91Ó°ÊÓ

Original angular speed is ${\mathrm{Ó\neg }}_1=0.40\mathrm{rev}/\mathrm{s}$.

Final angular speed is ${\mathrm{Ó\neg }}_2$.

Mass of arms is $m=8\mathrm{kg}$

Radius of cylinder is $R=25\mathrm{cm}$.

Length of arms outstretched is $L=1.8\mathrm{m}$.

Apply the law of conservation of angular momentum to a system whose moment of inertia changes gives:

$I_j{\mathrm{Ó\neg }}_j=I_f{\mathrm{Ó\neg }}_f=$constant

The total moment of inertia with arms outstretched is solved as:

$$\begin{gathered} I_1=I_{\text{body}}+I_{\text{arms}} \\ =0.40{\mathrm{kgm}}^2+\frac{1}{12}{\mathrm{mL}}^2 \\ =0.40{\mathrm{kgm}}^2+\frac{1}{12}×8\mathrm{kg}×(1.8\mathrm{m}{)}^2 \\ =2.56{\mathrm{kgm}}^2 \end{gathered}$$

Hence, the total moment of inertia when the arms are outstretched is $2.56{\mathrm{kgm}}^2$.

The total moment of inertia with arms wrapped around the body is solved as:

$$\begin{gathered} I_2=I_{\text{body}}+I_{\text{ams}} \\ =0.40{\mathrm{kgm}}^2+{\mathrm{mR}}^2 \\ =0.40{\mathrm{kgm}}^2+8\mathrm{kg}×(0.25\mathrm{m}{)}^2 \\ =0.9{\mathrm{kgm}}^2 \end{gathered}$$

Hence, the total moment of inertia when the arms are wrapped around is $0.9{\mathrm{kgm}}^2$.

Use the conservation of angular momentum to obtain angular speed:

$$\begin{gathered} I_1{\mathrm{Ó\neg }}_1=I_2{\mathrm{Ó\neg }}_2 \\ 2.56{\mathrm{kgm}}^2×0.40\mathrm{rad}/\mathrm{s}=0.9{\mathrm{kgm}}^2×{\mathrm{Ó\neg }}_2 \\ {\mathrm{Ó\neg }}_2=\frac{2.56{\mathrm{kgm}}^2×0.40\mathrm{rad}/\mathrm{s}}{0.9} \\ {\mathrm{Ó\neg }}_2=1.14\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the final angular speed is $1.14\mathrm{rev}/\mathrm{s}$