Chapter 5, Problem 97P

To determine

To Find: The safe range of speeds

Answer to Problem 97P

  $20.1\text{km/hr to 56}\text{.1 km/hr}$

Explanation of Solution

Given Information:

Radius of the curve is $30\text{m}$ .

Mass of the car, $m=950\text{kg}$ car

Speed of the car is $40.0\text{km/h}$

The coefficient of static friction between the road and the tires is $0.300$ .

Formula Used:

Newton’s second law of motion:

  ${\displaystyle \sum F}=ma$

Where, m is mass and a is acceleration.

Calculation:

  

From the Free body diagram is shown above. When the coefficient of static friction is zero along x-axis,

  $F_n\sin \theta =m\frac{v_{\min }^2}{r}.\mathrm{..}\left(1\right)$

And along y-axis

  $\begin{array}{l}{F}_n\cos \theta -mg=0\\ F_n\cos \theta =mg\mathrm{...}\left(2\right)\end{array}$

Along x-axis

  $F_n\sin \theta -f_s\cos \theta =m\frac{v_{\min }^2}{r}\mathrm{...}\left(3\right)$

And along y-axis

  $F_n\cos \theta +f_s\sin \theta -mg=0\mathrm{...}\left(4\right)$

Now the force of static friction is given as

  $f_s=f_{s,\max }={\mu }_s{F}_n\mathrm{...}\left(5\right)$

Substituting $\left(5\right)\text{ in }\left(3\right)\text{ and }\left(4\right)$

  $\begin{array}{l}{F}_n\left({\mu }_s\cos \theta -\sin \theta \right)=-m\frac{v_{\min }^3}{r}\mathrm{...}\left(6\right)\\ \text{and}{F}_n\left(\cos \theta +{\mu }_s\sin \theta \right)=mg\mathrm{...}\left(7\right)\end{array}$

Now dividing $\left(6\right)\text{ and }\left(7\right)$

  $\begin{array}{l}\frac{F_n\left({\mu }_s\cos \theta -\sin \theta \right)}{F_n\left(\cos \theta +{\mu }_s\sin \theta \right)}=-\frac{m\left(\frac{v_{\min }^2}{r}\right)}{mg}\\ \Rightarrow \frac{\left({\mu }_s\cos \theta -\sin \theta \right)}{\left(\cos \theta +{\mu }_s\sin \theta \right)}=\left(\frac{v_{\min }^2}{rg}\right)\\ \Rightarrow v_{\min }=\sqrt{\frac{rg\left(\sin \theta +{\mu }_s\cos \theta \right)}{\left(\cos \theta +{\mu }_s\sin \theta \right)}}\\ \Rightarrow v_{\min }=\sqrt{\frac{\left(\text{30 m}\right)\left(\text{9}{\text{.81 m/s}}^{\text{2}}\right)\left(\sin {22.8}^{\text{o}}-0.3\cos {22.8}^{\text{o}}\right)}{\left(\cos {22.8}^{\text{o}}+0.3\sin {22.8}^{\text{o}}\right)}}\\ \Rightarrow v_{\min }=5.59\text{m/s}\\ \Rightarrow v_{\min }=20.1\text{km/h}\end{array}$

Now when the car is traveling at maximum speed:

Along x-axis

  $F_n\sin \theta +f_s\cos \theta =m\frac{v_{\max }^2}{r}\mathrm{...}\left(8\right)$

And along y-axis

  $F_n\cos \theta -f_s\sin \theta -mg=0\mathrm{...}\left(9\right)$

Now the force of static friction is given as

  $f_s=f_{s,\max }={\mu }_s{F}_n\mathrm{...}\left(10\right)$

Substituting $\left(10\right)\text{ in }\left(8\right)\text{ and }\left(9\right)$

  $\begin{array}{l}{F}_n\left({\mu }_s\cos \theta +\sin \theta \right)=m\frac{v_{\max }^2}{r}\mathrm{...}\left(11\right)\\ \text{and }{F}_n\left(\cos \theta -{\mu }_s\sin \theta \right)=mg\mathrm{...}\left(12\right)\end{array}$

Now dividing $\left(11\right)\text{ and }\left(12\right)$

  $\begin{array}{l}\frac{F_n\left({\mu }_s\cos \theta -\sin \theta \right)}{F_n\left(\cos \theta +{\mu }_s\sin \theta \right)}=\frac{m\left(\frac{v_{\max }^2}{r}\right)}{mg}\\ \Rightarrow \frac{\left({\mu }_s\cos \theta +\sin \theta \right)}{\left(\cos \theta -{\mu }_s\sin \theta \right)}=\left(\frac{v_{\max }^2}{rg}\right)\\ \Rightarrow v_{\max }=\sqrt{\frac{rg\left({\mu }_s\cos \theta +\sin \theta \right)}{\left(\cos \theta -{\mu }_s\sin \theta \right)}}\\ \Rightarrow v_{\max }=\sqrt{\frac{\left(\text{30 m}\right)\left(\text{9}{\text{.81 m/s}}^{\text{2}}\right)\left(0.3\cos {22.8}^{\text{o}}+\sin {22.8}^{\text{o}}\right)}{\left(\cos {22.8}^{\text{o}}-0.3\sin {22.8}^{\text{o}}\right)}}\\ \Rightarrow v_{\max }=15.6\text{m/s}\\ \Rightarrow v_{\max }=56.1\text{km/h}\end{array}$

Conclusion:

The safe range of speed is $20.1\text{km/hr to 56}\text{.1 km/hr}$ . $$