Physics - Circular Motion Question with Solution | TestHub

Considering an element of tube of length $dx$ having mass $dm$ at a distance of $x$ from right end as shown in figure below.

Now, force acting of that element is $F=\int \left(dm\right){\omega }^{2}x$

Integrating the above relation for whole length of tube, we have 

 $F={\int }_0^L\left(\frac{m}{L}dx\right){\omega }^{2}x$

$=\frac{m}{L}{\omega }^2\frac{L^2}{2}$ $=\frac{m{\omega }^{2}L}{2}$

$\Rightarrow \omega =\sqrt{\frac{2}{mL}}\sqrt{F}$  $=\sqrt{\frac{2}{0.25\times 0.5}}\sqrt{F}$

$=\sqrt{16}\sqrt{F}$$=4\sqrt{\mathrm{F}} \mathrm{rad} {\mathrm{s}}^{-1}$

Hence, the value of $x=4$.