A  midpoint of a thin uniform rod AB of mass m and length l is rigidly fixed to a rotation axis  $OO\text{'}$ as shown in figure. The rod is set into rotation with a constant angular velocity  $\omega$. If the resultant moment of the centrifugal forces of inertia relative to the point C in the reference frame fixed to the axle $OO\text{'}$ and the rod is $\frac{1}{2}m{\omega }^2{\mathcal{l}}^2\sin \left(n\theta \right)$ Find n…..

Consider a small element of length dx at a distance x from the point C, which is rotating in a circle of radius  $r=x\sin \theta$
Now mass of the elements =  $\left(\frac{m}{l}\right)dx$
So centrifugal force acting on this element  $=\left(\frac{m}{l}\right)dx{\omega }^{2}x\sin \theta$

And moment of this force about C is  $\left|dN\right|=\left(\frac{m}{l}\right)dx{\omega }^{2}x\sin \theta .x\cos \theta =\frac{m{\omega }^2}{2l}\sin 2\theta x^{2}dx$
And hence, total moment  $N=2\underset{\upsilon }{\overset{l/2}{\int }}\frac{m{\omega }^2}{2l}\sin 2\theta x^{2}dx=\frac{1}{24}m{\omega }^2{l}^2\sin 2\theta$