Chapter 7, Problem 7.59P

To determine

(a)

An equation for the speed which Joe can pedal into the wind.

Answer to Problem 7.59P

$V^3+2V_w{V}^2+\left(V_w^2+\frac{2C_{rr}mg}{C_{D}A\rho }\right)V-\left(\frac{2C_{rr}mg}{C_{D}A\rho }{V}_b+V_b^3\right)=0$

Explanation of Solution

Given information:

Joe pedals his bike at $10m/s$ without the effect of wind

The rolling resistance is equal to $0.8N.s/m$

The drag area is $0.422m^2$

Weight of Joe is $80kg$

Weight of bike is $15kg$

The speed of headwind is $5m/s$

The rolling force $F_R$ is defined as,

$F_R=C_{rr}mg$

Where,

$C_{rr}$ - rolling resistance co-efficient

$m$ - Total mass acting normal to the plane

The air drag is defined as,

$F_{drag}=C_D\frac{\rho }{2}{V}^{2}A$

Where,

$C_D$ - Drag co-efficient

$\rho$ - Density of the fluid

$V$ - Velocity

$A$ - Characteristic area

The power is defined as,

$P=FV$

Calculation:

Evaluate the total force,

${\displaystyle \sum F=}{F}_R+F_{drag}$

For drag force the relative velocity is equal to,

$V_{relative}=V+V_{wind}$

Substitute for forces,

${\displaystyle \sum F=}{C}_{rr}mg+C_{D}A\frac{\rho }{2}{\left(V+V_{wind}\right)}^2$

Calculate the power,

$P=FV$

Substitute for total force,

$P=FV=\left\{C_{rr}mg+C_{D}A\frac{\rho }{2}{\left(V+V_{wind}\right)}^2\right\}V$

Evaluate the power required without head wind,

$P=FV=\left\{C_{rr}mg+C_{D}A\frac{\rho }{2}{V}_{bike}{}^2\right\}{V}_{bike}$

But, we can say the power output will be same at both occasions,

Therefore,

$\left\{C_{rr}mg+C_{D}A\frac{\rho }{2}{V}_{bike}{}^2\right\}{V}_{bike}=\left\{C_{rr}mg+C_{D}A\frac{\rho }{2}{\left(V+V_{wind}\right)}^2\right\}V$

Solve further,

$V^3+2V_w{V}^2+\left(V_w^2+\frac{2C_{rr}mg}{C_{D}A\rho }\right)V-\left(\frac{2C_{rr}mg}{C_{D}A\rho }{V}_b+V_b^3\right)=0$

Conclusion:

The equation for the speed at which Joe has to ride into the wind can be given as,

$V^3+2V_w{V}^2+\left(V_w^2+\frac{2C_{rr}mg}{C_{D}A\rho }\right)V-\left(\frac{2C_{rr}mg}{C_{D}A\rho }{V}_b+V_b^3\right)=0$.

To determine

(b)

Speed $V$.

Answer to Problem 7.59P

$V=9.64m/s$

Explanation of Solution

Given information:

Joe pedals his bike at $10m/s$ without the effect of wind

The rolling resistance is equal to $0.8N.s/m$

The drag area is $0.422m^2$

Weight of Joe is $80kg$

Weight of bike is $15kg$

The speed of headwind is $5m/s$

According to sub-part A,

We have found a cubic equation for speed $V$

$V^3+2V_w{V}^2+\left(V_w^2+\frac{2C_{rr}mg}{C_{D}A\rho }\right)V-\left(\frac{2C_{rr}mg}{C_{D}A\rho }{V}_b+V_b^3\right)=0$

In above equation,

$V_w$ - Speed of wind

$V_b$ - Speed of bike

Assume, the air at $20°C$ will have,

$\rho =1.2kg/m^3$

Calculation:

The cubic equation for speed $V$ is given as,

$V^3+2V_w{V}^2+\left(V_w^2+\frac{2C_{rr}mg}{C_{D}A\rho }\right)V-\left(\frac{2C_{rr}mg}{C_{D}A\rho }{V}_b+V_b^3\right)=0$

Substitute given values,

$\begin{array}{l}{V}^3+2\left(5m/s\right)V^2+\left({\left(5m/s\right)}^2+\frac{2\left(0.8N.s/m\right)\left(95kg\right)\left(9.81m/s^2\right)}{\left(0.422m^2\right)\left(1.2kg/m^3\right)}\right)V\\ -\left(\frac{2\left(0.8N.s/m\right)\left(95kg\right)\left(9.81m/s^2\right)}{\left(0.422m^2\right)\left(1.2kg/m^3\right)}\left(10m/s\right)+{\left(10m/s\right)}^3\right)=0\end{array}$

Solve further,

$V^3+10V^2+2970V-30445.5V=0$

Therefore,

$V=9.64m/s$

Conclusion:

Joe should ride at $9.64m/s$ into the headwind.

To determine

(c)

The reason the result is not simplified $10-5=5m/s$, as one might first suspect.

Answer to Problem 7.59P

Since, $F_{drag}\infty V^2$

We cannot able to estimate that,

$V=V_{bike}-V_{wind}=10m/s-5m/s$

Explanation of Solution

Given information:

Joe pedals his bike at $10m/s$ without the effect of wind

The rolling resistance is equal to $0.8N.s/m$

The drag area is $0.422m^2$

Weight of Joe is $80kg$

Weight of bike is $15kg$

The speed of headwind is $5m/s$

The drag force is defined as,

$F_{drag}=C_D\frac{\rho }{2}{V}^{2}A$

Therefore, we can say that,

$F_{drag}\infty V^2$

Hence, we cannot say that,

$V=V_{bike}-V_{wind}$

The rolling resistance will also have an effect on the velocity.

Conclusion:

Since, $F_{drag}\infty V^2$

We cannot able to estimate that,

$V=V_{bike}-V_{wind}=10m/s-5m/s$.