Title: Default for 15.2 Calculating Centres of Mass
Course: HELM Workbooks 2023-2024

Finding the centre of mass of a plane uniform lamina

We are now interested in finding the centre of mass of a thin sheet of material, such as a plane sheet of metal, called a lamina. The mass is now not located at individual points. Rather, it is distributed continuously over the lamina. In what follows, we assume that the mass is distributed uniformly over the lamina and you will see how integration as the limit of a sum is used to find the centre of mass.

The figure below shows a lamina where the centre of mass has been marked at point G with coordinates \((\bar{x}, \bar{y})\). If the total mass of the lamina is M then the moments about the \(y\)- and \(x\)-axes are respectively \(M\bar{x}\) and \(M\bar{y}\). Our approach to locating the position of G, i.e. finding \(\bar{x}\) and \(\bar{y}\), is to divide the lamina into many small pieces, find the mass of each piece, and calculate the moment of each piece about the axes. The sum of the moments of the individual pieces about the \(y\)-axis must then be equal to \(M\bar{x}\) and the sum of the moments of the individual pieces about the x-axis must equal \(M\bar{y}\).

[Image: Figure for section 15.2]
The centre of mass of the lamina is located at \(G(\bar{x}, \bar{y})\).

There are no formulae which can be memorized for finding the centre of mass of a general lamina because of the wide variety of possible shapes. Instead you should be familiar with the general technique for deriving the centre of mass. We will then see how in some cases, where the lamina can be described as the region bounded by simple curves, we can use the technique to find a standard formula.

An important preliminary concept is ‘mass per unit area’ which we now introduce.

Mass per unit area

Suppose we have a uniform lamina and select a piece of the lamina which has area equal to one unit. Let \(\rho\) stand for the mass of such a piece. Then \(\rho\) is called the mass per unit area. The mass of any other piece can be expressed in terms of \(\rho\). For example, an area of 2 units must have mass \(2\rho\), an area of 3 units must have mass \(3\rho\), and so on. Any portion of the lamina which has area \(A\) has mass \(\rho A\).

Key Point

If a lamina has mass per unit area, \(\rho\), then the mass of part of the lamina having area \(A\) is \(A \rho\).

In order to find the centre of mass of a lamina we will calculate \(M_x\) and \(M_y\), the moments of inertia about the \(x\)-axis and \(y\)-axis respectively. We do this in two ways.

First, acting as a point mass \(A\rho\) at its centre of mass \((\bar{x},\bar{y})\) we have \(M_y=A\rho \bar{x}\) and \(M_x=A\rho \bar{y}\). In other words \(\bar{x}=\dfrac{M_y}{A\rho}\) and \(\bar{y}=\dfrac{M_x}{A\rho}\).

Secondly, we take thin rectangular slices of the lamina and sum the contribution of each to the moments of inertia. Taking the limit as the width of these slices approaches \(0\) we will obtain an integral and solving this gives an expression for the moment of inertia which does not rely explicitly on \((\bar{x},\bar{y})\).

Finally we combine this data to find \((\bar{x},\bar{y})\).

Exercise

Consider the plane sheet, or lamina, shown below. In this exercise we will find the location of its centre of mass \((\bar{x}, \bar{y})\).

[Image: Section 15.2.2 exercise 1]

1. First, by noting the symmetry of the lamina we can already determine the \(y\) coordinate, \(\bar{y}\), of the centre of mass.
\(\bar{y} = \)
2. Let \(\rho\) stand for the mass per unit area of the lamina hence the total area is \(3\) units and the total mass is \(3\rho\). Its moment about the \(y\)-axis is \(3\rho\bar{x}\).

