Answer
$\omega_2=2.55~rev/s$
Work Step by Step
We can determine the required angular velocity as follows:
According to the conservation of angular momentum
$I_{z1}\omega_1=I_{z_2}\omega_2$ [eq(1)]
We know that
$I_{z1}=I_{body}+I_{arms}+I_{dumbbells}$
$\implies I_{z1}=\frac{1}{2}m_br_b^2+2(\frac{m_al_a^2}{12}+m_ar_a^2)+2(m_dr_d^2)$
We plug in the known values to obtain:
$I_{z1}=\frac{1}{2}(68)(0.2)^2+2[\frac{6(0.65)^2}{12}+6(\frac{0.65}{2}+0.2)^2]+2[10(0.65+0.2)^2]$
This simplifies to:
$I_{z1}=19.54Kg/m^2$
Similarly $I_{z2}=I_{body}+I_{dumbbells}$
$\implies I_{z2}=\frac{1}{2}m_br_b^2+2(m_dr_d^2)$
$\implies I_{z2}=\frac{1}{2}(80)(\frac{0.45}{2})^2+2(10)(0.3)^2=3.825Kg/m^2$
Now, we plug in the known values in eq(1) to obtain:
$19.54(0.5)=3.825\omega_2$
This simplifies to:
$\omega_2=2.55~rev/s$
