# Centrifugal force in rotating frame of reference

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## Main Question or Discussion Point

Can someone enlighten me why it is mentioned in the thread
that when the water is viewed from a rotating frame of reference it experiences an outward centrifugal force. This does not make sense to me.

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## Answers and Replies

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quasar987
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Gold Member
I'm going to attempt an explanation, but I'm not sure it's accurate. It would be nice if it were cuz it would make explaining centrifugal forces easy.

Let's focus on one molecule of water, instead of the whole cylinder.

This molecule is decribing uniform circular motion. Hence, there is a centripetal force towards the center acting on the molecule. But in a rotating frame of reference, the molecule is at rest. Hence, there is must be an equal and opposite force acting on it: the centrifugal force.

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Yes, the universe has a nasty way of making things difficult for us to understand. It is us humans who try to make the universe easier for us to comprehend. The problem with reconciling your explanation with what was said in the thread is that there would then be only gravity to control the shape of the surface of the water, and since it is shaped parabolically in the rotating frame your suggestion would not explain the effect, or am I understanding you incorrectly?

Doc Al
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Centrifugal force only appears when analyzing the situation from a rotating frame of reference. Since such a frame is non-inertial, Newton's law do not apply without modification. One modification is the introduction of "fictitious" forces (such as centrifugal force) which are entirely an artifact of the acceleration of the frame of reference. Some problems are much easier to analyze when viewed from a rotating frame.

Of course, the parabolic shape of the water does not depend on what frame you view it in. It is perfectly OK to analyze the problem from an ordinary inertial frame in which the water is rotating and thus centripetally accelerating. I give such an analysis in the referenced thread.

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In the rotating frame of reference the horizontal outwards force is a fictitious one (meaning it does not really exist), the centrifugal force, that we introduce in order to explain an effect that we know exist, even in the rotating frame of reference - is a manifestation of the accelerating frame of reference. The resultant force will then be pointing into the water, perpendicular to it's surface. Which explains everything well in this frame. In the inertial frame of reference the resultant force will however point somewhat along the surface of water due to the known required centripetal force?

Doc Al
Mentor
andrevdh said:
In the rotating frame of reference the horizontal outwards force is a fictitious one (meaning it does not really exist), the centrifugal force, that we introduce in order to explain an effect that we know exist, even in the rotating frame of reference - is a manifestation of the accelerating frame of reference.
Right.
The resultant force will then be pointing into the water, perpendicular to it's surface. Which explains everything well in this frame.
In the rotating frame, the resultant force (on an element of the water surface) is zero, since it does not accelerate in that frame. The combination of gravity plus centrifugal force will be normal to that parabolic surface.
In the inertial frame of reference the resultant force will however point somewhat along the surface of water due to the known required centripetal force?
In the inertial frame of reference, the resultant force is centripetal.

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And here I thought it would get easier as we go along... , but it is getting clearer! I forgot about the supporting force of the surrounding water on a water element. That is why I was thinking that the resultant force in the inertial frame points along the surface. I understand the zero resultant force in the rotating frame now - we are ignoring the fictitious force in this regard, since it does not really exist - but to explain the curvature of the water (which cannot appear out of nowhere) we use it to get the apparent gravity perpendicular to the surface. We cannot explain the curvature the same way in the inertial frame. We just say that the curvature is a result of the spinning.