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For example, you are in a box that is connected a distance $R$ from a clockwise spinning centre. If I understand correctly, the spinning box is a result of the resulting centrifugal force $F_{centrifugal}$ = $\frac{mv^2}{R}$. The person then would be pushed against the wall opposite to the direction of $F_{centrifugal}$

See this picture in top view: enter image description here

Where distance $R$, the direction of $F_{centrifugal}$ and v are indicated. The person in the box is looking in the opposite direction of the center and perpendicular to $v$ (speed).

If $R$ is small enough or $v$ large enough, then at a certain combination of $R$ and $v$, $F_{centrifugal}$ would be larger than $F_z$ = $mg$ (gravity).

Question

If $F_{centrifugal}$ > $F_z$ would it be possible to walk up the wall that is in front of him?

(The forces $F_{centrifugal}$ and $F_z$ are perpendicular on each other which mean they do not counteract each other. Similarly for example in a bus with the vertical gravity force and the horizontal acceleration of a person due to the acceleration of the bus.)

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  • $\begingroup$ Does anyone know how to write equations, subscripts? $\endgroup$
    – Faceb Faceb
    Commented Feb 10, 2017 at 18:24
  • $\begingroup$ A more common scenario is a large vertical cylinder which is spun about its axis. I have been in such a thing, and I am sure you could walk on its wall if you were brave enough. So yes, you can. $\endgroup$
    – user107153
    Commented Feb 10, 2017 at 18:49
  • $\begingroup$ If you consider the consequential effects on one's inner ear, I don't think a person would be able to walk unless the radius was very large (much larger than the person's height) so that $\omega$ could be small. The necessary friction might be there, but the muscular control due to poor balance would be questionable. When you put the question in the context of walking, you add a physiological dimension to the question. $\endgroup$
    – Bill N
    Commented Feb 10, 2017 at 19:31
  • $\begingroup$ You can. I did this in one of those enclosed fairground spinning rides! I got yelled at, but not before standing horizontally on the wall! $\endgroup$
    – user12029
    Commented Feb 10, 2017 at 19:35
  • $\begingroup$ @BillN That was the case for me. I wasn't actually sick from just 'lying down' (standing up initially) but I was extremely nauseous when I got out of the thing. I suspect that people differ though: isn't more-or-less this the sort of thing they tested astronauts for? $\endgroup$
    – user107153
    Commented Feb 10, 2017 at 22:12

3 Answers 3

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Centrifugal force greater than gravity should be one of the prerequisites for walking up the wall. However, it is the friction of the wall which is the deciding factor.

Suppose the man lies down on the floor with his head towards the centre of the circle and the feet on the wall . Under normal circumstances, he would simply slip off the wall as the friction is negligible compared to weight. But due to the centrifugal force, he is now pushing against the wall with significant force. Now, it is possible for friction to overcome the gravity. But it depends upon the coefficient of friction. If it is k, then k times the centrifugal force must be greater than the person's weight. Only then can he walk up the wall

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Partly agree with provided answers, but partly not, so here is my version.

TL;DR possible if speed and friction are big enough

Let's tie coordinate axis to box, which is a non-inertial reference frame. Man standing on the wall will be affected by four forces only:

  • inertial force of $\frac{mv^2}{R}$ directed to the wall
  • gravity
  • normal force from the wall
  • friction

Standing still (or moving without acceleration) means the following: $$F_n = F_{inertia}$$ $$F_{friction} = mg $$

$F_{friction}$ can't extend $\mu F_n = \mu\frac{mv^2}{R}$, so the only condition for standing is $\mu\frac{mv^2}{R} \geq mg$. So the only requirement for climbing is that centrifugal force is greater than gravity divided by $\mu$.

Ok, once we met this condition, how would it look from climber's perspective?

Climber is affected by force (sum of gravity and inertial) which pushes him towards wall and along it, the other stuff (friction and normal force) is quite common. This situation is full equivalent of man climbing a hill on planet with gravity greater than on Earth. So @Lz4's advice to not remove both his legs while climbing is wrong. You are free to jump forward if you are strong enough to cope with "hill" angle and increased "gravity".

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  • $\begingroup$ I like the "from the climbers perspective" drawing you made. This shows what is really going on: in the non-inertial frame of reference there is a new fictitious acceleration which adds to gravity. To the "walker" it is just like walking up a slope (with gravity a little larger than usual). This clearly illustrates that the coefficient of friction (which determines the minimum centrifugal force needed) plays a big role, and that a sufficiently large $\mu$ can make this quite practical. $\endgroup$
    – Floris
    Commented Feb 10, 2017 at 22:21
  • $\begingroup$ @nnovich-OK This a really interesting way of thinking about the problem. But please look carefully at my answer, especially the last lines before you say that my assumptions are inaccurate. I have made the correct assumption about coefficient of friction. The assumption in the first paragraph is simply because I felt that the coefficient in practical cases will be less than 1. Please update your answer accordingly. $\endgroup$
    – TheFool
    Commented Feb 11, 2017 at 6:55
  • $\begingroup$ @TheFool, Ok, removed reference to your answer. $\endgroup$
    – nnovich-OK
    Commented Feb 11, 2017 at 12:25
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In simple words: Depends on your velocity.

Because if we look from the frame of reference of the box we find the the centrifugal force would provide the normal reaction (assuming cofficient of friction to be constant) and hence the friction force to balance gravitational force acting on the man.

u*(mv^2/r) ≥ m*g(this is the minimum condition for the man to stand on wall. If you want to be extra safe then you should solve the above equation with strict inequality to find v.) Once this condition is achieved the man can move on the wall. But remember while walking he must not remove both his legs at the same time because the frictional force is a contact force.It would only act when at least one the man's leg is in contact with the wall.

Where u is coefficient of friction between man's shoes and the wall m is the mass of the man v is the speed with which the box is rotating

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  • $\begingroup$ Sorry, I meant the opposite wall. I've edited it $\endgroup$
    – Faceb Faceb
    Commented Feb 10, 2017 at 18:16
  • $\begingroup$ @FacebFaceb now check the answer $\endgroup$
    – Lz4
    Commented Feb 10, 2017 at 19:15

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