Tipping a container of water

  • #1
kbannister
17
1
Homework Statement
A weightless cylindrical tank of diameter D and height H rests on a nonslip surface on a flatcar going around a circular track of radius R. The tank contains water to depth h with a free surface, and we assume h < H such that water won't slop over the tank's edge. The flatcar's speed is gradually increased to V. Two possibilities exist: 1. Only part of the water occupies the "ungula" (the heel-shaped volume under the sloping free surface, and 2. All the water ends up in the Ungula, exposing some of the tank's floor. At what value of V will the tank begin to tip over?
Relevant Equations
Radial acceleration, a = V^2/R;
Angle of free surface, alpha = arctan(a/g), where g = acceleration of gravity
Not clear how to proceed. Does the cylindrical surface of the ungula need to be considered?
 
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  • #2
kbannister said:
Does the cylindrical surface of the ungula need to be considered?
Yes.
What attribute of the water volume do you need to determine in order to decide whether it will tip?
 
  • #3
haruspex said:
Yes.
What attribute of the water volume do you need to determine in order to decide whether it will tip?
Assume the water has a weight density of gamma.
 
  • #4
The question concerns determining the speed, V, around the circular track of radius R. A simpler version of the problem is the situation when the container is accelerated in a straight line at acceleration a.
 
  • #5
kbannister said:
Assume the water has a weight density of gamma.
No, that's an attribute of the water itself. I asked for an attribute of the volume (shape) the water occupies.
 
  • #6
haruspex said:
Yes.
What attribute of the water volume do you need to determine in order to decide whether it will tip?
The tank is cylindrical, diameter D, height H, water depth h where h < H so that the water won’t spill over.
 
  • #7
Will a diagram help? Does it need anything added?

1705270178121.png
 
  • #8
kbannister said:
The tank is cylindrical, diameter D, height H, water depth h where h < H so that the water won’t spill over.
For any object of given weight standing (not slipping) on a horizontal surface and subject to a sideways force of given magnitude, what determines whether it will tip over?
 
Last edited:
  • #9
kbannister said:
A simpler version of the problem is the situation when the container is accelerated in a straight line at acceleration a.
What (and how much of it) would make the container tip over in that simpler version?
 

1. What happens when you tip a container of water?

When you tip a container of water, the water inside will pour out due to the force of gravity. The water will flow in the direction of the tilt until the container is empty or until it reaches a surface that stops its flow.

2. Why does the water pour out when the container is tipped?

The water pours out when the container is tipped because of gravity. Gravity pulls the water downward, causing it to flow out of the container in the direction of the tilt.

3. Can you control the speed at which the water pours out when tipping a container?

Yes, you can control the speed at which the water pours out when tipping a container by adjusting the angle at which you tilt the container. A steeper tilt will result in faster pouring, while a gentler tilt will slow down the flow.

4. What factors can affect the way water pours out of a tipped container?

Factors that can affect the way water pours out of a tipped container include the size and shape of the container, the viscosity of the water, the angle at which the container is tilted, and any obstacles that may impede the flow of water.

5. Is there a limit to how much water can pour out when tipping a container?

There is no inherent limit to how much water can pour out when tipping a container. The amount of water that pours out will depend on the size of the container and the angle at which it is tilted. However, once the container is empty or reaches a surface that stops the flow, the pouring will naturally come to a halt.

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