Answer
The final angular speed is 1.14 rev/s.
Work Step by Step
We can find the moment of inertia $I_1$ with arms outstretched.
$I_1 = I_{body}+I_{rod}$
$I_1 = I_{body}+\frac{1}{12}ML^2$
$I_1 = (0.40~kg~m^2)+\frac{1}{12}(8.0~kg)(1.8~m)^2$
$I_1 = 2.56~kg~m^2$
We can find the moment of inertia $I_1$ with arms pulled in.
$I_2 = I_{body}+I_{cylinder}$
$I_2 = I_{body}+MR^2$
$I_2 = (0.40~kg~m^2)+(8.0~kg)(0.25~m)^2$
$I_2 = 0.90~kg~m^2$
We can use conservation of angular momentum to find the final angular speed.
$L_2 = L_1$
$I_2\omega_2 = I_1\omega_1$
$\omega_2 = \frac{I_1\omega_1}{I_2}$
$\omega_2 = \frac{(2.56~kg~m^2)(0.40~rev/s)}{0.90~kg~m^2}$
$\omega_2 = 1.14~rev/s$
The final angular speed is 1.14 rev/s.
