By symmetry, the
\(x\)-coordinate of the centre of mass will be
\(\bar x =0\text{;}\) that is, exactly in the middle, horizontally. To find the
\(y\)-coordinate of the centre of mass, we need to consider the origin of Equation
2.3.5.
We can make vertical strips or horizontal strips. A vertical strip of the circle has a density that varies from the bottom of the strip to the top, but a horizontal strip has a constant density (assuming the strip is very thin). So it seems that horizontal strips in this case will be the easier route.
Following the derivation of Equation
2.3.5, we model our circle as a vertical rod
\(R\text{,}\) filling the
\(y\)-interval
\([0, 6]\text{.}\) A portion of the rod with
\(a \le y \le b\) should have the same mass as the portion of the circle with
\(a \le y \le b\text{.}\) To achieve this, we slice the circle into thin horizontal strips of thickness
\(\dee{y}\text{,}\) calculate their mass, then use that to find
\(\rho(y)\text{,}\) the density of
\(R\text{.}\)
{First, let’s find a formula for the mass of a thin horizontal strip of the circle at position \(y\) with height \(\dee{y}\text{.}\)}
The circle with radius 3 centred at \((0,3)\) has equation \(x^2+(y-3)^2=9\text{.}\) So, the right half of the circle has equation \(x=\sqrt{9-(y-3)^2}\text{,}\) and the left half of the circle has equation \(x=-\sqrt{9-(y-3)^2}\text{.}\) So, the width of a strip at height \(y\) is \(2\sqrt{9-(y-3)^2}\) m. Its height is \(\dee{y}\) m, so its area is \(2\sqrt{9-(y-3)^2}\,\dee{y}\) m\(^2\text{.}\) Its density is \(2+y \frac{\text{kg}}{\text{m}^2}\) , so its mass is \(2(2+y)\sqrt{9-(y-3)^2}\,\dee{y}\) kg.
Now we can find \(\rho(y)\text{,}\) the density of \(R\) at position \(y\text{.}\) The mass of the section of \(R\) at position \(y\) with length \(\dee{y}\) is \(2(2+y)\sqrt{9-(y-3)^2}\,\dee{y}\) kg (the mass of the strip in the paragraph above), so its density is \(\frac{2(2+y)\sqrt{9-(y-3)^2}\,\dee{y} \text{kg}}{\dee{y} \text{m}} = 2(2+y)\sqrt{9-(y-3)^2} \frac{\text{kg}}{\text{m}} = \rho(y)\text{.}\)
Now, Equation
2.3.4 will tell us the centre of mass of
\(R\text{,}\) which is also the
\(y\)-coordinate of the centre of mass of the circle.
\begin{align*}
\bar y &= \frac{\int_a^b y\rho(y)\,\dee{y}}{\int_a^b \rho(y)\,\dee{y}} =
\frac{\int_0^6 y\times 2(2+y)\sqrt{9-(y-3)^2}\,\dee{y}}{\int_0^6 2(2+y)\sqrt{9-(y-3)^2}\,\dee{y}}\\
&=
\frac{\int_0^6 y(2+y)\sqrt{9-(y-3)^2}\,\dee{y}}{\int_0^6 (2+y)\sqrt{9-(y-3)^2}\,\dee{y}}\\
\end{align*}
To make things look a little cleaner, we use the substitution \(u=y-3\text{,}\) \(\dee{u}=\dee{y}\text{.}\) Then the limits of integration become \(-3\) and \(3\text{,}\) respectively, and \(y=u+3\text{.}\) (Geometrically, we’re re-centring the circle at the origin, instead of at the point (0,3).)
\begin{align*}
&=\frac{\int_{-3}^3 (u+3)(2+u+3)\sqrt{9-u^2}\,\dee{u}}{\int_{-3}^3 \big(2+u+3\big)\sqrt{9-u^2}\,\dee{u}}\\
&=\frac{\int_{-3}^3 \big(u^2+8u+15\big)\sqrt{9-u^2}\,\dee{u}}{\int_{-3}^3 \big(u+5\big)\sqrt{9-u^2}\,\dee{u}}
= \frac{N}{D}\tag{$*$}\\
\end{align*}
Let’s start by finding \(D\text{,}\) the integral of the denominator. If we break it into two pieces, we can use symmetry and geometry to evaluate it.
\begin{align*}
D&=\int_{-3}^3 u\sqrt{9-u^2}\,\dee{u}+5\int_{-3}^3 \sqrt{9-u^2}\,\dee{u}\\
\end{align*}
The left integrand is odd, so its integral over a symmetric interval is 0. (You can also evaluate this using the substitution \(w=9-u^2\text{,}\) \(\dee{w}=-2u\,\dee{u}\text{.}\)) The right integral represents the area underneath half a circle of radius 3, centred at the origin.
