Rod rotating with a ring

Lets clear out @rajdeep brahma 's solution to you.

For any infinitesimally small element $dx$ at a distance x from the Axis of rotation, mass of the element will be $dm= \dfrac{m}{L} dx$

So force acting on it is simply the centrepetal force, which is $dF_{cp}= dm \omega^2 x$

$\implies a_{cp}=\omega^{2}x$

$\implies v\dfrac{dv}{dx} = \omega^2 x$

$\implies v dv = \omega^2 x dx$

Integrating with limits $0 \rightarrow v$ and $\dfrac{L}{2} \rightarrow L$

$\displaystyle{\int_{0}^{v} vdv = \int_{0}^{L} \omega^2 x dx}$

$\implies \dfrac{v^2}{2} = \dfrac{\omega^2 \dfrac{3L^2}{4}}{2}$

$\implies v^{2} = \omega^2 \dfrac{3L^2}{4}$

$\implies v_{n}= \omega \dfrac{L\sqrt{3}}{2}$

Now this was the normal velocity.

But the tangential velocity will be $\omega L$

$\implies v^{2}=v_{t}^2 + v_{n}^2$

$\implies v^2 = (\dfrac{3}{4} + 1) \omega^2 L^2$

$\implies v^2 = \dfrac{7}{4} \omega^2 L^2$

$\implies k =\dfrac{7}{4}$

$\implies \boxed{k=1.75}$

In order to find velocity of the rod at the end,integrate m (w^2) x with x ranging from $\frac{L}{2}$ to L.Let radial velocity be v.So u get $\frac{3}{8}$ m (w^2) (L^2)= $\frac{1}{2}$ m (v^2) and tangential velocity is w l.So the net velocity is root( $\frac{7}{4}$ )w*L (radial and tangential velocity sum).Hence k^2= $\frac{7}{4}$ =1.75.