Centrifugal Force and shape of water

[amcavoy]

"Centrifugal" Force

Say you have a cylinder of radius r with πr²h liters of water in it. If you put it on some sort of turntable and spin it with an angular velocity ω, is there a way to tell what the shape of the water will be? I mean, if you look at the cross section will you be able to find out if it forms for example, a parabola?

This isn't a homework problem or anything, I'm just curious. Thanks for your help.

 

[Doc Al]

Actually, the rotating water surface will assume the form of a parabola. The way to figure that out is to examine the forces on a element of the water surface. The net force is centripetal.

 

[amcavoy]

Actually, the rotating water surface will assume the form of a parabola. The way to figure that out is to examine the forces on a element of the water surface. The net force is centripetal.

Relative to some origin (let's say the vertex), is there a way to find the equation of that parabola or is the shape the most that can be determined?

 

[Doc Al]

According to my calculation, the equation (using the vertex as the origin) should be:
$$y = \frac{\omega^2}{2 g} x^2$$

 

[Tide]

Said another way,the surface of the water will be an equipotential surface in a corotating frame of reference.

 

[dpsguy]

Are you rotating just the turntable or the cylinder as well? In the former case I think some water will fall out due to its inertia of rest.Tell me if I am wrong,and also if water would fall out in the latter case too.

 

[amcavoy]

According to my calculation, the equation (using the vertex as the origin) should be:
$$y = \frac{\omega^2}{2 g} x^2$$

Is this shown by the following?:

$$E_k=E_p\implies v=\sqrt{2gh}$$

$$y=h;\quad x=r$$

$$\frac{ds}{dt}=v=x\frac{d\theta}{dt}=x\omega$$

$$\sqrt{2gy}=x\omega\implies y=\frac{{\omega}^2}{2g}x^2$$

This was the only way I could derive the equation you posted above, and I figured that the kinetic energy would be the same as the potential energy because if you look at a single molecule of water, it is doesn't experience any friction because it is rotating with every other molecule. Also, I just said that the distance from the center of the cylinder r was x (since the origin is the center) and the height above that center is y. Am I correct with what I posted above?

Thanks again.

 

[Doc Al]

I can't say I understand your derivation or reasoning, but maybe I'm just dense today.

Here's how I would derive it (from an inertial frame). Take an element of mass on the surface. The force due to the water/air pressure must be perpendicular to the surface. Call that force F. The only other force is the weight, mg. Applying Newton's 2nd law to vertical and horizontal components gives:
$$F \cos \theta = mg$$
$$F \sin \theta = m\omega^2 x$$
where $\theta$ is the angle the tangent to the curve makes with the horizontal axis, thus:
$$\tan \theta = dy/dx = \frac{\omega^2}{g} x$$
$$y = \frac{\omega^2}{2g} x^2$$

Viewed from the rotating frame, the derivation is even simpler. Just realize that the surface of the fluid will be perpendicular to the direction of the apparent "gravity": the sum of real gravity plus the outward centrifugal force.

 

[Danger]

For an excellent practical application of this idea, check this out.

 

[amcavoy]

I can't say I understand your derivation or reasoning, but maybe I'm just dense today.

Here's how I would derive it (from an inertial frame). Take an element of mass on the surface. The force due to the water/air pressure must be perpendicular to the surface. Call that force F. The only other force is the weight, mg. Applying Newton's 2nd law to vertical and horizontal components gives:
$$F \cos \theta = mg$$
$$F \sin \theta = m\omega^2 x$$
where $\theta$ is the angle the tangent to the curve makes with the horizontal axis, thus:
$$\tan \theta = dy/dx = \frac{\omega^2}{g} x$$
$$y = \frac{\omega^2}{2g} x^2$$

Viewed from the rotating frame, the derivation is even simpler. Just realize that the surface of the fluid will be perpendicular to the direction of the apparent "gravity": the sum of real gravity plus the outward centrifugal force.

Ahh, ok that makes sense. I don't know what I was thinking when I derived it the other way :uhh:

Thanks for the help