The Thanjavur doll of Tamil Nadu, the Thalaiyatti Bommai, stands on a round base. Push it over and it rocks back and forth, and then it stands upright again. Push it harder and it still rises. Its secret lies in where its mass is placed. The heavy rounded base must pull its centre of mass low enough, and this lesson finds how low is enough.
The semicircular ring
A ring bent into a half circle is the first shape whose centre of mass needs a real integral. Place the ring of radius \(R\) and mass \(M\) with its diameter along the x axis and its centre at the origin O. The y axis is a line of symmetry, so \(x_{cm} = 0\). Only \(y_{cm}\) is left to find.
An element of the ring at angle \(\theta\) from the x axis has length \(R\,d\theta\). With \(\lambda = M/\pi R\), its mass is \(dm = \lambda R\,d\theta\), and its height is \(y = R\sin\theta\). The angle runs from \(0\) to \(\pi\) along the ring.
\[ y_{cm} = \dfrac{1}{M}\int y\,dm = \dfrac{1}{M}\int_0^{\pi} (R\sin\theta)(\lambda R\,d\theta) \]
Why. the definition, with \(y = R\sin\theta\) and \(dm = \lambda R\,d\theta\).
\[ y_{cm} = \dfrac{\lambda R^2}{M}\int_0^{\pi}\sin\theta\,d\theta = \dfrac{\lambda R^2}{M}\cdot 2 \]
Why. \(\lambda R^2\) is constant, and \(\int_0^{\pi}\sin\theta\,d\theta = [-\cos\theta]_0^{\pi} = 2\).
\[ y_{cm} = \dfrac{2\lambda R^2}{\lambda\pi R} = \dfrac{2R}{\pi} \]
Read it. put \(M = \lambda\pi R\), the density times the length of the half circle.
The centre of mass lies on the axis at \(2R/\pi \approx 0.64R\) above the diameter. It is in empty space, inside the curve, as the last lesson led us to expect for a curved wire.
A uniform wire is bent into a semicircle of radius \(14\,\text{cm}\). How far from the centre of the semicircle is its centre of mass? Take \(\pi = 22/7\).
The wire is a semicircular ring, so \(y_{cm} = 2R/\pi\) on the axis of symmetry.
\(y_{cm} = \dfrac{2 \times 14}{22/7} = \dfrac{28 \times 7}{22} \approx 8.9\,\text{cm}\).
The centre of mass is \(8.9\,\text{cm}\) from the centre along the axis, inside the curve, where there is no wire.
The semicircular disc
A semicircular disc, or half-disc lamina, is a set of semicircular rings one inside another. The ring of radius \(r\) and width \(dr\) has area \(\pi r\,dr\). With the surface mass density \(\sigma = M/(\tfrac12\pi R^2)\), its mass is \(dm = \sigma\pi r\,dr\). Each such ring has its own centre of mass at \(2r/\pi\), by the result just found. So
\[ y_{cm} = \dfrac{1}{M}\int_0^R \dfrac{2r}{\pi}\,\sigma\pi r\,dr = \dfrac{2\sigma}{M}\cdot\dfrac{R^3}{3} = \dfrac{4R}{3\pi} \]
after putting \(M = \tfrac12\sigma\pi R^2\). The disc's centre of mass, at \(0.42R\), is lower than the ring's, at \(0.64R\). The disc has mass near its centre as well as at its rim, and that inner mass pulls the average down.
The hemispherical shell
Stand a thin hemispherical shell of radius \(R\) on its rim, with its axis along z. Cut it by horizontal planes into thin bands. The band between heights \(z\) and \(z + dz\) is a narrow strip around the shell. A remarkable fact, known to Archimedes, makes this shell easy: every band of height \(dz\) has the same area, \(2\pi R\,dz\), whatever its height.
