I have an actual physics problem and also a conceptual question to go with it. From some test prep I was presented with this straight-forward question:

What is the minimum radius that a cyclist can ride around at 10 kilometers per hour without slipping if the coefficient of friction between her tires and the road is 0.5?

I assumed this is a F=$a_c$m = (m)(v^2/r) = $u_k$(mg), but upon solving I get r = 1.57, whereas the answer is r = 4.1 Does anyone see what I am doing wrong here?

Also, it seems the correct free-body diagram has both the centripetal acceleration and the force of friction pointing toward the center of the circle. How is this possible, shouldn't there be some other force in the diagram pointing out of the circle that centripetal force has to be equilibrium with?


1 Answer 1


Your answer is correct.

I would like to remove your confusion surrounding the "free-body" diagram you speak of. Firstly, because the cyclist is traveling in a circle, the net force is not zero, we have a net centripetal force. In other words the cyclist is not in equilibrium.

Where does the centripetal force come from? In the example given here the centripetal force is purely frictional. That is, the friction between the tires and the ground is the centripetal force. So when you speak of the centripetal force and frictional force pointing in the same direction, you're right, they are one and the same force.

Your example:

The maximum frictional force given coefficient of static friction, $\mu$, and normal force $N$ is $F_{\small{max}} = \mu~N$. In the case given, the normal force cancels the force due to gravity (because the cyclist is not accelerating up/down), leaving us with $F_{\small{max}} = \mu~mg$. The centripetal force $F_c = \frac{mv^2}{r}$ is this equal to the frictional force $F$ which gives us $$F_c = F \leq F_{\small{max}}\implies \frac{mv^2}{r}\leq \mu~mg\implies r \geq \frac{v^2}{\mu g} = \frac{\left(\frac{10}{3.6}\right)^2}{0.5\times9.8}\mbox{m} = 1.57~\mbox{m}$$ thus $r_{\small{min}} = 1.57~\mbox{m}$.


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