Unobstructed Toast soars through the air, requiring only a few parameters to describe its motion, as illustrated in the Figure. We start our stopwatch (t=0) when the Toast leaves the Table, noticing that the Toast may already be moving at this time. Based on my earlier discussion, any initial velocity was likely gained while tumbling off the table edge. It is useful to separate the velocity into two components: the vertical velocity vv, which is parallel the gravitational acceleration (g), and horizontal velocity vh.
The only thing vh determines is WHERE the Toast lands. Unless you drop your Toast while brunching on the side of a very steep hill, this does not influence the height or time of the fall, so it does not impact the Buttered Side Down condition. Thus, assuming you are not a mountain goat, ignore vh and just note that the Toast might land a few centimeters over from the Table Edge.
The other significant parameters are 1) the height (h0) above the Floor at t=0, 2) the uniform gravitational acceleration (g=9.8 m/s2 at Earth’s surface), 3) the angle of the Toast at t=0, θ0 (the angle it rotated through before leaving the Table) and 4) its rotation rate, ω. (Rotational speed is also known as angular velocity, which can be a bit tricky to think about if you haven't before. It helps to think of angular speed similar to how you think of regular speed. Instead of miles per hour or centimeters per second, it measures flips per second or angle per second.)
Now let’s find the equations that describe what happens. (If you are not used to interpreting equations, be sure to read the comments in blue!) The height of the Toast above the Floor as a function of time is
(4.1) h(t) = h0 - vv t - 1/2 g t2.
(Notice that at t=0, h(t) = h0, which we expected from the definition of h0. The second term, vv t, is just velocity times time. The last term takes care of the extra speed the toast gains as it falls and is accelerated by gravity, as discussed here. )
The Toast hits the Floor when h(t)=0. By setting the right hand side of Equation 4.1 to 0 and solving for t, we find the time of the fall (tf),
(4.2) tf = -[ vv - √ (vv2 + 2 h0 g) ] / g.
(This can be found using the quadratic formula and verified by substituting back into Equation 4.1. Of course, quadratic equations have two solutions, but the second solution produces a time from before the Toast fell off the Table (t<0). Clearly, Equation 4.1 was not valid before the Toast fell because the Table was supporting the Toast! Therefore, we throw this other mathematical solution away because it does not represent our physical situation.)
Equation 4.2 is exact but overly complicated because the initial velocity is very small. (The fact that vv starts off small was justified here, by assuming you have not hurled your toast energetically. I will also prove later that the amount it increases as the toast slides off the table is also quite small.) Neglecting vv (i.e., setting it to zero) in Equation 4.1 and solving for tf results in the much simpler expression
(4.3) tf = √ ( 2 h0 / g),
which we are much more likely to use later on. (When you drop an object, its time of flight only increases with the square root of distance because the longer it falls, the faster it goes at the end of the fall.)
Now that we know how long it takes for the Toast to fall, it is an easy matter to calculate how much it flips. The total flipping angle θ is
(4.4) θ(t) = θ0 + ω t.
(Notice again that at t=0, θ(0) = θ0, as we designed. Also notice the way θ(t) increases linearly with t, at a “speed” of ω. This will really drive home the physical meaning of ω!)
Equations 4.2 through 4.4 are the main results of this discussion, and we’ll refer back to them in later posts to predict the Buttered Side Down condition after approximating the Tumble from the Table.
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