A centrifuge accelerates uniformly from rest to 15,000 rpm in 220 s. Through how many revolutions did it turn in this time?

1 Answer
Oct 20, 2016

Answer:

I got: #2.75xx10^4#revolutions
BUT check my maths!

Explanation:

We can say that:
#1"rpm"=(2pi)/60(rad)/s#

[1 revolution is #2pi# radians
and 1 min=60 s]

In our case the object changes ANGULAR velocity and goes from #omega_0=0# to #omega_f=15,000"rpm"=15,000xx(2pi)/60(rad)/s# in #t=220s#.
This corresponds to an angular acceleration:
#alpha=(omega_f-omega_0)/t=15,000xx(2pi)/60xx1/220=7.14(rad)/s^2#
Let us use a (rotational) kinematic relationship to find tha ANGULAR distance #theta# described, as:
#theta=omega_0t+1/2alphat^2#
#theta=0+1/2*7.14*(220^2)~~1.72xx10^5rad# in total (during the #220s#)

But each revolution corresponds to #2pi# rad giving us:
#(1.7xx10^5)/(2pi)=2.75xx10^4#revolutions