Question

A bicycle rider has a drag area of 0.22 m2. She is riding at a velocity of 8.3 m/s along the road and the air density is 1.2 kg/m3. Total mass of the rider and bike is 93 kg. The coefficient of rolling resistance is 0.004. She is riding into a headwind with velocity 1.8 m/s. The road is horizontal. What power (watts) must the cyclist produce to maintain the speed of 8.3 m/s?

Answer #1

drag force is given by the drag equation=0.5*pho*v^2*Cd*A

where pho=density of air=1.2 kg/m^3

v=speed of the rider w.r.t. air=8.3+1.8 (as headwind means the air is moving in the opposite direction of the vehichle)

==>v=10.1 m/s

drag coefficient for bicylce=0.9

A=0.22 m^2

then drag force=0.5*1.2*10.1^2*0.9*0.22=12.119 N

rolling friction force=coefficient of rolling resistance*normal force

=0.004*mass*g

=0.004*93*9.8=3.6456 N

total force to be applied by the cyclist=3.6456+12.119=15.765 N

then power to be developed=force*speed

=15.765*8.3=130.85 W

12) There are several types of drag on a car other than air
resistance. Effects having to do with the squeezing of the tires
(rolling resistance) and frictional forces in the drivetrain (the
system that transfers energy from the engine to the rotation of the
wheels) also must be taken into account. Engineers use the
following equation to model the total force due to these different
effects
Fdrag=A+Bv+Cv2Fdrag=A+Bv+Cv2
For a Camry, these coefficients are estimated to be
A=117.130A=117.130 N, B=1.800...

12) There are several types of drag on a car other than air
resistance. Effects having to do with the squeezing of the tires
(rolling resistance) and frictional forces in the drivetrain (the
system that transfers energy from the engine to the rotation of the
wheels) also must be taken into account. Engineers use the
following equation to model the total force due to these different
effects
Fdrag=A+Bv+Cv2Fdrag=A+Bv+Cv2
For a Camry, these coefficients are estimated to be
A=117.130A=117.130 N, B=1.800...

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