First of all we need to convert everything into SI units.
Let's start with the initial angular speed, [tex]\omega _i = 3650 rev/min[/tex]. Keeping in mind that
[tex]1 rev = 2 \pi rad[/tex]
[tex]1 min=60 s[/tex]
we have
[tex]\omega _i = 3650 \frac{rev}{min} \cdot \frac{2 \pi rad/rev}{60 s/min} =382.0 rad/s [/tex]
And we should also convert the angle covered by the centrifuge:
[tex]\theta = 48.0 rev= 48.0 rev \cdot 2 \pi \frac{rad}{rev}=301.4 rad [/tex]
This is the angle covered by the centrifuge before it stops, so its final angular speed is [tex]\omega_f =0[/tex].
To solve the problem we can use the equivalent of
[tex]2aS = v_f^2 -v_i^2[/tex]
of an uniformly accelerated motion but for a rotational motion. It will be
[tex]2 \alpha \theta = \omega_f^2-\omega_i^2[/tex]
And by substituting the numbers, we can find the value of [tex]\alpha[/tex], the angular acceleration:
[tex]\alpha=- \frac{\omega_i^2}{2 \theta}=- \frac{(382 rad/s)^2}{2 \cdot 301.4 rad}=-242.1 rad/s^2 [/tex]
