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17

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 ...


15

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 ...


14

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

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 ...


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) ...


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 ...


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

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 ...


3

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: ...


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 ...


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

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

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 ...


2

I am not particularly an expert either, but my understanding is that shuttle flight is a very active process compared to ballistic motion, so any effects the Coriolis effect might have can just as well be considered as additional errors in the trajectory, which is being adjusted. There's an active feedback loop at work: "observe flightpath -> identify ...


2

You can think about it like this: It takes one day for the earth to perform a full rotation (about 86k seconds), on the other hand, it takes a few seconds for your sink to drain (lets say 10 seconds). So it takes 8600 times longer for the earth to do a full rotation than it takes the water to drain down the sink. It is not too hard to imagine that the ...


2

Rockets lean as they climb on purpose, in order to obtain the high orbital velocities needed to stay in space once they get there. For a nice explanation, see Orbital Speed at xkcd what-if, but the gist is the following. Being in orbit means going so fast that the Earth begins to curve away from you as you fall down towards it. The classic image to keep in ...


2

The Coriolis effect cannot possibly account for this. Other factors (e.g the shape of the basin and initial conditions for the water flow) should have a comparatively huger effect. This is intuitively obvious---if you are not convinced you should perform an experiment. Could be fun! You could also estimate characteristic scales (via dimensional analysis) ...


2

Well, first we know that the Coriolis force acting on the scale of a toilet is going to be a pretty small force. So the question can generally be tackled in the following way: Estimate the magnitude of the Coriolis force on the toilet water. Use this to estimate the magnitude of effect of the Coriolis force on the toilet water spin. Enumerate the ...


2

At any given moment, the pendulum is swinging in a certain plane. The driving force should be within this plane. If the earth weren't rotating, then such a force could never cause the plane of swing to rotate about a vertical axis, since by symmetry there would be no preferred direction for the rotation.


2

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 ...


2

Surprisingly, the answer is that yes you do, though the effect is very small. To see this consider the following (highly exaggerated) diagram of the lift shaft: The Earth rotates at a constant angular velocity of one rotation every 24 hours ($\omega = 7.27 \times 10^{-5}$ radians/sec). The tangential velocity of a part of the lift shaft at a distance $r$ ...


2

Look in Wikipedia http://en.wikipedia.org/wiki/Coriolis_effect. For understanding intuitively the Coriolis force effect, assume an object moving according to a static (inertial) frame of reference, in the plane perpendicular to the rotation axis, and along the radius, In the rotating frame, see the animation in Wikipedia, the Coriolis force imposes an ...


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 ...


1

Disclaimer: borrowed from wikipedia with minor modifications. The acceleration is the second time derivative of position, $$ \mathbf{a}_{\mathrm{i}} \ \stackrel{\mathrm{ }}{=}\ \left( \frac{d^{2}\mathbf{r}}{dt^{2}}\right)_{\mathrm{i}} = \left( \frac{d\mathbf{v}}{dt} \right)_{\mathrm{i}} = \left[ \left( \frac{d}{dt} \right)_{\mathrm{r}} + ...


1

You are right: the coriolis effect in the equator is converted to a kind of centrifugal force. But in many situations only the horizontal Coriolis forces are accountes, because in the vertical one the gravity is so much greater. The vertical coriolis and the centrifugal force may point parallel, but they are still different forces. Let's think of a ...


1

If you project the velocity of a water parcel traveling up the channel into radial coordinates in both the inertial frame (space frame) and the rotating frame of reference (earth frame) then I think the necessary effects will become apparent. In the earth's frame, the velocity is (use a ' to symbolize the rotating frame): $$\vec{v} = 0\cdot\hat{r}' + ...


1

Ok, let's see if I get this right. The circumference of the equator is 24902 miles. The Earth rotates 360° in 86,164.098 seconds. This means the surface of the Earth moves 1525.96 ft/sec at the equator. Adding 1000 ft to the height give a circumference of 24903 miles. This means the lateral speed is 1526.03 ft/sec; a difference of about 0.87 inches/sec. It ...



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