Centrifuge - find centripetal acceleration

[dragonladies1]

Homework Statement

You are designing a centrifuge to spin at a rate of 15,300 rev/min.
(a) Calculate the maximum centripetal acceleration that a test-tube sample held in the centrifuge arm 14.7 cm from the rotation axis must withstand.
377361.36 m/s2

(b) It takes 1 min, 16 s for the centrifuge to spin up to its maximum rate of revolution from rest. Calculate the magnitude of the tangential acceleration of the centrifuge while it is spinning up, assuming that the tangential acceleration is constant.

Homework Equations

Is the correct equation for the tangential acceleration a<sub>t[SUB=r*a?
If not, what is the correct formula(s)?

[h2]The Attempt at a Solution[/h2]
at=.147*377361.36=55472.12 m/s²</sub>

 

[Delphi51]

I get the 377361.36 m/s2 answer. You are supposed to show how you got yours so we can find out where you went wrong . . .
I started with v = 2πrn/T where n is the number of turns and T the time taken.
Then the standard formula for centripetal acceleration can be used to finish it.

 

[willem2]

Homework Equations

Is the correct equation for the tangential acceleration a<sub>t[SUB=r*a?
[/QUOTE]

No.
The tangential acceleration is the derivative with respect to time of the tangential
speed. It's also the component of acceleleration in a direction perpendicular to the radius.
For a constant angular acceleration, the acceleration is equal to

$$\frac { v_{final} - v_{initial} } { t_{final} - t_{initial} }$$

Just like linear acceleration.</sub>

 

[dragonladies1]

Thank you. So let me make sure I got this correctly...for my problem:

V<sub>final=(2*pi*.147)/(1/255)=235.53 m/s
Vinitial= 0
Tfinal=76 seconds
Tinitial=0

So...
235.53/76=3.099 m/s²


Is this correct?</sub>

 

[dragonladies1]

Thank you very very much. I got the answer.