Find the centroid of the arc of the catenary y=a cosh

Find the centroid of the arc of the catenary $y=a\cosh\dfrac xa$ from $x=-a$ to $x=a$.

Step 1: Arc-length element and total length

\[ds=\cosh\frac xa\,dx,\qquad L=\int_{-a}^a\cosh\frac xa\,dx = \left[a\sinh\frac xa\right]_{-a}^a=2a\sinh1\]

Step 2: Moment about the $y$-axis

The integrand $x\cosh\frac xa$ is odd, so \[M_y=\int_{-a}^ax\cosh\frac xa\,dx=0\]

Step 3: Moment about the $x$-axis

\[M_x=\int_{-a}^aa\cosh^2\frac xa\,dx=a\int_{-a}^a\frac{\cosh\frac{2x}a+1}2\,dx = \frac a2\left[\frac a2\sinh\frac{2x}a+x\right]_{-a}^a\]

\[=\frac a2\left[\left(\frac a2\sinh2+a\right)-\left(-\frac a2\sinh2-a\right)\right]=\frac a2\big[a\sinh2+2a\big]=\frac{a^2}2(\sinh2+2)\]

Step 4: Centroid

\[\bar x=0,\qquad \bar y=\frac{M_x}L=\frac{a^2(\sinh2+2)/2}{2a\sinh1}=\frac{a(\sinh2+2)}{4\sinh1}\]

Using $\sinh2=2\sinh1\cosh1$:

\[\bar y=\frac{a(2\sinh1\cosh1+2)}{4\sinh1}=\frac{a(1+\sinh1\cosh1)}{2\sinh1}\]

\[\boxed{\text{Centroid: }\left(0,\ \frac{a(1+\sinh1\cosh1)}{2\sinh1}\right)}\]

Step 5: Numerical value

With $\sinh1\approx1.175,\ \cosh1\approx1.543$: $\bar y\approx1.197a$.