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.

a. What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation?

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.

a. What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation?

Accepted Answer

Hi everyone. In this practice problem, we are being asked to calculate the container cent pedal acceleration. We will have a thin glass container placed 7.5 cm away from the axis of rotation of a spin coder device. The spin coder rotates at 2500 R PM. And we're being asked to calculate the container cent pedal acceleration. And the options given are a 2.0 times 10 to the power of one m per second squared. B 1.9 times 10 to the power of two m per second squared, C 2.6 times 10 to the power of two m per second squared and D 5.1 times 10 to the power of three m per second squared. So we will model the container as a particle and the center of mass of the particle is going to be placed 7.5 cm away from the axis of rotation. So that will mean the R is going to then be 7.5 cm or essentially equals to 0.075 m. Um The spin coder rotates at 2500 R PM. So the angular velocity is then going to be equals to 2500 R PM just like. So the centro pedal acceleration A is given by omega squared multiplied by R with omega squared or omega being the angular velocity and R being the radius, the angular velocity should be converted to radiant per seconds from R PM. In order for us to be able to get the cental acceleration in SI unit. So let's substitute everything in by also still converting the angular speed or angular velocity. So we have the omega squared omega squared is going to be 2500 R PM, which in this case, R PM will equals to revolution per minute. And in order to convert that into radiant for seconds for the SI unit, we wanna multiply it by two pi radiant per every rota revolution. And then also multiply that by one minute per every seconds And all of that is going to be squared and multiply the omega squared with the radius which is going to be 0.075 m. Calculating all of this, the result of the A a Centipede acceleration is going to equals to 5.1 times 10 to the power of three m per second squared. And that will essentially be the answer to this practice problem with a center pedal acceleration of 5.1 times 10 to the power of three m per second squared, which will correspond to option D in our answer choices. So option D is going to be the answer to this practice problem. And if you guys still have any sort of confusion on this particular practice problem, please make sure to check out our other lesson videos on similar topics and that'll be it for this one. Thank you.

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. a. What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation?

Verified Solution

Key Concepts

Centripetal Acceleration

Tangential Velocity

Rotational Motion