Explanation

(a) If no external torque acts on a system of particle then angular momentum of the system remains constant, that is,
$\tau = 0$
$\Rightarrow$ $\frac{dL}{dt} = 0$
$\Rightarrow$ $L = I\omega =$constant
$\Rightarrow$ $I_{1}\omega_{1} = I_{2}\omega_{2}$
$\therefore$ $\frac{1}{2}Mr^{2}\omega_{1} = \frac{1}{2}(M + 2m)r^{2}\omega_{2}$ …(i)
Here, $M = 2$kg, $m =$0.25 kg, $r =$0.2m.
$\omega_{1} = 30\ rads^{- 1}$
Hence, we get after putting the given values in Eq. (i)
$\frac{1}{2} \times 2 \times (0.2)^{2} \times 30 = \frac{1}{2} \times (2 + 2 \times 0.25)(0.2)^{2} \times \omega_{2}$
$\Rightarrow$ $60 = 2.5\omega_{2}$
$\therefore$ $\omega_{2} = 24\ rads^{- 1}$