To give an appropriate solution to this problem we require the concepts of Angular Moment and Inertia Moment,
The angular velocity is given by,
[tex]\omega_f = \frac{L}{I}[/tex]
Where,
L is the initial angular momentum, equal to
[tex]L =mvr[/tex]
And I the moment of inertia, which for a point rotating in a circle is given by
[tex]I=mr^2[/tex]
Here m is the mass and r the radius.
For our problem the values are,
[tex]m= 44kg\\v=5m/s \\r=2.6m[/tex]
Then the angular momentum is
[tex]L = mvr = (44)(5)(2.6)\\L = 576kgm^2/s[/tex]
In the other hand the moment of inertia of the system is,
[tex]I=I_0 +mr^2\\I = 857+(50)(2.6)^2\\I = 987kgm^2/s[/tex]
Finally we can calculate the angular momentum, which is given by,
[tex]\omega_f=\frac{L}{I}= \frac{576}{987}\\\omega _f= 0.584rad/s[/tex]
Therefore the angular velocity after the boy jumps on is 0.584rad/s
