Center of Mass & Centroids

Origin at the left end: $(2, 1)$, $(3, 4)$, $(5, 6)$.

$M = 2 + 3 + 5 = 10$ kg.

$2(1) + 3(4) + 5(6) = 44$.

$\bar x = \dfrac{44}{10} = 4.4$.

$(4, 0)$, $(6, 5)$, $(m, 8)$.

$M = 10 + m$.

$4(0) + 6(5) + 8m = 30 + 8m$.

$\dfrac{30 + 8m}{10 + m} = 4$, so $30 + 8m = 40 + 4m$ and $m = 2.5$.

$(1, -3)$, $(2, -1)$, $(4, 2)$; keep the minus signs.

$M = 1 + 2 + 4 = 7$.

$1(-3) + 2(-1) + 4(2) = 3$.

$\bar x = \dfrac{3}{7}$.

$\rho(x) = 1 + x$ on $[0, 4]$, denser on the right.

$M = \displaystyle\int_0^4 (1 + x)\,dx = 4 + 8 = 12$ kg.

$\displaystyle\int_0^4 (x + x^2)\,dx = 8 + \tfrac{64}{3} = \tfrac{88}{3}$.

$\bar x = \dfrac{88/3}{12} = \dfrac{22}{9} \approx 2.44$, right of the middle as expected.

$\rho(x) = e^{-x}$ on $[0, 2]$, denser on the left.

$M = \Big[-e^{-x}\Big]_0^2 = 1 - e^{-2}$.

By parts, $\displaystyle\int_0^2 xe^{-x}\,dx = \Big[-(x + 1)e^{-x}\Big]_0^2 = 1 - 3e^{-2}$.

$\bar x = \dfrac{1 - 3e^{-2}}{1 - e^{-2}} = \dfrac{e^2 - 3}{e^2 - 1} \approx 0.687$, left of the middle.

$\rho(x) = \dfrac{1}{x + 1}$ on $[0, 3]$.

$M = \Big[\log(x + 1)\Big]_0^3 = \log 4$.

$\dfrac{x}{x + 1} = 1 - \dfrac{1}{x + 1}$, so $\displaystyle\int_0^3 \dfrac{x}{x + 1}\,dx = 3 - \log 4$.

$\bar x = \dfrac{3 - \log 4}{\log 4} = \dfrac{3}{\log 4} - 1 \approx 1.16$.

They meet at $x = 0$ and $x = 1$; on $[0, 1]$, $T = x$ and $B = x^2$.

$A = \displaystyle\int_0^1 (x - x^2)\,dx = \tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6}$.

$\bar x = 6\displaystyle\int_0^1 (x^2 - x^3)\,dx = 6\left(\tfrac{1}{3} - \tfrac{1}{4}\right) = \tfrac{1}{2}$.

$\bar y = \dfrac{1}{2(1/6)}\displaystyle\int_0^1 (x^2 - x^4)\,dx = 3\left(\tfrac{1}{3} - \tfrac{1}{5}\right) = \tfrac{2}{5}$.

$T = 9 - x^2$, $B = 0$ on $[-3, 3]$. Symmetric about the $y$-axis.

$A = \displaystyle\int_{-3}^3 (9 - x^2)\,dx = 54 - 18 = 36$.

By symmetry, $\bar x = 0$: no integral needed.

$\bar y = \dfrac{1}{72}\displaystyle\int_{-3}^3 (9 - x^2)^2\,dx = \dfrac{1}{72}\cdot\dfrac{1296}{5} = \dfrac{18}{5}$.

$T = \sin x$, $B = 0$ on $[0, \pi]$. Symmetric about $x = \tfrac{\pi}{2}$.

$A = \Big[-\cos x\Big]_0^{\pi} = 2$.

By symmetry, $\bar x = \tfrac{\pi}{2}$.

$\bar y = \dfrac{1}{4}\displaystyle\int_0^{\pi}\sin^2 x\,dx = \dfrac{1}{4}\displaystyle\int_0^{\pi}\dfrac{1 - \cos 2x}{2}\,dx = \dfrac{1}{4}\cdot\dfrac{\pi}{2} = \dfrac{\pi}{8}$.

Density depends on $x$: vertical strips, each $2$ m tall.

Strip mass: $(1 + x)(2)\,dx$.

$M = \displaystyle\int_0^3 2(1 + x)\,dx = 15$ and moment $\displaystyle\int_0^3 2x(1 + x)\,dx = 9 + 18 = 27$.

$\bar x = \dfrac{27}{15} = \dfrac{9}{5}$. The density does not change with $y$, so $\bar y = 1$ by symmetry.

Density depends on $y$: horizontal strips. At height $y$ the strip has width $4 - y$.

Strip mass: $y(4 - y)\,dy$.

$M = \displaystyle\int_0^4 (4y - y^2)\,dy = \tfrac{32}{3}$ and moment $\displaystyle\int_0^4 (4y^2 - y^3)\,dy = \tfrac{64}{3}$.

$\bar y = \dfrac{64/3}{32/3} = 2$. Shape and density are symmetric about the $y$-axis, so $\bar x = 0$.

$m = 1000 \cdot \pi(1)^2(3) = 3000\pi$ kg.

Full cylinder: center of mass at $1.5$ m.

All the water ends at the top: $3$ m. So $\Delta h = 1.5$ m.

$W = 3000\pi(9.8)(1.5) = 44\,100\pi$, the same as slicing.

$m = 1000(2)(2)(1.5) = 6000$ kg.

Before: $5 + 0.75 = 5.75$ m.

$6$ m$^3$ over a $12$ m$^2$ base is $0.5$ m deep, so after: $0.25$ m. $\Delta h = 5.5$ m.

$W = 6000(9.8)(5.5) = 323\,400$.