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Question: If aboard a train being driven along a track by centripetal force, would it feel the same as if you were on a regular straight track?

I hope this isn't subjective, I think there is a right or wrong answer here. Either it does feel similiar or it doesn't so hopefully this question isn't closed. I don't see it as an opinion based answer.

I recently did a little khan academy course because I wanted to understand the physics of centripetal force. I was a little unsure of some things so I devised this thought experiment to help me clear things up in my own head. I was hoping you fine folks might be able to help clarify a few things for me.

The Scenario:

You board a train cart, sit down and make yourself comfortable inside your carriage. This train is special because it doesn't have an engine, the carriage is driven along a rail sitting on a rotating plane, your carriage starts close to the axis of rotation and is pushed towards the edge of the plane and away from the rotating center axis by centripetal forces. A lot like being inside a centrifuge!

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Details:

Now I know as your radius changes the centripetal force changes, so the speed of our train changes as given by the formula: Centripetal force = mass x velocity^2 / radius. Constant acceleration/deceleration isn't good for passengers so for this scenario were going to presume the rpm of the plane is being altered to induce a generally constant speed like a normal train.

The track in the picture is a little short so please use your imagination to extend it, it's just to get the general idea across.

What I don't understand:

What I'd like to know, is once the train is at it's correct speed and my body has caught up to that speed, could I hypothethically get up and walk around on the train like a person would in a normal train or passenger plane, drink a cup of tea or go to the bathroom. Because I'm moving forward I'm not dealing with the forces imposed by centripetal forces, that's an asuumption I'm making which I presume is correct but what I'm not sure about is because the angle of my velocity vector is changing constantly (The direction I would fly off in if were to jump of the train.) does my body have to be for lack of a better word consistently updated by the carriage, pinning me against the train window at a suffiently high rpm? I don't know if that force is there or not. So the real question can I have a cup of tea on board or not?

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  • $\begingroup$ Put your tea in a travel mug so you can be sure that you can drink it even if the ride is bumpy 👌 $\endgroup$ – Jonathan R Swift Feb 16 at 18:00
  • $\begingroup$ Not a full answer, but this scenario is very much related to Coriolis forces - things not connected to the train itself will move in a straight line as usual, but that path will appear curved from someone sitting on board the train, since the reference frame itself is undergoing constant acceleration. $\endgroup$ – Nuclear Hoagie Feb 16 at 18:00
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Your question is related to rotating frames of reference and the Coriolis Force.

As a passenger you are going to be experiencing a lateral force which will be greater the the faster you are travelling. The magnitude of the acceleration you'd feel would be

$$a_{coriolis} = -2\omega\times v_{train} $$

and the force would be :

$$F_{coriolis}= -2\cdot m\cdot \omega\times v_{train} $$

where:

  • m : is the mass of the object
  • $\omega$: is the angular velocity
  • $v$: is the velocity of the train in this example

To understand what happens you have to start from Newton's law. Basically, whenever there is no force applied on a mass (or if the net force is zero), the mass tends not to change its kinetic state. Therefore, (in the absence of gravity), it would tend to travel in a straight line. In order for the mass to change its course, you need to apply a force on it.

Once you are on the train and you are travelling with the train, if the train is not rotating then you wouldn't feel anything. However, if the carousel starts to rotate, a force would need to be applied on you to keep you up with the train. In particular example that force would be friction.

This is exactly the same example, with someone who is at the center on a merry-go-round and wants to travel to the edge. Try that when the merry-go-round is not rotating and everything is fine. Try it when its spinning, and there is a very good chance that you loose a couple of teeth.

There is a very nice video which gives a very nice explanation about this.

is the force there?

Well... that depends on your frame of reference. :-)

Coriolis is one of the pseudo-forces. If you are on an inertial frame of reference (Newtonian), then you don't need the Coriolis force. To make that clearer (this will make more sense if you watch the video), if you throw a ball from a rotating frame of reference, then you don't need Coriolis to describe its movement.

However, when you are observing from the rotating frame of reference, then Coriolis is necessary. Therefore the ball that was thrown in the video, will need the existence of a force in order to describe its trajectory.

Hope that makes sense!

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  • $\begingroup$ Thanks for the great answer but for further clarify, If the carosel was rotating at a high rpm and the train is moving along the rail, am I pinned against the window? $\endgroup$ – David Nolan Feb 16 at 18:45
  • $\begingroup$ yeap. Just like on the carousel or these guys in the centrifugal simulator $\endgroup$ – NMech Feb 16 at 18:49
  • $\begingroup$ But in the centrifugal similuator there are three forces the force pushing me into the seat which is variable depending on the speed, the force of earths gravity which is set and another force? The centripetal force which is the one they measure? $\endgroup$ – David Nolan Feb 16 at 18:54
  • $\begingroup$ Strike out my previous comment. I gave the example of the G simulator because of the mental image of a cheek under pressure. The coriolis would not play into that because there is no velocity on the rotating frame. $\endgroup$ – NMech Feb 16 at 19:05
  • $\begingroup$ Okay thanks, I think its clicked for me, took a minute. $\endgroup$ – David Nolan Feb 16 at 19:09

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