10 Week 10
We look at the important topics of centroids and centers of gravity, before a brief look at volumes and surfaces of revolution.
10.1 Centre of Gravity and Centroids
Many structures and mechanical systems act as if their masses were concentrated at a single point called the center of mass or centre of gravity. We wish to find out how to locate this point. First let us consider the situation in 1-dimension and then in 2-dimensions.
10.1.1 Masses along a line
Figure 10.1: mass along a line
Suppose that the \(x\)-axis is supported by a fulcrum at the origin and we have 2 masses, \(m_1\) and \(m_2\), situated at positions \(x_1\) and \(x_2\) from the origin, see Figure 10.1. This is similar to a ‘see-saw’. Does the system balance, or not? The answer, from experience, depends on both the values of the masses and the positions of the objects. The quantity \(m_1x_1\) is referred to as the moment of the mass \(m_1\) about the origin. Notice that in our figure, \(m_1x_1<0\) and \(m_2x_2>0\).
The principle of moments says that the system will balance providing \[ m_1x_1 + m_2x_2 = 0. \]
If we are given a system with more than 2 masses (see Figure 10.2)
Figure 10.2: mass along a line – many bodies
the principle says that the system will balance providing \[ \sum_i m_ix_i = 0. \] In principle, we should use the force acting on the body, which is \(m_ig\) where \(g\) is the gravitational acceleration. However the constant \(g\) will cancel out. Suppose however, the system does not balance. Can we find a point \(\overline x\) such that if we were to move the fulcrum to that position, the system would balance (see Figure 10.3)?
Figure 10.3: centre of mass
This would require that \[\begin{align*} \sum(x_i-\overline x)m_i& = 0\\ \sum x_i m_i-\sum{\overline x}m_i& = 0\\ \sum m_ix_i&=\overline x\sum m_i\\ \overline x&=\frac{\sum m_ix_i}{\sum m_i}=\frac{\text{system moment about $O$}}{\text{system mass}}. \end{align*}\]
The point \(\overline x\) is called the center of mass or center of gravity.
In order to generalise to an ‘infinite number of points’, we need to use integration, rather than a finite sum.
Consider a rod on Figure 10.4: we assume that its density varies with \(x\), i.e. it is given by a function \(\rho(x)\).
Figure 10.4: rod with variable density
The density (in general) is \[ \text{density} = \frac{\text{mass}}{\text{volume}}, \] Where ‘volume’ needs to be the appropriate one for the number of dimensions we work with: ‘length’ in 1 dimension, ‘surface area’ in 2 dimensions, and ‘volume’ in 3 dimensions.
If we imagine the rod in Figure 10.4 to be 2-dimensional, then the mass of the shaded region, of area \(\delta A\) (where the width \(\delta x\) is small), is \[ \rho(x)\delta A \] Summing over all these ‘small areas’ gives \[ \text{mass} = \int\rho(x)\ \mathrm{d}A. \] If we further suppose that the ‘height’ of the rod is constant ‘h’, then \(\mathrm{d}A=h\cdot\mathrm{d}x\) and \[ \text{mass} = h\cdot \int\rho(x)\ \mathrm{d}x. \]
Example 10.1 Suppose that the rod above (Figure 10.4) is 5 m long with mass per unit length at the point \(x\) given by \(2+x^2/2\ Kg/m\). Find the total mass of the rod.
Solution.
