Example 9.6.1 Suppose the beam is 10 meters long and that the density is $1+x$ kilograms per meter at location $x$ on the beam. To approximate the solution, we can think of the beam as a sequence of weights "on'' a beam. For example, we can think of the portion of the beam between $x=0$ and $x=1$ as a weight sitting at $x=0$, the portion between $x=1$ and $x=2$ as a weight sitting at $x=1$, and so on, as indicated in figure 9.6.2. We then approximate the mass of the weights by assuming that each portion of the beam has constant density. So the mass of the first weight is approximately $\ds m_0=(1+0)1=1$ kilograms, namely, $(1+0)$ kilograms per meter times 1 meter. The second weight is $\ds m_1=(1+1)1=2$ kilograms, and so on to the tenth weight with $\ds m_9=(1+9)1=10$ kilograms. So in this case the total torque is $$ (0-\bar x)m_0+(1-\bar x)m_1+\cdots+(9-\bar x)m_9= (0-\bar x)1+(1-\bar x)2+\cdots+(9-\bar x)10. $$ If we set this to zero and solve for $\ds \bar x$ we get $\ds \bar x=6$. In general, if we divide the beam into $n$ portions, the mass of weight number $i$ will be $\ds m_i=(1+x_i)(x_{i+1}-x_i)=(1+x_i)\Delta x$ and the torque induced by weight number $i$ will be $\ds (x_i-\bar x)m_i=(x_i-\bar x)(1+x_i)\Delta x$. The total torque is then $$\eqalign{ (x_0-\bar x)(1+x_0)\Delta x&+(x_1-\bar x)(1+x_1)\Delta x+\cdots+ (x_{n-1}-\bar x)(1+x_{n-1})\Delta x\cr &=\sum_{i=0}^{n-1} x_i(1+x_i)\Delta x- \sum_{i=0}^{n-1}\bar x(1+x_i)\Delta x\cr &=\sum_{i=0}^{n-1} x_i(1+x_i)\Delta x- \bar x\sum_{i=0}^{n-1}(1+x_i)\Delta x.\cr }$$ If we set this equal to zero and solve for $\ds \bar x$ we get an approximation to the balance point of the beam: $$\eqalign{ 0&=\sum_{i=0}^{n-1} x_i(1+x_i)\Delta x- \bar x\sum_{i=0}^{n-1}(1+x_i)\Delta x\cr \bar x\sum_{i=0}^{n-1}(1+x_i)\Delta x&=\sum_{i=0}^{n-1} x_i(1+x_i)\Delta x\cr \bar x&={\ds\sum_{i=0}^{n-1} x_i(1+x_i)\Delta x\over \ds\sum_{i=0}^{n-1}(1+x_i)\Delta x}.\cr }$$ The denominator of this fraction has a very familiar interpretation. Consider one term of the sum in the denominator: $\ds (1+x_i)\Delta x$. This is the density near $\ds x_i$ times a short length, $\Delta x$, which in other words is approximately the mass of the beam between $\ds x_i$ and $\ds x_{i+1}$. When we add these up we get approximately the mass of the beam.

Now each of the sums in the fraction has the right form to turn into an integral, which in turn gives us the exact value of $\ds \bar x$: $$ \bar x={\ds\int_0^{10} x(1+x)\,dx\over\ds\int_{0}^{10}(1+x)\,dx}. $$ The numerator of this fraction is called the moment of the system around zero: $$\int_0^{10} x(1+x)\,dx=\int_0^{10} x+x^2\,dx={1150\over 3},$$ and the denominator is the mass of the beam: $$\int_0^{10} (1+x)\,dx=60,$$ and the balance point, officially called the center of mass, is $$\bar x={1150\over 3}{1\over 60} = {115\over 18}\approx 6.39.$$ $\square$

Example 9.6.2 Suppose a beam lies on the $x$-axis between 20 and 30, and has density function $\sigma(x)=x-19$. Find the center of mass. This is the same as the previous example except that the beam has been moved. Note that the density at the left end is $20-19=1$ and at the right end is $30-19=11$, as before. Hence the center of mass must be at approximately $20+6.39=26.39$. Let's see how the calculation works out. $$\eqalign{ M_0&=\int_{20}^{30} x(x-19)\,dx=\int_{20}^{30} x^2-19x\,dx= \left.{x^3\over3}-{19x^2\over2}\right|_{20}^{30}={4750\over3}\cr M&=\int_{20}^{30} x-19\,dx=\left.{x^2\over2}-19x\right|_{20}^{30}=60\cr {M_0\over M}&={4750\over3}{1\over60}={475\over18}\approx 26.39.\cr }$$ $\square$

Example 9.6.3 Suppose a flat plate of uniform density has the shape contained by $\ds y=x^2$, $y=1$, and $x=0$, in the first quadrant. Find the center of mass. (Since the density is constant, the center of mass depends only on the shape of the plate, not the density, or in other words, this is a purely geometric quantity. In such a case the center of mass is called the centroid.)

This is a two dimensional problem, but it can be solved as if it were two one dimensional problems: we need to find the $x$ and $y$ coordinates of the center of mass, $\ds \bar x$ and $\ds \bar y$, and fortunately we can do these independently. Imagine looking at the plate edge on, from below the $x$-axis. The plate will appear to be a beam, and the mass of a short section of the "beam'', say between $\ds x_i$ and $\ds x_{i+1}$, is the mass of a strip of the plate between $\ds x_i$ and $\ds x_{i+1}$. See figure 9.6.3 showing the plate from above and as it appears edge on. Since the plate has uniform density we may as well assume that $\sigma=1$. Then the mass of the plate between $\ds x_i$ and $\ds x_{i+1}$ is approximately $\ds m_i=\sigma(1-x_i^2)\Delta x=(1-x_i^2)\Delta x$. Now we can compute the moment around the $y$-axis: $$M_y=\int_0^1 x(1-x^2)\,dx={1\over4}$$ and the total mass $$M=\int_0^1 (1-x^2)\,dx={2\over3}$$ and finally $$\bar x = {1\over4}{3\over2}={3\over8}.$$ Next we do the same thing to find $\ds \bar y$. The mass of the plate between $\ds y_i$ and $\ds y_{i+1}$ is approximately $\ds n_i=\sqrt{y}\Delta y$, so $$M_x=\int_0^1 y\sqrt{y}\,dy={2\over5}$$ and $$\bar y={2\over5}{3\over2}={3\over5},$$ since the total mass $M$ is the same. The center of mass is shown in figure 9.6.3. $\square$

Example 9.6.4 Find the center of mass of a thin, uniform plate whose shape is the region between $y=\cos x$ and the $x$-axis between $x=-\pi/2$ and $x=\pi/2$. It is clear that $\ds \bar x=0$, but for practice let's compute it anyway. We will need the total mass, so we compute it first: $$ M=\int_{-\pi/2}^{\pi/2} \cos x\,dx=\sin x\Big|_{-\pi/2}^{\pi/2}=2. $$ The moment around the $y$-axis is $$ M_y=\int_{-\pi/2}^{\pi/2} x\cos x\,dx= \cos x+x\sin x\Big|_{-\pi/2}^{\pi/2}=0 $$ and the moment around the $x$-axis is $$ M_x=\int_{0}^{1} y\cdot2\arccos y\,dy= \left.y^2\arccos y-{y\sqrt{1-y^2}\over2}+{\arcsin y\over 2} \right|_{0}^{1}={\pi\over4}. $$ Thus $$\bar x={0\over2},\quad \bar y={\pi\over8}\approx 0.393.$$ $\square$