9.25 a centrifuge takes up only 0.127 m of bench space

[karush]

$\textsf{An advertisement claims that a centrifuge takes up only $0.127 m$ of bench space}$
$\textsf{but can produce a radial acceleration of $4100 \, g$ at $6830 \, rev/min$}$

$\textsf{a. Calculate the requested radius of the centrifuge}$

OK the only thing I can guess here is $\frac{0.127}{2}$ as max

 

[skeeter]

$\textsf{An advertisement claims that a centrifuge takes up only $0.127 m$ of bench space}$
$\textsf{but can produce a radial acceleration of $4100 \, g$ at $6830 \, rev/min$}$

$\textsf{a. Calculate the requested radius of the centrifuge}$

OK the only thing I can guess here is $\frac{0.127}{2}$ as max

$a_r = r\omega^2 \implies r = \dfrac{a_r}{\omega^2}$

you'll need to convert $\omega$ given in rpm to rad/sec

 

[karush]

$a_r = r\omega^2 \implies r = \dfrac{a_r}{\omega^2}$

you'll need to convert $\omega$ given in rpm to rad/sec

$r = \dfrac{a_r}{\omega^2}\\$
$\textit {how do you cancel the units? }\\$
\begin{align}
\displaystyle
\frac{0.127 \, m}{2} &\ge \dfrac{4100 \, g}{(6830\cdot 2\pi\cdot 60)^2 \, rad/s}\\
0.0635&\ge \frac{4100 \, g}{6.629849 \, rad/s}\\
0.0635&\ge 6.184
\end{align}not!

 

[skeeter]

$\dfrac{6830 \, rev}{min} \cdot \dfrac{2\pi \,rad}{rev} \cdot \dfrac{1 \min}{60 \,sec} = \dfrac{683\pi}{3} \, \dfrac{rad}{sec}$

$r = \left(4100g \, m/sec^2\right) \left(\dfrac{3}{683\pi} \, \dfrac{sec}{rad} \right)^2 \approx 0.0785 \, m$

$d = 2r \approx 0.157 \, m > 0.127 \, m$

seems their "claim" is false ...

 

[karush]

$\dfrac{6830 \, rev}{min} \cdot \dfrac{2\pi \,rad}{rev} \cdot \dfrac{1 \min}{60 \,sec} = \dfrac{683\pi}{3} \, \dfrac{rad}{sec}$

where do you get

$$\dfrac{683\pi}{3}$$

 

[skeeter]

$\dfrac{6830}{1} \cdot \dfrac{2\pi}{1} \cdot \dfrac{1}{60} = \dfrac{6830 \cdot 2\pi}{60} = \dfrac{6830 \cdot \pi}{30} = \dfrac{683 \cdot \pi}{3}$