In many Indian villages, mangoes high on a tree are plucked with a long bamboo pole. The pole is thin at the top and thick at the bottom. Try to balance such a pole across one shoulder and you will not hold it at its middle. You slide it until the thick end is much nearer your shoulder than the thin end. The pole has no separate lumps of mass to average; its mass is spread along its whole length.
From a sum to an integral
The formula of the last lesson needs particles with definite masses and positions. A rod, a plate or a ball has no separate particles to count. Its mass is spread continuously through its volume. The way forward is to cut the body, in the mind, into a very large number of very small pieces.
Let the pieces have masses \(\Delta m_1, \Delta m_2, \ldots\) at positions \(x_1, x_2, \ldots\). Each piece is small enough to be treated as a point, so the particle formula applies to the pieces. Now let the pieces become smaller and more numerous without limit. Each sum then becomes an integral, and each small mass becomes an element of mass \(dm\).
\[ x_{cm} = \dfrac{\sum_i x_i\,\Delta m_i}{\sum_i \Delta m_i} \;\longrightarrow\; x_{cm} = \dfrac{1}{M}\int x\,dm \]
The denominator becomes \(\int dm = M\), the total mass of the body.
The linear mass density
An integral over \(dm\) cannot be done until \(dm\) is written in terms of a coordinate. For a thin rod, the right tool is the linear mass density \(\lambda\), the mass per unit length. A small element of length \(dx\) at position \(x\) then has mass \(dm = \lambda\,dx\). A rod is uniform when \(\lambda\) is the same everywhere, and then \(\lambda = M / L\).
\[ dm = \lambda\,dx, \qquad M = \lambda L \]
Why. the rod is uniform, so every element of length \(dx\) has the same mass per unit length \(\lambda\), as in Fig 2.1.
\[ x_{cm} = \dfrac{1}{M}\int_0^L x\,\lambda\,dx = \dfrac{\lambda}{\lambda L}\int_0^L x\,dx \]
Why. put \(dm = \lambda\,dx\) into the formula; the rod runs from \(x = 0\) to \(x = L\).
\[ x_{cm} = \dfrac{1}{L}\cdot\dfrac{L^2}{2} = \dfrac{L}{2} \]
Read it. \(\int_0^L x\,dx = L^2/2\), and \(\lambda\) cancels.
The answer, the middle of the rod, is what common sense expects. Its value lies in the method. The same three steps work for every rod, even one whose density changes along it. Write \(dm\) in terms of \(x\), set the limits, and integrate.
A rod that is heavier at one end
Now let the density grow along the rod. The simplest case is \(\lambda = kx\), where \(k\) is a constant: zero at the end \(x = 0\) and largest at \(x = L\). Both the mass and the numerator are integrals now, and \(k\) cancels at the end.
\[ M = \int_0^L kx\,dx = \dfrac{kL^2}{2} \]
Why. the total mass is the integral of \(dm = \lambda\,dx = kx\,dx\) over the rod.
\[ \int_0^L x\,dm = \int_0^L kx^2\,dx = \dfrac{kL^3}{3} \]
Why. each element contributes \(x\,dm = kx^2\,dx\).
\[ x_{cm} = \dfrac{kL^3/3}{kL^2/2} = \dfrac{2L}{3} \]
Read it. divide the numerator by the mass; \(k\) cancels, and so does \(L^2\).
The centre of mass is at \(2L/3\) from the light end, so it has moved from the middle towards the heavy end. This is the pattern for every rod. Where the mass is concentrated, the weighted average is pulled.
At \(b = 0\) the rod is uniform and C sits at \(L/2\). As \(b\) grows, the right end becomes heavier and C moves right: to \(0.556L\) at \(b = 1\), to \(0.600L\) at \(b = 3\), and to \(0.636L\) at \(b = 9\). It never reaches \(2L/3\). For a very large \(b\) the constant part of the density hardly matters, and the rod behaves like the \(\lambda = kx\) rod.
A rod \(2\,\text{m}\) long lies along the x axis from \(x = 0\). Its linear mass density is \(\lambda = (2 + 3x)\,\text{kg/m}\), with \(x\) in metres. Find its mass and the position of its centre of mass.
