A worker carries a long bamboo pole across a building site in Guwahati. The pole is thick at the base and thin at the tip. He does not hold it at the middle. He slides it on his shoulder until it balances, and the balance point is well towards the thick end. A rod of the same length and of the same thickness all along would balance at its middle. Where exactly is the balance point of the tapering pole? No formula for a single particle can answer this.
Cut the body into elements
A particle has one position, so its contribution to a sum is easy to write. A rod has mass spread along its whole length. The method of this lesson is to cut the rod into short pieces called elements. Each element is so short that all of its mass sits, very nearly, at one position \(x\). Then the rod behaves like a long row of particles, and every sum over particles becomes an integral.
The linear density \(\lambda\) of a rod at a point is its mass per unit length there, in kg/m. An element of length \(dx\) at position \(x\) has mass \(dm = \lambda(x)\,dx\). For a sheet, the mass per unit area \(\sigma\) gives \(dm = \sigma\,dA\). For a solid, the density \(\rho\) gives \(dm = \rho\,dV\).
For a uniform rod of mass \(M\) and length \(L\), the density is the same everywhere, \(\lambda = M/L\). For the bamboo pole, \(\lambda\) is larger near the base and changes with \(x\). The integral handles both cases in the same way. It adds the masses \(\lambda(x)\,dx\) of all the elements, from one end of the rod to the other. On a graph of \(\lambda\) against \(x\), each element's mass is one thin strip. So the mass of the rod is the area under the graph.
A rod of length \(L\) lies from \(x = 0\) to \(x = L\), and its linear density is \(\lambda = kx\). What is its mass?
The balance point of a rod
For a row of particles, the centre of mass is \(\sum m_i x_i / \sum m_i\). Each mass is weighted by its position. For a rod, each element \(dm = \lambda\,dx\) sits at \(x\), so the sum becomes \(\int x\,dm\). Divide by the total mass. The rod balances on a support placed at this point. For a uniform rod, \(\int_0^L x\,(M/L)\,dx = ML/2\), so the balance point is \(L/2\), the middle, as expected.
This is the worked example "A rod with a linear density" of Centre of Mass, lesson 2, done with the tool. 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 balances \(1.2\,\text{m}\) from its light end, past the middle and towards the dense end. These are the physics chapter's numbers, and the values that the figure's sums approach.
A rod \(1.5\,\text{m}\) long has linear density \(\lambda = kx\), zero at the end \(x = 0\). How far from that end is its centre of mass, in metres?
Choose the element to fit the shape
The element must fit the question. A moment of inertia needs \(r^2\,dm\), where \(r\) is the distance from the axis. So each element must have one distance from the axis. For a disc turning about its centre, a thin ring does this. Every point of a ring of radius \(r\) is the same distance \(r\) from the centre. The ring has circumference \(2\pi r\) and width \(dr\), so its area is \(dA = 2\pi r\,dr\).
The rings together make the area of a circle: \(\int_0^R 2\pi r\,dr = \pi R^2\). A sphere is built the same way from thin shells. A shell of radius \(r\) and thickness \(dr\) has volume \(4\pi r^2\,dr\), its surface area times its thickness. The shells together make the volume, \(\int_0^R 4\pi r^2\,dr = \tfrac43\pi R^3\). Two formulas from school geometry now follow from one method.
\[ dm = \sigma\,2\pi r\,dr, \quad \sigma = \dfrac{M}{\pi R^2} \]
Why. a ring of radius \(r\) and width \(dr\) has area \(2\pi r\,dr\), and the disc has mass \(\sigma\) per unit area.
\[ dm = \dfrac{2M}{R^2}\,r\,dr \]
Why. multiply out the constants: \(2\pi \times M/(\pi R^2) = 2M/R^2\).
\[ I = \int_0^R r^2\,dm = \dfrac{2M}{R^2}\int_0^R r^3\,dr \]
Why. every point of the ring is at the same distance \(r\) from the axis, so the ring contributes \(r^2\,dm\).
\[ I = \dfrac{2M}{R^2} \cdot \dfrac{R^4}{4} = \tfrac12 MR^2 \]
Read it. the reverse power rule gives \(\int_0^R r^3\,dr = R^4/4\).
A sphere is built from thin shells. What is the volume \(dV\) of the shell of radius \(r\) and thickness \(dr\)?
A uniform rod of mass \(0.6\,\text{kg}\) and length \(1.2\,\text{m}\) turns about one end. With \(dm = (M/L)\,dx\), its moment of inertia is \(I = \int_0^L x^2\,dm\). Find \(I\), in kg m².
An element must be small in the direction that matters. A thin straight strip across a disc fails for the moment of inertia about its centre. Its points lie at many different distances from the axis. A thin ring works, because all its points share one distance. Also keep the density and the element in the same variable: \(dm = \lambda(x)\,dx\), never \(\lambda\) at one point times the whole length.
The bamboo pole now has an answer. Model it as a rod whose density grows towards the base, and write \(dm = \lambda(x)\,dx\). The balance point is \(\int x\,dm\) divided by \(\int dm\). For the rod with \(\lambda = 2 + 3x\), it is \(1.2\,\text{m}\) from the thin end, not \(1\,\text{m}\). The worker's shoulder finds the same point by feel.
Integrating over a body is the method of every continuous body in the book. In Centre of Mass it gives the balance point of rods, arcs, discs, hemispheres and cones. In Rigid Bodies and Moment of Inertia it gives \(I\) for rods, rings, discs and spheres. In Gravitational Force and Field it adds the fields of a rod and a shell. In Electric Charge, Coulomb's Law and the Electric Field the same elements carry charge, \(dq = \lambda\,dx\). The stretch of a rope under its own weight and the emf of a rotating rod are found the same way.
- Multiplying the density at one end by the whole length.When \(\lambda\) changes, keep it inside the integral: \(M = \int \lambda(x)\,dx\).
- Forgetting to divide by the mass in the centre of mass.\(x_{cm} = \int x\,dm / M\). Check that the answer is a length inside the body.
- Using a strip where a ring is needed.Choose elements whose points all have the same distance from the axis or the point.