91Ó°ÊÓ

Angular momentum is equivalent to the linear momentum for rotation

$$\begin{gathered} L=/\mathrm{Ó\neg } \\ L=\mathrm{angular}\mathrm{momentum} \\ /=\mathrm{monent}\mathrm{of}\mathrm{inertia} \\ \mathrm{Ó\neg }=\mathrm{angular}\mathrm{velocity} \end{gathered}$$

For rod with axis per cu center$I_1=\frac{1}{12}M{L}^2$

Approximate the rings to point masses $I_2=m{L}^2$

$$\begin{gathered} \mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{rod}=0.03\mathrm{kg} \\ \mathrm{length}\mathrm{of}\mathrm{the}\mathrm{rod}=0.4\mathrm{m} \\ \mathrm{mass}\mathrm{of}\mathrm{ring}1\mathrm{and}2=0.2\mathrm{kg} \\ \mathrm{angulars}\mathrm{speed}=48\mathrm{rev}/\mathrm{m} \end{gathered}$$

Initial Instant: Before releasing the rings

$L_o=(I_1+2I_{2}o){\mathrm{Ó\neg }}_0$

Final instant: When the rings reach the end of the rod

$L_f=(I_1+2I_{2}f)\mathrm{Ó\neg }$

Angular momentum is conserved.

$$\begin{gathered} L_0=L_f \\ \left(I_1+2I_{20}\right){\mathrm{Ó\neg }}_0=\left(I_1+2I_{2f}\right)\mathrm{Ó\neg } \\ \mathrm{Ó\neg }=\frac{\left(I_1+2I_{20}\right)}{\left(I_1+2I_{2f}\right)}{\mathrm{Ó\neg }}_0 \\ {I}_1=\frac{1}{12}\mathrm{x}0.03\mathrm{x}0.4^2=4\mathrm{x}{10}^{-4}\mathrm{kg}.{\mathrm{m}}^2 \\ {I}_{20}=0.02\mathrm{x}0.{05}^2=5\mathrm{x}{10}^{-5}\mathrm{kg}.{\mathrm{m}}^2 \\ {I}_{2\mathrm{f}}=0.02\mathrm{x}0.2^2=8\mathrm{x}{10}^{-4}\mathrm{kg}.{\mathrm{m}}^2 \\ \mathrm{Ó\neg }=\frac{4\mathrm{x}{10}^{-4}+\left(2\mathrm{x}5\mathrm{x}{10}^{-5}\right)}{4\mathrm{x}{10}^{-4}+\left(2\mathrm{x}8\mathrm{x}{10}^{-4}\right)}\mathrm{x}48 \\ =12\mathrm{rpm} \end{gathered}$$

Hence the Angular speed of the system at the instant when the rings reach the ends of the rod is 12 rpm

Initial instant: With the rings reach the end of the rod

$L_o=(I_1+2I_{2}f){\mathrm{Ó\neg }}_f$

Final moment: without the rings

$$\begin{gathered} L_f=I_1\mathrm{Ó\neg } \\ {L}_0=L_f \\ \mathrm{Ó\neg }=\frac{\left(I_1+2I_{2f}\right)}{I_1}{\mathrm{Ó\neg }}_f \\ =\frac{4\mathrm{x}{10}^{-4}+\left(2\mathrm{x}8\mathrm{x}{10}^{-4}\right)}{4\mathrm{x}{10}^{-4}}\mathrm{x}12 \\ =15\mathrm{rpm} \end{gathered}$$

Hence the angular speed of the rod after the rings leave it is 15 rpm