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 Vectors I

A scalar quantity has only size or magnitude (i.e. all the information is just a single number). For example: mass, time, speed $\dots$

A vector quantity has magnitude and direction. For example: force, velocity, acceleration, moment, $\dots$

A vector is represented graphically by a straight line with an arrowhead showing its direction. Often the line is drawn to scale so that the magnitude of the vector may be determined by measuring its length. To refer to a vector we give it a name. There are several different conventions used to indicate that we are referring to a vector quantity. The vector above will often be referred to in books as $\underline{OP}$, or $\overrightarrow{OP}$, or $\underline p$, or $\vec{p}$. The point at which a vector starts is called the initial point. The point at which it ends is called the final point.

The magnitude of a vector, also sometimes called the length of a vector, is denoted by enclosing the name (any name) of a vector by vertical lines (“modulus lines”; it is the same notation as used for the absolute value of a number): for example $|\underline{OP}|$ or $|\underline{p}|$. It is a non-negative number.

We can multiply a vector by a non-negative scalar (i.e. a non-negative number). This has the effect of leaving the direction the same, but changing the magnitude. For example, below we have the vectors $\underline a$ and $3\underline a$. Note that $3\underline a$ is parallel to $\underline a$ (has the same direction) but is $3$ times as long. (At this point it may be helpful to also imagine that they have the same initial point, but for the sake of the picture they are drawn separately.) Multiplying a vector (say $\underline{p}$) by a negative scalar (say $a<0$) changes the direction of the vector to the opposite (imagine that it stays on the same line, the same initial point, but opposite direction). The magnitude of $a\cdot\underline{p}$ is $|a|\cdot |\underline{p}|$.

A position vector indicates the distance and direction that we have to travel from the origin to get to a point. We use the term displacement vector to refer to the same concept, but relative to any other point than origin.

In the diagram below, if $O$ is the origin then $\underline{OP}$ is the position vector of the point called $P$. $\underline{PQ}$ is a displacement vector from $P$ to $Q$. It shows the distance and direction that we have to travel from $P$ to get to $Q$. $\underline{OQ}$ is the position vector of the point $Q$.

10.2.1 Vector Algebra

We have already seen that a vector may be multiplied by a scalar. If the scalar is positive this changes the magnitude but not the direction of the vector. If the scalar is negative the vector is turned to the opposite direction, and the magnitude is changed.

We declare two vectors $\underline{a}$ and $\underline{b}$ to be equal, if they have the same magnitude and the same direction. They do not actually have to start and finish at the same point. In the above diagram, $\underline a= \underline b = \underline c$.

If a vector $\underline b$ has the same magnitude as vector $\underline a$, but points in the opposite direction then $\underline b = -\underline a$. Again it doesn’t matter whether they start and finish at the same point or not.

The sum (or resultant) of two vectors is obtained by placing the initial point of the second vector in the sum at the final point of the first vector in the sum. The sum is then the vector that completes the triangle (i.e. runs from the initial point of the first to the final point of the second). This is called the triangle law of vector addition.

The difference of two vectors is given by $\underline a + (-\underline b)$, where $-\underline b$ points in the opposite direction to $\underline b$. The sum of $\underline a + (-\underline b)$ is then found from the triangle law above. Equivalently, we define $\underline a-\underline b = \underline a + (-\underline b)$.

The sum of a set of vectors that form a closed loop is always zero. $\underline a + \underline b + \underline c + \underline d + \underline e = \underline 0$.

Example 10.3 For the diagram below, state the magnitude of the sum of the vectors $\underline{AB}$, $\underline{BC}$, $\underline{CD}$ and $\underline{DE}$.

Solution. The sum of all the vectors starts at the initial point of the first vector and ends at the final point of the last vector. This is the vector $\underline{AE}$. Its magnitude may be written $|\underline{AE}|$.

If a vector joins two points $A$ and $B$, then it is equal to the sum of any system of vectors that form a path from $A$ to $B$. For example the vector $\underline{AB}$ shown as a solid line below, is equivalent to the sum of the vectors shown dotted. This fact may be used to solve geometric problems using vectors.

Example 10.4 For the triangle $ABC$, if $D$ is the mid-point of $AB$ show that \[\begin{equation} 2\underline{AB}+3\underline{BC}+\underline{CA} = 2\underline{DC}. \tag{10.1} \end{equation}\]

Solution. From the triangle \[\begin{align} \underline{AB}&=2\underline{AD} \tag{10.2}\\ \underline{BC}&=\underline{BD}+\underline{DC} \tag{10.3}\\ \underline{CA}&=\underline{CD}+\underline{DA} \tag{10.4}\\ \end{align}\] Substituting the equations (10.2), (10.3), and (10.4) for the left-hand side of (10.1) gives \[\begin{equation} 2\underline{AB}+3\underline{BC}+\underline{CA}=4\underline{AD}+3\underline{BD}+3\underline{DC}+\underline{CD}+\underline{DA}. \tag{10.5} \end{equation}\] Now \[\begin{align*} \underline{BD}&=-\underline{AD}\\ \underline{CD}&=-\underline{DC}\\ \underline{DA}&=-\underline{AD}\\ \end{align*}\] Substituting these into the right-hand side of equation (10.5) gives \[ 2\underline{AB}+3\underline{BC}+\underline{CA}=4\underline{AD}-3\underline{AD}+3\underline{DC}-\underline{DC}-\underline{AD}. \] So \[ 2\underline{AB}+3\underline{BC}+\underline{CA} = 2\underline{DC}. \]