Take a small rectangular strip of width \(\delta x\) at the point \(x\) so that the mass of this rectangle is \[ \left(2+\frac{x^2}2\right)\delta x. \]
Figure 10.5: 2-dimensional rod
Now sum over all such rectangles and allow \(\delta x \to 0\) (in other words compute the integral) to get the mass \[ \text{mass} = \int_0^5\left(2+\frac{x^2}{2}\right)\ \mathrm{d}x = \left[2x+\frac{x^3}{6}\right]_0^5 = 30.8\text{ Kg}. \]
Suppose we wish to find the centre of mass of the rod in the previous Example. We use the same principle as in the discrete case when considering a finite number of points except we use an integral rather then a sum. The moment of the small rectangle of width \(\delta x\) is \(x(2+x^2/2)\delta x\). So \[ \overline x = \frac{\text{total moment}}{\text{total mass}} = \frac{1}{30.8}\int_0^5 x(2+\frac{x^2}{2})\ \mathrm{d}x = \frac{1}{30.8}\left[x^2+\frac{x^4}{8}\right]_0^5 = 3.35. \]
10.1.2 Centroid of a uniform lamina
A lamina is a solid body in the form of a thin sheet of uniform thickness, with uniform density. As a result, the weight is proportional to the area and the formula for the centre of gravity can have the \(m\) values replaced by \(a\) values where \(a\) represents the area of an element and \(A\) is the total area of the lamina. So \[ \overline x = \frac{\displaystyle\int xa\ \mathrm{d}x}{A},\;\;\; \overline y = \frac{\displaystyle\int ya\ \mathrm{d}y}{A} \] Then \(\overline x\) and \(\overline y\) are the coordinates of the centroid of the area or the centroid.
Suppose that the centroid of the area bounded by the curve \(y=f(x)\), the \(x\)-axis and the lines \(x=a\) and \(x=b\) is required.
Figure 10.6: lamina
Consider a narrow strip of width \(\delta x\) (see Figure 10.6). The strip can be considered as approximately a rectangle with height \(y\) and area \(y\delta x\). The centre of the area is at the midpoint with coordinate \((x,y/2)\). By summing the moments of the centre of area for each strip we can find the centroid.
Taking moments about the \(y\)-axis \[ \overline x = \frac{\displaystyle\int_a^bxy\ \mathrm{d}x}{\displaystyle\int_a^b y\ \mathrm{d}x}. \] Similarly taking moments about the \(x\)-axis \[ \overline y = \frac{\displaystyle\int_a^b \frac{y^2}2\ \mathrm{d}x}{\displaystyle\int_a^b y\ \mathrm{d}x}. \]
Example 10.2 Determine the centroid of area under the curve \(y=x^2\) between the coordinates \(x=1\) and \(x=3\).
Solution. \[ \overline x = \frac{\displaystyle\int_1^3xy\ \mathrm{d}x}{\displaystyle\int_1^3 y\ \mathrm{d}x} = \frac{\displaystyle\int_1^3x^3\ \mathrm{d}x}{\displaystyle\int_1^3x^2\ \mathrm{d}x} = \frac{\left[\frac{x^4}{4}\right]_1^3}{\left[\frac{x^3}{3}\right]_1^3} = \frac{20}{8.67} = 2.31 \] \[ \overline y = \frac{\displaystyle\int_1^3\frac{y^2}2\ \mathrm{d}x}{\displaystyle\int_1^3 y\ \mathrm{d}x} = \frac{\displaystyle\int_1^3\frac{x^4}2\ \mathrm{d}x}{\displaystyle\int_1^3x^2\ \mathrm{d}x} = \frac{\left[\frac{x^5}{10}\right]_1^3}{8.67} = \frac{24.1}{8.67} = 2.78 \] The centroid therefore has coordinates \((2.31,2.78)\).
10.2 Volume and Surface of Revolution
Consider a curve \(y = f(x)\) between the ordinates \(x = a\) and \(x = b\). The volume of revolution is the volume of the equivalent solid body obtained by rotating the curve around the \(x\) or \(y\) axes.
10.2.1 Rotation about the \(x\)-axis
Figure 10.7: volume of revolution
If the shaded strip in Figure 10.7 is rotated about the \(x\)-axis then, provided \(\delta x\) is small, it will trace out a disc with thickness \(\delta x\) and volume \(\delta V = \pi y^2\delta x\), where \(y\) is the radius of the disc, see Figure 10.8.
