= Solution
The uniform cylindrical torso contributes $I_c=Mr^2/2=50(0.20)^2/2=1.00\,\mathrm{kg\,m^2}$. Adding both arm <point masses> about the same vertical axis gives
$$
I_i=1+2(7)(0.70)^2=7.86\,\mathrm{kg\,m^2},\qquad
I_f=1+2(7)(0.30)^2=2.26\,\mathrm{kg\,m^2}.
$$
With initial <angular frequency> $\omega_i=2\pi\,\mathrm{rad\,s^{-1}}$, <conservation of angular momentum> gives $I_i\omega_i=I_f\omega_f$. Therefore
$$
\boxed{\omega_f=2\pi\frac{7.86}{2.26}=21.8521\,\mathrm{rad\,s^{-1}}.}
$$
This corresponds to $3.47788$ rotations per second; it is important not to confuse that <frequency> with the requested <angular frequency>.
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