A uniform solid square plate ABCD of mass m and side a is moving in x – y horizontal smooth plane. The velocity of centre of mass is ${\mathrm{v}}_0(2\widehat{\mathrm{i}}+4\widehat{\mathrm{j}})$ m/s. The end A of square plate is suddenly fixed by a pin, find the new velocity of centre of mass of square
 

${\overrightarrow{\mathrm{r}}}_{\mathrm{cm}}=\frac{\mathrm{a}}{2}\widehat{\mathrm{i}}+\frac{\mathrm{a}}{2}\widehat{\mathrm{j}}$(considering A to be origin AD to x-axis and AB to be y-axis)
$\begin{array}{r}{\overrightarrow{\mathrm{L}}}_{\mathrm{i}}=\mathrm{m}{\overrightarrow{\mathrm{r}}}_{\mathrm{cm}}\times {\overrightarrow{\mathrm{v}}}_{\mathrm{cm}}=\mathrm{m}\left(\frac{\mathrm{a}}{2}\widehat{\mathrm{i}}+\frac{\mathrm{a}}{2}\widehat{\mathrm{j}}\right)\times {\mathrm{v}}_0(2\widehat{\mathrm{i}}+4\widehat{\mathrm{j}})\\ ={\mathrm{mav}}_0(2\widehat{\mathrm{k}}-\widehat{\mathrm{k}})={\mathrm{mav}}_0\widehat{\mathrm{k}}\end{array}$
${\mathrm{L}}_{\mathrm{f}}=\left[\frac{\mathrm{m}}{12}\left({\mathrm{a}}^2+{\mathrm{a}}^2\right)+\mathrm{m}\frac{{\mathrm{a}}^2}{2}\right]\mathrm{\omega }=\left(\frac{{\mathrm{ma}}^2}{6}+\frac{{\mathrm{ma}}^2}{2}\right)\mathrm{\omega }$
From conservation of angular momentum about point A
$\begin{array}{l}{\mathrm{L}}_{\mathrm{f}}={\mathrm{L}}_{\mathrm{i}}\\ \frac{2}{3}{\mathrm{ma}}^2\mathrm{\omega }={\mathrm{mav}}_0\\ \Rightarrow \mathrm{\omega }=\frac{3{\mathrm{v}}_0}{2\mathrm{a}}\\ \left|{\overrightarrow{\mathrm{v}}}_{\mathrm{cm}}\right|=\mathrm{\omega }\frac{\mathrm{a}}{\sqrt{2}}=\frac{3{\mathrm{v}}_0}{2\mathrm{a}}\frac{\mathrm{a}}{\sqrt{2}}=\frac{3{\mathrm{v}}_0}{2\sqrt{2}}\end{array}$