Answer
The angular speed immediately after the collision is $~~1.5~rad/s$
Work Step by Step
Let $m_p$ be the mass of the putty.
We can find the rotational inertia of the rod-putty system after impact:
$I_f = (0.12~kg~m^2)+m_p~L^2$
$I_f = (0.12~kg~m^2)+(0.20~kg)(0.60~m)^2$
$I_f = 0.192~kg~m^2$
We can use conservation of angular momentum to find the angular speed immediately after the collision:
$L_f = L_i$
$I_f~\omega_f = I_i~\omega_i$
$\omega_f = \frac{I_i~\omega_i}{I_f}$
$\omega_f = \frac{(0.12~kg~m^2)(2.4~rad/s)}{0.192~kg~m^2}$
$\omega_f = 1.5~rad/s$
The angular speed immediately after the collision is $~~1.5~rad/s$
