January 28th, 2017, 02:57 PM  #1 
Member Joined: Oct 2016 From: Slovenia, Europe Posts: 52 Thanks: 5  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) 
January 29th, 2017, 03:28 AM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,833 Thanks: 649 Math Focus: Yet to find out.  
January 29th, 2017, 03:37 AM  #3 
Member Joined: Oct 2016 From: Slovenia, Europe Posts: 52 Thanks: 5 
nobody really helped me there, so i asked here too

January 29th, 2017, 03:47 AM  #4 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,833 Thanks: 649 Math Focus: Yet to find out. 
But you only posted 12 hours ago . But fair enough. Just whatever has been posted there may be of help to someone here.

January 29th, 2017, 05:55 AM  #5 
Member Joined: Oct 2016 From: Slovenia, Europe Posts: 52 Thanks: 5 
Joppy, you do not know the answer?

January 29th, 2017, 05:59 AM  #6 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,833 Thanks: 649 Math Focus: Yet to find out.  
January 29th, 2017, 08:22 AM  #7 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587  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 $y$. 
January 29th, 2017, 08:31 AM  #8 
Member Joined: Oct 2016 From: Slovenia, Europe Posts: 52 Thanks: 5  The shape of water is not a parabola then??? 
January 29th, 2017, 08:55 AM  #9 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587  
January 29th, 2017, 11:23 AM  #10 
Member Joined: Oct 2016 From: Slovenia, Europe Posts: 52 Thanks: 5 
ok, i got it. Thanks for your effort. This forum is really helpful


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centrifugal, force, glass, water 
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