If 2 ~kg mass is rotating on a circular path of radius 0.8 ~m with...

Mass $(m)=2 \mathrm{~kg}$; initial radius of the path $\left(r_1\right)=0.8 \mathrm{~m}$; initial angular velocity $\left(\omega_1\right)=44 \mathrm{rad} / \mathrm{sec}$ and final radius of the path \left(r_2\right)=1 \mathrm{~m}$Moment of inertia,$I_1=m r_1^2=2 \times(0.8)^2 =1.28 \mathrm{~kg} \mathrm{~m}^2 and I_2=m r_2^2=2 \times(1)^2=2 \mathrm{~kg} \mathrm{\textrm {m } ^ { 2 }}$Therefore from the law of conservation of angular momentum$I_1 \omega_1=I_2 \omega_2 or \omega_2=\frac{I_1 \times \omega_1}{I_2}=\frac{1.28 \times 44}{2} \omega_2=28.16 \mathrm{rad} / \mathrm{sec}