Tides

Antimoon! Antimoon!

I was saying, you tip a bowl of water to one side, the water flows to that side due to gravity acting on it. If there is no external gravity, and the USA accelerates the whole earth the same amount, how does the earth "Tippin" cause tides??

Equivalence Principle.   

This is where I gathered you were referring to tides - my mistake I guess.

C-me!
John

He asked how the tilting can cause the water to 'slosh' to one side if there is no 'gravity' but only acceleration.  Therefore, EP.

but that would mean one side of the FE has significantly less "gravity" than the other.

but that would mean one side of the FE has significantly less "gravity" than the other.

No, I don't think it would.

Edit:

Actually, i suppose it would have more downwards pull in some areas, but barely any since the tilt is supposedly very small.

So when you tilt a bowl of soup, you've just altered the gravitational properties of the RE? 

He asked how the tilting can cause the water to 'slosh' to one side if there is no 'gravity' but only acceleration.  Therefore, EP.

Yes, that's what I meant by my mistake. I misread his question to include EP (or the basic idea of it) as an assumption within it the way it was worded. Reading it more closely, I see your point in having referenced EP.

So when you tilt a bowl of soup, you've just altered the gravitational properties of the RE? 

No, but you have altered the relative direction the RE gravity acts in (i.e. before the tilt, gravity pulls things towards the bottom of the bowl. After the tilt, RE gravity pulls things towards the edge of the bowl, which is now the lowest point.)

No, but you have altered the relative direction the RE gravity acts in (i.e. before the tilt, gravity pulls things towards the bottom of the bowl. After the tilt, RE gravity pulls things towards the edge of the bowl, which is now the lowest point.)

Same thing for acceleration.  Glad we got that cleared up.

I think we are responding to the wrong things or something.

Yes, we got off topic onto whether or not EP / acceleration would have the same effects on tilted water (basically) as to the RE idea of gravity. I think that's pretty much a done deal - how either applies to tides, well I gave my opinion on that already via EP.

 

Quote from: Moon squirter on November 13, 2007, 09:36:29 AM

Except, in this case, you are wrong.  For the bucket, ω²r is the correct equation, as the angular velocity (radians per second) is constant for the all radial points along the swinging bucket.  For this constant angular velocity, a is proportional to r, as I originally stated.

Yeah, what about it? I thought we're talking about centripetal (inward) acceleration...

Quote from: wikipedia

However, an object moving in a circle at constant speed has a changing direction of motion. The rate of change of the object's velocity vector is the centripetal acceleration.

The centripetal acceleration varies with the radius r of the circle and speed v of the object, becoming larger for greater speed and smaller radius. ω = v / r is the magnitude of angular velocity. The negative sign indicates that the direction of this acceleration is towards the center.

ω = v/r
 
a = -ωr²

a = -v²/r

Thus, inward acceleration.

Quote from: Moon squirter on November 13, 2007, 09:36:29 AM

The other equation, v²/r, deals with linear velocity (v), which varies for the all points along the swinging bucket.

Sure, it is one of the equations that deals with uniform circular motion. a = v²/r is valid for this bucket situation. Velocity varies in direction along the points.

Quote from: Moon squirter on November 13, 2007, 09:36:29 AM

Wikipedia Quote:  The quote from about acceleration applies only to a constant linear speed (v), which does not apply in this case because it is going to be different for points along the bucket

Mind giving me a source of where you get this? Or did you just pull this out of your ass?

Quote from: Moon squirter on November 13, 2007, 09:36:29 AM

Nice try too!

Thanks, since I've been winning the whole time.

No, you are still wrong, because you have misused the equations for this example.  Angular velocity (ω) in constant for the whole bucket system (e.g. from the bottom of the bucket to the top). 

The equation you derived from a = -ωr2 and ω = v/r, is indeed from centripetal (inward) acceleration.  However, v is linear velocity (velocity at tangents to the rotation system).

In the bucket example, this linear velocity changes with radius; it is small near the centre of the system and larger nearer the outside.  Therefore the linear velocity is not the same in the bottom of the bucket as it is at the top.  Therefore the Wikipedia quote is not relevant.

However the angular velocity is constant.  Therefore for the bucket example it is sensible to apply the equation a = -ωr², thus avoiding linear velocity.

Another example may be:  If you are on a child's roundabout with a radius of 4 feet, rotation at a speed of 1 revolution per second, you would feel the acceleration reasonably easily.  However, if the roundabout was radius 40 feet, at a speed of 1 revolution per second, you would have to hang on for dear life and would probably be forced off.

I did not pull this out of my ass, you complete f@cking ignoramus.  I learned it at school and passed an exam, so there are no sources.  However if you need more information, this link may be helpfully.

Ok, not sure if theengineer got what I meant. The water in a dish thing is an example. I'm asking, if the tides were caused by water flowing to the lower side of the tilting FE, (which it isn't, because of our new friend the antimoon), what causes the water to flow downwards, when there is no external "downward" gravity.

