밀도·질량과 질량중심

Note $-x^2+2x+1 = 2-(x-1)^2$. So, both parabolas are symmetric about the line $x=1$, and the $x$-coordinate of the centroid is $x=1$.

The parabolas meet when:

$$\begin{aligned} (x-1)^2&=2-(x-1)^2\\ 2(x-1)^2&=2\\ |x-1|&=1\\ x=0,\quad x&=2 \end{aligned}$$

At both these points, $y=1$, so we see the figure is symmetric about the line $y=1$. Then the $y$-coordinate of the centroid is $1$.

Therefore, the centroid is at $(1,1)$.

The circle and the cut-out rectangle are symmetric about the $x$-axis, and about the $y$-axis, so the centroid is the origin.

Remark: the centroid of a region doesn't have to be a point in the region!

(a) If we were to set this figure on a pencil lined up along the vertical line $x=a$, it seems pretty clear that it would fall to the left. So, the centre of mass is to the left of the line $x=a$. The same is true in (b): the added density on the left makes it only more lopsided. However, in (c), the right side is denser than the left, which could counterbalance the left. Without knowing more about the dimensions and the density, we can't say where the centre of mass is in relation to the line $x=a$.

(d) Consider a section of the figure, consisting of all points $(x,y)$ in the figure with $b\le x\le c$, and its “mirror" section on the other side of the line $x=a$. These two sections, which are drawn in red in the sketches below, will have the same area, at the same distance from $x=a$. Since we only care about the $x$-coordinate of the centre of mass, it doesn't matter that the two halves are at different $y$-coordinates. The centre of mass falls along the line $x=a$.

(e) There is the same amount of area to the left and right of the line $x=a$, as in part (d). However, the area to the right is “stretched out" more, so that it occupies space farther away from the line $x=a$. So, the centre of mass will be to the right of the line $x=a$.

• The volume of water in Tank A is $\frac{4}{3}\pi(1)^3 = \frac{4}{3}\pi$ cubic metres.
• The mass of water is $\frac{4000}{3}\pi$ kg.
• By symmetry, the centre of mass of the water when it fills Tank $A$ is exactly in the centre of the sphere, at height $\bar y_1=4$ metres above the ground (one metre above the bottom of Tank $A$, which is three metres above the ground).
• When the water is entirely in Tank $B$, its height is $\frac{2}{3}\pi$ metres. (The base of Tank $B$ has area 2 m$^2$, and the volume of water is $\frac{4}{3}\pi$ m$^3$.) By symmetry, the centre of mass is exactly halfway up, at height $\bar y_2=\frac{1}{3}\pi$ metres.
• So, the point mass in our model is moved from $\bar y_1=4$ to $\bar y_2=\frac{1}{3}\pi$, a distance of $4-\frac{1}{3}\pi$ metres, by gravity.
• The work involved is: $$\left(\frac{4000}{3}\pi~\text{kg}\right)\times\left(4-\frac{1}{3}\pi~\text{m}\right)\times\left(9.8~\frac{\text{m}}{\text{sec}^2}\right) = \frac{39200\pi}{9}(12-\pi)\approx 121,212~\text{J}$$

(a) Here is a sketch of the specified region, which we shall call $R$.

The top of $R$ has equation $y=T(x)=e^x$, the bottom has equation $y=B(x)=-1$ and $x$ runs from $0$ to $1$. So, using vertical strips, we see that $R$ has area

$$\begin{aligned} A & = \int_0^1 \big[T(x)-B(x)\big]\,\mathrm{d}x = \int_0^1 \big[e^x-(-1)\big]\,\mathrm{d}x = \int_0^1 \big[e^x+1\big]\,\mathrm{d}x = \big[e^x+x\big]_0^1 = e \end{aligned}$$

and

$$\begin{aligned} \bar y &=\frac{1}{2A} \int_0^1 \big[T(x)^2-B(x)^2\big]\,\mathrm{d}x \\ &= \frac{1}{2e}\int_0^1 \big[ e^{2x} - 1\ \big] \,\mathrm{d}x = \frac{1}{2e} \bigg[\frac{e^{2x}}{2} - x \bigg]_0^1 \\ &=\frac{1}{2e} \bigg(\frac{e^2}{2} - 1 - \frac{1}{2} \bigg) = \frac{e}{4} - \frac{3}{4e} \end{aligned}$$

(b) To compute the volume when $R$ is rotated about the line $y=-1$

• we use thin vertical strips of width $\mathrm{d}x$ as in the figure above.

• When we rotate about the line $y=-1$, each strip sweeps out a thin disk

• whose radius is $r=T(x)-B(x)=e^x+1$ and

• whose thickness is $\mathrm{d}x$ and hence

• whose volume is $\pi r^2\,\mathrm{d}x = \pi(e^x+1)^2\,\mathrm{d}x$.

• As our leftmost strip is at $x=0$ and our rightmost strip is at $x=1$, the total volume is

$$\begin{aligned} \pi \int_0^1 (e^x+1)^2\,\mathrm{d}x &=\pi \int_0^1 (e^{2x}+2e^x+1)\,\mathrm{d}x =\pi\bigg[\frac{e^{2x}}{2}+2e^x + x\bigg]_0^1 \\ &=\pi\bigg[\Big(\frac{e^{2}}{2}+2e + 1\Big) -\Big(\frac{1}{2}+2 + 0\Big)\bigg] \\ &=\pi\Big(\frac{e^{2}}{2}+2e -\frac{3}{2}\Big) \end{aligned}$$

(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.

