Fizyx for Felines: A Physics Textbook for the Curious Cat

Chapter 2 – Excerpt 7

Posted in Uncategorized by skonabrittain on 7 May, 2010

Yet another short excerpt, including a physics approach to analyzing catfood.

A Highly Recommended Lab Exercise

We rarely even suggest very involved laboratory experiments, as we generally prefer the thought experiments of theoreticians to the actual ones of experimentalists, but the following has significant redeeming value. We think that by performing it, you have a lot to gain, dietetically as well as pedagogically.

It is a well-known fact that the more expensive the canned catfood, the moister the contents. In your previous lives, it was easy to evaluate quality by price. But now that price tags have been replaced by bar codes, how can you determine the moisture content without opening a can? This experiment will enable you to rank cans of catfood by their levels of moisture.

Lab Exercise – Predicting Catfood Liquidity

Gather several different kinds of canned catfood and set up an inclined plane. For the plane, we particularly recommend a rectangular cat scratching board. If you don’t already own one, tell your person you need one for a scientific experiment. The kind with embedded catnip is particularly well-suited to our purpose, as will be explained below. (And afterwards, you can scratch it and enjoy the embedded catnip.)

Conduct individual trial runs as follows. Place a can at the top of the inclined plane on its side and let go. Be careful to not add any external force by pushing it, so that its initial speed is zero. Using a stopwatch, time how long it takes for it to reach the bottom of the plane.

Note that the longer the runs, the easier they are to time and obtain statistically significant timing differences. The length of time can be increased two ways: increasing the distance, i.e. the length of the board; or decreasing the speed of the rolls. In turn, the speed can be decreased two ways: decreasing the angle of elevation; or increasing the coefficient of friction. The best way to achieve a high coefficient of friction is to use a board with a rough texture. That’s why we recommended a cat scratching board, and the embedded catnip may add even more texture. If necessary to slow the motion down further, just adjust the angle by lowering the top of the board.

Now that you know which can rolls down the fastest and which the slowest, you need to interpret your results to determine which is best. Note that the more liquid the contents are, the more they will slosh around within the can, rather than just rolling along with it. Thus you can determine, from their relative speeds, which can contains the moistest, hence most delectable, treats.

Hint: If you are having trouble analyzing the data theoretically, add a can of evaporated milk to the experiment. This is an example of a general technique in both mathematics and theoretical physics, as well as in experimental physics: It is often useful to consider extreme values, or boundary cases, whenever deriving principles.

After all that work, you deserve to eat at least one can of catfood. Select the one that you have experimentally determined to be the best-tasting one and ask your person to open it for you. For extra credit, convince your person to open all of the cans so that you can verify all of your results.

If your person is hesitant to open many cans, here’s some extra ammunition. Say that you are also studying the can opener and wish to observe how the torque applied to the rotating handle leads to the rotation of the can in a different plane. Many instances of observation will be required to fully comprehend this phenomenon, even for your person, who will, we hope, become intrigued enough by the can opener’s operation to repeatedly engage it. If so, be sure to give credit to its inventor, William W. Lyman of Meriden, CT, who patented it in 1870.


Figure 2-26. Cartoon by Randy Glasbergen © 2005


Rotational Energy

The kinetic energy of rotation for an extended object is found by adding up the kinetic energy amounts of the individual pieces of rotating mass Δm. From Chapter 1, we know that they each contribute an amount of ½(Δm)v2 to the sum. Applying equations (2) and (5) yields the rotational analog of this expression:
KE = Σ½(Δm)v2 = Σ½(Δm)(rω)2 = ½[Σ(Δm)r22 = ½Iω2

Motion can be separated into its translational and rotational components, with each part analyzed separately, and then the results combined to produce the actual motion. For example, consider the motion of a spinning ping pong ball: the center of mass moves in a parabola while the body rotates around the center of mass. Thus the total kinetic energy is the translational kinetic energy of the center of mass plus the rotational kinetic energy around the center of mass, i.e.
KEtotal = KEtrans + KErot = ½mv2 + ½Iω2

Energy considerations provide the simplest way to calculate the speeds obtained by rolling objects, just as they did for purely translational motion. At the end of chapter 1, we used this approach to calculate the speed of a freely falling object when it hit the ground, which was, of course, independent of its mass: v = (2gh)½, where h was the height where it started at rest. Now consider a rounded solid object, such as a sphere or a cylinder, rolling down a ramp of height h. Assume its initial speed at the top, is zero and that it rolls without slipping down the ramp.

Since only shape determines the rotational inertia per mass, all solids of a particular shape, regardless of their size or mass (as long as their density is uniform) will roll down with the same acceleration due to gravity, and hence the same speeds. This is analogous to the fact that all objects undergo free fall with the same acceleration due to gravity. However, it may be surprising because the rotational inertia does depend on size, being proportional to the square of the radius, and hence is greater for larger objects. It turns out that this leads to differences in angular speed, but not linear speed, as the following energy calculation shows.

For uniform solid cylinders, the rotational inertia is I = ½MR2,

so the rotational kinetic energy is KErot = ½Iω2 = ½(½MR22 = ¼M(RW)2 = ¼mv2

and hence the total kinetic energy is KEtotal = KEtrans + KErot = ½mv2 + ¼mv2 = ¾mv2.

At the top of the ramp, all the energy is potential energy and when the bottom is reached it’s all kinetic. So, relying on conservation of energy, we set our expression for the final kinetic energy equal to the initial potential energy and then solve for v: mgh = ¾mv2 → v = ((4/3)gh)½.

Suppose instead that the cylinder were hollow and very thin. To make the calculation much simpler, let’s assume it has neither top nor bottom and hence more symmetry, such as a catfood can with both the top and bottom cut off. Then it’s basically a thin hoop, with I = MR2 and we have

KErot = ½Iω2 = ½(MR2)ω2 = ½M(Rω)2 = ½ mv2
KEtotal = KEtrans + KErot =½mv2 + ½mv2 = mv2
and mgh = mv2 → v = (gh)½.

All the cases in the lab experiment suggested above are between these two extreme cases of a solid cylinder and an empty cylinder, with the cheaper brands of catfood being closer to the former case than the moister brands.

Q. A uniform solid ball rolls down an inclined plane of height h, starting from rest, without slipping. What is its linear speed the moment it touches the bottom?
A. KE = KEtrans + KErot = ½mv2 + ½Iω2 = ½mv2 +½((2/5)MR2)ω2 = ½mv2 + 1/5 M(Rω)2 = ½mv2 + 1/5 mv2 = 7/10 mv2.
So PEinitial = KEfinal → mgh = (7/10)mvfinal2 → vfinal = ((10/7)gh)½.

Note that all the speeds being obtained are a fraction of the final speed for free falls from the same height. They’re slower because some of the energy is being taken up by the spinning. And although the linear speeds are independent of size, the smaller objects are spinning faster when they reach the bottom, since v = rω.

Furthermore, these results are also independent of the angle of inclination of the ramp. For a shallower inclined plane, the trip will just take longer. The same final linear speed is gained more slowly, due to the smaller component of gravitational force directed down the ramp.


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