# Kinematics - centrifuging a solution with bacteria

1. Jan 29, 2014

### kenji1992

Kinematics -- centrifuging a solution with bacteria

1. The problem statement, all variables and given/known data

In the lab, we have to separate the bacteria from the solution by centrifuging the solution with bacteria. Because the bacteria are more massive than the media molecules, they settle to the bottom of the container.
For this question, we will neglect friction or resistance from the buffer (we will see it later).
If the bacteria are spun at 100 xg, how long will it take for the bacteria to pellet to the bottom of a 10 cm centrifuge bottle?

2. Relevant equations

3. The attempt at a solution

So I know that I have to find the time (t)

I have no idea what 100 xg is--what is the unit 'xg'?

Is 10 cm relevant to the question?

If possible, can I be referred to another version of this problem or tutorial? I'm not sure what to do.

2. Jan 29, 2014

### voko

100 xg should be written as $100 \times g$, or just $100 g$, where $g$ is the acceleration due to gravity at sea level.

3. Jan 29, 2014

### kenji1992

So, t=?
Acceleration=100g
final velocity=10cm/s
initial velocity=0cm/s

acceleration=velocity/time
100g=10 cm/s-0cm/s / t
100g = 10 cm/s /t
t = 10 cm/s / 100 g
t=10 s
Is that right?

4. Jan 29, 2014

### Simon Bridge

Welcome to PF;
 beat me to it :)
the xg value is the "relative centrifugal force", usually written "100 x g" (a la voko) and read: "one hundred times gravity"
You'll see it in centrifuge specs all the time and it's one of the ways biologists annoy physicists who think that "100g" is clear enough.
http://en.wikipedia.org/wiki/Laboratory_centrifuge

The 10cm is relevant to the question.

Basically: where do the bacteria start out?
where do they end up?
how do they get there (what are the forces)?
how long do they take?

5. Jan 29, 2014

### kenji1992

Ok.

I'll assume that 10cm refers to displacement.
d1 = 0 cm
d2 = 10cm
acceleration=100 * 9.8 m/s^2
time=?

So would I use this equation?
vf=vi + a*t

but i'm not sure how that would work with displacement, as opposed to velocity?

Last edited: Jan 29, 2014
6. Jan 29, 2014

### voko

You need an equation that relates displacement with acceleration and time.

7. Jan 29, 2014

### kenji1992

This equation then: d = vi*t + 1/2at^2?

10 cm=0.1m

0.1m=0m/s*t + 1/2*980m/s^2 * t^2
0.1m=1/2*980m/s^2 * t^2
0.1 m = 490 m/s^2 * t^2
0.1m/490 m/s^2 = t^2
0.1/490 s^2 = t^2
sqrt 0.1/490 = t

8. Jan 29, 2014

### voko

Good so far.

9. Jan 29, 2014

### kenji1992

@voko What else is missing? Do I only have to simplify what I wrote?

10. Jan 29, 2014

### voko

I do not see the final number for time.

11. Jan 29, 2014

### kenji1992

t = 14.3 seconds

12. Jan 29, 2014

### voko

No. Not even close. Many orders of magnitude off.

13. Feb 4, 2014

### kenji1992

Magnitude means measurement, right?

14. Feb 5, 2014

### Simon Bridge

No.
In general: "magnitude" is a generalization of the concept of "length" as in "the magnitude of a vector"
... but in this specific case the term "order of magnitude" means "power of ten".

So if I said the answer was 5 and it was really 5000 ... I'd be three orders of magnitude off.
If I'd said the answer was 3000 ... I'd have the right order of magnitude but still the wrong answer.
(Order-of-magnitude calculations are often used for exploring an idea.)
This is different from if I'd have said the answer was 3 ... when I'd just be completely wrong.

Voko is saying you need to check your arithmetic.

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