An element \(dx\) at \(x\) has mass \(dm = (2 + 3x)\,dx\). The rod runs from \(x = 0\) to \(x = 2\,\text{m}\).
\(M = \int_0^2 (2 + 3x)\,dx\) and \(\int x\,dm = \int_0^2 (2x + 3x^2)\,dx\).
\(M = [2x + 1.5x^2]_0^2 = 4 + 6 = 10\,\text{kg}\). The numerator is \([x^2 + x^3]_0^2 = 4 + 8 = 12\,\text{kg m}\). So \(x_{cm} = 12 / 10 = 1.2\,\text{m}\).
The rod has a mass of \(10\,\text{kg}\), and its centre of mass is \(1.2\,\text{m}\) from the light end. That is past the middle, towards the denser end, as it must be.
A rod \(3\,\text{m}\) long lies along the x axis from \(x = 0\), with linear mass density \(\lambda = (1 + x)\,\text{kg/m}\). Find the position of its centre of mass, in metres.
The linear mass density of a rod of length \(L\) grows as the square of the distance from one end, \(\lambda = kx^2\). Where is its centre of mass, measured from that end?
Symmetry does the integral
Many bodies need no integral at all. Suppose a uniform body is symmetric about the y axis, so that for every element \(dm\) at \(x\) there is an equal element at \(-x\). In the integral \(\int x\,dm\) each pair contributes \(x\,dm - x\,dm = 0\). So \(x_{cm} = 0\): the centre of mass lies on the axis of symmetry.
If the mass of a body is distributed symmetrically about a line or a plane, its centre of mass lies on that line or plane. If there are two lines of symmetry, it lies at their crossing.
Fig 2.2 shows the consequences. A uniform ring, disc, sphere or rectangular plate has two or more lines of symmetry, and they cross at the geometric centre. For the ring, that centre is empty: the centre of mass lies where there is no material at all. A bangle, a tyre and a hollow ball all have their centres of mass in empty space.
A triangle has no line of symmetry in general, yet a symmetry argument still works. Cut a uniform triangular plate into thin strips parallel to one side. Each strip is a thin uniform rod, so its centre of mass is at its own middle. The middles of all the strips lie on one line, the median from the opposite corner. So C lies on that median, and by the same argument on the other two medians. The three medians meet at the centroid, one third of the height above each side.
Symmetry of shape is not enough. The mass must also be symmetric. A cricket ball with a heavy lump on one side has a round shape, but its centre of mass is not at its centre. Use the symmetry shortcut only for bodies that are uniform, or whose density is itself symmetric.
Where is the centre of mass of a uniform circular bangle?
For which of these uniform bodies does the centre of mass lie outside the material of the body?
A uniform rectangular plate has sides \(a\) and \(b\). Without any integral, explain why its centre of mass is at its geometric centre.
The plate is symmetric about the line through its middle parallel to side \(b\). Every element on one side of that line has an equal partner at the same distance on the other side. Their contributions to the weighted average cancel, so C lies on the line. The same argument puts C on the second line of symmetry, parallel to side \(a\). The two lines cross only at the geometric centre, so C is there.
At the 1968 Olympic Games in Mexico City, the American high jumper Dick Fosbury went over the bar head first and on his back. His body arched around the bar like a horseshoe. When a body bends like that, its centre of mass can lie outside the body, below the bar. So a jumper can clear a bar while his centre of mass passes just under it, which saves the energy of lifting it higher. Almost every high jumper now uses the Fosbury flop.
Return to the bamboo pole of the opening. Take a simple model in which the mass per metre grows in proportion to the distance from the thin tip, \(\lambda = kx\). A pole \(3\,\text{m}\) long then balances at \(2L/3 = 2\,\text{m}\) from the tip, only \(1\,\text{m}\) from the thick end. That is why you hold it so far from its middle.
- Integrating \(x\,dm\) but taking \(M\) as length times the density at one end.When \(\lambda\) changes along the rod, the mass is itself an integral, \(M = \int \lambda\,dx\).
- Using symmetry for a body whose density is not symmetric.The shortcut needs the mass to be symmetric, not only the shape.
- Thinking that the centre of mass must be a point of the body.It is a weighted average of positions. For a ring, a hollow ball or a bent body it lies in empty space.