I have tried to do the following circular motion question but I don't understand an assumption in the mark scheme. The specifics of the question are not necessary but I have linked the paper and mark scheme below anyway (question 5a):

Question paper: https://www.bpho.org.uk/user/pages/05.past-papers/01.round-1/_general/Round_1_Section_2_Nov_2020.pdf

Mark scheme: https://www.bpho.org.uk/user/pages/05.past-papers/01.round-1/_general/Mark%20scheme%20v4%20Round%201%20Nov%202020.pdf

The question essentially says: A hump-back bridge that crosses a river has a circular cross section. What is the maximum speed at which a car can drive over the bridge whilst still remaining in contact with it?

To solve the question you can resolve forces using F=mv^2/r then set the reaction force between the car and bridge to zero. However, the problem I have is that the mark scheme says that at the maximum velocity, the reaction force is zero at the top of the bridge. By drawing a force diagram, I would have thought that in the case of a bridge, the minimum reaction force would be at either end of the bridge as this is where the smallest component of the cars weight acts towards the centre of the circle. Thus, if you are travelling at a velocity which results in a reaction force of zero at the top of the bridge (like the mark scheme says), this will result in a negative reaction force at either end which is impossible. Therefore, when I tried the problem, I set the reaction force to zero at either the end of the bridge.

So my question is, is the mark scheme is wrong, or where does my reasoning fail?


3 Answers 3


The question is vague. The given answer implies that the problem is asking for "the fastest speed at any single point on the bridge such that the car remains in contact with the bridge in that moment." If the car has to remain in contact with the bridge the whole way across, then the answer would be determined (for the reasons given in your question) by the car's speed at the ends of the bridge where gravity provides the least amount of centripetal force.

In the extreme, imagine a bridge that was semicircular where the ends of the bridge were vertical. At the ends, even a stopped car will fall off the bridge because it cannot stay on a vertical road.

I would contact the Olympiad and tell them about this problem.


Suppose you have normal force $N$ with the object of mass $m$. At the top, $\frac{mv^2}{r}$+$N$=$mg$. At the bottom, $mg+\frac{mv^2}{r}=N$. In order to maximize our speed, set the normal force to zero at the top. Thus, $\frac{v^2}{r}=g$ and $v=\sqrt{rg}$.


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the radius of the bridge

with: $$(r-h)^2+\left(\frac{b}{2}\right)^2=r^2\Rightarrow\\ r=\frac{1}{8}\,\frac{4\,h^2+b^2}{h}=\frac{97}{8}=12.12\,\text{[m]}$$

Normal force


$$\hat r= \begin{bmatrix} \cos(\varphi) \\ \sin(\varphi) \\ \end{bmatrix}$$

thus $$N-\frac{m\,v^2}{r}+m\,g\,\hat y\cdot \hat r=0\quad \Rightarrow\\ N=\frac{m\,v^2}{r}-m\,g\,\sin(\varphi)$$


$$N=0\,, v^2=r\,g\,\sin(\varphi)\\ v_{\text{max}}(\varphi=\pi/2)=\sqrt{g\,r}$$

the maximum speed accrue at the middle of the bridge


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