To solve this problem it is necessary to apply the concepts related to the conservation of angular momentum. This can be expressed mathematically as a function of inertia and angular velocity, that is:
[tex]L = I\omega[/tex]
Where,
I = Moment of Inertia
[tex]\omega[/tex]= Angular Velocity
For the given object the moment of inertia is equivalent to
[tex]I = \frac{mr^2}{12}[/tex]
Considering that the moment of inertia varies according to distance, and that there are two of these without altering the mass we will finally have to
[tex]L_i = L_f[/tex]
[tex]I_i \omega_1 = I_f \omega_2[/tex]
[tex](\frac{mr_{initial}^2}{12})(\omega_1)=(\frac{mr_{final}^2}{12})(\omega_2)[/tex]
[tex](r_{initial}^2})(\omega_1)=(r_{final}^2)(\omega_2)[/tex]
Our values are given as,
[tex]r_{initial} = 3m\\\omega_1 = 0.05rad/s \\r_{final}=1.5m[/tex]
Replacing we have,
[tex](3^2})(0.05)=(1.5^2)(\omega_2)[/tex]
[tex]\omega_2 = 0.2rad/s[/tex]
Therefore the angular speed after the catch slips is 0.2rad/s
