# Centrifuge question

1. Oct 28, 2004

### matchboxdude

ok, you are using a centrifuge. the centripetal acceleration of the sample is 5.05E3 times larger than the accelleration due to gravity. how many revolutions/minute is the sample making if it is located at a radius of 4.62 cm from the axis of rotation?

i think i use: centripetal accellearion = (v^2)/r
so... 5.05E6*9.8m/s = v^2 / 4.62cm
v = 47.8167 rev/min. but i get it incorrect.. any hints?

2. Oct 28, 2004

### Galileo

Why take $5.05^6$ times g, instead of the given $5.05^3$?

Edit: Nevermind, that was a typo, you used the correct value.

But v is the velocity, not the number of revolutions per minute.
The time it takes (in seconds) for one revolution is $$T=\frac{2\pi r}{v}$$

Last edited: Oct 28, 2004
3. Oct 28, 2004

### matchboxdude

oops sorry. i meant 5.05^3 in my calculations. i used it, but just typed it here wrong.

4. Oct 28, 2004

### matchboxdude

ok, in my first equation. when i have the radius, should that be 4.62cm or .0462m ? because i used .0462m even though i wrote 4.62cm

Last edited: Oct 28, 2004
5. Oct 28, 2004

### Pseudo Statistic

You want angular velocity, you're using the linear velocity formula!
Just use simple substitution...
Because v = rw (Where w is angular velocity)
Centripetal Acceleration = (rw)^2/r Which gives:
(r^2w^2)/r = rw^2
Just solve for w.

Last edited: Oct 28, 2004
6. Oct 28, 2004

### matchboxdude

so everything i did was wrong? ok.
ok so i took:
5.05E3*9.8 = .0462w^2
w=1034.993778 and is that revolution/sec?
is that better?

7. Oct 28, 2004

### matchboxdude

anyone

8. Oct 29, 2004

### Galileo

That would radians per second. You traverse $2\pi$ radians in one revolution, so dividing $\omega[itex] by [itex]2\pi$ will give the number of revolutions per second.

And no, it's not true that everything you did was wrong. You could take
$$f=\frac{1}{T}=\frac{v}{2\pi r}$$ which also gives the number of revolutions per second.