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Merry go Round on a train

chancerph

New Blood
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Apr 26, 2011
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You are riding a Merry-go-Round (roundabout in the UK), which itself is on a train with no windows.

The Merry-go-Round is rotating at a constant speed of 30 mph at the circumference.

Would you be able to detect any difference when:

a) The train is travelling in a straight line at a constant speed.
b) The train is stationary


Can someone please explain the physics of it?
 
If the train's not accelerating, then you wouldn't notice a difference.


Steve S.
 
chancerph said:
You are riding a Merry-go-Round (roundabout in the UK), which itself is on a train with no windows.

The Merry-go-Round is rotating at a constant speed of 30 mph at the circumference.

Would you be able to detect any difference when:

a) The train is travelling in a straight line at a constant speed.
b) The train is stationary


Can someone please explain the physics of it?
Click to expand...

Heh. I assume you mean that one should not look out the windows.

Since the train is enclosed, there is no wind to affect the roundabout. Put it another way, assuming good soundproofing and suspension, how would you know if the train is moving or not? The general answer is that you can't. At least not in the sense of measuring forces on dynamic bodies.

And before some folks start picking nits, yes there are effects which can in principle be measured. Depending on direction, objects in the train can get heavier or lighter (changes in centrifugal force). And again, depending on direction, gyroscopes will behave slightly differently (angular rates change a little from local earth rotation). And if you have a very tall train, a Foucalt pendulum will behave a little differently. And the measured frequency of radio and television stations will vary slightly due to Doppler effect. By all means, jump on.

None of these affect the thought experiment.
 
Dr.Sid said:
I guess by 'straight' in thought experiments it's usually meant 'ignoring Earth curvature'.
Click to expand...

Yeah, but even then you get angular rate effects, unless you mean "ignoring Earth curvature and rotation and orbital velocity". Which is what the OP probably meant.
 
The big answer is "no difference" -see post for nuance!

chancerph said:
You are riding a Merry-go-Round (roundabout in the UK), which itself is on a train with no windows.

The Merry-go-Round is rotating at a constant speed of 30 mph at the circumference.

Would you be able to detect any difference when:

a) The train is travelling in a straight line at a constant speed.
b) The train is stationary


Can someone please explain the physics of it?
Click to expand...

With very very precise measuring instruments one could tell a difference because the train is on the earth which is rotating and therefore isn't a perfect Inertial frame of reference (but nearly a perfect one.)

For all practical purposes and for this forum, The train, still or moving, is linearly translating with respect to the initial reference frame (associated with the ground). Assuming no acceleration, all inertial frames are equivalent so there would, in general, be no difference or way to detect a difference in the spirit of the posed question. This is why tennis players can play on a smoothly moving cruise ship and never notice a difference from their ground game!

Regards,

WindGrins:D
 
Last edited:
chancerph said:
You are riding a Merry-go-Round (roundabout in the UK), which itself is on a train with no windows.

The Merry-go-Round is rotating at a constant speed of 30 mph at the circumference.

Would you be able to detect any difference when:

a) The train is travelling in a straight line at a constant speed.
b) The train is stationary


Can someone please explain the physics of it?
Click to expand...

Well, I should think that one should be able to tell if the train is in motion or not regardless of the fact that one cannot look out of the windows. At least all the trains that I have ever been on, one can always tell if the train is in motion or not because a moving train is a bit noisy and rocky; in other words, it is always obvious when the train is in motion.

But assuming that the train travel is perfectly smooth, straight, quiet, and otherwise unaccelerated, and that one cannot obtain any outside data by looking out of the windows, using a GPS device, asking the conductor, or other such thing, then one could not tell if the train was moving or not.

This is due to the fact that when the train is in motion, then everything else that one is dealing with has the exact same motion. And if the train is stationary, then everything else that one is dealing is also stationary.

Therefore, one would not be able to tell the difference between the two cases.

Does that help?
 
There are actually several real rides that come close to simulating this effect. They're called Disk'Os. You sit on a round platform, on the edge, facing out, with about 25 other people. The whole platform sits on a U shaped track. As the platform starts to spin, it moves back and forth on the track.

A similar ride, by the same manufacturer, is called a Rocking Tug. Similar layout, except you sit in rows and face toward the center of the car.

After riding one of these, I think your body would notice changes in speed and direction of the train, but it is a nice thought experiment.
 
I think it's a signal-noise-ratio problem, so detection would be dependent on the two speeds and the sensitivity of the sensor.

A box car is 9'+ wide, so call the merry-go-round 8'. Circumference = 3.14x8 = 25' (...and inches. Ignore them and they'll go away.) At 10 rpm that's 250'/min /60 = 4'/second.

30 mph train is moving 44'/second. Thus the difference between the forward and reverse portion of the rotation is 8'/second, just under 20% speed difference. So the next question is how little difference in velocity can the inner ear detect, and how quickly? Would the sinusoidal transition between forward and reverse blur the difference enough to prevent detection?
 
ApolloGnomon said:
I think it's a signal-noise-ratio problem, so detection would be dependent on the two speeds and the sensitivity of the sensor.

A box car is 9'+ wide, so call the merry-go-round 8'. Circumference = 3.14x8 = 25' (...and inches. Ignore them and they'll go away.) At 10 rpm that's 250'/min /60 = 4'/second.

30 mph train is moving 44'/second. Thus the difference between the forward and reverse portion of the rotation is 8'/second, just under 20% speed difference. So the next question is how little difference in velocity can the inner ear detect, and how quickly? Would the sinusoidal transition between forward and reverse blur the difference enough to prevent detection?
Click to expand...

No. The inner ear cannot detect velocity, it can detect acceleration.
 
Exactly, the inner ear would need to detect the change in acceleration from 44+4 fps to 44-4 fps, while whirling around in a circle. That's a lot of noise to get signal out of.
 
ApolloGnomon said:
Exactly, the inner ear would need to detect the change in acceleration from 44+4 fps to 44-4 fps
Click to expand...

Stop right there. fps (which I presume is feet per second) is a velocity, not an acceleration. Velocity and acceleration are different quantities. You can have large accelerations and small velocities, or small accelerations and large velocities. Telling us the velocity tells us nothing about the acceleration.

Don't mix your dimensions. That gets you into trouble.

while whirling around in a circle. That's a lot of noise to get signal out of.
Click to expand...

No, it's not. That's like saying it's harder to hear during the daytime because the light swamps out the sound. Velocity and acceleration are different quantities. A higher velocity doesn't change your sensitivity to acceleration.
 
ApolloGnomon said:
Exactly, the inner ear would need to detect the change in acceleration from 44+4 fps to 44-4 fps, while whirling around in a circle. That's a lot of noise to get signal out of.
Click to expand...

Nope. A merry-go-round on a moving train experiences velocities from 40 to 48 fps, and you say this happens in 3 seconds (1/2 rev at 10rpm)' for an acceleration of 8/3 feet/s/s. A merry-go-round on a stationary train goes from -4 to +4 f/s in 3s, also an acceleration of 8/3 f/s/s. These accelerations are exactly identical, and no accelerometer can tell them apart. The same thing would be true if you replace the 40 f/s train with a 1000 f/s rocket or a 4 f/s rowboat. Velocity is really, precisely relative.
 
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