To find \(\bar{x}\) first divide the lamina into a large number of thin vertical slices. In the figure below a typical slice has been highlighted. Note that the slice has been drawn from the point \(P\) on the line \(y = 3x\). The point \(P\) has coordinates \((x, 3x)\). The thickness of the slice is \(\delta x\). A typical slice of this sheet has been shaded in the left image below.
[Image: Slice of lamina and rectangular approximation with P on left]
Approximate the area of the slice by a rectangle as indicated in the right of the image above.
Approximate area of slice:
3. Writing \(\rho\) as the mass per unit area, approximate the mass of the slice using the area from the previous part.
Note: write rho for \(\rho\).
Approximate mass of slice:
4. The centre of mass of this slice lies on the \(x\)-axis. We will take the limit as the thickness approaches zero so the slice can be assumed to be a point mass located a distance \(x\) from \(O\). Write down the approximate moment of the mass of the slice about the \(y\)-axis using the mass above.
5. Now to consider the moment of the lamina we add up contributions from all such slices in the lamina. Following which we take a limit by letting \(\delta x \to 0\) to obtain the integral:
\(3\rho\bar{x}=\)

\(\displaystyle \int\) |  |  |  |  |  | \(~\mathrm{d}x\) |  |  |

6. By evaluating the integral we can find
\(\bar{x}=\)
Feedback: 1. \(\bar{y}=0\) since the lamina is symmetric through reflection in the \(x\)-axis. 2. The width is \(\delta\) and height \(3x+3x=6x\) so we have \(6x\delta\). 3. The mass is uniform so we simply have \(\rho \times 6x\delta\). 4. We are told to take the distance from the \(y\)-axis to the centre of mass of this rectangle to be \(x\) so the moment of inertia is \(x \times 6x\rho\delta = 6x^2\rho\delta\). 5. The lamina sits above the interval \([0,1]\), taking the limit as \(\delta \rightarrow 0\) give the integral \(\int_{0}^{1}6\rho x^2 dx\) 6. Evaluating the integral \(\int_{0}^{1}6\rho x^2 dx = \left[2\rho x^3 \right]_{0}^{1} = 2\rho\). This must equal the moment of the total mass acting at the centre of mass so \(3\rho\bar{x} = 2\rho\) giving \(\bar{x} =2/3\). Now the coordinates of the centre of mass are \((2/3, 0)\).

Exercise

Find the position of the centre of mass of a uniform semi-circular lamina of radius \(a\), as shown below.

[Image: Semi circle of radius a centered at the origin, fully contained in upper half plane.]

1. \(\bar{x}=\)
2. \(\bar{y}=\)
Feedback: 1. By symmetry we have \(\bar{x}=0\) 2. The lamina is bounded by \(y=0\) and \(y=\sqrt{a^2-x^2}\). The area is half that of semicircle of radius \(a\) so we have area \(\dfrac{a^2\pi}{2}\) Then \(bar{y}= \dfrac{\int_{-a}^a a^2-x^2 \mathrm{d}x}{a^2\pi}=[value]\).

Exercise

This is the first in a series of three exercises to find the centre of mass of the plane lamina shown below.

[Image]

The coordinates of \(\bar{x}\) and \(\bar{y}\) must be calculated separately. In this exercise we calculate \(\bar{x}\) as we did before.
1. Let \(\rho\) equal the mass per unit area. The total mass \(m\) is then

\(m=\)
Enter rho for \(\rho\).

Its moment about the \(y\)-axis is then \(m\bar{x}\).
2. To calculate \(\bar{x}\) the lamina is divided into thin slices; a typical slice is shown in the figure above. We approximate the area of the slice with the largest rectangle of width \(\delta x\) which fits within the slice. Calculate the mass of the rectangle.
Mass of rectangle:
3. Write down the moment about the \(y\)-axis of the typical strip, assuming the mass is a point mass on the leftmost edge of the rectangle.
Moment of rectangle:
4. We sum the moments of all rectangular strips and take the limit as \(\delta x \to 0\) to obtain the integral:
\( \displaystyle{\int_{0}^{2}}\)\(\ \textrm{d}x\).
5. Evaluate the integral from the previous part to find the moment:
6. Equating the moments from parts (a) and (e) we find \(\bar{x}=\)

The next questions will illustrate two alternative ways of calculating \(\bar{y}\).
Feedback: 1. \(2 \rho \bar{x}\) 2. \(\rho \times (y \delta x)\) 3. \((\rho x^2 \delta x) \) 4. \(\displaystyle{\int_0^2 x^2 \rho \mathrm{d}x = \left[\dfrac{\rho x^3}{3}\right]_0^2 = \dfrac{8 \rho}{3}-0=\dfrac{8 \rho}{3}.}\) 5. \(2 \rho \bar{x}=\dfrac{8 \rho}{3}\) so that \(\bar{x} = 4/3\).