Figure 10.8: volume of revolution
We may sum the volumes generated by all the strips between \(x = a\) and \(x = b\) to obtained the volume of the solid of revolution for the section of curve about the \(x\)-axis. Thus the total volume \(V\) is given by \[ V = \int_a^b\pi y^2\ \mathrm{d}x \]
10.2.2 Rotation about the \(y\)-axis
By a similar method it can be shown that for a rotation about the \(y\)-axis, the volume the solid of revolution will be \[ V = 2\pi\int_a^bxy\ \mathrm{d}x. \] This is for the volume of the solid created by rotating (about the \(y\)-axis) the same region as before (i.e. bounded by two vertical lines \(x=a\) and \(x=b\) on the sides, and the \(x\)-axis and the graph of a function \(y(x)\) from the bottom and the top) is rotated about the \(y\)-axis.
10.2.3 Surface area of revolution
The surface area of revolution is the surface area of the body generated by rotating a curve about the \(x\) or \(y\) axes. For a strip as shown in the figure above, we approximate the ‘arc’ (a part of the graph of the function) by a straight line. This yields a right triangle, from which we get an approximation for the length of arc to be \(\delta l=\sqrt{1+(\mathrm{d}y/\mathrm{d}x)^2}\cdot\delta x\).
If the strip is rotated about the \(x\)-axis, the surface area generated will be approximately \(\delta S = 2\pi y\delta l\). The surface area of the whole solid generated by rotating the curve around the \(x\)-axis between \(x = a\) and \(x = b\) is obtained by summing the contributions to the surface area from each of the elementary strips and we can eventually deduce that \[ S = 2\pi\int_a^b y\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\ \mathrm{d}x. \]
Example 10.3
The straight line \(y=mx\) between \(x=0\) and \(x=h\) is rotated about the \(x\)-axis thus generating a right circular cone. Calculate its volume in terms of \(h\) and \(R\) where \(R\) is the \(y\) value at \(x=h\). Find also the surface area of the resultant cone.
Figure 10.9: right-circular-cone
Solution. See Figure 10.9.
The volume of the solid of revolution for \(y = mx\) is given by \[ V = \pi\int_0^hy^2\ \mathrm{d}x = \pi\int_0^h (mx)^2\ \mathrm{d}x = m^2\pi\int_0^hx^2\ \mathrm{d}x = m^2\pi\left[\frac{x^3}3\right]_0^h = m^2\pi\frac{h^3}3. \] Now \(m = R/h\) and so \[ V = \frac13\frac{R^2}{h^2}h^3 = \frac13\pi R^2h. \] The surface area is given by \[\begin{align*} S&= 2\pi\int_0^hy\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\ \mathrm{d}x = 2\pi\int_0^h mx\sqrt{1+m^2}\ \mathrm{d}x\\ &= 2m\sqrt{1+m^2}\pi\left[\frac{x^2}2\right]_0^h = m\sqrt{1+m^2}\pi h^2 = \pi R h\sqrt{1+m^2}\\ &= \pi Rl. \end{align*}\] where \(l\) is the length of the inclined line.
Example 10.4 A curve is given parametrically as \(x=4t^2, y = 6t-t^2\). Determine the volume generated when the figure bounded by the curve, the \(x\)-axis and the ordinates corresponding to \(t=0\) and \(t=2\) is rotated about the \(x\)-axis.
Solution. \[ V = \pi\int_a^b y^2\ \mathrm{d}x. \] Since \(x=4t^2\) then \(\mathrm{d}x/\mathrm{d}t = 8t\) and so we can write \(\mathrm{d}x = 8t\ \mathrm{d}t\). Hence \[\begin{align*} V& = \pi\int_0^2(6t-t^2)^28t\ \mathrm{d}t = 8\pi\int_0^2(36t^3-12t^4+t^5)\ \mathrm{d}t\\ & = 8\pi\left[9t^4-\frac{12t^5}5+\frac{t^6}6\right]_0^2\\ &= 8\pi\times77.87 = 1957 \text{ units}^3. \end{align*}\]