A rod of length ‘$l$’ is traveling on a smooth horizontal surface with velocity ${\mathrm{u}}_{\mathrm{CM}}$ and rotating with an angular velocity $\mathrm{\omega }$ such that ${\mathrm{u}}_{\mathrm{CM}}$$=\frac{\mathrm{\omega l}}{2}.$ The distance covered by the point B in multiples of ‘$l$’ when the rod completes one full rotation is :

Here the point B is following a curvilinear path whose velocity at any instant is given by vector sum of $u_{CM}$ and $\frac{\omega l}{2}$. When rod rotates by an angle θ , the situation is as shown in the figure. 

The speed of end point at this instant is given as,

$v_B=\sqrt{v^2+{\left(\frac{\omega L}{2}\right)}^2+2v\times \frac{\omega L}{2}\times \cos (\pi -\theta )}$

$=\sqrt{2v}(\sqrt{1-\cos \theta }=2v \sin \frac{{\displaystyle \theta }}{2}$

$\Rightarrow v_B=2\times \frac{\omega l}{2}\times \sin \frac{{\displaystyle \theta }}{2}$

$\Rightarrow \frac{ds}{dt}=l \sin \frac{{\displaystyle \theta }}{2}\frac{d\theta }{dt}$

$\Rightarrow ds=l \sin \frac{{\displaystyle \theta }}{2}d\theta$

 $\Rightarrow {\int }_0^{s}ds=l{\int }_0^{2\mathrm{\pi }} \sin \frac{{\displaystyle \theta }}{2}d\theta$

$\Rightarrow s=-2l{\left|\cos \frac{{\displaystyle \theta }}{2}\right|}_0^{2\mathrm{\pi }}=4l$