(Before scaring readers away, I’ll warn you that the math in the following sections makes it easier (NOT HARDER) to understand what’s going on. An equation is worth a thousand words, or something like that. The math will be nothing more than 9th grade algebraic substitution and rearrangement. Equations will also be followed by a colored section in parenthesis that briefly discusses its meaning and can be skipped if considered obvious.)
What follows is a simple calculation based on some pretty coarse approximations, which may be relaxed in future calculations. However, these simple approximations will provide quite a bit of insight about rigid pieces of Toast as they topple. The initial condition is drawn in Figure 5.1a, representing
Toast positioned with its center of mass (marked with an x) overhanging a Table Ledge by a small distance d. As discussed at length previously, the force of gravity will cause the Toast to rotate, slip, and fall from the Ledge.My first approximation is that the Toast cannot slide while on the Table Ledge. Thus, rotation occurs as drawn in Figure 5.1b with the Toast rotating around the line of contact between its bottom surface and the Table until the Toast is vertical (θ = 90°, Figure 5.1c), at which point it falls. Using this approximation, I have already constrained one of my parameters; θ0 must be 90° when the free fall starts.
In Figure 5.1a, the Toast overhangs the Ledge by a distance d. Because there is no sliding, the Toast’s center of mass has been lowered by d when it reaches the position in Figure 5.1c. Since d is much smaller than the length of the Toast (l), approximately all the gravitational energy (Eg) released by lowering the Toast’s center of mass by d is deposited in the Toast’s rotation (call this Erot). Mathematically,
(5.1a) Erot ≅ -Eg
(5.1b) = m g d
(Here, m is the mass of the Toast, and g is the acceleration due to gravity. The energy released in lowering the Toast increases with d, m, and the “strength” of gravity, g. Therefore, it makes sense that these three values are multiplied together in Equation 5.1. Don't worry too much about the negative sign; it just makes sure that decreasing Eg, as the Toast is lowered, becomes increasing Erot, as it spins faster.)
I use an approximately equal sign (≅) in Equation 5.1 because as the Toast topples (on the Ledge), rotation occurs along the line of contact between the Toast and the Ledge. Once it enters free fall, rotation occurs around an axis through the center of mass. Since the displacement between these two axes is small (≅d), I have ignored the detailed transformation between them and assumed all the energy becomes rotation about the proper freely rotating axis. This is another approximation to be addressed in detail another time.
It turns out that this approximation immediately determines another of my parameters. Since all the energy goes into rotation, none can be in translation and vv must be zero. Another parameter can be constrained using a little bit of math because Erot determines the speed of rotation (also called the angular velocity, ω) according to
(5.2) Erot = 1/2 I ω2 .
(I is the moment of inertia, which increases as the Toast becomes heavier or larger, and therefore harder to spin. It plays a similar role as mass does in linear [non rotating] motion, where the energy Elinear = 1/2 m v2 is isomorphic with Equation 5.2.)
By substituting the expression on the right side of Equation 5.1 in Equation 5.2 and solving for ω, we find the expression for the Toast’s angular velocity as it leaves the Table:
(5.3) ω = √(2m g d / I).
(It is helpful to have a physical picture for an equation. For Equation 5.3, it helps if you are familiar with how a quantity scales with the square root of a variable. Plotting a graph of the square root of x versus x can help. That is how ω increases with d and 1/I, i.e. it doubles when they go up by four, triples when they increase by nine, and so on. More importantly, Equation 5.3 reveals that the Toast’s rotation (ω) increases as we increase the terms related to the gravitational energy: m, g, and d. On the other hand, it decreases as I gets bigger because I is dividing the other terms. Such a decrease makes sense because bigger I means it is harder to spin the Toast.)
Now all we need to know is how far the Toast falls, h0. One likely height is that of a Table or countertop. A very low Table, such as a Coffee Table, is probably the lowest the Toast is likely to fall from. (Perhaps you sometimes put food on the Floor, in which case, no, I won't join you for dinner.) An upper limit of the initial height might be the height of a clumsy person’s mouth while standing. In this model, h0 is actually these heights MINUS d. However, d is very small and h0 is not very well defined, so I will ignore this very minor correction. In fact, I’ll keep h0 as a variable, so we can see how the results change as it is varied over the range just defined.
(On that note, I’m also neglecting another term: the change in h0 caused by the angle of the Toast bringing its edge in contact with the Floor before the center of mass. That is, when the Toast is parallel to the Floor (θ = 0° or 180° as in Figure 5.1a) when it lands, the entire Toast has fallen by h0. However, when the Toast has rotated 90° (as in Figure 5.1c), the edge that hits first has fallen h0 MINUS l/2, because that edge of the Toast is l/2 closer to the Floor when θ = 90°. Again, l/2 is much smaller than h0, so I’ll ignore this for now and include it explicitly later.)
HOW MUCH DOES THE TOAST ROTATE BEFORE HITTING THE GROUND?
We answer this question by substituting t = tf via Equation 4.3 ( tf = √ ( 2 h0)and Equation 4.4 (θ(t) = θ0 + ω t) from my last post, and then expressing ω according to Equation 5.3, above, to find
(5.4a) θ = θ0 + √(2m g d / I) √ ( 2 h0 / g)
(5.4b) = θ0 + 2 √(m d h0 / I) .
This tells us how far the Toast rotates after falling a distance h0. Before getting into the “numbers” of the result, we can learn more about the situation by examining how Equation 5.4 scales with various quantities. θ increases with the height of the fall, h0, and the amount the Toast initially hangs over the Ledge, d. Although there seems to be some mass dependence, this actually cancels because I is proportional to m, and the ratio of the two is a term that only depends on the shape of the Toast and other geometric factors to be worked out in detail some other time.
