# Tag Info

52

Summary Centrifugal force and Coriolis force exist only within a rotating frame of reference and their purpose is to "make Newtonian mechanics work" in such a reference. So your teacher is correct; according to Newtonian mechanics, centrifugal force truly doesn't exist. There is a reason why you can still define and use it, though. For this reason, your ...

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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 ...

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The trick is, centrifugal force is a fictitious force. Centrifugal force exists! To everyone denying it, do this to them: xkcd.com/123. However it is a fictitious force. To quote wikipedia: A fictitious force is an apparent force that acts on all masses whose motion is described using a non-inertial frame of reference, such as a rotating reference ...

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The error is that you assume that the density distribution is "nearly spherically symmetric". It's far enough from spherical symmetry if you want to calculate first-order subleading effects such as the equatorial bulge. If your goal is to compute the deviations of the sea level away from the spherical symmetry (to the first order), it is inconsistent to ...

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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 ...

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The effective gravity inside the ISS is very close to zero, because the station is in free fall. The effective gravity is a combination of gravity and acceleration. If you're standing on the surface of the Earth, you feel gravity (1g, 9.8 m/s2) because you're not in free fall. Your feet press down against the ground, and the ground presses up against your ...

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The surprising answer is that the stability of the modern bicycle has little or nothing to do with centrifugal force or gyroscopes or any of that. Look up "bicycle stability" on Google. Experiments show that the sloped angle of the front fork is very important, e.g. If the fork pointed backwards it is very difficult to stay upright at any speed. At ...

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Suppose you are at a red light in your car. You apply Newton's second law on the street light. $$F=ma$$ $$F=0N, a=0ms^{-2}$$$$0N=0N$$ It works!! Now the light turns green and you start accelerating. Suppose your acceleration is $1ms^{-2}$. According to you, you are at rest. Do you see your nose moving? Apparently not. It means your body is at rest wrt you. ...

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Great photo! Edit: My language is "sloppy" (I like talking physics in "lay person" terms so anybody can understand) but @dcmkee made really nice comment clarifying my answer for the more advanced people. Thanks @dcmkee! Since the plane is in a loop there is significant g's due to centripetal acceleration. The water was being accelerated upward$^{1}$ with ...

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A report appeared in Science today which addresses this exact question: Kooijman et al., Science 332 (6027): 339-342, "A Bicycle Can Be Self-Stable Without Gyroscopic or Caster Effects." The abstract reads: A riderless bicycle can automatically steer itself so as to recover from falls. The common view is that this self-steering is caused by gyroscopic ...

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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, $$\dot{\vec{r}} = \vec{\omega} \times ... 14 By definition an orbit occurs when gravity balances with the "centrifugal" force. It is essentially a free fall situation. So the answer is the same reason why you don't get stuck to the ceiling of a free falling elevator. Both the spacecraft and the occupants are moving in-sync. 13 Here we would like to calculated analytically Lubos Motl's solution to the first order in the flatness parameter f,$$0<f:=1-\frac{b}{a}\approx\frac{a}{b}-1 \ll 1, $$where a and b are the equatorial and polar radius of the Earth, respectively, and a>b. (The \approx symbol will from now on mean equality up to higher-order terms in f.) We ... 13 It doesn't actually have anything to do with the plane being upside down, or even changing from a vertical direction to a horizontal one. It's purely the vertical velocity that's at play here. Imagine water being thrown upward. You know what, imagine a fountain, a really big fountain. As soon as the water leaves the underground pump, it starts falling back ... 12 As I disagree with all the answers I am going to try to explain some of the fundamentals of science: Science in its very essence can not explain why things happen the way they do, it simply tries to model reality based on observations in the past to predict events in the future. In other words, defining a centrifugal force is possible as for example your ... 11 These are two different effects. Satellites don't fall down because they are moving on a circular orbit. Actually, they are falling down all the time, since circular motion is accelerated (though the velocity doesn't change absolute value, it changes direction!), so it is kind of "falling around the earth". The second question is, why doesn't an astronaut ... 10 There is definitely a huge difference. Centripetal force is a real force that causes objects to move in a circular path or curved path that points to the center of the circle or curvature respectively. Centrifugal force is not a real force. It is an inertial force established so that Newtonian laws are valid when observing motion in an accelerated frame of ... 10 You will have two forces that act on an elementary mass element dm on the surface. The force in the x-direction will be dF_{x}=\omega^{2}xdm and in the y-direction dF_{y}=gdm. Also, we know that the slope of a curve is \tan{\alpha}=dy/dx. However, the tangent is equal also to \tan{\alpha}=dF_{x}/dF_{y}. So from this you have that ... 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 Centrifugal force is a particular example of a fictitious force. It is introduced so that Newton's second law holds in a rotating reference frame. Newton's second law says$$F = ma This means that whenever we find an object accelerating (speeding up, slowing down, turning, or some combination), we can look around and find a physical reason why this ...

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There was some doubt about Lubos' answer (which I've accepted), so this is just a verification. I copied the method Lubos described and found the potential difference for an ellipsoid with different eccentricities. Sure enough, for an oblate spheroid, if you make the center-equator distance a fraction $e$ larger than the center-pole distance, the ...

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The real force at work is centripetal force, or a force pushing inwards. Imagine you have a bucket on a string, and you swing that around in a circle: As you swing the bucket, it travels in a circle. The red line shows the path the bucket takes. In order to make it swing like this, you have to apply a constant force on the rope -- this is the green arrow ...

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Turn on your rocket engines or configure your space craft into a rotating cylinder.

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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, ...

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Actually, the astronaut would only float completely free in the middle of the space station. Elsewhere, he will stick slightly to whatever side is closer to the Earth than is the middle, or farther from the Earth than is the middle. The reason is the tidal force from the Earth, which will be very small but probably detectable. If the acceleration from ...

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Because it's effect is smaller than the variation in $g$ due to earth's bulge (caused by the same centrifugal force) or the local geology - when you use $9.8m/s^2$ that's just an approximation. The effect of the bulge and centrifugal force mean that $g$ at the equator is about 0.5% lower than $g$ at the poles edit: velocity at equator $40,000 km / 24 h = ... 7 It sounds like you don't want the normal rotating spaceship like in "2001" because you get motion sickness. No one really gets "motion sickness" just from moving, though. That's impossible because moving with constant velocity is physically the same as being stationary. What you get is "acceleration sickness". You feel the bumpiness of a car ride. Even ... 7 The force you feel when you round a corner in your car is the friction force of the car seat on your behind, and perhaps the pushing force of the door on your shoulder. These are very real forces that occur when your car tries to turn while your body tries to continue moving in a straight line. But from your point of view in the car, with the windows ... 6 At the moment the picture was taken the plane, with mass$M$was performing an inside loop, and was almost exactly inverted. It was moving at a speed$V$in a vertical circle with radius$R\$; both of these are chosen by the pilot as he execute the loop. The physics of circular motion requires that the plane experience a force towards the centre of the ...

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