Hi there!
[tex]\boxed{\omega = 0.38 rad/sec}[/tex]
We can use the conservation of angular momentum to solve.
[tex]\large\boxed{L_i = L_f}[/tex]
Recall the equation for angular momentum:
[tex]L = I\omega[/tex]
We can begin by writing out the scenario as a conservation of angular momentum:
[tex]I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)[/tex]
[tex]I_m[/tex] = moment of inertia of the merry-go-round (kgm²)
[tex]\omega_m[/tex] = angular velocity of merry go round (rad/sec)
[tex]\omega_f[/tex] = final angular velocity of COMBINED objects (rad/sec)
[tex]I_b[/tex] = moment of inertia of boy (kgm²)
[tex]\omega_b[/tex]= angular velocity of the boy (rad/sec)
The only value not explicitly given is the moment of inertia of the boy.
Since he stands along the edge of the merry go round:
[tex]I = MR^2[/tex]
We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:
[tex]\omega = \frac{v}{r}[/tex]
[tex]L_b = MR^2(\frac{v}{R}) = MRv[/tex]
Plug in the given values:
[tex]L_b = (20)(3)(5) = 300 kgm^2/s[/tex]
Now, we must solve for the boy's moment of inertia:
[tex]I = MR^2\\I = 20(3^2) = 180 kgm^2[/tex]
Use the above equation for conservation of momentum:
[tex]600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}[/tex]
