Okay so there's a car again around a roundabout, radius $8.5\ m$ at a speed of $\frac{100}{9} ms^{-1}$. There is a man of mass $72.5\ kg$ who is going to hold onto the back of the car. What force does he need to hold onto the car with?

I have been told in my physics lesson to use $\frac{mv^2}{r}$(which is the centripetal force) and that this gives you the force with which the man holds onto the car with ($1053\ N$). But I don't understand why it would be this force?

I get that the normal reaction force of the car on the man will provide the centripetal force for this scenario (I am assuming that this is correct?) but I don't understand why he needs to hold onto the car with a force equal to the centripetal force. Is it something to do with Newton's third law?


1 Answer 1


It has to do more with second law. The normal forces between car and stuntman are the same in opposite directions, so we normally don't take them into account. So now, you have to use second law to formulate the equilibrium condition, i.e. $$\sum_i \vec{F_i}=0$$ which means that all the forces involved should compensate so that the man isn't moved in any direction. Then, you have the force he has to do to hold on to the car and the centripetal force, which are going in opposite directions so we have $$F_{man}-\frac{mv^2}{r}=0 \Rightarrow F_{man}=\frac{mv^2}{r}$$

You can think at this point: well, the man has a normal acceleration as he is spinning with the car. That may seem different, but it is exactly the same approach. In this case, we would write Newton's second law as $$\sum_i \vec{F}_i=m\vec{a}$$ where $\vec{a}$ is the normal acceleration. In this case, you would not take into account the centripetal force, as its information is contained within the acceleration term. Now, the only other force acting is the force that the man has to do to hold on to the car. You will get the same result as above. The forces need to balance out because the man is not displacing with respect to the car. This connects with Newton's first law.

  • $\begingroup$ This is very similar to the explanation that my teacher gave, but why do the forces need to balance out? The man is moving in a circle so there must be a resultant force acting towards the centre. $\endgroup$
    – Zeazealia
    Jun 15, 2020 at 9:30
  • $\begingroup$ That is another approach to go. Let me update my answer. $\endgroup$ Jun 15, 2020 at 9:32

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