

0 0.2 0.4 0.6 0.8 1 1.2 1.40

0.2

0.4

0.6

0.8

1

1.2

f(x) = 0.224102928041391 ln(x) + 0.956165805589757f(x) = 0.981458167467079 x^0.341889476070784

velocity

Fric

tiona

l for

ce

Intimate contact

Fast Passover

Page 13: Decelerating cars - KYPT

Handicapso Experiments at lower velocity

o To determine the linear character of the graph

oMeasuring apparatus’ low accuracy

o Experiments at higher velocityo The change of slope at

extreme velocitiesoMechanical failure of cheap

elements

Page 14: Decelerating cars - KYPT

Conclusion• Conventional physics• Coefficient is constant regardless of situations• Horizontal line graph

• NOT a horizontal line

• 3 experimental solutions:• Linear• Decreasing slope• Decreasing, and then increasing slope