Chapter 8, Problem 80GP

(a)

To determine

To Explain: The reason why the angle $\theta$ must be given by $\tan \theta =\frac{F_{fr}}{F_N}$ .

Answer to Problem 80GP

The angle $\theta$ must be given by $\tan \theta =\frac{F_{fr}}{F_N}$ , if the cyclist is to maintain balance.

Explanation of Solution

Given:

Bicycle traveling speed is, $\text{v}\text{=}\text{4}\text{.2}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$

The radius of road turn is, $\text{r}\text{=}\text{6}\text{.4m}$

The normal force and friction force are ${\text{F}}_{\text{N}}$ and ${\text{F}}_{\text{fr}}$ that are exerted by the road on the tires respectively and $m\overrightarrow{g}$ is the total weight of the cycle and cyclist.

Formula used:

The sum of torques about the center of the mass is given as,

  ${\displaystyle \sum \text{τ}}\text{=}{\text{F}}_{\text{N}}\text{x}\text{-}{\text{F}}_{\text{fr}}\text{y}\text{=}\text{0}\mathrm{.....}\left(\text{1}\right)$

Calculation:

The total torque on the cyclist about the axis through the center of mass and align to the ground will be zero in order not to fall over. Now, according to the given free-body diagram, the summation of torques about their center of mass with clockwise as positive and its sum equals to zero such that,

  $\begin{array}{l}{\displaystyle \sum \text{τ}}\text{=}{\text{F}}_{\text{N}}\text{x}\text{-}{\text{F}}_{\text{fr}}\text{y}\text{=}\text{0}\\ \Rightarrow {\text{F}}_{\text{N}}\text{x}={\text{F}}_{fr}y\\ \Rightarrow \frac{x}{y}=\frac{{\text{F}}_{fr}}{{\text{F}}_{\text{N}}}\\ \Rightarrow \tan \theta =\frac{{\text{F}}_{fr}}{{\text{F}}_{\text{N}}}\end{array}$

Conclusion:

Hence, the angle $\theta$ must be given by $\tan \theta =\frac{F_{fr}}{F_N}$ .

(b)

To determine

To Calculate: The values of $\theta$ .

Answer to Problem 80GP

The angle for the given values is ${16}^{°}$ .

Explanation of Solution

Given:

Bicycle traveling speed is, $\text{v}\text{=}\text{4}\text{.2}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$

The radius of road turn is, $\text{r}\text{=}\text{6}\text{.4m}$

The normal force and friction force are ${\text{F}}_{\text{N}}$ and ${\text{F}}_{\text{fr}}$ exerted by the road on the tires respectively and $m\overrightarrow{g}$ is the total weight of the cycle and cyclist.

And, the angle $\theta$ is given as,

  $\tan \theta =\frac{F_{fr}}{F_N}$

Formula used:

The value of $\theta$ is calculates as,

  $\tan \theta =\frac{F_{fr}}{F_N}$

Where, ${\text{F}}_{\text{N}}$ is the normal force and ${\text{F}}_{\text{fr}}$ is the frictional force.

Calculation:

The normal force is equal to weight because the cycle is not accelerating vertically, such that ( ${\text{F}}_{\text{N}}\text{=}\text{mg}$ )

The frictional force must be supplying the centripetal force due to cyclists who traveling horizontally as he is moving in a circular path. Therefore, ${\text{F}}_{fr}\text{=}\frac{m{v}^2}{r}$

Now, by substituting the given values in this formula, the value of $\theta$ is calculated as,

  $\begin{array}{l}\text{tan}\text{θ}\text{=}\frac{{\text{F}}_{\text{fr}}}{{\text{F}}_{\text{N}}}\\ \text{tan}\text{θ}\text{=}\frac{\left(\frac{{\text{mv}}^{\text{2}}}{\text{r}}\right)}{\text{mg}}\\ \text{tan}\text{θ}\text{=}\frac{{\text{v}}^{\text{2}}}{\text{rg}}\\ \Rightarrow \text{θ}\text{=}{\text{tan}}^{\text{-1}}\left(\frac{{\text{v}}^{\text{2}}}{\text{rg}}\right)\\ \Rightarrow \text{θ}\text{=}{\text{tan}}^{\text{-1}}\left(\frac{{\left(\text{4}\text{.2}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.\right)}^{\text{2}}}{\left(\text{6}\text{.4m}\right)\text{×}\left(\text{9}\text{.8}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.\right)}\right)\\ \Rightarrow \text{θ}\text{=}\text{15}{\text{.71}}^{°}\simeq {16}^{°}\end{array}$

Conclusion:

The angle is ${16}^{°}$ .

(c)

To determine

To Calculate: The minimum turning radius.

Answer to Problem 80GP

The minimum turning radius is $\text{2}\text{.6m}$ .

Explanation of Solution

Given:

The given force-diagram is

  

Bicycle traveling speed is, $\text{v}\text{=}\text{4}\text{.2}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$

The radius of road turn is, $\text{r}\text{=}\text{6}\text{.4m}$

The normal force and friction force are ${\text{F}}_{\text{N}}$ and ${\text{F}}_{\text{fr}}$ exerted by the road on the tires respectively and $m\overrightarrow{g}$ is the total weight of the cycle and cyclist.

And, the angle $\theta$ is given as,

  $\tan \theta =\frac{F_{fr}}{F_N}$

Formula used:

• The value of $\theta$ is calculates as,

  $\tan \theta =\frac{F_{fr}}{F_N}$

Where ${\text{F}}_{\text{N}}$ is the normal force and ${\text{F}}_{\text{fr}}$ is the frictional force.

• The static friction force is calculated as,

  ${\text{F}}_{\text{fr}}\text{=}{\text{μF}}_{\text{N}}$

Where, $\text{μ}$ is the frictioncoefficient

Calculation:

Due to the smallest turning radius, there is a maximum force according to the friction force formula, ${\text{F}}_{fr}\text{=}\frac{m{v}^2}{r}$ and the static friction force is ${\text{F}}_{\text{fr}}\text{=}{\text{μF}}_{\text{N}}$ . Now using these two values of force, the radius is calculated such that

  $\begin{array}{l}{\text{F}}_{\text{fr}}\text{=}{\text{μF}}_{\text{N}}\\ \Rightarrow \frac{{\text{mv}}^{\text{2}}}{{\text{r}}_{\text{min}}}\text{=}\text{μmg}\\ \Rightarrow {\text{r}}_{\text{min}}\text{=}\frac{{\text{mv}}^{\text{2}}}{\text{μmg}}\text{=}\frac{{\left(\text{4}\text{.2m}\right)}^{\text{2}}}{\left(\text{0}\text{.70}\right)\left(\text{9}\text{.80}\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{sec}}^{\text{2}}$}\right.\right)}\text{=}\text{2}\text{.6m}\end{array}$

Conclusion:

The minimum radius is $\text{2}\text{.6m}$ .