Here is why. A band at polar angle \(\theta\) from the axis has radius \(R\sin\theta\) and slant width \(R\,d\theta\), so its area is \(2\pi R^2\sin\theta\,d\theta\). Its height is \(z = R\cos\theta\), so \(|dz| = R\sin\theta\,d\theta\). The area is therefore \(2\pi R\,|dz|\). Near the top the band is small around but wide along the slope, and near the rim it is large around but steep. The two effects cancel exactly.
So the mass of the shell is spread evenly in height, exactly like the mass of a uniform rod along its length. Its centre of mass is at half the height: \(z_{cm} = R/2\) above the flat face.
The solid hemisphere
A solid hemisphere is a stack of discs. The disc at height \(z\) has radius \(\sqrt{R^2 - z^2}\) and thickness \(dz\). With density \(\rho\), its mass is \(dm = \rho\pi(R^2 - z^2)\,dz\). Large discs near the flat face carry most of the mass, so the centre of mass must be lower than \(R/2\).
\[ M = \int_0^R \rho\pi(R^2 - z^2)\,dz = \rho\pi\left(R^3 - \dfrac{R^3}{3}\right) = \dfrac{2}{3}\rho\pi R^3 \]
Why. add the discs from the flat face \(z = 0\) to the top \(z = R\); the result is half the volume of a sphere times \(\rho\).
\[ \int z\,dm = \rho\pi\int_0^R (R^2 z - z^3)\,dz = \rho\pi\left(\dfrac{R^4}{2} - \dfrac{R^4}{4}\right) = \dfrac{\rho\pi R^4}{4} \]
Why. each disc contributes its height \(z\) times its mass.
\[ z_{cm} = \dfrac{\rho\pi R^4/4}{2\rho\pi R^3/3} = \dfrac{3R}{8} \]
Read it. divide; \(\rho\pi\) cancels, and so do three powers of \(R\).
The solid cone
A solid cone of height \(h\) and base radius \(R\), standing on its base, is also a stack of discs. The disc at height \(z\) has radius \(R(1 - z/h)\), shrinking to zero at the tip. Fill in the missing step of the derivation below.
\[ dm = \rho\pi R^2\left(1 - \dfrac{z}{h}\right)^2 dz \]
Why. a disc of radius \(R(1 - z/h)\) and thickness \(dz\).
?
Your turn. add the discs from the base to the tip. Which line gives the mass of the cone?
\[ \int z\,dm = \rho\pi R^2\int_0^h z\left(1 - \dfrac{z}{h}\right)^2 dz = \dfrac{\rho\pi R^2 h^2}{12} \]
Why. expand the bracket: \(\int_0^h (z - 2z^2/h + z^3/h^2)\,dz = h^2/2 - 2h^2/3 + h^2/4 = h^2/12\).
\[ z_{cm} = \dfrac{h^2/12}{h/3} = \dfrac{h}{4} \]
Read it. divide the numerator by the mass.
The cone's centre of mass is at \(h/4\) above its base, or \(3h/4\) below its tip. It is far below the middle of the height, because the wide part of a cone near its base holds most of its volume.
Step through the chips of Fig 3.1 and compare the heights. Mass far from the flat side raises C, and mass close to it lowers C. The ring, with all its mass on the curve, has the highest centre of mass. The shell spreads its mass evenly in height and sits at \(R/2\). The disc and the solid hemisphere hold mass near their flat sides, so their centres of mass are lower.
Four uniform bodies have the same radius \(R\). Rank them by the distance of the centre of mass from the centre of the flat side, largest first.
A uniform semicircular plate has a radius of \(15\,\text{cm}\). How far is its centre of mass from the centre of its straight edge, in cm?
Why is the centre of mass of a semicircular ring farther from the diameter than that of a semicircular disc of the same radius?
Archimedes of Syracuse found the centres of gravity of a triangle and other plane figures in his book On the Equilibrium of Planes. In a letter now called The Method, he also found the centre of gravity of a solid hemisphere. It divides the axis in the ratio five to three, the longer part towards the curved surface. That is our \(3R/8\). The Method was lost for centuries. In 1906 the Danish scholar Johan Heiberg recognised it under the prayers of a monk's prayer book in Constantinople.
