91Ó°ÊÓ

$v=29\mathrm{km}/\mathrm{hr}$,

role="math" localid="1654085678255" ${\mathrm{μ}}_s=0.32$

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. Also, this problem deals with the Newton’s second law of motion. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Frictional force $\mathbf{â‰\yen }$Frictional force

Formula:

$R_{min}=\frac{v^2}{{\mathrm{μ}}_{s}g}$

where,$v$is the velocity,$g$is an acceleration due to gravity,${\mathrm{μ}}_s$is the coefficient of friction and$R_{min}$ is the minimum radius.

The maximum value of static friction is,

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

If the bicycle does not slip,role="math" localid="1654086660834" ${}^{f≤{\mathrm{μ}}_{s}mg.}$This means,

$$\begin{gathered} \frac{v_2}{R}≤{\mathrm{μ}}_{s}g \\ Râ‰\yen \frac{v^2}{{\mathrm{μ}}_{s}g} \end{gathered}$$

Consequently, the minimum radius with which a cyclist moving at role="math" localid="1654087040889" ${}^{29\mathrm{km}/\mathrm{h}=8.1\mathrm{m}/\mathrm{s}}$can round the curve without slipping is,

$$\begin{gathered} R_{min}=\frac{v^2}{{\mathrm{μ}}_{s}g} \\ =\frac{{(8.1\mathrm{m}/\mathrm{s})}^2}{(0.32)(9.8\mathrm{m}/{\mathrm{s}}^{2)}} \\ =21m \end{gathered}$$

Therefore, the smallest radius of an unbanked (flat) track around which a bicyclist can travel is${}^{21\mathrm{m}}$.