Exercise

In this exercise we use the same tall and narrow rectangles as in the previous exercise, but we now consider the moment about the \(x\)-axis in order to find \(\bar{y}\).

[Image]

1. Referring to the figure again, the centre of mass of the rectangular slice must lie half way along its length, that is, its \(y\)-coordinate is \(\dfrac{y}{2}=\dfrac{x}{2}\). Assume that all the mass of the slice, \(x \rho \delta x\), acts at this point, the moment about the \(x\)-axis is:

Enter rho for \(\rho\).

2. Adding contributions from all slices gives the sum and letting \(\delta x \to 0\) we obtain the integral:
\( \displaystyle{\int_{x=0}^{2}}\) \(\textrm{d}x\)
3. Evaluate the above integral.
4. Similar to the previous exercise the moment of the total mass about the \(x\)-axis is \(2 \rho \bar{y}\). Setting this equal to the value in part (c) determine \(\bar{y}\).
\(\bar{y}=\)
Feedback: 1. \(\displaystyle [value]\) 2. \(\displaystyle \int_{x=0}^{2} \frac{\rho y^2}{2} \textrm{d}x\) 3. \({4 \rho}{3}\) 4. \(2 \rho \bar{y} = \dfrac{4 \rho}{3} \) from which \(\bar{y} = \dfrac{2}{3}\). The coordinates of the centre of mass are \((\frac{4}{3}, \frac{2}{3})\).

Exercise

In this exercise we divide the lamina into a number of horizontal slices; a typical slice is shown below.

[Image]

1. In this case we will take a rectangle of width \(2 - x\) and height \(\delta y\). Find the mass \(m\) of this rectangle in term of \(\rho\) and \(y\).
\(m=\)
Enter rho for \(\rho\).
The moment about the \(x\)-axis is then \(m\bar{y}\).
2. We take the sum of all such moments and letting \(\delta \to 0\) we obtain the integral:
\(\displaystyle{\int_{y=0}^{y=2}}\) \(\textrm{d}y\)
3. Evaluate the integral above.
4. Hence find \(\bar{y}=\)
Feedback: 1. \(\displaystyle (2 - x) \delta y, \rho (2 - x) \delta y, \rho (2 - x)y \delta y\) 2. \(\displaystyle \int_{y=0}^{y=2} \rho (2 - y)y \textrm{d}y\) 3. \(\displaystyle \frac{4 \rho}{3}\) 4. As before the total mass is \(2 \rho\), and its moment about the \(x\)-axis is \(2 \rho \bar{y}\). Hence \( 2 \rho \bar{y} = \frac{4 \rho}{3}\) from which \(\bar{y} =\frac{2}{3}\) which was the result obtained in the previous exercise.

The centre of mass of a collection of point masses

Suppose we have a collection of masses located at a number of known points along a line. The centre of mass is the point where, for many purposes, all the mass can be assumed to be located.

For example, if two objects each of mass \(m\) are placed at distances 1 and 2 units from a point \(O\), as shown in part (a) of the Figure below, then the total mass, \(2m\), might be assumed to be concentrated at distance 1.5 units as shown in part (b). This is the point where we could imagine placing a pivot to achieve a perfectly balanced system.

[Image]
Equivalent position of the centre of mass of the objects in (a) is shown in (b)

To think of this another way, if a pivot is placed at the origin \(O\), as on a see-saw, then the two masses at \(x = 1\) and \(x = 2\) together have the same turning effect or moment as a single mass \(2m\) located at \(x = 1.5\). This is illustrated in the Figure below.

[Image]
The single object of mass \(2m\) has the same turning effect as the two objects each of mass \(m\)

Before we can calculate the position of the centre of mass of a collection of masses it is important to define the term ‘moment’ more precisely. Given a mass \(M\) located a distance \(d\) from \(O\), as shown in the Figure below, its moment about \(O\) is defined to be moment \( = M\times d\)

[Image]
The moment of the mass \(M\) about \(O\) is \(M \times d\)

In words, the moment of the mass about \(O\), is the mass multiplied by its distance from \(O\). The units of moment will therefore be kg m if the mass is measured in kilogrammes and the distance in metres. (N.B. Unless specified otherwise these will be the units we shall always use.)

