Discussion
A cyclist speeding at a velocity v on a level road takes a sharp circular turn of radius R. If μ is the static friction between the tyres and road, then the condition for the cyclist not to slip is
Answer: B
The condition is \(v^{2} \leq \mu R g\).
On a level road the only horizontal force on the cycle is static friction, and it must supply the centripetal force for the turn. The largest friction available is \(\mu m g\):
\[ \begin{aligned} \frac{m v^{2}}{R} &\leq \mu m g \\ v^{2} &\leq \mu R g \end{aligned} \]
The mass cancels. The inequality must run this way, since a higher speed needs more friction; options with \(\geq\) describe the condition for slipping. \(\mu R\) and \(\frac{\mu R}{g}\) do not have the units of \(v^{2}\), and \(v \leq \mu R g\) leaves \(v\) unsquared.
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