A uniform rod of mass M is moving in a plane and has a kinetic energy of $\frac{4}{3}{\mathrm{MV}}^2$ where V is speed of its centre of mass. Find the maximum and minimum possible speed of the end point of the rod.

If $\varpi$ = angular speed 
 

$$\begin{gathered} \frac{1}{2}{\mathrm{MV}}^2+\frac{1}{2}\left(\frac{1}{12}{\mathrm{ML}}^2\right){\mathrm{\omega }}^2=\frac{4}{3}{\mathrm{MV}}^2 \\ =\frac{1}{24}(\mathrm{L\omega }{)}^2=\frac{5}{6}{\mathrm{MV}}^2\therefore (\mathrm{\omega L}{)}^2=20{\mathrm{V}}^2 \\ \Rightarrow \mathrm{\omega L}=2\sqrt{5}\mathrm{V}\Rightarrow \frac{\mathrm{\omega L}}{2}=\sqrt{5}\mathrm{V} \end{gathered}$$

The velocity of one end of the rod is given by vector sum of its velocity of COM and  $\frac{\mathrm{\omega L}}{2}=\sqrt{5}\mathrm{V}$ (that is perpendicular to length)
Speed of end point can range from $(\sqrt{5}\mathrm{V}-\mathrm{V})\text{ to }(\sqrt{5}\mathrm{V}+\mathrm{V})$depending on the direction of velocity of the COM of the rod.