*dp/dr=ρ**w*^{2}*r (1)*

Where *ρ *is the fluid density,* **w* is the angular velocity and *r* is the distance from the center of rotation.

This is easily demonstrated by spinning a cup of water and showing the paraboloid surface that forms with the low point at the center of pressure. However, the pressure gradient exists regardless of the free surface and, just as with the non-rotating hydrostatic pressure variation, an immersed object will experience a net force in toward the region of lower pressure. This can be demonstrated using a sealed container

**Equipment**

- A small cork
- A marble
- A Lazy Susan or some other cheap turntable
- A mason jar
- Some Velcro strips or something else to secure the Mason jar to the turntable.

**Demonstration**

- Place the cork and marble in the jar and fill it with water
- Seal the jar so that there are no bubbles (or at least no bubbles that are large compared to the size of the cork) and attach it to the turntable so that its long axis is horizontal and it is centered on the turntable (see figure above).
- Shake the jar until the marble and the cork are near the center of the jar (this is so that when the marble moves it is clearly due to the rotation of the jar).
- Rapidly spin the turntable. The marble should be pushed to one end of the jar while the cork should remain centered.

**Analysis**

The horizontal hydrostatic pressure gradient (equation above) means that any submerged object will experience a net pressure force acting toward the center of rotation. For a rectangular object of width *s *in the radial direction and area *A *normal to the radial direction located a distance *r * from the center of rotation, the net pressure force toward the center of rotation is given by

*F _{p}=ρ*

See figure below.

Expanding leads to

*F _{p}=ρAw^{2}rs= ρ∀w^{2}r (3)*

where *∀* is the volume of the object. Therefore, the net force toward the center of rotation is the angular acceleration multiplied by the mass of water displaced by the object. This is directly analogous to the buoyancy force in a stationary fluid. Therefore, if the object displacing the fluid has a lower density than the fluid then the centripetal pressure force will exceed that needed to maintain the angular velocity of the object and it will be pushed toward the center of rotation (as in the cork). If the object is denser than the fluid the centripetal pressure force will not be enough to maintain rotation and the object will move radially outward (note that this is a somewhat simplified linearized analysis and the integration constant is left out of the pressure equation used in (2) but it gets the appropriate result).

An index of all the demonstrations posted on this blog can be found here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (nbkaye@clemson.edu). I also welcome comments (through the comments section or via email) on improving the demonstrations.

Filed under: angular momentum, buoyancy, Density, Hydrostatic pressure, Hydrostatics ]]>

Dropping the coins into the cup showing that 9 quarters displaces 50 ml of water.

Dumping the coins out showing that, when submerged, they displace a lot less than 50 ml.

An index of all the demonstrations posted on this blog can be found here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (nbkaye@clemson.edu). I also welcome comments (through the comments section or via email) on improving the demonstrations.

Filed under: buoyancy, Hydrostatic pressure, Hydrostatics ]]>

This is a simple scaled down version that clearly demonstrates the puzzle result but can also be used to weigh the heavy object (in this case coins).

**Equipment**

- Small plastic cup
- measuring cup
- syringe (for small adjustments in the volume of water in the measuring cup
- a hand full of identical coins (I used 9 US quarters).

**Demonstration**

- fill the measuring cup so that the water level is one line below the top volume measurement when the empty cup is floating in the water
- drop the coins in one by one until the water level rises up to the top line (record the number of coins needed)
- pour the coins into the water and put the plastic cup back on the water.
- note that the water level is below what it was when the coins were floating in the cup.

**Analysis**

The puzzle answer is fairly straightforward. However,the demonstration can also be used to estimate the weight of the individual coins (hence the need to use all the same coins). The change in volume recorded as the coins are added is equal to the volume displaced by the coins (see figure below). Therefore, the volume change multiplied by the density of water will equal the mass of the coins added. I measured a 50 ml change in volume when I added 9 quarters. The density of water is approximately 1 g/ml so the 9 coins have a mass of 50 g. The weight of the individual quarters is, therefore 50/9=5.56 g (which is very close to the actual standard mass of a quarter of 5.67 g). It is also possible to work out the volume of the coins by measuring the volume they displace when they sink (see figure). My measuring cup did not have enough resolution to get a good volume. The total volume of the 9 coins would be 7.5 ml based on US Mint dimensions.

Figure showing (from left to right) the cup displacing the coins weight, the empty cup, and the coins displacing their mass with the water level lower than when the coins are floating in the cup.

The other great thing about the demonstration is that it is easy to ask a lot of questions around the demonstration. Obviously you can pose the original question about the water level going up or down. There are also simple calculations that can be asked such as “what is the mass (or weight) of the coins?”, “what is the approximate density of the coins?”. The US mint website has information about size and metal content that can be used to calculate the actual density for comparison.

Ii is also really easy to have this as an activity in class. I used a version of it on the last day of class one semester using pennies.I used small and large plastic cups and gave the students the dimensions of the small cup and the equation for the cup volume (here). In this case you count how many pennies it takes to sink the cup so that the mass of pennies matches the density of water multiplied by the cup volume. You will also need a roll of paper towel for the clean up. I did not explain the demonstration I just gave them the equipment and told the to calculate the weight of a penny. It was well received by the students though I had to give some groups a hint or two as to how it worked.

An index of all the demonstrations posted on this blog can be found here. Don’t forget to follow @nbkaye on twitter for updates to this blog. If you have a demonstration that you use in class that you would like to share on this blog please email me (nbkaye@clemson.edu). I also welcome comments (through the comments section or via email) on improving the demonstrations.

Filed under: buoyancy, Hydrostatic pressure, Hydrostatics ]]>

Filed under: Energy equation, head loss in pipes, Pump performance, pumps ]]>

Cup partially full with ball drawn to the edge of the cup

Cup over full with ball drawn tot he middle of the cup.

Filed under: surface energy, Surface tension, surfactant ]]>