A uniform rod of mass m and length l is struck at one end by a...

The angular impulse about the centre of mass is given by the product of torque and time.

Torque $\tau = F \times \dfrac{l}{2}$

Angular impulse $= \dfrac{F l t}{2}$

Change in angular momentum $\Delta L = I \omega$

The moment of inertia of a uniform rod about an axis passing through its centre of mass and perpendicular to its length is $I = \dfrac{m l^2}{12}$

Equating the angular impulse to the change in angular momentum:

$\dfrac{m l^2}{12} \omega = \dfrac{F l t}{2}$

$\omega = \dfrac{12 F l t}{2 m l^2}$

$\omega = \dfrac{6 F t}{m l}$