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39

Because the rotation of the earth is very smooth and doesn't change, the centripetal acceleration we feel is very nearly constant. This means that the (small) centrifugal force from the rotation gets added to gravity to make up the "background force" we don't notice. Earthquakes are not at all smooth and the accelerations involved are large and change ...


26

Dan's answer is essentially good, but miss one effect : the Coriolis effect. You can imagine a planet spinning much more rapidly than the earth, but at a constant angular speed. On that quickly rotating planet, the explanation of Dan would still stand, but as soon as on moves, we would feel a lateral Coriolis force. The Coriolis acceleration is ...


19

Method 1: Foucault Pendulum As user Rob asks, what is wrong with a short Foucault pendulum? There is a problem, but it can be overcome inexpensively, but with some DIY effort at home. The problem is that, by dint of imperfections in the suspending fibre and bob, no pendulum will swing in a plane even if the Earth were not rotating. Instead, the bob will ...


16

Ok, here is my (hopefully rigorous) demonstration of the origin of these forces here, from first principles. I've tried to be pretty clear what's happening with the maths. Bear with me, it's a bit lengthy! Angular velocity vector Let us start with the principal equation defining angular velocity in three dimensions, $$\dot{\vec{r}} = \vec{\omega} \times ...


13

The whirl is due to the net angular momentum the water has before it starts draining, which is pretty much random. If the circulation were due to Coriolis forces, the water would always drain in the same direction, but I did the experiment with my sink just now and observed the water to spin different directions on different trials. The Coriolis force is ...


10

I personally think the descriptions on Wikipedia are rather confusing, so I'm going to write a self-contained derivation in my own words; hopefully this helps. Note: I'll use Einstein summation notation throughout. In order to understand what's really going on in the derivation, I'm going to attempt to separate pure mathematics from physics. In ...


10

The answer: the ball appears to be deflected ~10 cm. The calculation: For simplicity, say we tee off at the north pole. The effects are a bit weaker at more typical locations, you multiply by sin(latitude) = 0.64 for a 40 degree (central california or washington DC) latitude. The Coriolis effect exists because the Earth rotates while the ball is in ...


9

Yes, the ball would land in front of you. If you watch from outside the space station, the ball moves in a straight line at constant speed while you move in a circle at constant speed. That means the distance the ball takes to get from point A (where you release it) to point B (where it hits the floor) is shorter than the distance you take. Further, ...


8

The Coriolis acceleration goes like $-2\omega \times v$, which for the sake of an order of magnitude estimate we can take to be $a\sim \omega v$. But in order to get an observable effect, we don't just need an acceleration, we need a difference in acceleration between the two ends of the tub, which are separated by some distance $L\sim 1$ m. The ...


5

Since you want to explain it to your daughter, take a plastic bottle, cut the bottom open, turn it upside town, hold the top closed and fill it with water. Give her that bottle and have her release the top (which is on the bottom now, sorry for the bad phrasing). The water will whirl in different orientations whenever you repeat this (if it whirls at all) ...


5

I know it's very late in the game for this question, but this is partly a biology question. We don't feel the rotation of the earth because our brains are biased, they evolved that way. It's not useful to experience/be aware of this rotation day by day, in the same way it isn't useful to be aware of gravity. This is also why this optical illusion works: ...


5

Of course that there would be forces that would try to bend the track but they would be tiny. Each segment of the track would be under the action of $-2m \Omega \times v$ Coriolis force. Note that the Coriolis force only depends on velocities, not accelerations as you stated! In other words, there is the Coriolis acceleration, $-2\Omega\times v$, and you ...


4

Indirectly, yes. It is called a gyro compass. When you constrain a gyroscope to spin about an axis in the horizontal plane it will experience a Coriolis like force unless it is pointing due North-South - if it isn't it will slowly align, then stay there, so you can use this as a compass - in fact it is widely used as it doesn't suffer from magnetic ...


4

An alternate derivation for a due-north ball, ignoring the diminishing effect of latitude, that confirms Kevin's order of magnitude: Acceleration due to the Coriolis effect: $a_C = -2 \, {\Omega \times v}$ $\Omega = 2 \pi/day$ $v = 45 m/s$ $a_C = -0.00654498469 m/s^2$ Horizontal displacement $d$ is given by $d = 1/2a_C\,t^2 $ Using earlier estimate of ...