More important is what Equation 5.4 does NOT depend upon. Notice that there is NO dependence on g, the acceleration due to gravity. What does this mean? Well, g has a particular value on the surface of Earth, another one on the moon, and another one in an accelerating spaceship. It could take almost any value depending on where you are in the universe. Yet, tumbling Toast does not seem to care. If you drop your Toast while visiting Mars, whether it lands Buttered Side Down is determined by the size and position of the dropped Toast, according to Equation 5.4, and not on the value of g on Mars! Of course, it will fall more slowly on Mars, (because of smaller g) so you’ll have extra time to try to catch it. However, left to its own devices, the Toast will go through the same number of rotations as it falls off a Table on Mars as it will on Earth.
We have discovered something fundamental about the universe! Within the approximations I have made over my last few posts, Toast (or any rigid body) will fall and tumble based on its height, h0, distance over the edge, d, and mass distribution, m / I, without regard for the value of the acceleration constant, g.
Ultimately, what does Equation 5.4 actually predict for the number of rotations? Using typical values for height h0 = 0.75 meters (about 2.5 feet) and overhang d = 1 mm (0.04 inches). The moment of inertia for a 40 g piece of Toast that is shaped like a square with a length, l = 10 cm, and negligible thickness is I = m l2 / 12 = 3.3×10-5 kg m2. Via Equation 5.4, we find θ is 3.0 radians or 200°.
This angle is only a little above the perfect Buttered Side Down condition at 180°. The small difference is illustrated in Figure 5.2. The Toast position in this illustration reveals that impact with the Floor will result in a Buttered Carpet.
(A simple approximation for the Landing Toast is that θ values between 90° and 270° result in the Buttered Side facing Down.)
Of course, the calculated result depends quite a bit on these chosen conditions. For instance, I just mentioned that Equation 5.4 is valid on any planet, but who is to say that aliens have the same size bread as we do? Since we don't know what Alien Bread is like, I'll compare across cultures on Earth, where a 10 cm piece of square Toast could be replaced with a 15 cm round object with the same mass, representing a tortilla (in Latin America) or pita (in the Middle East). Do Toasted Pitas suffer the same Buttered Side Down fate as Toast, or is Western European culture to blame for Greased Kitchen Floors? The moment of inertia of a thin, 40 g, 15 cm diameter circular Toast-Like Object is 5.63x10-5 kg m2. With the same initial conditions as used for the Square Toast, θ = 174°, also landing squarely Buttered Side Down.
Of course, the exact starting conditions chosen above seem a bit arbitrary. If I had used d = 4 mm or h0 = 3 m, the Toast would have rotated twice as far and landed Buttered Side Up. 3 m is about 10 feet, considerably higher than most people position their breakfast, so we can ignore this case. However, the exact distance that the Toast overhangs the Table IS likely to vary quite a bit. Who is to say I will miss the Table by 1 mm? Perhaps I am clumsy enough to miss by 10 mm, resulting in more than 3 times as much rotation as I predicted above.
This trend is presented graphically in Figure 5.3. In this figure, I've plotted Equation 5.4a with the values of m and I that I determined above, but leaving the product of d and h0 as the independent coordinate. (the x-axis) (Here, the rotation angle increases with the square root of d h0, starting from θ0 = 90°. The y-axis wraps around at 360°, which is one full rotation, bringing the Toast back to the perfect Buttered Side Up condition, which is also θ = 0°.)Several calculated points are marked in Figure 5.3. The circle corresponds to a gentle tumble off a Coffee Table, d = 1 mm and h0 = 30 cm. The triangle is the calculation worked above, with d = 1 mm and h0 = 75 cm. The square data point corresponds to an exceptionally clumsy person who drops the Toast while standing, so d = 10 mm and h0 = 1.5 m, and the diamond is an extreme of this using d = 25 mm (out of a maximum of missing the table at d = 50 mm!) and h0 = 200 cm (6.5 feet above the Floor!). The Buttered Side Down region has been shaded pink, revealing that there are many initial conditions that lubricate your carpet. However, as d h0 increases, there are competing regions which do not.
Clearly, for small overhang (d) and low drop height, the non-slipping model predicts a Buttered Side Down landing. The universe really does hate you, and your Toast will land in the messiest possible position when it falls from a low position and only slightly unbalanced.
It is easy to seek limiting conditions where this prediction breaks down. For instance very Small Toast (such as a Buttered Cracker or Dipped Chip) will spin more easily, and a “low” falling height will be correspondingly lower. Thus, the scaling of Figure 5.3’s x-axis is much smaller and the Small Toast is likely to go through many rotations as it falls from a human-sized height. Since there are so many rotations on the way down, the number it goes through will be very sensitive to the exact h0, and the result will appear to be “random.” In the other extreme, Large Toast (perhaps a Rigid Pizza) will not really fall or topple as described. In the extreme case, the Large Toast has such a large length that l/2 > h0 and the Floor stops its fall before it even finishes toppling from the Ledge!
These trends will hold, even as I relax the approximations inherent to the non-slipping model. Thus, the propensity for Toast to land Buttered Side Down is entirely determined by the ratio of the size of our Toast to the height from which we drop it. That is to say, it depends on the size of our food relative to ourselves. More modest morsels (Small Toast) result in random landings, and Very Large Toast does more of a tip than a tumble from the Table.
Is there another factor, not included here, which cancels out the amount of overhang? My biggest simplification has been neglecting sliding. If the Toast were allowed to slide, the effective overhang, d, would be changing as it fell, and this might compensate for the exact starting conditions.
This is a postulated answer I will try to address (experimentally!) next time.
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