decelerating cars - kypt
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DECELERATING CARS
KangSan Kim혜움나래
Problem Statement
1. Simple skid2. reverse
Existing theoretical background𝜇𝑘
velocity
fric
tion
Experiment setup - 1
Experiment setup - 2
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
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
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
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
Explanation #1
time
forc
e
Δ 𝑡 Δ 𝑡 ′
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
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
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
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
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