Question

A circular road course track has a radius of $500\text{m}$ and is banked to ${10}^{\circ }$. If the coefficient of friction between the road and tyre is $0.25$, compute (i) the maximum speed to avoid slipping (ii) optimum speed to avoid wear and tear of the tyres. [Take $\tan {10}^{\circ }=0.1763$]

• A. $46.74\text{m/s}$, $29.39\text{m/s}$
• B. $30\text{m/s}$, $24\text{m/s}$
• C. $74.46\text{m/s}$, $23.39\text{m/s}$
• D. $50\text{m/s}$, $44\text{m/s}$

Answer 1

The correct option is A $46.74\text{m/s}$, $29.39\text{m/s}$

Given: radius of curve = $r=500\text{m}$, Angle of banking $=\theta ={10}^{\circ }$,
and coefficient of friction $=\mu =0.25,g=9.8{\text{m/s}}^2$,
To find safe velocity = $v=?$, To avoid wear and tear, $v_{sp}=?$

Maximum speed to avoid slipping:
In this case, centripetal force is provided by horizontal component of both normal reaction and friction (i.e there will be wear and tear)
$v_{max}=\sqrt{\frac{rg(\tan \theta +{\mu }_s)}{(1-{\mu }_s\tan \theta )}}$
$=\sqrt{\frac{500\times 9.8(\tan {10}^{\circ }+0.25)}{(1-0.25\tan {10}^{\circ })}}$
$=\sqrt{\frac{4900(0.1763+0.25)}{(1-0.25\times 0.1763)}}=\sqrt{\frac{4900\left(0.4263\right)}{1-0.0441}}=\sqrt{\frac{2089}{0.9559}}$
$\therefore v_{max}=\sqrt{2185}=46.74\text{m/s}$

Maximum speed to avoid wear and tear:
In order to avoid wear and tear, centripetal force for making the turn has to provided by the horizontal component of normal reaction alone. (i.e no friction)
$v_{opt}=\sqrt{rg\tan \theta }=\sqrt{500\times 9.8\times \tan {10}^{\circ }}$
$=\sqrt{500\times 9.8\times 0.1763}=29.39\text{m/s}$