Center of mass and moment of inertia of catenary

[brkomir]

Homework Statement

A homogeneous catenary $z=acosh(x/a)$, $y=0$ and $x\in \left [ -a,a \right ]$ is given. Calculate the center of mass and moment of inertia

Homework Equations

The Attempt at a Solution

I started with $x=at$, for$t\in \left [ -1,1 \right ]$, therefore y remains $y=0$ and $z=acosh(ta/a)=acosh(t)$.
Btw, a is constant greater than zero, not a sign of arcus or any other inverse function.

Ok, now $r(t)=(at,0,acosh(t))$ and $\dot{r}(t)=(a,0,asinh(t))$ and $\left |\dot{r} \right |=acosh(t)$

So the length of the catenary is $l=\int_{-1}^{1}acosh(t)dt=2asinh(1)$

The catenary is symmetrical on x axis, but point z is given as $z=\frac{\int_{-1}^{1}a^2cosh^2(t)dt}{\int_{-1}^{1}acosh(t)dt}=\frac{a^2}{2asinh(1)}\int_{-1}^{1}cosh^2(t)dt=\frac{asinh(2)}{4sinh(1)}=\frac{a}{2}cosh(1)$

I would just like to know if this is ok, before I start calculating moment of inertia?

 

[vela]

Your method looks fine, but I'm getting a different result for the integral of cosh² t.

 

[brkomir]

Hmmm, let's take a look..

$a^2\int_{-1}^{1}cosh^2(t)dt=\frac{a^2}{2}\int_{-1}^{1}(1+cosh(2t))dt=\frac{a^2}{2}(t+\frac{1}{2}sinh(2t))$ for $t$ form $0$ to $2\pi$. (I don't know how to type bar with integrating borders in latex...

$\frac{a^2}{2}(t+\frac{1}{2}sinh(2t))=\frac{a^2}{2}(1+\frac{1}{2}sinh(2)-1-\frac{1}{2}sinh(-2))=\frac{a^2}{2}sinh(2)$

OR is it not true that $cosh^2(t)=\frac{1}{2}(1+cosh(2t)$ or do I have some other problems? :/

 

[vela]

You messed up the constant term. It works out to 1-(-1) = 2, not 1-1=0.

 

[brkomir]

True... Thank you vela!