Question

A typical laboratory centrifuge rotates at 4000 rpm. 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 10 cm from the axis of rotation? For comparison, what is the magnitude of the acceleration a test tube would experience if dropped from a height of 1.0 m and stopped in a 1.0-ms-long encounter with a hard floor?

Answer 1

Answer:

1. $a_{rad}=17545.2\frac{m}{s^{2}}$

2. $a=4429.45 \frac{m}{s^{2}}$

Explanation:

Radial acceleration is:

$a_{rad}=\frac{v^2}{r}$ (1)

With r the radius respects the axis of rotation and v the tangential velocity that is related with angular velocity (ω) by:

$v= \omega r$ (2)

By (2) on (1):

$a_{rad}= \frac{(\omega r)^2}{r}= (\omega )^2r=(418,88\frac{rad}{s})^2(0.1m)$

$a_{rad}=17545.2\frac{m}{s^{2}}$

To find the acceleration of the tube with the fall, we can use the expression:

$\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t)$ (3)

Due impulse-momentum theorem:

$\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ (4)

with p the momentum and J the impulse. By (4) on (3):

$\overrightarrow{p}_{f}-\overrightarrow{p}_{i}=\overrightarrow{F}_{avg}(\varDelta t)$

And using Newton's second law (F=ma) and that (P=mv):

$mv_f-mv_i=(ma)(\varDelta t)$ (5)

Final velocity is the velocity just after the encounter with hard floor, and initial momentum us just before that moment so the first one is zero and the second one can be found sing conservation of energy:

$\frac{mv_i}{2}=mgh$

$v_i=\sqrt{2gh}=\sqrt{2(9.81)(1.0)}=4.43\frac{m}{s}$

So (5) is:

$-m(4.43)=(ma)(\varDelta t)$

solving for a:

$a=\frac{4.43}{\varDelta t}=\frac{4.43}{1.0\times10^{-3}}=-4429.45$

It’s negative because is opposed to the tube movement.