Question

For a touring bicyclist the drag coefficient C (wherefair=12CAρv2) is 1.06, the frontal area A is 0.460 m2 , the air density ρ is 1.20 kg/m3 ,and the coefficient of rolling friction is 4.10×10⁻³. The rider has mass 53.0 kg , and her bike has mass 12.4 kg .

Part A

To maintain a speed of 12.3 m/s on a level road, what must the rider's power output to the rear wheel be?

P =

W

Part B

For racing, the same rider uses a different bike with a coefficient of rolling friction 2.50×10⁻³ and mass 9.00 kg . She also crouches down, reducing her drag coefficient to 0.860 and reducing her frontal area to 0.360 m2 . What must her power output to the rear wheel be then to maintain a speed of 12.3 m/s ?

P =

W

Part C

For the situation in part B, what power output is required to maintain a speed of 6.30 m/s ? (For more on aerodynamic speed limitations for a wide variety of human-powered vehicles see "The Aerodynamics of Human-Powered Land Vehicles," Scientific American, December, 1983.)

P =

W

Answer 1

C = 1.06,
The frontal area A = 0.460 m2
The air density ρ = 1.20 kg/m3 ,
The coefficient of rolling friction = 4.10×10−3.
The rider has mass 53.0 kg ,
Bike has mass 12.4 kg .

The formulas are:
fa = .5 *a * cd * rho * v^2 (fa=aero force, a=area, cd=drag coeff, rho=density, v=velocity)
ff = m * g * cf (ff=friction force, m=mass, g=grav accel, cd=friction coeff)


(a)
Power= v * (fa + ff)
Power= 12.3 * (.5 *a * cd * rho * v^2 + m * g * cf)
Power= 12.3 * (.5 *0.460 * 1.06 *1.20 * 12.3^2 + (53+12.4) * 9.8 * 4.10*10^-3)
Power = 576.7 Watt

(b)
Power= v * (fa + ff)
Power= 12.3 * (.5 *a * cd * rho * v^2 + m * g * cf)
Power= 12.3 * (.5 *0.360 * 0.860 *1.20 * 12.3^2 + (53+9) * 9.8 * 2.50*10^-3)
Power = 364.35 Watt

(c)

Power= v * (fa + ff)
Power= 6.30 * (.5 *a * cd * rho * v^2 + m * g * cf)
Power= 6.30 * (.5 *0.360 * 0.860 *1.20 * 6.30^2 + (53+9) * 9.8 * 2.50*10^-3)
Power = 56.0 Watt


a. 40.0032 2.736055 512.8711
b. 27.82771 1.735777 354.7619
c. 6.956928 1.735777 52.15623