Let us imagine some amount of mercury $\left(\text{Hg}\right)$ contained in a cylinder. It is being rotated about the central axis at a constant angular velocity $ω$, and the system has moment of inertia $I$.  I want to find an expression for its radius of curvature. Let us place the cross section of the system on the $\left(x,\,y\right)$-plane such that the $y$-axis passes through the center of the surface and the $x$-axis tangents the surface in the following way:$\hspace{200px}$.  Now, if we can express this curvature as $y=f\left(x\right)$, then we can say that$$ r=\left| \frac{\left[1+\left(f^{'}\left(x\right)\right)^2 \right]^{3/2}}{f^{''}\left(x\right)} \right| \,.$$   Now, if we can find $f\left(x\right)$ in terms of the $I$ and/or $ω$ and any physical quantity like acceleration due to gravity, we would replace it in the previous equation and find $r$.   Please help me to proceed.  EDIT:  I read the page tagged. My Question is how did the get $y= \dfrac{(x^2 \cdot \omega^2)}{2g}$?
I was going through fluid dynamics and I came upon this section on the shape of a liquid surface in a rotating cylindrical vessel. The working in the book is something like this:   It considers a mass $m$ of water on the surface of the rotating liquid at a distance $x$ from its axis of rotation .It takes into account a pseudo force ($m \cdot x \cdot \omega^2$) radially outwards and its weight force ($mg$).It then proceeds to calculate the net force and the angle it makes with the vertical.From there through basic integration it gets the equation of the surface of liquid  i.e. $y= \dfrac{(x^2 \cdot \omega^2)}{2g}$ .  I get this working.  But if I were to view the same rotation through an inertial frame,how would I proceed? I mean, there wouldn't be any pseudo force, rather a centripetal acceleration of same magnitude radially inwards.This is where I am facing certain doubts.  The forces on the liquid would be  the centripetal force inwards and the weight force downwards.The net force would therefore be in a different direction,and consequently the slope of the surface at that point would be in a different direction.How is it possible for the same surface to have two different shapes in two different frames?   I would be obliged if someone could help me out by pointing out the flaw in my understanding.
How can I delete a (null) app in usage? In my usage list(settings>general>storage &amp; iCloud usage>manage storage), there is an app that registers as (null) with no icon, a version number that doesn't match anything I've previously downloaded, and it's taking up storage space. I first noticed it about a week or two ago. And no, I can't download recent apps to find the one that it is, because I downloaded multiple apps recently, and didn't notice the (null) app until my storage was critically low and I was looking to delete something. I am running iOS 9.2.1 on an iPhone 5C. This is question 2.9 #9 from Topics in Algebra by Herstein:     If $o(G)$ is $pq$ where $p$ and $q$ are distinct prime numbers and if $G$ has a normal subgroup of order $p$ and a normal subgroup of order $q$, prove that $G$ is cyclic.   Let $N$ denote the normal subgroup of order $p$ and $M$ denote the normal subgroup of order $q$. 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