Answer:
The coefficient of static friction between the tires and the road should be at least [tex]0.29[/tex], assuming that the traffic circle is level, the tires are not skidding, and that [tex]g \approx 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
When the vehicle travels in a circular path, the vehicle would be in a centripetal motion, and the net force on the vehicle would point towards the center of the circle. If the circle is of radius [tex]r[/tex], the tangential velocity of the vehicle is [tex]v[/tex], and the mass of the vehicle is [tex]m[/tex], the net force on the vehicle would be [tex](m\, v^{2} / r)[/tex].
Forces on the vehicle are:
Thus, friction from the ground would be the only force on the vehicle that can supply this force in the horizontal direction towards the center of the circle. Since the net force on the vehicle would be [tex](m\, v^{2} / r)[/tex], the friction on the vehicle should also be [tex](m\, v^{2} / r)[/tex].
There are two types of friction: kinetic and static. When the two surfaces in contact are sliding across each other, the friction would be kinetic; otherwise, the friction between the two surfaces would be static. Assuming that the vehicle isn't sliding, the friction between the tire and the ground would be static friction.
The coefficient of static friction is the ratio between the magnitude of the maximum possible static friction and the magnitude of normal force. Assuming that the surface is level, the normal force on the vehicle would be equal in magnitude to weight: [tex]m\, g[/tex].
Since the static friction on the vehicle is [tex](m\, v^{2} / r)[/tex] when normal force is [tex]m\, g[/tex], the coefficient of static friction [tex]\mu_{\text{s}}[/tex] would be at least:
[tex]\displaystyle \mu_{\text{s}} \ge \frac{(\text{static friction})}{(\text{normal force})} = \frac{m\, v^{2} / r}{m\, g} = \frac{v^{2}}{g\, r}[/tex].
Ensure that all quantities are in standard units before substituting the values into this expression:
[tex]\displaystyle v = 50\; {\rm km \cdot h^{-1}} \times \frac{10^{3}\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} = 13.9\; {\rm m\cdot s^{-1}}[/tex].
[tex]\begin{aligned} \mu_{\text{s}} & \ge \frac{v^{2}}{g\, r} \\ &\approx \frac{(13.9\; {\rm m\cdot s^{-1}})^{2}}{(9.81\; {\rm m\cdot s^{-2}})\, (70\; {\rm m})} \\ &\approx 0.29\end{aligned}[/tex].
(Rounded up.)
In other words, the coefficient of static friction between the tire and the ground should be at least [tex]0.29[/tex].