\begin{problem*}{3.24}
Suppose a copper sleeve of inner radius $r_1=2.10\U{cm}$ and outer
radius $r_2=2.20\U{cm}$ is to be cast.  To eliminate bubbles and give
high structural integrity, the centripetal acceleration of each bit of
metal should be at least $a_{c\U{min}}=100g$.  What rate of rotation
is required?  State the answer in revolutions per minute (rpm).
\end{problem*} % problem 3.24

\begin{solution}
For circular motion
\begin{align}
 a_c &= v^2/r \\
 v &= \sqrt{a_c r}
\end{align}
And if the circular motion is uniform, the time $T$ taken to complete 
one revolution is given by
\begin{align}
 \Delta x &= v T \\
 T &= \Delta x / v = \frac{2 \pi r}{v}
\end{align}
So the frequency $f = 1/T$ of rotation is
\begin{equation}
 f = 1/T = \frac{v}{2 \pi r}
	= \frac{\sqrt{a_c r}}{2 \pi r}
	= \frac{1}{2 \pi}  \sqrt{ \frac{a_c}{r} }
\end{equation}

Because for a given frequency $a_c \propto r$,
 we must use the inner radius $r_1$ to compute the $f$ required
 to create an acceleration $a_c \geq 100g$.
\begin{equation}
 f = \frac{1}{2 \pi} \sqrt{ \frac{100 * 9.8\U{m/s}^2}{0.0210\U{m}} }
		\cdot \frac{60\U{s}}{1\U{min}}
	= \ans{2060\U{rpm}}
\end{equation}

To reassure ourselves that this frequency creates enough centripetal 
acceleration $a_{c2}$ at $r_2$, we can compute $a_{c2}$.
\begin{align}
 f &= \frac{1}{2 \pi}  \sqrt{ \frac{a_c}{r} } \\
 2 \pi f &= \sqrt{ \frac{a_c}{r} }
\end{align}
The left hand side is constant, so applying this equation to both $r_1$ 
and $r_2$ we get,
\begin{align}
 \sqrt{ \frac{a_{c\U{min}}}{r_1} } &= \sqrt{ \frac{a_{c2}}{r_2} } \\
 a_{c2} &= a_{c\U{min}} \frac{r_2}{r_1}
	= a_{c\U{min}} \frac{2.20\U{cm}}{2.10\U{cm}}
	= 1.05 \cdot a_{c\U{min}}
\end{align}
So the acceleration is indeed $\geq 100g$ throughout the cylinder.
\end{solution}
