# Finding the radius of curvature of mercury surface in a rotating cylinder [duplicate]

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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 $y= \dfrac{(x^2 \cdot \omega^2)}{2g}$?

## marked as duplicate by John Rennie newtonian-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 18 '18 at 15:53

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## 2 Answers

The answer is pretty easy if you choose a non-inertial frame.

If you choose a reference frame such that it is rotating at the same angular speed as the fluid, then this observer sees the fluid at rest, so the observer can apply the hidrostatic equation:

$P+\rho\cdot U= const$, with P=pressure, $\rho$=density, $\phi$=field potential.

The thing is that, in this case, the potential is not only the gravitational one: you must add the centrifugal force.

$F_c=-\omega^2 R$

which is conservative, and its potential is $-\frac{1}{2} \omega^2 R^2$.

So your equation is

$$P + \rho gz + \frac{1}{2}\rho \omega^2 R = const.$$

You'll find that the shape is parabolic.

• I have edited my Q, please see the edit. – SeaDog May 18 '18 at 17:05
• Okay, I'll add one mroe clue: the top surface is all at the same pressure (the atmospehric one). You can use that to find the constant. Once you do that, you only have to rearrange the terms, you'll get $z(R)$. You justh ave to work it out. – FGSUZ May 18 '18 at 18:14

I don't have enough reputation otherwise I would have added this as a comment.

You need to find the shape of the liquid. So simply write out the forces acting on a liquid particle on the surface. You want a differential equation so ask yourself: what is dy/dx (for any curve)? You will then find out with a bit of geometry that you can directly link dy/dx to the forces that you wrote down. Integrate and you get a nice parabola equation

• Hi, weclome to Physics! I think this is perfectly fine as a (partial) answer rather than as a comment, especially since we discourage complete answers to homework-like questions. Looking forward to seeing more from you. – rob May 18 '18 at 16:45
• I have edited my Q, please see the edit. – SeaDog May 18 '18 at 17:05
• By doing what I wrote above. Have you tried any of that? If you don't show us what you've tried so far we can't really help you – physicss May 18 '18 at 17:52