7.3 Centre of mass of continuous bodies

(a) Definition

A body with a continuous distribution of mass may be divided into \(N\) small elements of masses \(m_j\) at positions \(\mathbf r_j\). The formula of Section 7.1 then applies only approximately, because the elements are not true particles, but the approximation becomes exact in the limit \(N\to\infty\) in which the size of each element tends to zero. The limit of the sum defines an integral,

\[\lim_{N\to\infty}\sum_{j=1}^{N}m_j\mathbf r_j=\int\mathbf r\,dm,\]

where \(dm\) is a differential mass element. Hence

\[\boxed{\mathbf R_{\text{CM}}=\frac1M\int\mathbf r\,dm},\qquad X_{\text{CM}}=\frac1M\int x\,dm,\quad Y_{\text{CM}}=\frac1M\int y\,dm,\quad Z_{\text{CM}}=\frac1M\int z\,dm.\]

To evaluate the integral we regard \(dm\) as the mass in a volume element \(dV\) at position \(\mathbf r\) (Figure 7.2, right). If the density there is \(\rho\), then \(dm=\rho\,dV\) and \(\mathbf R_{\text{CM}}=\frac1M\int\mathbf r\,\rho\,dV\), a volume integral. For a thin plate with surface density \(\sigma\) we write \(dm=\sigma\,dA\), and for a thin wire or rod with linear density \(\lambda\) we write \(dm=\lambda\,dl\). In practice the integral is set up in the coordinates that suit the symmetry of the body: plane polar coordinates for discs and rings, with \(dA=r\,dr\,d\theta\), and spherical polar coordinates for spheres and hemispheres, with \(dV=r^2\sin\theta\,dr\,d\theta\,d\phi\).

(b) Bodies with cavities

If a portion of mass \(M_1\) with centre of mass \(\mathbf r_1\) is removed from a body of mass \(M\) whose centre of mass is \(\mathbf r\), the remaining portion and the removed portion together make up the original body. Hence \(M\mathbf r=(M-M_1)\mathbf R+M_1\mathbf r_1\), where \(\mathbf R\) is the centre of mass of the remainder, and

\[\boxed{\mathbf R=\frac{M\mathbf r-M_1\mathbf r_1}{M-M_1}}.\]

The removed part behaves as a body of negative mass. The same device applies to moments of inertia (Section 7.6).

(c) Worked examples

Example 7.3 A rod of length \(L\) lies along the \(x\)-axis between \(x=0\) and \(x=L\). Its linear density varies with the distance from the origin as \(\lambda=Rx\), where \(R\) is a positive constant. Find the position of its centre of mass.

Solution. The element of length \(dx\) at \(x\) has mass \(dm=\lambda\,dx=Rx\,dx\) and centre of mass \((x,0,0)\). Hence

\[x_{\text{CM}}=\frac{\int_0^Lx\,dm}{\int_0^Ldm}=\frac{R\int_0^Lx^2\,dx}{R\int_0^Lx\,dx}=\frac{L^3/3}{L^2/2}=\frac{2L}{3}.\]

Since \(y=z=0\) for every element, \(y_{\text{CM}}=z_{\text{CM}}=0\), and the centre of mass is at \((2L/3,0,0)\). It lies beyond the midpoint because the rod is heavier towards \(x=L\).

Example 7.4 Find the centre of mass of a thin rectangular plate with sides \(a\) and \(b\) whose mass per unit area varies as \(\sigma=\sigma_0xy/ab\), where \(\sigma_0\) is a constant.

Solution. Place the plate with corners at \((0,0)\) and \((a,b)\). Its mass is

\[M=\int_0^b\int_0^a\sigma_0\frac{xy}{ab}\,dx\,dy=\frac{\sigma_0}{ab}\cdot\frac{a^2}{2}\cdot\frac{b^2}{2}=\frac14\sigma_0ab.\]

The \(x\)-coordinate of the centre of mass is

\[\begin{aligned}X&=\frac1M\int_0^b\int_0^ax\,\sigma_0\frac{xy}{ab}\,dx\,dy=\frac1M\cdot\frac{\sigma_0}{ab}\cdot\frac{a^3}{3}\cdot\frac{b^2}{2}\\ &=\frac{4}{\sigma_0ab}\cdot\frac{\sigma_0a^2b}{6}=\frac23a,\end{aligned}\]

and in the same way \(Y=\frac23b\). Hence

\[\mathbf R=\frac23\left(a\hat{\mathbf i}+b\hat{\mathbf j}\right).\]

Example 7.5 Find the centre of mass of a uniform semicircular ring of mass \(M\) and radius \(R\).

