Riding a roundabout on a train

[chancerph]

You are riding a roundabout which is itself on a train traveling in a straight line at a constant speed.

Would you notice any different inertial effects or forces compared to when the train is stationary? I think not but am discussing this with a friend who says this:

"If the train was traveling at 15mph and the circumference of the roundabout was also traveling at that speed you could exit the roundabout safely on one side of the train at 0mph or exit the other side at 30mph far less safely! You would accelerate 0-30mph in 180º, this is going to be very noticeable indeed, a real gut churning, vomit inducing experience."

Yes, if we were getting off the train I agree, but we are talking about the effect (forces) the passenger would feel on the ride.

Can someone explain the physics of it?

 

[russ_watters]

Welcome to PF!

What's a roundabout?

In any case, any time you're in circular motion, you do feel a constant acceleration. But that's completely decoupled from the motion of the train due to the principle of relativity which holds that there is no self-contained experiment you can do while at constant velocity which will allow you to identify that motion.

 

[Doc Al]

"If the train was traveling at 15mph and the circumference of the roundabout was also traveling at that speed you could exit the roundabout safely on one side of the train at 0mph or exit the other side at 30mph far less safely! You would accelerate 0-30mph in 180º, this is going to be very noticeable indeed, a real gut churning, vomit inducing experience."

That's a true statement, but ask your friend what the change in velocity would be if the train were stationary? Or moving at 100 mph? (You'll find it doesn't matter.)

What's a roundabout?

What we'd call a Merry-go-Round.

 

[rcgldr]

Compare the speed of the train to the speed of the surface of the earth, which is moving at over 1500 kph to the east if you're close enough to the equator. Do you notice any effect from the Earth's surface speed? Then what about the orbital speed of the earth, 107,000 kph, or the speed that the solar system orbits the center of the Milky Way, 792,000 kph?

As far as the change in speed experienced by the rider, from the roundabout's axis frame of reference in any direction, the riders go from -15 mph to + 15 mph, the same 30 mph change in speed they'd experience with the roundabout on a train.

 

[chancerph]

Ok, thanks for the replies. I thought I had enough info there, so I emailed my adversary with the following:

Yes, if we were getting off the train I agree with the first sentence, but we are talking about the effect (forces) the passenger would feel on the ride. The ground is irrelevant, so let’s take it away and call it a flying train. All you would feel is the constant acceleration from the centripetal force, whatever speed the train was moving. (not sure about near light speeds here).

So let’s say you are on a stationary train, and you have a 1lb weight on a short length of rope. You start to swing it around your head in a constant circular motion. You will feel the constant force on the rope, agreed?

Now continue twirling and increase the speed of the (flying) train to 15 mph. You think you will notice a difference. How about at 100 mph? how about 1000mph? How about you twirl it on a 747 at 600mph?

Back to the rider on the roundabout. Any time you're in circular motion, you do feel a constant acceleration. But that's completely decoupled from the motion of the train due to the principle of relativity which holds that there is no self-contained experiment you can do while at constant velocity which will allow you to identify that motion.

That thar hole’s getting bigger.

_________________________________________

He replied with the following comments interjected in my text using red type. As I can't use red text here, I have separated his comments for clarity.

My words in quotes, his without:

"Yes, if we were getting off the train I agree with the first sentence, but we are talking about the effect (forces) the passenger would feel on the ride. The ground is irrelevant, so let’s take it away and call it a flying train. All you would feel is the constant acceleration from the centripetal force, whatever speed the train was moving. (not sure about near light speeds here)."

The ground is totally relevant, the force we are talking about is weight, the mass however remains constant. The centripetal force is indeed a good way of explaining what some would normally refer to as centrifugal force, ie one is constantly drawn toward the centre of the radius the wing is taking but we are talking about the vector so centripetal can mislead here as it only applies when there is no “velocity to the circling”, so, can be used but not on its own ie it must be added to another variable degree or rotation or velicity."

"So let’s say you are on a stationary train, and you have a 1lb weight on a short length of rope. You start to swing it around your head in a constant circular motion. You will feel the constant force on the rope, agreed?"

The weight feels constant agreed.

"Now continue twirling and increase the speed of the (flying) train to 15 mph. You think you will notice a difference. How about at 100 mph? how about 1000mph? How about you twirl it on a 747 at 600mph?"

