Shape of fastest spinning rod A one-meter steel rod of variable thickness is attached at one end to a spinning hub. The cross-sectional area of the rod is a function $f(x)$ of the distance $x$ in meters from the hub, x ranging from 0 to 1. My question is: how can I choose the function $f(x)$ to maximize the speed at which the rod can spin without flying apart?
Additional constraints: the rod has a minimum cross-section of 1 cm$^2$ everywhere, and the rod weighs 10 kg. Density of steel = $\rho$ = 8 g/cm$^3$, and the ultimate tensile strength is $F_{tu}$ = 800 MPa.
What I have: assume that at each distance c from the hub, the rod's cross-section at that distance has just enough tensile strength to support the rest of the rod. By setting $F_{tu} f(c)$ equal to the sum of centripetal forces needed for the rest of the rod, with $\omega$ angular velocity, I get
$$F_{tu} f(c) = \int_c^1 \rho \cdot x \cdot f(x) \cdot \omega^2  dx$$
But I do not know how to solve for f.
 A: start with this equation ?
$$F_{tu} f(x) = \int \rho \cdot x \cdot f(x) \cdot \omega^2  dx$$
heche
$$F_{tu} \frac{df(x)}{dx}=\rho \cdot x \cdot f(x) \cdot \omega^2$$
with $~f(0)=f_0~$ you obtain
$$f(x)=f_0\,e^{\frac{\rho\omega^2\,x^2}{2\,F_{tu}}}$$
A: Assume that you know the primitive F(x) of the integrand.$$ \frac{dF(x)}{dx}= \rho. x.f(x). \omega ^{2}$$ Then your equation reads:$$ F_{tu} f(c)=F(1)-F(c)$$ Differentiate both sides relative to c to get:$$F_{tu}  \frac{df(c)}{dc} =- \frac{dF(c)}{dc}=-\rho. c.f(c). \omega ^{2}$$ From which you get:$$ f(c)=k. e^{- \frac{ \rho  c^{2}  \omega ^{2}  }{2 F_{tu} } }$$ The negative sign indicates the rod is getting thiner as the distance from the axis increases. It makes sense because the centrepital force increases with x.  k is a constant to be found with the other constraints.The mass of the rod is given by:$$ m= \int_0^1  \rho. f(x)dx $$
The integral involves the erf function, all calculus done, one finds:
$$ k= \frac{m. \omega  \sqrt{ \frac{2 \rho }{ F_{tu} } }  }{ \rho . \sqrt{ \pi }.erf \big( \omega. \sqrt{ \frac{ \rho }{2. F_{tu} } }  \big)} $$