Exercise

Calculate the moments of mass about \(O\) in each of the following cases:

1. [Image]
\(\textrm{kg} \, \textrm{m}\)
2. [Image]
\(\textrm{kg} \, \textrm{m}\)
Feedback: The moment is given by \(\text{moment } = M \times d\) so we get: 1. \(5 \times 8 = 40 \, \textrm{kg} \, \textrm{m}\) 2. \( 5 \times 10 = 50 \, \textrm{kg} \, \textrm{m}\)

Exercise

Objects of mass \(m\) and \(3m\) are placed at the locations shown in diagram (a). Find the distance \(\bar{x}\) of the centre of mass from the origin \(O\) as illustrated in diagram (b).

[Image]

First calculate the sum of the individual moments: \( \textrm{kg}\, \textrm{m}\)

The moment of the total mass about \(O\) is \(4 m \times \bar{x}\).

The moment of the total mass is equal to the sum of the moments of the individual masses. What equation do we get by equating the two moments?

\(\times \; m = \) \(\times \; m\bar{x}\)

So what is the distance \(\bar{x}\)?
\(\bar{x}=\)
Feedback: The sum of the individual moments is: \(6\times m + 10 \times 3m = 36m.\) Equating the two moments gives: \(36m = 4m\bar{x}\). And thus \(\bar{x}=\frac{36m}{4m}=9.\) So the centre of mass is located a distance 9 units along the \(x\)-axis. Note that it is closer to the position of \(3m\) mass than to the position of \(1m\) mass (actually in the ratio 3:1).

Example

Obtain an equation for the location of the centre of mass of two objects of masses \(m_1\) and \(m_2\):

1. located at distances \(x_1\) and \(x_2\) respectively, as shown part (a) of the Figure below.
2. positioned on opposite sides of the origin as shown in part (b) of the Figure below.

[Image]

Referring to part (a) of the above Figure, we first write down an expression for the sum of the individual moments: The total mass is \(m_1 + m_2\) and the moment of the total mass is \((m_1 + m_2) \times \bar{x}\)

The moment of the total mass is equal to the sum of the moments of the individual masses. The equation satisfied by \(\bar{x}\) is

\((m_1+m_2)\bar{x} = m_1x_1+m_2x_2\), so \(\bar{x}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}\)

However, this is precisely what would have been obtained if, when working out the moment of a mass, we use its coordinate (which takes account of sign) rather than using its distance from the origin.

For the second case, as depicted in part (b) of the figure, the mass \(m_1\) positioned on the left-hand side has a turning effect opposite to that of the mass \(m_2\) positioned on the right-hand side. To take account of this difference we use a minus sign when determining the moment of \(m_1\) about the origin. This gives a total moment

\(\qquad -(m_1x_1)+(m_2x_2)\)

leading to

\(\qquad (-m_1x_1+m_2x_2)=(m_1+m_2)\bar{x}\qquad\)so \(\qquad\bar{x}=\dfrac{-m_1x_1+m_2x_2}{m_1+m_2}\)

However, this is precisely what would have been obtained if, when working out the moment of a mass, we use its coordinate (which takes account of sign) rather than using its distance from the origin.

The above formulae can be generalised very easily to deal with the general situation of \(n\) masses, \(m_1, m_2, \dots, m_n\) located at coordinate positions \(x_1,x_2,\dots,x_n\) and is given in the Key Point below.

Key Point

The centre of mass of individual masses \(m_1, m_2, \dots, m_n\) located at positions \(x_1,x_2,\dots,x_n\) is \[\bar{x}=\dfrac{\sum_{i=1}^n m_i x_i}{\sum_{i=1}^n m_i}\]

Exercise

Calculate the centre of mass of the 4 masses distributed as shown below

[Image]

Give your answer as a rational number (fraction)

\(\bar{x}=\)
Feedback: \(\bar{x}=\dfrac{(9)(-1)+(1)(2)+(5)(6)+(2)(8)}{9+1+5+2}=\dfrac{39}{17}\). The centre of mass is located a distance \(\frac{39}{17} \approx 2.29\) units along the \(x\)-axis from \(O\).

Distribution of particles in a plane

If the particles are distributed in a plane then the position of the centre of mass can be calculated in a similar way.

