# Homework Help: HELP! Circular motion problem.

1. Jan 12, 2008

### Bensky

[SOLVED] HELP! Circular motion problem.

1. The problem statement, all variables and given/known data
A test tube in a centrifuge is pivoted so that it swings out horizontally as the machine builds up speed. If the bottom of the tube is 165 mm from the central spin axis, and if the machine hits 59000 revolutions per minute, what would be the centripetal force exerted on a giant amoeba of mass 1.0 x 10^-8 kg at the bottom of the tube?

t=period=1/frequency
V=velocity
m=mass=1.0 x 10^-8 kg

2. Relevant equations

V = (2*pi*r)/T
F=(mv^2)/r

3. The attempt at a solution

r=.165m (converted to m from mm by dividing by 1000)
m=1.0 X 10^-8
f = 983.3 revolutions per second (I changed 59000 rpm to revolutions per sec. by dividing by 60)
T (period) = 1/983.3 ~= .001

V=(2*pi*r)/T
V= ((2*pi)*.165)/0.001
V= 1036.725576 m/s

F=(mv^2)/r
F=((1 x 10^-8)(1036.725^2)) / .165
F=.0651393 N

Now, I've checked my math several times, so I don't think there's a problem with that - I also think I have the correct formulas. I'm thinking this is either a rounding error or I have the period wrong.

The answers I have tried so far: .065, .0651, and .07 - all are wrong somehow.

Can anyone explain what I have done wrong?

Last edited: Jan 12, 2008
2. Jan 12, 2008

### hage567

I get a different velocity than you. I think the ~0.001 s you are using for T is causing you trouble. Since T = 1/983.3 s just multiply your v = 2*pi*r by 983.3 instead of dividing by 0.001. This will keep more figures.

3. Jan 12, 2008

### Bensky

So you're saying do 2*pi*r*f instead of 2*pi*r/T?

I would think you meant divide by f, but why would I divide by f if the formula tells me to divide by the period?

4. Jan 12, 2008

### hage567

No, you're not dividing by f. You can either divide by T or multiply by f, since f = 1/T.

v = 2*pi*r*(1/T) (the way you did it)

which is the same as v = 2*pi*r*f

It's just one way to avoid the 0.001 s. Which I think is you're problem because you rounded it.

Hopefully I'm not confusing you.

5. Jan 12, 2008

### Bensky

Thank you! Turns out it was a rounding error and when I did it the way you did it it worked fine. I will use that formula from now on.

6. Jan 12, 2008

### hage567

There's nothing wrong with the way you were doing it. Either way works and both can be useful. Just be aware of rounding when you're using small numbers.