Centrifugal force on water in a glass

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Homework Help Overview

The problem involves a cylindrical glass filled with water that is rotating around its vertical axis. The objective is to determine the height difference of the water level between the rim and the center of the glass due to the effects of centrifugal force. The discussion revolves around the dynamics of fluids in a rotating reference frame.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the parabolic shape of the water surface and the implications of rotational motion on pressure distribution. There are attempts to derive relationships involving the height difference and the forces acting on a small parcel of water. Some participants express uncertainty about specific concepts, such as free body diagrams and the parameters involved in the equations.

Discussion Status

The discussion includes various attempts to formulate the problem mathematically, with some participants providing equations relating to the slope of the water surface and its height. There is a mix of understanding and confusion regarding the application of free body diagrams and the physical principles at play. While some calculations have been presented, there is no explicit consensus on the final answer, and the discussion remains open for further exploration.

Contextual Notes

Some participants express uncertainty about the definitions and applications of certain concepts, such as free body diagrams, which may affect their ability to engage fully with the problem. Additionally, there are variations in the values used for gravitational acceleration, indicating a need for clarification on assumptions.

srecko97
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Homework Statement


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.

Homework Equations


p=ρah
ω=2πγ
a=ω2r

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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srecko97 said:
All I know is that the shape of water is parabolic.

That's right. If we let the vertex be at (0,0) then you just need to find the Y value when X=0.05m.
Start by drawing a free body diagram for a piece of water on the surface.
 
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probably is y=Cx2 ... but I do not know what is C, maybe a/g ... a=17.8 m/s^2, g=10 m/s^2 i do not know, honestly just guessing, as i do not know how to solve it
 
billy_joule said:
Start by drawing a free body diagram for a piece of water on the surface.

Post your FBD and we'll see where you've gone wrong.
 
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I do not know what free body diagram is. Sorry. Please explain to me and I will try
 
srecko97 said:
I do not know what free body diagram is. Sorry. Please explain to me and I will try
Consider a small parcel of water at the surface, at some radius x. If the surface shape is described by y=f(x), what is the slope there? Draw a diagram of the parcel of water and the forces acting on it. Treat the rest of the water as a frictionless solid.
 
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20170129_121122.jpg
 
srecko97 said:
20170129_121122.jpg
Ok. You left out a normal force, and you need to answer this question
haruspex said:
If the surface shape is described by y=f(x), what is the slope there?
I.e., in terms of x and y.
 
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tanφ =ω2r / g
 
  • #10
and dy/dx = ω2r / g
so: y= ω2r x /g ... for r=x I get y= ω2x2 /2g
 
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  • #11
and I have y if i put r=x!
 
  • #12
srecko97 said:
and I have y if i put r=x!
Right. So what answer do you get?
 
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  • #13
4,4cm
 
  • #14
srecko97 said:
4,4cm
Using g=10m/s2, yes, but 4,5cm would be a bit closer.
Well done.
 
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