[Image]
These masses are distributed throughout the \(xy\) plane

Now we must consider the moments of the individual masses taken about the \(x\)-axis and about the \(y\)-axis. For example, in the Figure above, mass \(m_i\) has a moment \(m_iy_i\) about the \(x\)-axis and a moment \(m_ix_i\) about the \(y\)-axis. Now we define the centre of mass at that point \((\bar{x}, \bar{y})\) such that the total mass, \(M = m_1 + m_2 + \dots + m_n\), placed at this point would have the same moment about each of the axes as the sum of the individual moments of the particles about these axes.

Key Point

The centre of mass of \(m_1, m_2, \dots, m_n\) located at \((x_1,y_1), (x_2,y_2), \dots, (x_n,y_n)\) has coordinates \((\bar{x},\bar{y})\) where \[ \bar{x}=\dfrac{\sum_{i=1}^n m_i x_i}{\sum_{i=1}^n m_i}, \quad \quad \bar{y}=\dfrac{\sum_{i=1}^n m_i y_i}{\sum_{i=1}^n m_i}. \]

Exercise

Masses of \([value] \, \mathrm{kg}, [value] \, \mathrm{kg}\) and \([value] \, \mathrm{kg}\) are located at the points \(([value],[value]), ([value],[value])\) and \(([value],[value])\) respectively. Find the coordinates of their centre of mass.

Give your answer to 2 decimal places.

\(\bar{x}=\)

\(\bar{y}=\)
Feedback: \[\bar{x}=\dfrac{\sum_{i=1}^3 m_i x_i}{\sum_{i=1}^3m_i}=\dfrac{[value]\times([value]) + [value]\times([value]) + [value]\times([value])}{[value]+[value]+[value]} \approx [value]\] \[\bar{y}=\dfrac{\sum_{i=1}^3 m_i y_i}{\sum_{i=1}^3m_i}=\dfrac{[value]\times([value]) + [value]\times([value]) + [value]\times([value])}{[value]+[value]+[value]} \approx [value]\] Hence the centre of mass is located at the point \(([value],[value])\).

Exercise

Find the \(x\) coordinate of the centre of mass of 5 identical masses placed at \(x=[value], x=[value], x=[value], x=[value], x=[value].\)

Give your answer as a rational number (fraction).

\(\bar{x}= \)
Feedback: We are asked to find the \(x\) coordinate of the centre of mass of 5 identical masses placed at \(x=[value], x=[value], x=[value], x=[value], x=[value].\) We compute: \[\bar{x} = \dfrac{\sum_{i=1}^{5} x_i}{5} = [value]\]

Application: Suspended cable

Introduction

A cable of constant line density is suspended between two vertical poles of equal height such that it takes the shape of a curve, \(y = 6 \cosh(\dfrac{x}{6})\). The origin of the curve is a point mid-way between the feet of the poles and \(y\) is the height above the ground. If the cable is 600 metres long show that the distance between the poles is 55 metres to the nearest metre. Find the height of the centre of mass of the cable above the ground to the nearest metre.

Mathematical statement of the problem

We can draw a picture of the cable, as below, where \(A\) and \(B\) denote the end points.

[Image]

For the first part of this problem we use the result that the distance along a curve \(y = f(x)\) from \(x = a\) to \(x = b\) is given by

\(\displaystyle{s = \int_{a}^{b} \sqrt{ 1 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 } \mathrm{d}x}\),

where in this case we are given \(y = 6 \cosh\left(\dfrac{x}{6}\right)\), and therefore \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \sinh\left(\dfrac{x}{6}\right)\). If we take the distance between the poles to be \(2d\) then the values of \(x\) in this integration go from \(-d\) to \(+d\). So we need to find \(d\) such that:

\(\displaystyle{600 = \int_{-d}^{+d} \sqrt{1 + \left(\sinh\left(\dfrac{x}{6}\right)\right)^2} \mathrm{d}x}.\)

For the second part of this problem we need to find the centre of mass of the cable. From the symmetry of the problem we know that the centre of mass must lie on the \(x\)-axis. To find the height of the centre of mass we need to take each section of the cable and consider the moment about the x-axis through the origin. A section of the cable has mass \(\rho \delta s\) where \(\rho\) is the line density of the cable and \(\delta s\) is the length of a small section of the cable.

so the moment about the \(x\)-axis will be \(\displaystyle{\sum_{x=-d}^{x=+d} \rho y \delta s}\),

taking the limit as \(\delta s \to 0\) and using the fact that \(\delta s = \sqrt{1 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2} \delta x\).

