Luc, who is 1.80 m tall and weighs 950 N, is standing at the center of a playground merry-go-round with his arms extended, holding a 4.0 kg dumbbell in each hand. The merry-go-round can be modeled as a 4.0-m-diameter disk with a weight of 1500 N. Luc’s body can be modeled as a uniform 40-cm-diameter cylinder with massless arms extending to hands that are 85 cm from his center. The merry-go-round is coasting at a steady 35 rpm when Luc brings his hands in to his chest. Afterward, what is the angular velocity, in rpm, of the merry-go-round?

Height of Luc, h=1.8 m
Weight of the Luc, W=950 N
Mass of the dumbbell, m=4 kg
Diameter of the merry-go-round, D=4m
Weight of merry-go-round, W_m=1500 N
Diameter, of Luc, d=40cm =0.4 m
Length of the arms, l=85 cm= 0.85 m
Steady speed=ω=35 rpm

First calculate mass

Mass of Luc $m_{Luc}=\frac{W}{g}=\frac{950N}{10m/s^2}=95\mathrm{kg}$ ... taken g=10

Mass of the merry-go-round $m_m=\frac{W_m}{g}=\frac{1500N}{10m/s^2}=150\mathrm{kg}$

Calculate initial moment of inertia of the system

$$\begin{gathered} I_{\text{initial}}=I_{\text{merry}}+I_{\text{Luc}}+2I_{\text{dumbbell}} \\ =\frac{m_m(D/2{)}^2}{2}+\frac{m_{\text{Luc}}(d/2{)}^2}{2}+2\times m{l}^2 \end{gathered}$$

Substitute values we get

$=\frac{\left(150kg\right)(4m/2{)}^2}{2}+\frac{\left(95kg\right)(0.4m/2{)}^2}{2}+2\times \left(4kg\right)\times {(0.85m)}^2=307.6\mathrm{kg}.{\mathrm{m}}^2$

When arm is closed moment of inertia changes say final inertia

$$\begin{gathered} I_{\text{final}}=I_{\text{merry}}+I_{\text{Luc}} \\ =\frac{m_m(D/2{)}^2}{2}+\frac{m_{\text{Luc}}(d/2{)}^2}{2} \\ =\frac{\left(150kg\right)(4m/2{)}^2}{2}+\frac{\left(95kg\right)(0.4m/2{)}^2}{2} \\ =301.9\mathrm{kg}.{\mathrm{m}}^2 \end{gathered}$$

Use the law of energy conservation and solve for final angular velocity

$$\begin{gathered} I_{\text{final}}\times {\omega }_{\text{final}}=I_{\text{initial}}\times {\omega }_{\text{initial}} \\ 301.9\mathrm{kg}.{\mathrm{m}}^2\times {\omega }_{\text{final}}=307.6\mathrm{kg}.{\mathrm{m}}^2\times 35\mathrm{rpm} \\ {\omega }_{\text{final}}=35.66=36\mathrm{rpm} \end{gathered}$$