# For a rotating fluid in a beaker, why does the depth of the parabola increase as the angular velocity increases (in terms of forces)? [closed]

I am writing a highschool paper on this and wanted to know the relationship between the parabolic depth and the angular velocity of the rotating fluid in terms of the forces.

This is called Newton's bucket, and is most commonly solved by transforming into a reference frame which rotates at the same angular velocity as the water, such that the water appears stationary. When we do this, we must introduce a fictitious centrifugal force term of magnitude $$mr\omega^{2}$$ acting away from the axis of rotation!

Consider a small mass element $$dm$$ at the surface of the water, $$x$$ units to the right of the vertex of the parabola and $$y$$ units above the vertex. It's weight, $$gdm$$ acts downward, and the centrifugal force, $$mx\omega^{2}$$, acts to the right. If we add these two vectors, we obtain an effective weight.

The surface of the water is necessarily perpendicular to the effective weight, so if the "gradient of the weight vector" is $$\frac{-mg}{mx\omega^{2}} = -\frac{g}{x\omega^{2}}$$, then using standard geometry the gradient of the water at this point has to be $$\frac{x\omega^{2}}{g}$$!

So finally we write down the relationship $$\frac{dy}{dx} = \frac{x\omega^{2}}{g}$$, we can separate variables

$$\int dy = \int \frac{x\omega^{2}}{g} dx \implies y = \frac{\omega^{2}x^{2}}{2g}+c$$

As you should now be able to see, increasing $$\omega$$ will result in the parabola being stretched vertically (the arbitrary constant $$c$$ can just be taken to be the height of the vertex above the bottom of the bucket). This will make the well deeper relative to the top of the parabola at the edge of the bucket!

• Are there any academic papers that have this information present? If you can suggest an academic paper, it would be great as I will have to cite the information in the research paper I am writing. Commented Jan 9, 2020 at 5:04
• How about the same analysis from the Pressure point of view? ∆P=(ρω^2 r^2)/2 This show that increasing the angular velocity increases the change in pressure. Does the increaasing pressure show that the fluid particles are futher forced down? Commented Jan 9, 2020 at 5:13
• Moreover, what exactly takes place in terms of physical changes as it changes its shape between the two angular velocities such that the fluid takes place of a new parabola with a greater depth? Commented Jan 9, 2020 at 5:17
• @AmanMehta Where did you come across that pressure equation? Commented Jan 9, 2020 at 11:00
• informit.com/articles/article.aspx?p=2832417&seqNum=7 this is the source @JamesWirth Commented Jan 10, 2020 at 5:20