Question

A centrifuge in a medical laboratory rotates at an angular speed of 3,650 rev/min. when switched off, it rotates through 48.0 revolutions before coming to rest. find the constant angular acceleration (in rad/s2) of the centrifuge.

Answer 1

First of all we need to convert everything into SI units.

Let's start with the initial angular speed, $\omega _i = 3650 rev/min$. Keeping in mind that
$1 rev = 2 \pi rad$
$1 min=60 s$
we have
$\omega _i = 3650 \frac{rev}{min} \cdot \frac{2 \pi rad/rev}{60 s/min} =382.0 rad/s$

And we should also convert the angle covered by the centrifuge:
$\theta = 48.0 rev= 48.0 rev \cdot 2 \pi \frac{rad}{rev}=301.4 rad$

This is the angle covered by the centrifuge before it stops, so its final angular speed is $\omega_f =0$.

To solve the problem we can use the equivalent of
$2aS = v_f^2 -v_i^2$
of an uniformly accelerated motion but for a rotational motion. It will be
$2 \alpha \theta = \omega_f^2-\omega_i^2$
And by substituting the numbers, we can find the value of $\alpha$, the angular acceleration:
$\alpha=- \frac{\omega_i^2}{2 \theta}=- \frac{(382 rad/s)^2}{2 \cdot 301.4 rad}=-242.1 rad/s^2$