밀도·질량과 질량중심

$(0,3)$

$\left(0,\frac{4}{\pi }\right)$

$\left(\frac{5}{8},\frac{1}{3}\right)$

$\left(\frac{4}{3\pi },\frac{4}{3\pi }\right)$

$\left(\frac{1}{2},\frac{2}{5}\right)$

$\left(0,\frac{28}{9\pi }\right)$

(a) To find mass we slice the rod horizontally into uniform slices of length $\Delta x$ and approximate the density on the $i$th slice with the constant density $\rho(x_i)$. Then the approximate mass of the $i$th slice is $\rho(x_i) \Delta x$. Adding the approximations and taking the limit as the number of slices goes to infinity gives us the definite integral $$\int_0^a \rho(x) \ dx$$ to represent the mass of the rod. Using the substitution $u = -0.1x$ and $du = -0.1 \ dx$ we have $$\begin{aligned}\int_0^a \rho(x) \ dx &= \int_0^a 10e^{-0.1x} \ dx \\ &= \int_0^{-0.1a} -100 e^u \ du \\ &= -100e^u \biggm|_0^{-0.1a} \\ &= -100\left(e^{-0.1a} - 1\right) \\ &= 100\left(1-e^{-0.1a}\right)\end{aligned}\,.$$ So if the mass is $30$ g, then $100\left(1-e^{-0.1a}\right) = 30$, so $1-e^{-0.1a} = 0.3$, and $e^{-0.1a} = 0.7$. It follows that $-0.1a = \ln(0.7)$, and thus, $a = -10 \ln(0.7) \approx 3.567 \text{ cm}$. (b) Since the density is a decreasing function, we should expect that the center of mass of the rod will lie to the left of the midpoint. (c) To find the center of mass, we slice the rod horizontally into uniform slices of length $\Delta x$ and approximate the mass on the $i$th slice with the constant density $\rho(x_i)$ to be $\rho(x_i) \Delta x$ as in (a). The position of the $i$th slice is approximately $x_i/$. Treating the rod as a collection of $n$ discrete masses (the slices), the center of mass $\overline{x}$ is approximately $$\overline{x} \approx \frac{\sum_{i=1}^n x_i \rho(x_i) \Delta x}{\sum_{i=1}^n \rho(x_i) \Delta x}\,.$$ Letting the number of slices go to infinity gives us the exact location of the center of mass of the rod, $$\overline{x} = \frac{\int_0^a x \rho(x) \ dx}{\int_0^a \rho(x) \ dx}\,.$$ The mass of the rod is $30$ g, so now we just need to evaluate $\int_0^a x \rho(x) \ dx = \int_0^a 10xe^{-0.1x} \ dx$. Integrating by parts with $u = x$, $du = dx$, $dv = e^{-0.1x} \ dx$, and $v = -10e^{-0.1x}$ gives us $$\begin{aligned}\int_0^a 10xe^{-0.1x} \ dx &= 10\left(-10xe^{-0.1x}\biggm|_0^a + \int_0^a -10e^{-0.1x} \ dx \right) \\ &= 10\left(-10ae^{-0.1a} + 100(1-e^{-0.1a})\right) \\ &\approx 10\left(-10(3.567)ae^{-0.1(3.567)} + 100(1-e^{-0.1(3.567)})\right) \\ &\approx 50.3338\end{aligned}\,.$$ So the center of mass is $\overline{x} \approx \frac{50.3338}{30} \approx 1.687$, which is less than midway across the $3.567$ cm rod as expected. (d) Each piece would have mass $15$ grams. Let $q$ be a value of $x$ between 0 and $a$ so the the mass of the rod on the interval $[0, q]$ is 15. Using the substitution $u = -0.1x$ and $du = -0.1 \ dx$ we have that the mass of the rod on the interval $[0, q]$ is $$\begin{aligned}\int_0^q \rho(x) \ dx &= \int_0^q 10e^{-0.1x} \ dx \\ &= \int_0^{-0.1q} -100e^{u} \ du \\ &= -100e^u\biggm|_0^{-0.1q} \\ &= 100\left(1-e^{-0.1q}\right)\end{aligned}$$ We want this mass to be $15$ grams, so we find $q$ such that $100\left(1-e^{-0.1q}\right) = 15$. Solving for $q$ in the usual way, we find $q = -10\ln(0.85) \approx 1.625$ cm. Note that this is not the same as the center of mass. The center of mass takes into account the torque (determined by distance from the center of mass), which is not incorporated into the calculation of the point to split the object into two pieces of equal mass.