The top of the wheel rim, will be travelling forward, at double the speed of the bottom of the wheel rim - but I cannot think of anything which actually goes backwards, apart from the opposing force, to its movement forward.
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The top of the wheel rim, will be travelling forward, at double the speed of the bottom of the wheel rim - but I cannot think of anything which actually goes backwards, apart from the opposing force, to its movement forward.
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almost double the speed of the train, not the bottom of the wheel rim
The wheel rim in relation to the axle? It's all relative (isn't everything?)
Well, for a start, as bernard has said, the accelerations and decelerations concerned are far too small to be perceptible by human beings. If I've got my very rough 'back of a ciggie packet' calculations right ...
... the circumference of the earth is roughly 40,000 km, so the rotational speed at the equator is roughly 460 m/sec (a bit over 1,000 mph). That means that, relative to any distant object (e.g. the sun) the velocity of the surface of earth at equator will vary from about -460 m/s to +460 m/s, a change of about 920 m/sec over 12 hours. Assuming the rate of change of velocity is constant, that seems to equate to an acceleration of about 0.02 m/s², which is only about 0.002g - hence, as said, far too small to be perceived.
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The outer wheel flange rim (largest diameter is larger than the wheel surface that sits on the rail surface) the bottom of the rim flange will be travelling at X and the bottom of the wheel Y will be travelling ever so slightly slower according to wheel diameter- pi times D in each case. So the b two diameters rotate but their linear length around the circumference is not quite the same. The train travel (smaller diameter) is less than the outermost rim (large diameter ) so it goes very very fractional backwards (but never quite makes it to get to Huddersfield for Albert Einstein.
It reminds me of A a piece of string around the Earth is one length and another piece 1 metre above it all the way round. Length difference in string is 6 and a bit metres ( Pi x D1 - Pi x D2) , not much is it?
It reminds me of A a piece of string around the Earth is one length and another piece 1 metre above it all the way round. Length difference in string is 6 and a bit metres ( Pi x D1 - Pi x D2) , not much is it?
Is it not the other way round - for both rotation and orbit?
err no. the rate of spin is constant day and night with an infinitesimal slowing caused by the tides sloshing about, 1.7 milliseconds every hundred years.
As you are stood on a thing that rotates constantly there is no acceleration due to spin, however our planet does accelerate and decelerate around its orbit of the sun as its ever so slightly eccentric, max velocity is 30.29 km/s and lowest at closet approach in 29.29
At different points on the globe pole and equator , you will be travelling from an outsiders perspective at different speeds as speed is just distance / time, but your velocity would be the same, stand on the pole looking one way say due south and it will take 24 hours ( roughly) to complete the full circle.
As for the train wheel, you are viewing it as the outside observer passing a point , it only accelerates / decelerates as a mass however, you need to change your point of reference to being on the wheel, then at a constant velocity there would be no acceleration or declaration phase, just constant motion until an outside force is applied.
One can avoid most of these discussions/arguments by talking in terms of angular, rather than linear velocity - since, foir example, all parts of a train wheel have the same angular velocity.
Admittedly, even determining 'angles' (through which something has rotated) is not totally straightforward, since it relies on some 'external reference', but in terms of things happening on (or near) the surface of our planet, the direction that gavity is pulling will usually suffice at that reference.
No, not much at all - but, proportionately, exactly the same (very small) difference as is that 1 metre compared with the radius of the earth - go up about 6.37 km (rather than 1 metre) above the surface and the piece of string will be double what it was at the surface!
Kind Regards, John
The rate of rotation is constant, but the rate of linear movement of a point on the surface relative to some external object (e.g. the sun) will vary (and, indeed, reverse in direction) during the rotation ...
As I wrote, at one extreme a point on the (equator of the) surface of the earth is moving towards , say, the sun at about 460 m/s, and at the other extreme it is moving away from the sun at that same speed - with all rates of linear movement relative to the sun (including zero) in-between.
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Aye but that surface speed of the earth at the equator of about 1,000 mph just pales into insignificance when you consider the speed of our Galactical Orbit (around the milky way) of over 500,000 mph
The first thing to realise is that we don't feel acceleration or gravity directly, we only feel the forces that cause acceleration or oppose gravity.
In particular someone in free-fall feels "weightless". Gravity is causing acceleration but they don't feel either of them.
When we construct newton's laws into a rotating reference frame, we get "fictitious forces". Since these fictitious forces represent accelerations in the global reference frame, just like acceleration and gravity we can't feel them directly, only feel the forces opposing them.
So yes, we absolutely do feel the reaction force from the ground, but we mentally just attribute it to "gravity" and don't notice the (relatively small) contributions from the fictitious forces.
Quite so, and I presume the speed of our orbit around the centre of the Milky Way pales into insignificance in relation to the speed at which the Milky Way galaxy moves within the universe'.
I suppose that is in many senates true, but, as I think you go on to acknowledge, we can nevertheless 'feel the effects of acceleration - just ask any fighter pilot, 'stunt pilot' or astronaut etc. - or, indeed, anyone who has ever been in a vehicle which drove into something very solid (and survived)!
Yes - but, as you say, someone in free fall 'feels weightless', which they don't when not in free fall - so they are 'feeling' something as a result of that unaccustomed acceleration, even if it is a sort-of 'negative' something!.
As I see it, human beings (and presumably most/all other land animals) are designed/programmed to regard 'the norm' as being standing on the earth (and experiencing the reaction to the force of gravity), but anything which increases or decreases that reaction (e.g. going up or down in a high-speed lift) will be perceived as something 'different from normal' - so, again, they do 'feel it'.
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