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broad set of circumstances and are easy to implement and tune. Another
commonly used control system is a simple first order lag, x = kx, under
which the error decays exponentially. This is similar to a critically damped
second-order system, but with a bit jerkier response to a sudden change
in the set point. Another important and common technique is to “chase”
the set point at a fixed velocity. A filter is another broad class of control
system, in which the output is computed by taking some linear combination
of set points or values on previous frames.
12.3
Momentum
Let's say that Moe's box from Section 12.1.3 has a mass m, and at a certain
instant we observe it moving with a velocity v . Coming in late to the story,
we cannot tell what magnitude of forces were used to achieve that motion,
or how long the forces were applied, or what the history of the box's velocity
was. For example, it could have been that the box was accelerated as a
result of a constant net force f being applied over a duration ∆t. But we
have no way of knowing the values of f and ∆t. Was a large force used for
a small duration, or a small force for a longer duration? In fact, we have no
reason to assume that the force was constant at all! Moe could have given
the box a good shove and set it in motion, and then gave it another shove
to speed it up.
While we don't know the exact history of Moe's pushes, we do know
what the “total” was, in the sense about to be described. Assume that
Moe did make one push with a constant force f applied for a duration ∆t.
Then according to Newton's second law, the acceleration was a = f /m.
Assuming the initial velocity was zero, we know that
v = a ∆t.
Substituting a = f /m and rearranging, we get
v = ( f /m) ∆t,
m v = f ∆t.
Two ways to think of
momentum
(12.16)
The left- and right-hand sides of Equation (12.16) illustrate two different
ways of thinking about the important concept of momentum. Momentum
is the correct quantity to track in order to quantify the “total amount of
pushing.”
Let's do dimensional analysis on Equation (12.16), first just to verify
that it makes physical sense—it isn't intuitively obvious that these two
products would bear the same physical meaning—and also to see what the
 
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