Chapter 5, Problem 25P

To determine

The coefficient of static friction would be necessary between the tires and the road to provide this acceleration with no slipping or skidding.

Answer to Problem 25P

Solution:

Tangential acceleration $\left(a_t\right)$ and radial acceleration $\left(a_R\right)$ :

  $\begin{array}{l}\left(a_t\right)=4.07{\text{m/s}}^2\\ \left(a_R\right)=12.78{\text{m/s}}^2\end{array}$

The coefficient of static friction to avoid skidding or slipping must be greater than or equal to 1.32.

Explanation of Solution

Given:

Initial speed =0 m/s

Final speed

  $\begin{array}{l}=270\text{km/h }\\ \text{=}\frac{270\times 1000}{60\times 60}\\ =75\text{m/s}\end{array}$

Radius of semi-circular arc, $r=\text{220m}$

Tangential acceleration is constant. $\left(a_t\right)=$ constant

Formula Used:

Radial acceleration, $a_R=\frac{v^2}{R}$

Where, v is the velocity and R is the radius.

Total acceleration, $a=\sqrt{a_t{}^2+a_R{}^2}$

From third equation of motion:

  $v_f{}^2=v_i{}^2+2as$

Where, $v_f$ is the final velocity, $v_i$ is the initial velocity, a is the acceleration and s is the displacement.

Calculation:

Tangential acceleration is constant, so

  $\begin{array}{l}s=\pi R\\ s=\pi \times 220\text{m}\\ s=220\pi \text{m}\\ v_i=0\\ v_f=75\text{m/s}\\ v_f{}^2=v_i{}^2+2as\\ {\left(75\right)}^2={\left(0\right)}^2+2\left(a\right)\left(220\pi \right)\\ a=\frac{{\left(75\right)}^2}{440\pi }\\ a=4.07\mathrm{m}/s^2\end{array}$

Hence, tangential acceleration $=4.07\mathrm{m}/s^2$

Calculation of radial acceleration when it is halfway:

  $\begin{array}{l}{a}_t=4.07\text{m/s}\\ {\left(v_f\right)}^2={\left(v_i\right)}^2+2as\end{array}$

  $\begin{array}{l}{v}_i=0\\ s\text{=}\frac{220\pi }{2}\\ s=110\pi \text{m}\end{array}$

  $\begin{array}{l}{\text{v}}_f{}^2=2\left(\frac{{\left(75\right)}^2}{440\pi }\right)\times 110\pi \\ {\text{v}}_f{}^2=\frac{{75}^2}{2}\end{array}$

  $\begin{array}{l}{a}_R=\frac{v^2}{R}\\ =\frac{{75}^2}{2}\times \frac{1}{220}\end{array}$

  $a_{{}_R}=12.78{\text{m/s}}^{\text{2}}$

Hence tangential acceleration $\left(a_t\right)=4.07\text{m/s}$ and radial acceleration $\left(a_R\right)=12.78{\text{m/s}}^2$

Calculation of coefficient of static friction

Total acceleration:

  $a=\sqrt{a_t{}^2+a_R{}^2}$

  $a=\sqrt{{12.78}^2+{4.07}^2}$

  $a=12.94{\text{m/s}}^2$

This total acceleration is provided by static friction force and static friction

  ${\mu }_{s}N={\mu }_{S}mg$

  ${\mu }_s=\frac{a}{g}$

So, coefficient of static friction is:

  $\begin{array}{c}{\mu }_s=\frac{12.94}{9.80}\\ =1.32\end{array}$

Conclusion:

Hence, to avoid skidding and slipping static friction coefficient must be greater than or equal to 1.32.