(a) To find the mass of the first bar, we use the fact that $M = \int_0^a \rho(x) \, dx$, with $a = 10$ and $\rho(x) = \frac{1}{1+x^2}$. Thus, the mass of the first bar is $$M_1 = \int_0^{10} \frac{1}{1+x^2} \, dx = \left. \arctan(x) \right|_0^{10}\,,$$ so $M_1 = \arctan(10) \approx 1.47113$. Similar computations for the second bar shows that its mass is $$M_2 = \int_0^{10} e^{-0.1x} \, dx = \left. -10e^{-0.1x} \right|_0^{10}\,,$$ so $M_2 = 10 - 10e^{-1} \approx 6.32121$. (b) To find the center of mass of each bar, we use the standard formula: $$\overline{x} = \frac{\int_0^{10} x \rho(x) \, dx}{\int_0^{10} \rho(x) \, dx}\,.$$ Having already evaluated the denominator for each mass distribution, we simply need to compute the numerator. It is straightforward to show that $$\int_0^{10} x \frac{1}{1+x^2} \, dx = \frac{1}{2} \ln(101) \approx 2.30756\,,$$ so the first bar's center of mass is $$\overline{x_1} = \frac{\arctan(10)}{\frac{1}{2}\ln(101)} \approx \frac{2.30756}{1.47113} \approx 1.56857\,.$$ For the second bar, $$\int_0^{10} x e^{-0.1x} \, dx = 100-200e^{-1} \approx 26.42411\,,$$ so its center of mass is $$\overline{x_2} = \frac{10 - 10e^{-1}}{100-200e^{-1}} \approx \frac{26.42411}{6.32121} \approx 4.18023\,.$$ (c) Consider a new 10 cm bar whose mass density function is $f(x) = \rho(x) + p(x)$. (i) The mass of this new bar is $M = \int_0^{10} (\rho(x) + p(x)) \, dx$. Using the additive property of the definite integral, $$M = \int_0^{10} \rho(x) \, dx + \int_0^{10} p(x) \, dx \approx 1.47113 + 6.32121 = 7.79234\,.$$ (ii) Applying the same idea to $\int_0^{10} xf(x) \, dx$, we have $$\int_0^{10} x(\rho(x) + p(x))) \, dx = \int_0^{10} x \rho(x) \, dx + \int_0^{10} x p(x) \, dx \approx 2.30756 + 26.42411 = 28.73167\,.$$ (iii) False. The center of mass of the new bar is $$\overline{x} = \frac{\int_0^{10} x(\rho(x) + p(x)) \, dx}{\int_0^{10} (\rho(x) + p(x)) \, dx} \approx \frac{2.30756 + 26.42411}{1.47113 + 6.32121} = 3.68717$$ while the average of the two bars' centers of mass is $$\frac{\overline{x_1} + \overline{x_2}}{2} \approx \frac{\frac{2.30756}{1.47113} + \frac{26.42411}{6.32121}}{2} = 2.8744\,.$$ Although the computations prove the conclusion, the result is also intuitive: the average treats the two bars as equal contributors and doesn't take into account that one bar has much more mass than the other. Indeed, a weighted average is needed to correctly compute the center of mass, which is what our earlier computations in (i) and (ii) accomplish.

(a) The plot shown above (with the curve reflected across the $x$-axis to generate the solid of revolution) shows why this solid of revolution is a reasonable approximation to a baseball bat (if perhaps a bit thin between $x = 3$ and $x = 9$. (b) We use the disk method to find the volume of the solid of revolution. In the usual way, the volume of a typical slice is $$V_{\text{slice}} \approx \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \triangle x$$ and thus using a definite integral to sum the volumes of the slices, we get $$V = \int_0^{30} \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \, dx\,.$$ Using technology to evaluate the integral, we find that $V \approx 52.0666$ cubic inches. (c) Given that the bat has (constant) weight density $0.6$ ounces per cubic inch, it follows that the total weight of the bat is $W \approx 0.6 \cdot 52.0666 = 31.23996$ ounces. (d) If we slice the baseball bat perpendicular to the horizontal axes, we get thin cylindrical disks as cross-sections. At a given $x$-location, the amount of weight concentrated there is approximately the weight density ($0.6$ ounces per cubic inch) times the volume of the slice, which is $V_{text{slice}} \approx \pi f(x)^2$. Thus, we see that the amount of weight concentrated at a location $x$ along the bat is given by $\rho(x) = 0.6 \pi f(x)^2$, which we can therefore view as being the weight density function for how the weight of the baseball bat is distributed from $x = 0$ to $x = 30$. (e) Even though we are dealing with “weight” instead of “mass”, we can compute the center of mass in the usual way since weight differs from mass only by the gravitational constant, and this constant is present in both the numerator and denominator of the center of mass. Thus, using the standard formula, $$\overline{x} = \frac{\int_0^{30} x \rho(x) \ dx}{\int_0^{30} \rho(x) \ dx}$$ so $$\begin{aligned}\overline{x} &= \frac{\int_0^{30} x \cdot 0.6 \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2) \ dx}{\int_0^{30} 0.6 \pi (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \ dx} \\ &= \frac{\int_0^{30} x \cdot (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2) \ dx}{\int_0^{30} (2xe^{-1.25x} + (30-x) e^{-0.25(30-x)})^2 \ dx} \\ &\approx \frac{384.735}{16.5733} = 23.21415\end{aligned}\,,$$ so the center of mass is just over 2/3 along the length of the bat.