May 30th, 2010

Introducing the new series of ‘Question of the Week’ (QotW) where I pose physics and math questions to myself and others. These QotW are meant to explore concepts and generally have a little fun. I am no expert in these so it may very well be that my solutions are wrong and I invite criticism and scrutiny.

**QotW #1**

When I clean dishes (I do!) and encounter a dirty jam jar, milk bottle or similar container I occasionally consider how much water to put in the container before shaking it. The idea being that the more water (mass) I have inside the more force the water will hit the dirty corners with. The catch is that the more water I put in the jar the less distance it will have to accelerate when I shake the jar and hit the other end. So the question is how much of the container should be filled with water to maximize the force with which the water hits the other side of the jar.

An assumption:

**My Solution**

I am assuming that since force (**F**) is what dislodges the crusties in the jar and **F=ma**. Therefore the greater the mass and (de)acceleration will cause the greatest force. Calculating the deceleration of a liquid is pretty difficult, but since I know that the greater deceleration will be caused by the greatest velocity I can simply say that I am trying to maximize mass and velocity, or momentum (**M=mv**).

To generalize I will ignore the shape of the container (it could be a cylinder, rectangle) and simply say the mass is the length of the container containing water:

**m = h-x**

where:

**m** is the mass of the liquid (water),

**h** is the height of the container, and

**x** is the height of the emptiness in the container.

Calculating the velocity is based on gravity and the distance, **x**, the liquid will travel. Consider the (reduced) Newtonian mechanics equation:

**v² = 2ad**

where:

**v** is the velocity at impact with the bottom of the jar,

**a** is the acceleration from when the water starts falling, and

**d** is the distance the water falls.

Since we know the acceleration is gravity in this case and the abstract value of the distance is **x** we can modify the equation to:

**v = (2gx)^½**

We now have two equations showing the relationship between the jar proportion of water and its velocity and mass. We are trying to maximize momentum (**M = mv**) so we combine the equations accordingly and get:

**M = (h-x)(2gx)^½**

In order to find the maximum we need to find the derivative of the function. Using the product and chain rules we get:

**M’ = -1((2gx)^½) + (h-x)(1/2)((2gx)^-½)*(2g))**

Cleaned up:

**M’ = g(h-x)/((2gx)^½) – (2gx)^½**

Since we want to maximize we make M’ = 0 and solve for x:

**x = h/3**

**Conclusion**

There we have it. In order to maximize the impact we should leave a third of the bottle/container empty. Also note that the **g** variable was cancelled out. This has important implications. This means that regardless of the acceleration with which the water moves towards the other end of the bottle this proportion will always be the most effective. So we have more than solved the original scenario of simply letting the water volume fall. This means when I shake the bottle hard back and forth I should still fill the container two-thirds full.

September 15th, 2010 at 04:29

I love googling.

I also love that you have similar obsessions about such things, and that you not only devote time to thinking about how to do it better, but to blog about it! Tell me your secret.

p.s. I fill half way. I figure I’ve got my bases covered.