No, you are still wrong, because you have misused the equations for this example.  Angular velocity (ω) in constant for the whole bucket system (e.g. from the bottom of the bucket to the top). 

Right, which is why you only need to use this very simple centripetal acceleration equation: a = -ω²/r (negative means inward) with ω = v/r.
                     

The equation you derived from a = -ωr2 and ω = v/r, is indeed from centripetal (inward) acceleration.  However, v is linear velocity (velocity at tangents to the rotation system).

That's why I use the equation, because we're talking about inward acceleration. Unless you want to argue about other things...

In the bucket example, this linear velocity changes with radius; it is small near the centre of the system and larger nearer the outside.  Therefore the linear velocity is not the same in the bottom of the bucket as it is at the top.  Therefore the Wikipedia quote is not relevant.

Eh, the velocity doesn't change with radius; acceleration changes with radius. The velocity is constant and varies in direction in a circular path. Hence, inward acceleration.     

a = v²/r refers to uniform circular motion; constant velocity.

However the angular velocity is constant.  Therefore for the bucket example it is sensible to apply the equation a = -ωr², thus avoiding linear velocity.

Eh, it's a = -ω²r, not a = -ωr².

Another example may be:  If you are on a child's roundabout with a radius of 4 feet, rotation at a speed of 1 revolution per second, you would feel the acceleration reasonably easily.  However, if the roundabout was radius 40 feet, at a speed of 1 revolution per second, you would have to hang on for dear life and would probably be forced off.

You feel inward acceleration; however, you always go in a straight line, thus you feel being forced outwards. Direction of velocity changes faster in smaller radius, which means the rate of change in velocity is quicker. Thus, acceleration is greater. In a spinning table, put a penny near the center and another one near the edge. Which one spins faster (changing direction of velocity)?

I did not pull this out of my ass, you complete f@cking ignoramus.  I learned it at school and passed an exam, so there are no sources.  However if you need more information, this link may be helpfully.

Luckily I don't go to your school...

Imagine there is a river with banks running paralell to the way the earth 'tilts'
this means, according to your FE theory, that on one bank of the river the tide would be high, while the other would be really low. But don't rivers have equal tides on both banks? (ie gravitational pull?)

Could an FE explain it to me simply, and if you really can't, then please direct me to a link. But don't just direct me to a link first, because if its so simple you really should be able to explain it.

Ok, not sure if theengineer got what I meant. The water in a dish thing is an example. I'm asking, if the tides were caused by water flowing to the lower side of the tilting FE, (which it isn't, because of our new friend the antimoon), what causes the water to flow downwards, when there is no external "downward" gravity.

I don't understand what the complication is.  Acceleration is responsible. 

Quote from: Moon squirter on November 13, 2007, 02:19:04 PM

However the angular velocity is constant.  Therefore for the bucket example it is sensible to apply the equation a = -ωr², thus avoiding linear velocity.

Eh, it's a = -ω²r, not a = -ωr².

Ok, that was a typing error, should have been a = -ω²r.

Quote from: Moon squirter on November 13, 2007, 02:19:04 PM

Another example may be:  If you are on a child's roundabout with a radius of 4 feet, rotation at a speed of 1 revolution per second, you would feel the acceleration reasonably easily.  However, if the roundabout was radius 40 feet, at a speed of 1 revolution per second, you would have to hang on for dear life and would probably be forced off.

You feel inward acceleration; however, you always go in a straight line, thus you feel being forced outwards. Direction of velocity changes faster in smaller radius, which means the rate of change in velocity is quicker. Thus, acceleration is greater. In a spinning table, put a penny near the center and another one near the edge. Which one spins faster (changing direction of velocity)?

The table/coin example further illiterates my point, and proves you do not understand what "v" is in a=v²/r.  v is the tangential velocity, v_t=rω (or could be considered straight line speed):

From this diagram we can see that the tangential velocity will be smaller towards the centre of the system.

Both coins have the same rotational velocity ω. Therefore, according to a = -ω²r, the coin on edge of the table will experience more inward (centripetal) acceleration, and hence centripetal force (a = -mω²r) that the one near to centre.  Therefore the coin on the edge will overcome friction first, flying off the table at a tangent, which makes sense if you think about it.

I will stop trading insults.

How about we put an end to this. 

The position of any point on a rotating bar at any angle is given by:
P = r*e^j*theta   (1)
where r is the distance from the center of rotation and theta is the angle.

Taking the derivative of (1) yields the velocity of any point at any angle:
V = j*w*r*e^j*theta  (2)
with w being the angular velocity.

Taking the derivative of (2) gives us the acceleration of any point at any angle:
A = (-w² +j*a)r*e^j*theta   (3)
with a being the angular acceleration of the bar.

As is evident from (3), increasing the distance from the center of rotation increases the acceleration. 

There.  Can we put this to rest now?