Solution. The length of the ring is \(\pi R\), so its mass per unit length is \(M/\pi R\). The arc subtending \(d\theta\) at the centre has length \(R\,d\theta\) and mass \(dm=(M/\pi)\,d\theta\), and is located at \(x=R\cos\theta\), \(y=R\sin\theta\). Then

\[X_{\text{CM}}=\frac1M\int_0^\pi R\cos\theta\,\frac{M}{\pi}\,d\theta=0,\qquad Y_{\text{CM}}=\frac1M\int_0^\pi R\sin\theta\,\frac{M}{\pi}\,d\theta=\frac R\pi\Big[-\cos\theta\Big]_0^\pi=\frac{2R}{\pi}.\]

Hence

\[\boxed{\text{semicircular ring: } Y_{\text{CM}}=\frac{2R}{\pi}}.\]

Example 7.6 Find the centre of mass of a uniform semicircular disc of mass \(M\) and radius \(R\).

Solution. The surface density is \(M/(\frac12\pi R^2)=2M/\pi R^2\). With \(x=r\cos\theta\), \(y=r\sin\theta\) the mass element is \(dm=(2M/\pi R^2)\,r\,dr\,d\theta\), with \(0\le r\le R\) and \(0\le\theta\le\pi\). By symmetry about the \(y\)-axis \(X_{\text{CM}}=0\), as the integral \(\int_0^\pi\cos\theta\,d\theta=0\) confirms. Then

\[Y_{\text{CM}}=\frac1M\int y\,dm=\frac{2}{\pi R^2}\int_0^Rr^2\,dr\int_0^\pi\sin\theta\,d\theta=\frac{2}{\pi R^2}\cdot\frac{R^3}{3}\cdot2,\]

that is,

\[\boxed{\text{semicircular disc: } Y_{\text{CM}}=\frac{4R}{3\pi}}.\]

Example 7.7 Find the centre of mass of a uniform hollow hemisphere (a hemispherical shell) of radius \(R\) and mass \(M\).

Solution. Use spherical polar coordinates, \(x=R\sin\theta\cos\phi\), \(y=R\sin\theta\sin\phi\), \(z=R\cos\theta\), with \(0\le\theta\le\pi/2\) and \(0\le\phi\le2\pi\). The area of the shell is \(2\pi R^2\), so

\[dm=\frac{M}{2\pi R^2}R^2\sin\theta\,d\theta\,d\phi=\frac{M}{2\pi}\sin\theta\,d\theta\,d\phi.\]

The \(x\)- and \(y\)-coordinates vanish because \(\int_0^{2\pi}\cos\phi\,d\phi=\int_0^{2\pi}\sin\phi\,d\phi=0\). For the \(z\)-coordinate,

\[Z_{\text{CM}}=\frac1M\int z\,dm=\frac{1}{2\pi}\int_0^{\pi/2}R\cos\theta\sin\theta\,d\theta\int_0^{2\pi}d\phi=R\int_0^{\pi/2}\frac12\sin2\theta\,d\theta=\frac R2.\]

Hence

\[\boxed{\text{hemispherical shell: } Z_{\text{CM}}=\frac R2}.\]

The same result follows by noting that the shell between heights \(z\) and \(z+dz\) is a ring of radius \(R\sin\theta\) and width \(R\,d\theta\), whose area \(2\pi R\sin\theta\cdot R\,d\theta=2\pi R\,dz\) is independent of \(z\); the mass is therefore spread uniformly in height, and its mean height is \(R/2\).

Example 7.8 Find the centre of mass of a uniform solid hemisphere of radius \(R\) and mass \(M\).

Solution. By symmetry the centre of mass lies on the \(z\)-axis, at a height \(Z=\frac1M\int z\,dm\) above the equatorial plane. In spherical polar coordinates the volume of the hemisphere is \(\frac23\pi R^3\), so

\[dm=\frac{M}{\frac23\pi R^3}r^2\sin\theta\,dr\,d\theta\,d\phi=\frac{3M}{2\pi R^3}r^2\sin\theta\,dr\,d\theta\,d\phi.\]