We get that the moment about the \(x\)-axis to be \(\displaystyle{\rho \int_{-d}^{d} y \sqrt{ 1 + \left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2 } \mathrm{d}x}\).

This must equal the moment of a single point mass, equal to the total mass of the cable, placed at its centre of mass. As the length of the cable is 600 metres then the mass of the cable is \(600 \rho\) and we have

\(\displaystyle{600 \rho \bar{y} = \rho \int_{-d}^{d} y \sqrt{ 1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 } \mathrm{d}x}.\)

Dividing both sides of this equation by \(\rho\) we get:

\(\displaystyle{600 \bar{y} = \int_{-d}^{d} y \sqrt{ 1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 } \mathrm{d}x}.\)

where we have already established the value of \(d\) from Equation (1) so we can solve this equation to find \(\bar{y}\).

Mathematical analysis

We need to find \(d\) so that \(\displaystyle{600 = \int_{-d}^{+d}\sqrt{1 + \left(\sinh\left(\frac{x}{6}\right) \right)^2}\mathrm{d}x}.\)

Rearranging the hyperbolic identity \(\cosh^{2}(u) - \sinh^{2}(u) \equiv 1\) we obtain \(\sqrt{1 + (\sinh(u))^2} = \cosh(u)\)

so the integral becomes \(\displaystyle{\int_{-d}^{d} \cosh\left(\dfrac{x}{6}\right)\mathrm{d}x = \left[6 \sinh\left(\frac{x}{6}\right)\right]^{+d}_{−d} = 6 \left(\sinh\left(\frac{d}{6}\right) - \sinh\left(-\frac{d}{6}\right)\right)}.\) so \(12 \sinh\left(\dfrac{d}{6}\right) = 600\) and \({d = 6 \sinh^{−1}(50)}.\)

Using the log identity for the \(\sinh^{-1}\) function:

\(\sinh^{-1}(x) \equiv \ln(x + \sqrt{x^2 + 1})\)

we find that \(d = 27.63\) m so the distance between the poles is \(55\) m to the nearest metre. To find the height of the centre of mass above the ground we use

\(\displaystyle{600 \bar{y} = \int_{-d}^{+d} y \sqrt{ 1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 }\ \mathrm{d}x}\)

Substituting \(y = 6 \cosh\left(\dfrac{x}{6}\right)\) and therefore \(\displaystyle{\sqrt{1 + (\frac{\mathrm{d}y }{\mathrm{d}x})^2} = \sqrt{1 + \left(\sinh\left(\frac{x}{6}\right)\right)^2} = \cosh\left(\frac{x}{6}\right)}\) we get \(\displaystyle{\int_{-d}^{d} 6 \cosh\left(\frac{x}{6}\right) \cosh\left(\frac{x}{6}\right) \ \mathrm{d}x = \int_{-d}^{d} 6 \cosh^{2}\left(\frac{x}{6}\right)\ \mathrm{d}x}. \)

From the hyperbolic identities we know that \(\cosh^{2}(x) \equiv \dfrac{(\cosh(2x) + 1)}{2}\)

so this integral becomes \(\displaystyle{\int_{-d}^{d} 3\left(\cosh\left(\frac{x}{3}\right) + 1\right) \ \mathrm{d}x = \left[9 \sinh\left(\frac{x}{3}\right) + 3x\right]_{-d}^{+d} = 18 \sinh\left(\frac{d}{3}\right) + 6d}\).

So we have that \(600\bar{y} = 18 \sinh\left(\dfrac{d}{3}\right) + 6d\)

From the first part of this problem we found that \(d = 27.63\) so substituting for \(d\) we find \(\bar{y} = 150\) metres to the nearest metre.

Interpretation

We have found that the two vertical poles holding the cable have a distance between them of 55 metres and the height of the centre of mass of the cable above the ground is 150 metres.