\begin{align*}
D&=0+5\cdot\frac{1}{2}\pi\cdot 3^2 = \frac{45}{2}\pi
\end{align*}
Now, let’s evaluate our numerator integral from (\(*\)), \(N=\int_{-3}^3 \big(u^2+8u+15\big)\sqrt{9-u^2}\,\dee{u}\text{.}\) If we break it into three pieces, we can simplify the integration somewhat.
\begin{align*}
N&=\int_{-3}^3 u^2\sqrt{9-u^2}\,\dee{u}+
8\int_{-3}^3 u\sqrt{9-u^2}\,\dee{u}+
15\int_{-3}^3 \sqrt{9-u^2}\,\dee{u}\\
\end{align*}
The first integrand is even, with a symmetric interval of integration, so we can simplify its limits of integration a little bit. The middle integrand is odd, so its integral over the symmetric interval \([-3,3]\) is zero. The last integral is the area of half a circle of radius 3.
\begin{align*}
N&=2\int_{0}^3 u^2\sqrt{9-u^2}\,\dee{u}+0+15\cdot\pi\cdot3^2\\
&=\frac{135}{2}\pi+2\int_{0}^3 u^2\sqrt{9-u^2}\,\dee{u}\\
\end{align*}
The remaining integral has a quadratic function underneath a square root with no obvious substitution, so we use a trigonometric substitution. Let \(u=3\sin\theta\text{,}\) \(\dee{u}=3\cos\theta\,\dee{\theta}\text{.}\) Note \(3\sin(0)=0\) and \(3\sin(\pi/2)=3\text{,}\) so the limits of integration become 0 and \(\frac{\pi}{2}\text{.}\)
\begin{align*}
N&=\frac{135}{2}\pi+2\int_{0}^{\pi/2} \big(3\sin\theta\big)^2\sqrt{9-\big(3\sin\theta\big)^2}\cdot 3\cos\theta\,\dee{\theta}\\
&=\frac{135}{2}\pi+2\int_{0}^{\pi/2} 9\sin^2\theta \cdot \sqrt{9-9\sin^2\theta}\cdot 3\cos\theta\,\dee{\theta}\\
&=\frac{135}{2}\pi+54\int_{0}^{\pi/2} \sin^2\theta \cdot \sqrt{9\cos^2\theta}\cdot \cos\theta\,\dee{\theta}\\
&=\frac{135}{2}\pi+54\int_{0}^{\pi/2} \sin^2\theta \cdot 3\cos \theta\cdot \cos\theta\,\dee{\theta}\\
&=\frac{135}{2}\pi+162\int_{0}^{\pi/2} \sin^2\theta \cdot \cos^2 \theta\,\dee{\theta}\\
\end{align*}
Using the identity \(\sin(2\theta)=2\sin\theta\cos\theta\text{,}\) we see \(\sin^2\theta\cos^2\theta =\big(\sin\theta\cos\theta\big)^2 = \frac{1}{4}\sin^2(2\theta)\)
\begin{align*}
N&=\frac{135}{2}\pi+162\int_{0}^{\pi/2} \frac{1}{4}\sin^2(2\theta)\,\dee{\theta}\\
\end{align*}
Now, we use the identity \(\sin^2 x = \frac{1}{2}(1-\cos(2x))\text{,}\) with \(x=2\theta.\)
\begin{align*}
N&=\frac{135}{2}\pi+162\int_{0}^{\pi/2} \frac{1}{8}\big(1-\cos(4\theta)\big)\,\dee{\theta}\\
&=\frac{135}{2}\pi+\frac{81}{4}\int_{0}^{\pi/2} 1-\cos(4\theta)\,\dee{\theta}\\
&=\frac{135}{2}\pi+\frac{81}{4}\left[\theta-\frac{1}{4}\sin(4\theta)\right]_{0}^{\pi/2}\\
&=\frac{135}{2}\pi+\frac{81}{4}\left(\frac{\pi}{2}\right)\\
&=\frac{621}{8}\pi\\
\end{align*}
Now, using equation (\(*\)), we find \(\bar y\text{:}\)
\begin{align*}
\bar y &=\frac{N}{D} = \frac{\frac{621}{8}\pi}{\frac{45}{2}\pi} = \frac{69}{20}=3.45
\end{align*}
Let’s quickly check that this makes sense: if the circle has uniform density, its centre of mass would lie at \((0,3)\text{.}\) Since it’s denser at the top, the centre of mass should be higher, and indeed \(3.45\) is higher than 3 (without being so high it’s above the entire circle).
