MHB 9.25 a centrifuge takes up only 0.127 m of bench space

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The discussion centers on calculating the radius of a centrifuge that claims to occupy 0.127 m of bench space while producing a radial acceleration of 4100 g at 6830 rev/min. Participants note the need to convert the rotational speed from rpm to rad/sec for accurate calculations. Using the formula for radial acceleration, they derive that the calculated radius is approximately 0.0785 m, leading to a diameter of about 0.157 m. This diameter exceeds the claimed bench space, suggesting the advertisement's claim is inaccurate. The calculations highlight the importance of unit conversion and proper application of physics formulas in verifying claims.
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$\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
 
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karush said:
$\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
 
skeeter said:
$a_r = r\omega^2 \implies r = \dfrac{a_r}{\omega^2}$

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

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

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

seems their "claim" is false ...
 


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

where do you get

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