Bodies built from pieces
Most real objects are built from simple pieces: a plate and a rod, a cone on a hemisphere, two rectangles in an L. Such a composite body needs no new integral. Each piece can be replaced by a point mass, equal to the mass of the piece, at the centre of mass of that piece. Then the particle formula of the first lesson finishes the job.
This works because the integral \(\int x\,dm\) over the whole body is the sum of the integrals over its pieces. For each piece, \(\int x\,dm = m_{\text{piece}}\,x_{\text{piece}}\) by the definition of its own centre of mass. So the whole sum is \(\sum m_{\text{piece}}\,x_{\text{piece}}\), exactly the particle formula.
A uniform L-shaped plate is made of two rectangles, as in Fig 3.2. One, \(40\,\text{cm}\) by \(10\,\text{cm}\), lies along the x axis; the other, \(10\,\text{cm}\) by \(30\,\text{cm}\), stands on its left end. Find its centre of mass.
The plate is uniform, so each piece's mass is proportional to its area: \(A_1 = 400\,\text{cm}^2\) at \(C_1 = (20, 5)\) and \(A_2 = 300\,\text{cm}^2\) at \(C_2 = (5, 25)\).
\(x_{cm} = \dfrac{400 \times 20 + 300 \times 5}{700}\) and \(y_{cm} = \dfrac{400 \times 5 + 300 \times 25}{700}\).
\(x_{cm} = 9500/700 \approx 13.6\,\text{cm}\) and \(y_{cm} = 9500/700 \approx 13.6\,\text{cm}\).
C is at \((13.6, 13.6)\,\text{cm}\). That point has \(x > 10\) and \(y > 10\), so it lies in the empty corner of the L, outside the material.
A uniform rod of mass \(2\,\text{kg}\) and length \(1\,\text{m}\) has a small \(3\,\text{kg}\) ball fixed at one end. How far is the centre of mass of the pair from the other, free end of the rod, in metres?
The doll that rights itself
Now return to the doll. A simple model is a solid cone of height \(h\) standing on a solid hemisphere of radius \(R\), both of the same wood. The doll rests on its curved base. Tilt it a little, and the point of contact moves along the curve. The weight then turns the doll back upright only if the centre of mass lies below O. O is the centre of the flat face and of the curved base.
A solid cone of height \(h\) stands with its base on the flat face of a solid hemisphere of radius \(R\). Both are made of the same material. For what heights \(h\) does the centre of mass lie below O, the centre of the flat face?
Take O as the origin and the axis as y, upward. The cone has mass \(\tfrac13\pi R^2 h\rho\) at \(+h/4\). The hemisphere has mass \(\tfrac23\pi R^3\rho\) at \(-3R/8\).
\(y_{cm} = \dfrac{\tfrac13\pi R^2 h\rho \cdot \tfrac{h}{4} - \tfrac23\pi R^3\rho \cdot \tfrac{3R}{8}}{M}\), where \(M\) is the total mass.
The numerator is \(\pi\rho R^2\left(\dfrac{h^2}{12} - \dfrac{R^2}{4}\right)\). It is negative when \(h^2 < 3R^2\), that is when \(h < \sqrt{3}\,R\).
The centre of mass lies below O, and the doll rights itself, when \(h < \sqrt3\,R \approx 1.73R\). A taller cone lifts C above O, and the doll topples.
Real Thanjavur dolls go further. Their bases are not solid wood but hollow, with a heavy lump of clay at the bottom. That lowers the centre of mass far below O. So the doll rises even from a large tilt, and its body can be tall and slender.
- Mixing up the ring and the disc results.A thin wire or ring gives \(2R/\pi\); a flat plate gives \(4R/3\pi\), which is smaller.
- Measuring the cone's \(h/4\) from the tip.\(h/4\) is measured from the base. From the tip the distance is \(3h/4\).
- Treating a piece of a composite body as a point at its geometric middle when it is not symmetric.Place each piece at its own centre of mass, from the table of standard results where needed.