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Water in rotating cylinder

by Pi-Bond
Tags: cylinder, rotating, water
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Pi-Bond
#1
Mar19-13, 06:00 PM
P: 304
A question I was doing asked to find the equation describing the surface of water placed in a cylinder rotating about its central axis. The question asserts that in the rotating frame, the water is at rest, and centrifugal force and the gravitational force of a volume element are perpendicular to each other. I don't see why this is the case - can anyone explain?
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tiny-tim
#2
Mar19-13, 07:02 PM
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Hi Pi-Bond!
Quote Quote by Pi-Bond View Post
The question asserts that in the rotating frame, the water is at rest, and centrifugal force and the gravitational force of a volume element are perpendicular to each other. I don't see why this is the case - can anyone explain?
Which part don't you get?

In the rotating frame, obviously the water is at rest.

And obviously the gravity is vertical, and the centrifugal force is horizontal.
MrAnchovy
#3
Mar19-13, 07:54 PM
P: 541
The geometrical axis of the cylinder, which is also the axis of rotation, must be vertical so the centrifugal force must be horizontal: this is clearly at right angles to gravity.

The water must have reached equilibrium so it is rotating with the cylinder (friction forces and viscosity will eventually achieve this). "Rotating with" means that the water is at rest wrt a rotating frame of reference about the axis of rotation of the cylinder.

Pi-Bond
#4
Mar19-13, 08:49 PM
P: 304
Water in rotating cylinder

Tiny-tim, what I was confused about is the thing MrAnchovy explained. I think I was getting confused by "at rest in the rotating frame" somehow. But now since both of you have written it out, it makes more sense. Thanks!

As a follow on, I was wondering if something similar to this method can be used to find the equation of the surface if the cylinder is rotated around a point at some angle to the vertical. For example, rotating a bucket of water by using a rope. The question previously mentioned used

dh/dr = Centrifugal Force/Gravitational Force
h: height of water w.r.t centre
r: distance from centre
olivermsun
#5
Mar19-13, 08:55 PM
P: 844
You should work out the bucket/rope problem. It might be helpful for your understanding of this type of problem. But first ask yourself what the angle to the vertical will be for the rope.
tiny-tim
#6
Mar20-13, 04:19 AM
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Hi Pi-Bond!

(just got up )
Quote Quote by Pi-Bond View Post
I was wondering if something similar to this method can be used to find the equation of the surface if the cylinder is rotated around a point at some angle to the vertical. For example, rotating a bucket of water by using a rope. The question previously mentioned used

dh/dr = Centrifugal Force/Gravitational Force
h: height of water w.r.t centre
r: distance from centre
Yes, so long as the water keeps the same shape inside the bucket, you can use a rotating frame whose axis is the vertical line through the top of the rope.

Since the surface has a shape such that an object placed on top of the surface will not move relative to the water, it will be in equilibrium in the rotating frame.

Since the only three forces on it are gravity centrifugal and normal, you can take tangential components and get dh/dr = Centrifugal Force/Gravitational Force
Pi-Bond
#7
Mar20-13, 04:24 PM
P: 304
After thinking it through, I think I understand it now. Thanks for the help!


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