Question

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

Answer 1

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