Intuition tells us that a large moment corresponds to a large turning effect. A mass placed 8 metres from the origin has a smaller turning effect than the same mass placed 10 metres from the origin. This is, of course, is why it is easier to rock a see-saw by pushing it at a point further from the pivot. Our intuition also tells us the side of the pivot on which the masses are placed is important. Those placed to the left of the pivot have a turning effect opposite to those placed to the right.

For a collection of masses, the moment of the total mass located at the centre of mass is equal to the sum of the moments of the individual masses. This definition enables us to calculate the position of the centre of mass. It is conventional to label the \(x\) coordinate of the centre of mass as \(\bar{x}\), pronounced ‘x bar’.

Key Point

The moment of the total mass located at the centre of mass is equal to the sum of the moments of the individual masses.

Exercise

Find the centre of mass of a lamina bounded by \(y^2 = [value]x\), for \(0\leq x \leq [value]\).

1. \(\bar{x} =\)
2. \(\bar{y} =\)
Feedback: We can follow the approach above, noting that the region is bounded by \(x=0\), \(x=[value]\), \(y=\sqrt{[value]x}\) and \(y=-\sqrt{[value]x}\).Hence \[ \bar{x} = \frac{\int_0^[value] 2x \sqrt{[value]x} \mathrm{d}x}{\int_0^[value] 2\sqrt{[value]x} \mathrm{d}x} = \frac{ [value] }{ [value]} =[value]. \] Then either by symmetry or the formula we can see that \(\bar{y}=0\). Alternatively we can work with respect to \(y\). First we compute the total mass of the lamina taking vertical strips. Rearranging the equation gives \(y_{max} = \sqrt{[value]\times[value]}= [value]\), \(y_{min} = -[value]\). The total mass is then given by: \[ mass = \rho \int_{-[value]}^{[value]} ([value] - \frac{y^2}{[value]})dy = 2\rho\times[value]\times[value] - \frac{2\rho}{3\times[value]}\times\frac{[value]^3}{[value]} = [value]\rho\] Similarly we compute the total moment to be: \[ \begin{align*} moment &= \rho\int_{-[value]}^{[value]} x([value] - \frac{y^2}{[value]})dy\\ &= \rho\int_{-[value]}^{[value]} \left(\frac{y^2}{[value]}+\frac{[value]}{2}\right) \left([value] - \frac{y^2}{[value]}\right)dy\\ &= [value]\rho \end{align*} \] Taking the ratio gives \[\bar{x}= [value] \] By symmetry, \(\bar{y} = 0\).

Formulae for the centre of mass

Now we consider the case of a lamina bounded above by some function \(f(x)\), bounded below by \(g(x)\) and bounded left and right by the lines \(x=a\) and \(x=b\) respectively (here we assume \(f(x) \geq g(x)\) for all \(x \in [a,b]\). Although this case is quite restrictive it does allow the process we have used before to be used to obtain a formula for each of \(\bar{x}\) and \(\bar{y}\).

[Image: Lamina as described in text]

Finding \(\bar{x}\)

As before, we find the moment of inertia about the \(y\)-axis in two ways. Assuming again that the lamina has an area \(A\) and mass per unit area of \(\rho\) we have the moment of inertia is \(A\rho \bar{x}\).

We slice the lamina in thin vertical slices of width \(\delta x\). We can approximate the area of the slices by the area of the rectangle of width \(\delta x\) and height \(f(x)-g(x)\). The moment of inertia of this rectangle is \( x(f(x)-g(x)) \delta x\). By adding all of these and letting \(\delta x \to 0\) we obtain the integral giving the moment of inertia of lamina so that

\[ \int_a^b x\rho (f(x)-g(x)) \mathrm{d}x = A\rho \bar{x}.\]

We can simplify this to obtain

\[ \bar{x}= \frac{1}{A} \int_a^b x (f(x)-g(x)) \mathrm{d}x.\]

Note that we can calculate \(A=\int_a^b f(x)-g(x) \mathrm{d}x\) to find

\[ \bar{x}= \frac{\int_a^b x (f(x)-g(x)) \mathrm{d}x}{\int_a^b f(x)-g(x) \mathrm{d}x}.\]

Finding \(\bar{y}\)

