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quarto/_freeze/integrals/center_of_mass/execute-results/html.json

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+    "markdown": "# Center of Mass\n\n\n\nThis section uses these add-on packages:\n\n``` {.julia .cell-code}\nusing CalculusWithJulia\nusing Plots\nusing Roots\nusing QuadGK\nusing SymPy\n```\n\n\n\n\n---\n\n![ A silhouette of two children on a seesaw. The seesaw can be balanced only if the distance from the central point for each child reflects their relative weights, or masses, through the formula $d_1m_1 = d_2 m_2$. This means if the two children weigh the same the balance will tip in favor of the child farther away, and if both are the same distance, the balance will tip in favor of the heavier. ](../integrals/figures/seesaw.png)\n\n\n\nThe game of seesaw is one where children earn an early appreciation for the effects of distance and relative weight. For children with equal weights, the seesaw will balance if they sit an equal distance from the center (on opposite sides, of course). However, with unequal weights that isn't the case. If one child weighs twice as much, the other must sit twice as far.\n\n\nThe key relationship is that $d_1 m_1 = d_2 m_2$. This come from physics, where the moment about a point is defined by the mass times the distance. This balance relationship says the overall moment balances out.  When this is the case, then the *center of mass* is at the fulcrum point, so there is no impetus to move.\n\n\nThe [center](http://en.wikipedia.org/wiki/Center_of_mass) of mass is an old concept that often allows a possibly complicated relationship involving weights to be reduced to a single point. The seesaw is an example: if the center of mass is at the fulcrum the seesaw can balance.\n\n\nIn general, we use position of the mass, rather than use distance from some fixed fulcrum. With this, the center of mass  for a finite set of point masses distributed on the real line, is defined by:\n\n\n\n$$\n\\bar{\\text{cm}} = \\frac{m_1 x_1 + m_2 x_2 + \\cdots + m_n x_n}{m_1 + m_2 + \\cdots + m_n}.\n$$\n\n\nWriting $w_i = m_i / (m_1 + m_2 + \\cdots + m_n)$, we get the center of mass is just a weighted sum: $w_1 x_1 + \\cdots + w_n x_n$, where the $w_i$ are the relative weights.\n\n\nWith some rearrangment, we can see that the center of mass satisfies the equation:\n\n\n\n$$\nw_1 \\cdot (x_1 - \\bar{\\text{cm}}) + w_2 \\cdot (x_2 - \\bar{\\text{cm}}) + \\cdots + w_n \\cdot (x_n - \\bar{\\text{cm}}) = 0.\n$$\n\n\nThe center of mass is a balance of the weighted signed distances. This property of the center of mass being a balancing point makes it of intrinsic interest and can be - in the case of sufficient symmetry - easy to find.\n\n\n##### Example\n\n\nA set of weights sits on a dumbbell rack. They are spaced 1 foot apart starting with the 5, then the 10-, 15-, 25-, and 35-pound weights. Where is the center of mass?\n\n\nWe begin by letting $m_1=5$, $m_2=10$, $m_3=15$, $m_4=25$ and $m_5=35$. Our positions will be labeled $x_i = i-1$, so the five-pound weight is at position $0$ and the $35$-pound one at $4$. The center of mass is then given by:\n\n\n\n$$\n\\frac{5\\cdot 0 + 10\\cdot 1 + 15 \\cdot 2 + 25 \\cdot 3 + 35\\cdot 4}{5 + 10 + 15 + 25 + 35} = \\frac{255}{90} = 2.833\n$$\n\n\nIf the $25$-pound weight is removed, how does the center of mass shift?\n\n\nWe could add the terms again, or just reduce from our sum:\n\n\n\n$$\n\\frac{255 - 25\\cdot 3}{90 - 25} = \\frac{180}{65} = 2.769...\n$$\n\n\nThe center of mass shifts slightly, but since the removed weight was already close to the center of mass, the movement wasn't much.\n\n\n## Center of mass of figures\n\n\nConsider now a more general problem, the center of mass of a solid figure. We will restrict our attention to figures that can be represented by functions in the $x-y$ plane which are two dimensional. For example, consider the region in the plane bounded by the $x$ axis and the function $1 - \\lvert x \\rvert$. This is triangle with vertices $(-1,0)$, $(0,1)$, and $(1,0)$.\n\n\nThis graph shows that the figure is symmetric:\n\n::: {.cell hold='true' execution_count=5}\n``` {.julia .cell-code}\nf(x) = 1 - abs(x)\na, b = -1.5, 1.5\nplot(f, a, b)\nplot!(zero, a, b)\n```\n\n::: {.cell-output .cell-output-display execution_count=6}\n![](center_of_mass_files/figure-html/cell-6-output-1.svg){}\n:::\n:::\n\n\nAs the center of mass should be a balancing value, we would guess intuitively that the center of mass in the $x$ direction will be $x=0$.\n\n\nBut what should the center of mass formula be?\n\n\nAs with many formulas that will end up involving a derived integral, we start with a sum approximation. If the region is described as the area under the graph of $f(x)$ between $a$ and $b$, then we can form a Riemann sum approximation, that is a choice of $a = x_0 < x_1 < x_2 \\cdots < x_n = b$ and points $c_1$, $\\dots$, $c_n$. If all the rectangles are made up of a material of uniform density, say $\\rho$, then the mass of each rectangle will be the area times $\\rho$, or $\\rho f(c_i) \\cdot (x_i - x_{i-1})$, for $i = 1, \\dots , n$.\n\n::: {.cell holLine truncated
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