RPM of centrifuge sample with 7.80×10³g centripetal acceleration

[Doug Desatnik]

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 7.80 x 10^3 times as large as the acceleration due to gravity. How many revolutions per minute is the sample making, if it is located at a radius of 8.00 cm from the axis of rotation?
--------------------------------------------------------------------------

Suppose the centripetal acceleration of the sample is 7.80 x 10^3 times as large as the acceleration due to gravity.

Question 1
Does this simply mean that the centripetal force is 7.8x10^3 x 9.80? IF so then 7.8x10^3 x 9.80 = 76400.

Question 2
I worked through the solution but the answer I get doesn't seem right to me. Can somebody maybe please point out where it is that I went wrong?

1. First I found the velocity

[tex]Ac=V^2/r[\tex] --> [tex]76400=V^2/.08[\tex] --> [tex]v=78.20 m/s[\tex]<br /> <br /> 2. Then I used the following equation<br /> <br /> [tex]V=2\Pir/T[\tex] --> [tex]78.20=(2)(3.14)(.08)/T[\tex] --> [tex]T=.006s[\tex] --> $$.0001 RPM[\tex]<br /> <br /> .0001 Revolutions per Minute just seems incorrect for some reason :)<br /> <br /> Sorry for the screwy latex code, I must be doing something wrong here to ...<br /> <br /> Thanks in advance for the help!<br /> <br /> - Doug$$[/tex][/tex][/tex][/tex][/tex][/tex]

 

[krab]

LaTeX: your [\tex] should be forward slash.

You've interpreted a time T directly as a circular rate. Wrong. T is the time of one cycle. So How many cycles per second?

 

[M98Ranger]

Hi Doug,

To solve this problem, we first need to understand what is meant by "centripetal acceleration of the sample is 7.80 x 10^3 times as large as the acceleration due to gravity." This means that the centripetal acceleration (Ac) is 7.80 x 10^3 times the acceleration due to gravity (g). In other words, Ac = 7.80 x 10^3 * g.

Now, let's use the formula for centripetal acceleration: Ac = V^2/r, where V is the velocity and r is the radius. We can rewrite this as V = √(Ac * r). Plugging in the given values, we get:

V = √(7.80 x 10^3 * 9.80 * 0.08) = 78.20 m/s

Using the formula for velocity in circular motion, V = 2πr/T, where T is the period (time for one revolution), we can solve for T:

T = (2π * 0.08) / 78.20 = 0.016 s

Finally, we can convert this period to revolutions per minute (RPM):

T = 0.016 s = 0.016/60 min = 0.000267 min

Therefore, the sample is making 0.000267 RPM or approximately 0.0003 RPM.

I hope this helps clarify the solution for you. Keep in mind that the answer may seem small because the radius is in centimeters and the acceleration is in terms of g, which is a very small value. But the calculations are correct and the answer is in fact very low, indicating that the sample is rotating at a slow speed.