Centrifugal/Centripetal Acceleration

[thebigeis]

This question is actually posted up, but not under a very descriptive title so I'm reposting it and hoping the admin will just close my other one... sorry. Anyway, here is my problem-

#1) A typical lab centrifuge rotates at 4000rpm. Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. What is the acceleration at the end of a test tube that is 10cm from the axis of rotation in?

What I've already done is find the circumference of the circle, multiplied that by how many revs in a minute the lab does, divide that by 60 to get it in how many revs it does per second, then plug it into the centripetal acceleration equation a=(v^2)/r.

Work:
Known- 4000rpm, r=5cm
Find-a

4000rpm * (2pi(r=5)) = 1256.6m/min
1256.6/60 = 20.94m/s
a = ((v=20.94)^2)/(r=.05m) = 8772.98m/s^2

Obviously, this isn't right; so what do I need to do?

 

[Andrew Mason]

Work:
Known- 4000rpm, r=5cm
Find-a

4000rpm * (2pi(r=5)) = 1256.6m/min
1256.6/60 = 20.94m/s
a = ((v=20.94)^2)/(r=.05m) = 8772.98m/s^2

Obviously, this isn't right; so what do I need to do?

Why are you using 5 cm as the radius? Should it not be 10 cm?

AM

 

[kyle8921]

I understand your confusion and I will be happy to help you with this problem. First, let's clarify the difference between centrifugal and centripetal acceleration. Centrifugal acceleration is the apparent force felt by an object moving in a circular path, while centripetal acceleration is the actual force acting on the object that keeps it moving in a circular path.

In this problem, we are dealing with centripetal acceleration since we are looking at the acceleration of the test tube as it moves in a circular path in the centrifuge. The equation for centripetal acceleration is a = (v^2)/r, where v is the tangential velocity and r is the radius of the circular path.

In your work, you have correctly calculated the tangential velocity at the end of the test tube, but you have used the incorrect radius. The radius in this case is not 5cm, but rather 10cm (since the test tube is 10cm from the axis of rotation). So, the correct calculation would be:

v = 20.94m/s
r = 0.1m
a = ((20.94)^2)/(0.1) = 4384.92m/s^2

Therefore, the acceleration at the end of the test tube is approximately 4384.92m/s^2. This is a very large acceleration, which is why it is important to handle the test tube carefully when placing it in the centrifuge.

I hope this helps clarify the problem for you. Keep up the good work in your lab experiments!