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{\bf Question}

A smooth straight tube rotates in a plane with constant angular
velocity $\omega$ about a perpendicular axis through a point O.
Inside the tube is a particle of mass $m$ joined to O by a spring
of stiffness $k = 5m\omega^2$ and natural length $a$. The particle
is released from its position of relative rest, with the spring at
its natural length.  Show the the particle describes oscillations
with period $\pi/\omega$ and amplitude $a/4$ relative to the tube.
What is the largest reaction on the tube?

\vspace{.25in}

{\bf Answer}

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Newton's 2nd law: In a frame of reference stationary with respect
to the tube:

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gives $m\ddot x {\bf i} = -T {\bf i} - mg {\bf k} + {\bf R} -
\underbrace{2m \mbox{\boldmath $\omega$} \times {\bf v}} -
\underbrace{m \mbox{\boldmath $\omega$} \times (\mbox{\boldmath
$\omega$} \times {\bf r})}$

\hspace{2.2in} Coriolis \hspace{.2in} Centrifugal

$\ds {\bf v} = {\bf i} \dot x$ and ${\bf R}$ is perpendicular to
$\bf i$ as the tube is smooth.

$\ds \mbox{\boldmath $\omega$} = \omega {\bf k},$ therefore

\begin{eqnarray*} m \ddot x &amp; = &amp; -5m\omega^2 (x - a) + m \omega^2
x \\ \ddot x &amp; = &amp; -4\omega^2x + 5 \omega^2 a \\ &amp; = &amp; -4\omega
\left( x - \frac{5}{4}a \right) \\ x - \frac{5}{4} a &amp; = &amp; A \cos
(2 \omega t + \alpha) \hspace{.2in} A, \alpha {\rm \ \ are\
constant} \\ &amp; &amp; {\rm at\ }\ t = 0,\ x = a,\ \dot x = 0
\Rightarrow A = - \frac{a}{4} \alpha = 0 \\ x &amp; = &amp; a \left[
\frac{5}{4} - \frac{1}{4} \cos 2 \omega t \right]\end{eqnarray*}

The period of oscillation is $\ds \frac{2\pi}{2\omega} =
\frac{\pi}{\omega}$

The amplitude of oscillation is $\ds \frac{a}{4}$

${\bf R} = m g {\bf k} + 2m \mbox{\boldmath $\omega$} \times {\bf
v} = mg {\bf k} + 2m \omega \frac{a}{2} \omega \sin 2 \omega t
{\bf j}$

Therefore $R_{\rm max} = m \sqrt{g^2 + a^2 \omega^4}$




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