4

Coriolis force is not an actual force, but rather an effect observed in rotating frame of reference. The light path is not actually bent, so it doesn't matter that the photon has no mass, the Earth's rotation will have an affect on the photon's apparent path. This does not contradict your calculation of $F_{Coriolis}=0$, because you have to put this force ...


4

I don't believe the Coriolis force has much effect on a tsunami because it does only affect moving masses. Coriolis force in fact isn't a force but a movement pattern looking as though a force were involved. It is a result of inertia "driving" the moving masses towards a constant direction in space and at the same time the earth's rotation taking place. ...


4

Firstly, is that correct? Yes your intuitive understanding for this part of the Coriolis effect is correct. The second part, that is, why wind in the East direction is deflected South, is a bit trickier, and involves the use of centripetal force. this is given by the equation: $F = \frac{mv^2}{r}$ If we re-arrange the above equation, we can find ...


3

The rotation of your coordinate system causes the Coriolis effect. Things move in straight lines, but if your coordinate system is rotating, then the straight lines look curved from the perspective of your coordinate system. If you want to insist that objects move in straight lines in your coordinate system, then you must invent a fictitious reason why ...


3

Well, there is a partial yes that is a direct result of the Coriolis force: If you go up in a hot air balloon, you will be subject to various winds which will move you. And these winds are a result of the Earth spinning. In principle you should be able to navigate to most places on the globe by choosing height etc. in reality it is much too complex to do ...


3

It does have an effect. Also see this paper about modelling tsunami propagation. As noted in the paper, the Coriolis force only becomes important over large distances. Here's an article on MathWorld including many references.


3

What you indicate appears to be the case. You want to minimize the Coriolis effect in any volume of space, which requires making the rotating station large.


3

The whirl happens in the draining tube, whose optimal solution to drain the bathtub is a laminar flow allowing for some rotation in the tube. What you see in the surface is the match between the solution of flow in the tube and the solution of flow in the surface. Angular momentum of the flow gets modified a lot as the tube twists and twists, sometimes even ...


3

could the Coriolis effect on snowing be so dramatic...? No. The Coriolis effect is only noticeable for objects traveling long distances with respect to Earth's surface for significant periods of time. For example, a ballistic missile fired hundreds of miles or a hurricaine that is hundreds of miles in diameter and lasts for days. Across the street is ...


3

The Coriolis force on the equator indeed does point outwards, if you are moving west to east. This is not the same as the centrifugal force, because the centrifugal force is present always - even if you are not moving. But when you move (west to east), there is an additional force on top of the centrifugal force - the Coriolis force. If you travel east to ...


3

The Coriolis force $\vec F_{\text{coriolis}} = -2m \, \vec \omega \times \vec v$ only depends on velocity. The centrifugal force $\vec F_{\text{centrifugal}} = -m \, \vec \omega \times (\vec \omega \times \vec r)$ only depends on position. Finally, if the object is not rotating uniformly ($\dot {\vec \omega} \ne 0$), then yet another fictitious force comes ...


2

I believe you want to replace mass with charge and angular velocity with the magnetic induction. The Coriolis effect is an apparent force due to the fact that the observer is measuring with respect to a rotating frame of reference. There is no actual force acting on the body, so this can be made to disappear by changing the frame of reference. Classical ...


2

Why do airplanes experience negligible Coriolis force while bullets experience the Coriolis force in long range shooting? You are confusing the force with the consequence of the force. Consider a powered parafoil whose total mass is a mere 100 kg (motor+parafoil+pilot) and is moving at a mere 25 km/h and a 50 caliber bullet whose mass is 50 grams and is ...


2

Just to make things simple, suppose you are standing at the north pole, and you shoot a bullet south at some speed, aiming for a target 1 km away. In the time it takes the bullet to get there, the target has moved east a certain distance, because the target travels in a complete circle around the north pole in 24 hours. From the viewpoint of the shooter, who ...


2

Note that Fitzpatrick states towards the beginning, The following solution method exploits the fact that the Coriolis force is much smaller in magnitude that the force of gravity: hence, $\Omega$ can be treated as a small parameter Generally, when statements like that are made, powers (greater than 1) of the term in question are considered to be zero: ...


2

Ahh, Richard Fitzpatrick. Great guy. Ok, If you start with the second set of expressions, use the appropriate double-angle-formula and then assume the "angle" $2\Omega \sin\lambda t$ is small (note that the $t$ is not within the sin function!), you get the first expressions, e.g. $$\cos(\theta+\phi) = \cos\theta\cos\phi - \sin\theta\sin\phi,$$ and then ...



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