- #1
brkomir
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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?