Two discs of moments of inertia $\mathrm{I}_1$ and $\mathrm{I}_2$ about their transverse symmetry axes, respectively rotating with angular velocities to $\omega_1$ and $\omega_2$, are brought into contact with their rotation axes coincident. Find the angular velocity of the composite disc.
Q 109
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We assume that the initial angular momenta $\left(\vec{L}_1\right.$ and $\left.\vec{L}_2\right)$ of the discs are either in the same direction or in opposite directions. Then,
the total initial angular momentum $=\vec{L}_1+\vec{L}_2=I_1 \overrightarrow{\omega_1}+I_2 \overrightarrow{\omega_2}$
After they are coupled, the total moment of inertia, i.e., the moment of inertia of the composite disc is $I=I_1+I_2$ and the common angular velocity is $\vec{\omega}$. Assuming conservation of angular momentum,
$
l \vec{\omega}=\left(I_1+I_2\right) \vec{\omega}=I_1 \vec{\omega}_1+I_2 \vec{\omega}_2
$
$\therefore \vec{\omega}=\frac{I_1 \vec{\omega}_1+I_2 \vec{\omega}_2}{I_1+I_2}$
If $\vec{\omega}_1$ and $\vec{\omega}_2$ are in the same direction, $\omega=\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}$. If $\vec{\omega}_1$ and $\vec{\omega}_2$ are in opposite directions, $\omega=\left|\frac{I_1 \omega_1-I_2 \omega_2}{I_1+I_2}\right|$.
the total initial angular momentum $=\vec{L}_1+\vec{L}_2=I_1 \overrightarrow{\omega_1}+I_2 \overrightarrow{\omega_2}$
After they are coupled, the total moment of inertia, i.e., the moment of inertia of the composite disc is $I=I_1+I_2$ and the common angular velocity is $\vec{\omega}$. Assuming conservation of angular momentum,
$
l \vec{\omega}=\left(I_1+I_2\right) \vec{\omega}=I_1 \vec{\omega}_1+I_2 \vec{\omega}_2
$
$\therefore \vec{\omega}=\frac{I_1 \vec{\omega}_1+I_2 \vec{\omega}_2}{I_1+I_2}$
If $\vec{\omega}_1$ and $\vec{\omega}_2$ are in the same direction, $\omega=\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}$. If $\vec{\omega}_1$ and $\vec{\omega}_2$ are in opposite directions, $\omega=\left|\frac{I_1 \omega_1-I_2 \omega_2}{I_1+I_2}\right|$.
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