We can use the same vertical slices to find the moment of inertia about the \(x\)-axis, and thus find \(\bar{y}\). We found the area of one of our rectangles to be  \((f(x)-g(x)) \delta x\) but the \(y\)-coordinate of the centre of the rectangle is the average of \(f(x)\) and \(g(x)\), namely \(\frac{1}{2}(f(x)+g(x))\) so that the moment of inertia about the \(x\)-axis is

\[ \frac{1}{2}(f(x)+g(x))(f(x)-g(x)) \delta x=\frac12 (f(x)^2-g(x)^2) \delta x.\]

By adding all of these and letting \(\delta x \to 0\) we obtain the integral giving the moment of inertia of lamina so that

\[ \int_a^b \frac12 (f(x)^2-g(x)^2)  \mathrm{d}x = A\rho \bar{y}.\]

We can simplify this to obtain

\[ \bar{y}= \frac{1}{2A} \int_a^b f(x)^2-g(x)^2 \mathrm{d}x.\]

Note that we can calculate \(A=\int_a^b f(x)-g(x) \mathrm{d}x\) to find

\[ \bar{y}= \frac{\int_a^b f(x)^2-g(x)^2 \mathrm{d}x}{2\int_a^b f(x)-g(x) \mathrm{d}x}.\]

Key Point

Given functions \(f(x),\, g(x)\) such that \(g(x)\leq f(x)\) on the interval \([a,b]\) the centre of mass of the lamina bounded by \(x=a\), \(x=b\), \(y=f(x)\) and \(y=g(x)\) is
\[ (\bar{x},\bar{y}) = \left( \frac{\int_a^b x (f(x)-g(x)) \mathrm{d}x}{\int_a^b f(x)-g(x) \mathrm{d}x},\frac{\int_a^b f(x)^2-g(x)^2 \mathrm{d}x}{2\int_a^b f(x)-g(x) \mathrm{d}x}\right).\]

In the special case that \(g(x)=0\) is the \(x\)-axis we have

\[ (\bar{x},\bar{y}) = \left( \frac{\int_a^b x f(x) \mathrm{d}x}{\int_a^b f(x) \mathrm{d}x},\frac{\int_a^b f(x)^2 \mathrm{d}x}{2\int_a^b f(x) \mathrm{d}x}\right).\]

Remember that the integral in the denominator is related to the area of the lamina and may be possible to calculate without the integral.

Exercise

Find the position of the centre of mass of a uniform semi-circular lamina of radius \(a\), shown below.

[Image]

The equation of a circle centre the origin, and of radius \(a\) is \(x^2 + y^2 = a^2\). By symmetry \(\bar{x} = 0\). However it is necessary to calculate \(\bar{y}\).

1. The lamina is divided into a number of horizontal strips and a typical strip is shown. Assume that each strip is rectangular. Writing the mass per unit area as \(\rho\), state the area and the mass of the strip:
Note: write \(\delta\) and \(\rho\) as 'delta' and 'rho'
Area:
Mass:
2. Write down the moment of the mass about the x-axis:
3. Write down the expression representing the sum of the moments of all strips and the corresponding integral obtained as \(\delta y \to 0\):
\(\displaystyle{\sum_{y=0}^{y=a}}\)
\(\displaystyle{\int_{0}^{a}}\) \(dx\)
4. Evaluate the above integral by making the substitution \(u^2 = a^2 - y^2\) to obtain the total moment. Note that since \(x^2 + y^2 = a^2\) we have \(x = \sqrt{ a^2 - y^2}\).
5. The total area is half that of a circle of radius \(a\), that is \(\dfrac{\pi a^2}{2}\). The total mass is \(\dfrac{\pi a^2 \rho}{2}\) and its moment is \(\dfrac{\pi a^2 \rho \overline{y}}{2}\).
Deduce \(\bar{y}\):
Feedback: Solution 1. \(2x \delta y,\ 2 x \rho \delta y\) 2. \(2x \rho y \delta y\) 3. \(\sum_{y=0}^{y=a}{2x \rho y \delta y},\ \int_{0}^{a} 2x \rho y \textrm{d}y\) 4. \(\frac{2 \rho a^3}{3}\) 5. \(\frac{\pi a^2 \rho \bar{y}}{2} = \frac{2 \rho a^3}{3}\) from which \(\bar{y} = \frac{4a}{3\pi}\)