How Do You Calculate Centripetal Acceleration in a Blood Centrifuge?

AI Thread Summary
To calculate centripetal acceleration in a blood centrifuge, the key formulas are a = v²/r and v = d/t, where 'v' is the tangential velocity and 'r' is the radius from the rotation axis. The period of rotation can be determined from the centrifuge's speed, converting revolutions per minute to seconds per revolution. It's important to consider the angle of the test tube when determining the effective radius for calculating centripetal acceleration. After applying these concepts, a participant calculated the centripetal acceleration to be 10749 m/s², confirming the accuracy of their solution. Understanding the relationship between the angle and radius is crucial for correct calculations.
Jpyhsics
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Homework Statement


A centrifuge is a laboratory device used for spinning samples of material. In a blood centrifuge, a test tube is inserted at an angle θ=32.0° with respect to the vertical and the whole sample is spun at high speed. For a typical test that is l= 15.9 cm long and is spun in a centrifuge at a rate of 4316 rotations per minute, what is the centripetal acceleration experienced by the sample at a point x=9.93 cm from the rotation axis, as measured along the test tube? Express your answer in units of the standard gravity g. Thus, if your answer is 19.62 m/s2 = 2.00×9.81 m/s2, you would enter 2.00 into the box.

Homework Equations


a=v2/r
T=2πr/f
v=d/t

The Attempt at a Solution


First, I found the period
T=2π(15.9)/(4316/60)=1.3888
v=2π(9.93)/1.3888=71.9
a=(71.9)2/9.93

I don't think I did this correctly, as I didnt use the angle...
I have included a photo as well,
 

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Jpyhsics said:

Homework Equations


a=v2/r
T=2πr/f
v=d/t
The equation T=2πr/f cannot be correct as you can tell by doing a unit analysis of the equation.

3. The Attempt at a Solution
First, I found the period
T=2π(15.9)/(4316/60)=1.3888
v=2π(9.93)/1.3888=71.9
a=(71.9)2/9.93
You need to be clear on the meaning of r in the equations. You want the centripetal acceleration of a particular point in the test tube. Mark this point in your diagram and try to draw a representation of r in the diagram.
 
TSny said:
The equation T=2πr/f cannot be correct as you can tell by doing a unit analysis of the equation.You need to be clear on the meaning of r in the equations. You want the centripetal acceleration of a particular point in the test tube. Mark this point in your diagram and try to draw a representation of r in the diagram.
So how what formulas would I have to use?
I would assume ac=v2/r? but what about the period (T)?
 
Jpyhsics said:
So how what formulas would I have to use?
Would I use the 15.9cm?
 
Jpyhsics said:
So how what formulas would I have to use?
You should be able to solve it using your two equations

a=v2/r
v=d/t

But if you have learned how to write centripetal acceleration in terms of angular velocity ω, you might find that to be a quicker way to the answer.
 
Jpyhsics said:
but what about the period (T)?
If the test tube makes 4316 revolutions in one minute, how much time does it take for it to make one revolution?
 
TSny said:
If the test tube makes 4316 revolutions in one minute, how much time does it take for it to make one revolution?
0.0139s per rotation?
 
Jpyhsics said:
0.0139s per rotation?
Oh I see how the angle relates now. I got an answer of 10749 m/s2. Thank you so much for your help! :)
 
Jpyhsics said:
Oh I see how the angle relates now. I got an answer of 10749 m/s2. Thank you so much for your help! :)
I think your answer is correct. Good work.
 
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