A uniform rod AB of mass 2kg and Length 30cm at rest on a smooth...

Impulse $J = 0 . 2Ns$

$⇒ J = ∫ Fdt = 0 . 2Ns$

Now, angular impulse $( \overset{\to}{M} )$ will be

${M}_{c} = ∫ τdt$

$= ∫ F\dfrac{L}{2}dt$

$= \dfrac{L}{2} ∫ Fdt = \dfrac{L}{2} \times J$

$= \dfrac{0 . 3}{2} \times 0 . 2$

$= 0 . 03$

Moment of inertia of the rod about the centre of mass, ${I}_{cm} = \dfrac{M{L}^{2}}{12} = \dfrac{2 \times ( 0 . 3{)}^{2}}{12} = \dfrac{0 . 09}{6}$

Angular impulse will be equal to the change in angular momentum.

$M = {I}_{cm}\left({\omega}_{f} - {\omega}_{i}\right)$

$⇒ 0 . 03 = \dfrac{0 . 09}{6}\left({\omega}_{f}\right)$

$⇒ {\omega}_{f} = 2rad{s}^{- 1}$

For angular displacement, we can write $\theta = \omegat$

$⇒ t = \dfrac{\theta}{\omega} = \dfrac{\pi}{2 \times 2} = \dfrac{\pi}{4}s$.

Therefore, $x = 4$.