The integrals for \(X_{\text{CM}}\) and \(Y_{\text{CM}}\) vanish, and

\[\begin{aligned}Z_{\text{CM}}&=\frac{3}{2\pi R^3}\int_0^Rr^3\,dr\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta\int_0^{2\pi}d\phi\\ &=\frac{3}{2\pi R^3}\cdot\frac{R^4}{4}\cdot\frac12\cdot2\pi=\frac38R.\end{aligned}\]

The integral is three-dimensional, but the symmetry allows it to be reduced to a single integral. We divide the hemisphere into thin discs parallel to the base (Figure 7.3). The disc at height \(z\) with thickness \(dz\) has radius \(r\), where \(r^2=R^2-z^2\), and volume \(dV=\pi r^2\,dz\). Its mass is \(dM=(M/V)\,dV\) with \(V=\frac23\pi R^3\), so

\[Z=\frac1V\int_0^R\pi r^2z\,dz=\frac\pi V\int_0^Rz(R^2-z^2)\,dz=\frac\pi V\left(\frac12R^4-\frac14R^4\right)=\frac{\frac14\pi R^4}{\frac23\pi R^3}=\frac38R.\]

Hence

\[\boxed{\text{solid hemisphere: } Z_{\text{CM}}=\frac38R}.\]

Example 7.9 Find the centre of mass of a uniform triangular plate of mass \(M\) with vertices at \(O\), \((l,h)\) and \((l,-h)\), as shown.

Solution. The area of the plate is \(\frac12(2h)l=hl\), so \(dm=(M/hl)\,dx\,dy\). At a given \(x\) the plate extends from \(y=-hx/l\) to \(y=hx/l\). Hence

\[X_{\text{CM}}=\frac{1}{hl}\int_0^lx\,dx\int_{-hx/l}^{hx/l}dy=\frac{1}{hl}\cdot\frac{2h}{l}\int_0^lx^2\,dx=\frac{2}{l^2}\cdot\frac{l^3}{3}=\frac{2l}{3},\]

and

\[Y_{\text{CM}}=\frac{1}{hl}\int_0^ldx\int_{-hx/l}^{hx/l}y\,dy=0,\]

since the integrand is odd in \(y\). The centre of mass is at \((2l/3,0)\), which is the intersection of the medians, as it must be for any uniform triangle.

Example 7.10 Find the centre of mass of the uniform lamina shown, which is a disc of radius \(a\) from which a disc of radius \(a/2\) touching the rim and passing through the centre \(O\) has been removed.

Solution. Let the complete disc, of area \(\pi a^2\), have mass \(M\) and centre of mass \((0,0)\). The removed disc has area \(\pi(a/2)^2=\pi a^2/4\), so its mass is \(M_1=M/4\), and its centre is at \((a/2,0)\). By Section 7.3(b),

\[X_{\text{CM}}=\frac{0\cdot M-\frac a2\cdot\frac M4}{M-\frac M4}=-\frac a6,\qquad Y_{\text{CM}}=0.\]

The centre of mass of the lamina is at \((-a/6,0)\).

Example 7.11 A square plate of side \(l\) consists of two halves, the left half of density \(\rho_1=2\rho\) and the right half of density \(\rho_2=\rho\) (mass per unit area). A circular hole of radius \(l/8\) is cut from the right half, with its centre at the centre of that half. Find the centre of mass of the remaining plate.

Solution. Take the origin at the corner \(O\). The left half (1) has area \(l^2/4\) and centre \((l/4,l/2)\); the right half (2) has area \(l^2/4\) and centre \((3l/4,l/2)\); the hole (3) has area \(\pi(l/8)^2=\pi l^2/64\) and centre \((3l/4,l/2)\). Treating the hole as a negative mass of density \(\rho\),

\[\begin{aligned}x_{\text{CM}}&=\frac{2\rho\frac{l^2}{4}\cdot\frac l4+\rho\frac{l^2}{4}\cdot\frac{3l}{4}-\rho\frac{\pi l^2}{64}\cdot\frac{3l}{4}}{2\rho\frac{l^2}{4}+\rho\frac{l^2}{4}-\rho\frac{\pi l^2}{64}}\\ &=\frac{\frac18+\frac3{16}-\frac{3\pi}{256}}{\frac34-\frac{\pi}{64}}\,l=\frac{80-3\pi}{192-4\pi}\,l\approx0.39\,l.\end{aligned}\]

Every part has its centre at height \(l/2\), so \(y_{\text{CM}}=0.5\,l\). The centre of mass is at \((0.39\,l,\ 0.5\,l)\), displaced to the left of the centre of the square because the left half is the denser.