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# Physics Question?

A centrifuge is a device in which a small container of material is rotated at a high speed on a circular path. Such a device is used in medical laboratories, for instance, to cause the more dense red blood cells to settle through the less dense blood serum and collect at the bottom of the container. Suppose the centripetal acceleration of the sample is 6.45 × 103 times as large as the acceleration due to gravity. What is the linear speed of the sample, if it is located at a radius of 5.30 cm from the axis of rotation?

### 2 Answers

- NCSLv 74 years agoFavorite Answer
You made a couple of errors here.

On the LHS, you have "6450" instead of "6450*g" ("6.45 × 10³ times as large as the acceleration due to gravity").

On the RHS you (at first) have a nonsensical term -- v²/t. What is that? It's certainly not an acceleration, since the units are m²/s³. "v²/t" has no meaning, ever. The gravitational force is simply m*g and has nothing to do with the velocity. And you only need to consider the gravitational force (or just acceleration) if the question asks for the "net" or "total" or "resulting" acceleration, which wouldn't happen in a problem like this where the object isn't accelerating vertically.

But your last question is correct: it's as simple as v² / r.

So, your equation should be

6450 * 9.8m/s² = v² / 0.053m

and in the end you are only off by a factor of "g."

Note that if you had kept track of your units, you'd have probably known it.

m²/s = (m²/s²) / m

Hope this helps!

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- 4 years ago
I tried relating centripetal force and gravitational force so,

ac=v^2/r and ag=v^2/t

6450=(v^2/r)/(v^2/t)

but I don't know the time...

Or is it as simple as 6450=v^2/.053

v=18.5 ?

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