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A person stadns on a platform, initially at rest, that can rotate freely without friction. The moment of inertia of the person plus the platform is [itex]I_p[/itex]. The person holds a spinning bicycle wheel with axis horizontal. The wheel has momemnt of inertia [itex]I_w[/itex] and angular velocity [itex]\omega_w[/itex]. What will the angular velocity [itex]\omega_p[/itex] of the platform if the person moves the axis of the wheel so that it points vertically upward.
Using conservation of momentum:
According to my book though,
Is it just me, or is my book on crack?
Using conservation of momentum:
[tex]\vec{L}_w = \vec{L'}_w + \vec{L}_p[/tex]
where[tex]\begin{align*}
\vec{L}_w &= (I_w\omega_w, 0)\\
\vec{L'}_w &= (0, I_w\omega_w)\\
\vec{L}_p &= I_p\vec{\omega}_p = I_p(\omega_1, \omega_2)
\end{align}[/tex]
Solving for [itex]\omega_1[/itex] and [itex]\omega_2[/itex] gives\vec{L}_w &= (I_w\omega_w, 0)\\
\vec{L'}_w &= (0, I_w\omega_w)\\
\vec{L}_p &= I_p\vec{\omega}_p = I_p(\omega_1, \omega_2)
\end{align}[/tex]
[tex]\begin{align*}
\omega_1 &= \frac{I_w\omega_w}{I_p}\\
\omega_2 &= -\frac{I_w\omega_w}{I_p}
\end{align}[/tex]
And since [itex]\omega_p = \sqrt{\omega_1^2 + \omega_2^2}[/itex] then\omega_1 &= \frac{I_w\omega_w}{I_p}\\
\omega_2 &= -\frac{I_w\omega_w}{I_p}
\end{align}[/tex]
[tex]\omega_p = \frac{\sqrt{2}I_w\omega_w}{I_p}[/tex]
According to my book though,
[tex]\omega_p = \frac{I_w\omega_w}{I_p}[/tex]
Is it just me, or is my book on crack?