91影视

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

Final angular speed is ${蝇}_2$.

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:

$I_j{蝇}_j=I_f{蝇}_f=$constant

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

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

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:

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

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:

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

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