[![enter image description here][1]][1]

The "centripetal force" on a small mass element  is

$$dF=dm\,\omega^2 \,x\\
\text{with}\quad dm=A(x)\,\rho\,dx=A\,\rho\,dx $$

 

you obtain

$$F=A\,\rho\,\omega^2\int x\,dx=\frac 12 A\,\rho\,\omega^2\,x^2+c\\
\text{with}\quad F(L/2)=0\quad\Rightarrow c=-\frac 18\,A\rho\,{\omega}^{2}L^2$$ 

thus
$$F(x)=\frac 12\,A\,\rho\,\omega^2\left(x^2-\frac{L^2}{4}\right)
\quad, 0\le x\le \frac L2$$

you can obtain the angular velocity $~\omega~$ from the EOM

$$I_{\text{CM}}\,\dot\omega=\tau\quad,I_{\text{CM}}=\frac{m\,L^2}{12}$$


  [1]: https://i.sstatic.net/Rh7c1.png