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January 28th, 2017, 02:57 PM   #1
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Question centrifugal force on water in a glass

1. The problem statement, all variables and given/known data
There is a cyllindrical glass (r=0.05 m) full of water. It is rotating around its vertical geometric axis with 3 turns per second. How many centimetres higher is water at the rim than in the centre of the glass . Water rotates together with the glass.

2. Relevant equations
p=ρah
ω=2πγ
a=ω2r

3. The attempt at a solution
All I know is that the shape of water is parabolic. In the centre there is no additional pressure from rotation, the pressure on the glass is p=ρah, a=ω2r
total pressure on any point is px + py (vectorial sumation)
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January 29th, 2017, 03:28 AM   #2
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http://https://www.physicsforums.com...-glass.901899/
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January 29th, 2017, 03:37 AM   #3
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nobody really helped me there, so i asked here too
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January 29th, 2017, 03:47 AM   #4
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But you only posted 12 hours ago . But fair enough. Just whatever has been posted there may be of help to someone here.
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January 29th, 2017, 05:55 AM   #5
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Joppy, you do not know the answer?
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January 29th, 2017, 05:59 AM   #6
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Quote:
Originally Posted by srecko View Post
Joppy, you do not know the answer?
I do not.
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January 29th, 2017, 08:22 AM   #7
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Been a long time since I've dealt with fluid mechanics, but here is a go at it ...

Referencing the diagram, the red dot represents a fluid particle on the surface of mass $m$. The free body diagram to the right is a representation of the forces acting on that particle with the blue line representing a line tangent to the curve formed by the fluid surface.

fluid particle is in translational equilibrium ...

$F_c = mr \omega^2 = N\sin{\theta}$

$mg = N\cos{\theta}$

$\dfrac{N\sin{\theta}}{N\cos{\theta}} = \dfrac{mr\omega^2}{mg}$

$\tan{\theta} = \dfrac{r \omega^2}{g}$

since $r = x$ from the coordinate system ...

$\tan{\theta} = \dfrac{x \omega^2}{g}$

slope of the tangent line, $\tan{\theta} = \dfrac{dy}{dx}$ ...

$\dfrac{dy}{dx} = \dfrac{x \omega^2}{g} \implies y = \dfrac{x^2 \omega^2}{2g}$

sub in the radius of the cylinder for $x$ and $\omega = 6\pi \, rad/sec$ to determine $y$.
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January 29th, 2017, 08:31 AM   #8
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The shape of water is not a parabola then???
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January 29th, 2017, 08:55 AM   #9
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Quote:
Originally Posted by srecko View Post
The shape of water is not a parabola then???

yes ... look again at the equation for $y$

$y = kx^2$ where $k = \dfrac{\omega^2}{2g}$
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January 29th, 2017, 11:23 AM   #10
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ok, i got it. Thanks for your effort. This forum is really helpful
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