The resistance to motion of a good bicycle on smooth pavement is nearly all due

Question

The resistance to motion of a good bicycle on smooth pavement is nearly all due to
aerodynamic drag. Assume that the total mass of rider and bike is M = 100 kg. The frontal area measured from a photograph is A = 0.46 m2. Experiments on a hill, where the road grade is 8 percent, show that terminal speed is Vt = 15 m/s. From these data, drag coefficient is estimated as CD = 1.2. a) Verify this calculation of drag coefficient. B) Estimate the distance needed for the bike and rider to decelerate from 15 to 10 m/s while coasting after reaching level road.

Explanation / Answer

Road grade = Tan = 0.08.

Hence, = 4.57 degrees. Sin = 0.08.

(a) For terminal velocity while going downhill, forward force due to gravitation = backward force due to drag.

Hence, MgSin = Cd*1/2*AV2

Taking air density = 1.2 kg/m3 and putting values we get,

100*9.8*0.08 = Cd*1/2*1.2*0.46*152

This gives, Cd = 1.26

This is approximately the same as mentioned in the question. Hence verified.

(b) Drag force on level road, F = Cd*1/2*AV2

Hence, deceleration = F/m = Cd*1/(2m)*AV2

dV/dt = Cd*1/(2m)*AV2

(dV/ds)(ds/dt) = Cd*1/(2m)*AV2 (Here s denotes the distance.)

V(dV/ds) = Cd*1/(2m)*AV2

dV/V = Cd*1/(2m)*A*ds

Integrating it on both the sides, ln (V2/V1) = Cd*1/(2m)*A*(s2-s1)

Hence, s2-s1 = 2m/Cd * ln (V2/V1) / A

Putting values, Distance = 2*100/1.2 * ln(10/15) / (1.2*0.46)= -122.5 m

Ignoring the negatie sign, distance = 122.5 m

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