Where is the centre of mass of half this cylinder?

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The volume of the lower half can be thought of as the sum of the volumes of thin vertical slices parallel to the \(y\)-axis.

Each slice has a shape approximating a cuboid, with thickness \(\delta x\).

To find the height of a slice we look at the front view of the cylinder. The slice at distance \(x\) from the origin has height \(z=x+a\).

For the width of a slice, we look at the plan view. The base is a circle of radius \(a\) and Pythagoras’ theorem tells us that \(x^2+y^2=a^2\). The width of the slice is twice the \(y\) value, namely \(2\sqrt{a^2 - x^2}\).

The volume of each slice, then, is \[\begin{equation*} 2\sqrt{a^2 - x^2} (x+a)\delta x, \end{equation*}\] and so the total volume of the bottom half (as in the limit, the sum becomes an integral) is \[\begin{equation*} \int_{-a}^a 2\sqrt{a^2 - x^2} (x+a) \:dx, \end{equation*}\]

as required.

Take \(x = a \sin \theta\), so that \(dx = a \cos \theta\: d\theta\). Then \[\begin{align*} V = \int_{-a}^a 2\sqrt{a^2 - x^2} (x+a) \:dx &= \int_{-\pi/2}^{\pi/2} 2 \sqrt{a^2 - a^2 \sin^2 \theta} (a \sin \theta + a) a \cos \theta \:d\theta \\ &= 2a^3\int_{-\pi/2}^{\pi/2} \cos \theta (\sin \theta + 1) \cos \theta \:d\theta \\ &= 2a^3\int_{-\pi/2}^{\pi/2} \cos^2 \theta \sin \theta + \cos^2 \theta \:d\theta . \end{align*}\] The first term can be integrated by substituting \(u=\cos\theta\), \(du=-\sin\theta \:d\theta\). The second term can be expanded using a trig identity. \[\begin{align*} V &= 2a^3\left\{\int_{0}^{0} -u^2 \: du + \int_{-\pi/2}^{\pi/2} \dfrac{\cos 2\theta+1}{2} \:d\theta \right\} \\ &= 2a^3\left\{ 0 + \left[\dfrac{\sin 2\theta}{4} +\dfrac{\theta}{2}\right]_{-\pi/2}^{\pi/2} \right\}\\ &= 2a^3\left\{ 0+0+\dfrac{\pi}{2} \right\}\\ &= \pi a^3 \end{align*}\]

Since the object has reflective symmetry in the \(x-z\) plane, the centre of mass must lie in this plane, i.e. the plane \(y = 0\).

The \(x\)-coordinate, \(x_0\), of the centre of mass must be such that the mass would balance around it. Taking moments we have that \[\begin{equation*} \int_{-a}^a x \times 2\sqrt{a^2 - x^2} (x+a) \:dx = \pi a^3 \times x_0. \end{equation*}\]

To find this integral we’ll use the substitution \(x = a \sin \theta\) again and trig identities for \(\sin 2\theta\) and \(\cos 2\theta\). We have

\[\begin{align*} \int_{-a}^a x \times 2 \sqrt{a^2 - x^2} (x+a) \:dx &= \int_{-\pi/2}^{\pi/2} a\sin \theta \times 2 \sqrt{a^2 - a^2 \sin^2 \theta} (a \sin \theta + a) a \cos \theta \:d\theta \\ &= 2a^4\int_{-\pi/2}^{\pi/2} \sin \theta \cos \theta ( \sin \theta + 1) \cos \theta \:d\theta \\ &= 2a^4\int_{-\pi/2}^{\pi/2} \cos^2 \theta \sin^2 \theta + \cos^2 \theta\sin \theta \:d\theta \\ &= 2a^4\int_{-\pi/2}^{\pi/2} \dfrac{\sin^2 2\theta}{4} + \cos^2 \theta\sin \theta \:d\theta \\ &= 2a^4\int_{-\pi/2}^{\pi/2} \dfrac{1-\cos 4\theta}{8} + \cos^2 \theta\sin \theta \:d\theta \\ &= 2a^4\left[\dfrac{\theta}{8}-\dfrac {\sin 4\theta}{32} - \dfrac{\cos^3 \theta}{3}\right]_{-\pi/2}^{\pi/2} \\ &= \dfrac{\pi a^4}{4}. \end{align*}\]

Thus \(\dfrac{\pi a^4}{4} = \pi a^3 x_0 \implies x_0 = \dfrac{a}{4},\) as required, and so the centre of mass of the lower half lies directly above the point \((\frac{1}{4}a,0)\).