Excluding the effects that only become apparent nearing the speed of light, the effective force we are examining is given by the difference in acceleration along a changing vector. When the train is at rest all is balanced. As you begin to accelerate along a rail or fly above it the effect will be that derived from the difference in velocity about the rotation or moment. The max effect would be obtained when traveling at the circumference speed. At 1000 mph the difference in ground speed if he circumference is traveling at 15mph, would be 30mph. the same was true at 100mph, or 650mph.

"Back to the rider on the roundabout."

This rider is very, very sick by now.

"Any time you're in circular motion,

(relative to what?)

"you do feel a constant acceleration."

True IF relative to the ground

"But that's completely decoupled from the motion of the train"

You get this bit perhaps I think, but, instead of decoupling, ADD in the motion of the train: you do understand it only decouples when the train is stationary? And bingo “WE NO LONGER HAVE CONSTANT VELOCITY” making the rest of this sentence irrelevant

"due to the principle of relativity which holds that there is no self-contained experiment you can do while at constant velocity which will allow you to identify that motion."

"That thar hole’s getting bigger."

Any and all holes Repaired methinks.Please help!



 

[Doc Al]

Excluding the effects that only become apparent nearing the speed of light, the effective force we are examining is given by the difference in acceleration along a changing vector. When the train is at rest all is balanced. As you begin to accelerate along a rail or fly above it the effect will be that derived from the difference in velocity about the rotation or moment. The max effect would be obtained when traveling at the circumference speed. At 1000 mph the difference in ground speed if he circumference is traveling at 15mph, would be 30mph. the same was true at 100mph, or 650mph.

This is confusing. Are you talking about a roundabout on an accelerating train? In that case, the acceleration felt by the rider will be the total acceleration, including that of the train. But once the train is moving at a constant velocity--whether 0 or 100 mph--the acceleration of the roundabout rider is completely independent of the train's velocity.

You get this bit perhaps I think, but, instead of decoupling, ADD in the motion of the train: you do understand it only decouples when the train is stationary? And bingo “WE NO LONGER HAVE CONSTANT VELOCITY” making the rest of this sentence irrelevant

Nonsense. Have your friend actually do the calculation for a train with constant velocity. The change in velocity of someone on the roundabout--and thus the acceleration--is the same regardless of the train's speed.

(And why can't you use red font?)

 

[chancerph]

Thanks again.

We were talking about a train moving at constant velocity - my friend has apparently forgotten that. I'll remind him.

I also think it is nonsense, and I told him so, but neither of us knows enough physics to do the calculation, could you be kind enough to give me an example?

Our original discussion was about aviation and gliders.

When a pilot turns the glider in circles to stay in lift and there is wind, the glider will move in relation to the ground (ground-speed).

My friend thinks he feels a difference at different parts of the turn, assuming constant turning and wind velocity. Obviously the ground-speed will change throughout the turn. But I say we are in a moving block of air Independent of the ground.Sorry I'm new to this. How do you get red text?

 

[jtbell]

How do you get red text?

In the toolbar above the message-composition window, click on the icon that looks like an artist's palette, next to the "size" box.

 

[Doc Al]

We were talking about a train moving at constant velocity - my friend has apparently forgotten that. I'll remind him.

I also think it is nonsense, and I told him so, but neither of us knows enough physics to do the calculation, could you be kind enough to give me an example?

Sure. Let the tangential speed of the rider to be 15 mph. Let's calculate the rider's speed with respect to the ground when the train is moving with speed v:

When the rider is moving in the same direction as the train his groundspeed is: v + 15
When the rider is moving opposite to the train his groundspeed is: v - 15

The change in velocity for the 180 degree turn is: (v + 15) - (v - 15) = 30 mph. Note that the speed of the train drops out.

Note further that the change in rider velocity will be the same when viewed from any inertial frame, the ground or the train--it doesn't matter.

Our original discussion was about aviation and gliders.

When a pilot turns the glider in circles to stay in lift and there is wind, the glider will move in relation to the ground (ground-speed).

My friend thinks he feels a difference at different parts of the turn, assuming constant turning and wind velocity. Obviously the ground-speed will change throughout the turn. But I say we are in a moving block of air Independent of the ground.

Sounds like a similar situation. If you are in a glider in a moving block of air (with constant velocity), then the acceleration you'll feel will be the same regardless of the velocity of the air mass. (As long as your speed with respect to the air remains the same, of course.)

Sorry I'm new to this. How do you get red text?

In the edit menu you'll see something that looks like an artist's palette. Click it to change the font color.

 

[chancerph]

Great, I'll try that. :-)

 

[russ_watters]

Btw, Einstein's Relativity incoporates the effect of travel near the speed of light so no, nothing changes at very high speed.