91Ó°ÊÓ

$$\begin{gathered} {\mathrm{μ}}_s=0.60 \\ R=30.5\mathrm{m} \end{gathered}$$

This problem is based on the concept of uniform circular motion. Uniform circular motion is a motion in which an object moves in a circular path with constant velocity.

Formula:

The velocity in uniform circular motion is given by,

$v=\frac{2\mathrm{Ï€}R}{T}$

where, v is the velocity, R is the radius and T is the time period

The magnitude of the acceleration of the car as it rounds the curve is given by ${}^{v^2/R}$,where v is the speed of the car and R is the radius of the curve. Since, the road is horizontal, only the frictional force of the road on the tires makes this acceleration possible. The horizontal component of Newton’s second law is

$f=\frac{m{v}^2}{R}$

If ${}^{F_N}$is the normal force of the road on the car and m is the mass of the car, the vertical component of Newton’s second law leads to ${}^{F_N=mg}$. Thus, the maximum value of static friction is,

$$\begin{gathered} f_{amax}={\mathrm{μ}}_s{F}_N \\ ={\mathrm{μ}}_{s}mg \end{gathered}$$

If the car does not slip,${}^{f≤{\mathrm{μ}}_{s}mg}$.This means,

$\frac{v^2}{R}≤{\mathrm{μ}}_{s}g$

$v≤\sqrt[]{{\mathrm{μ}}_{s}Rg}$

Consequently, the maximum speed with which the car can round the curve without slipping is,

$$\begin{gathered} v_{max}=\sqrt[]{{\mathrm{μ}}_{s}Rg} \\ =\sqrt[]{\left(0.60\right)\left(30.5\mathrm{m}\right)\left(9.8\mathrm{m}/\mathrm{s}\right)} \\ =13.4\mathrm{m}/\mathrm{s} \\ ≈48\mathrm{km}/\mathrm{h} \end{gathered}$$

Hence, the speed that will put the car on the verge of sliding as it rounds a level curve of${}^{30.5\mathrm{m}}$radius is${}^{48\mathrm{km}/\mathrm{h}}$.