A uniform rod of mass m = 2 kg and length l = 0.5 m is sliding along two mutually perpendicular smooth walls with the two ends P and Q having velocities v_p = 4 m/s and v_Q = 3 m/s as shown in figure. Then,

$$\begin{gathered} x^2+y^2=l^2=(0.25)m^2 \\ y\omega =4 ...\left(i\right) \\ x\omega =3 ...\left(ii\right) \end{gathered}$$

Squaring and adding, we get

$$\begin{gathered} (x^2+y^2){\omega }^2=25 or (0.25){\omega }^2=25 \\ \therefore \omega =10 rad/s \\ OC=\frac{l}{2}=0.25 m \\ {v}_{cm}=\left(OC\right)\omega =2.5 m/s \\ {K}_{Total}=\frac{1}{2}m{v}_{CM}^2+\frac{1}{2}{I}_{CM}{\omega }^2 \\ =\frac{1}{2}\times 2\times {(2.5)}^2+\frac{1}{2}\left(\frac{2\times 0.25}{12}\right){\left(10\right)}^2 \\ =\frac{25}{3}J \end{gathered}$$