Finding Constant Angular Acceleration

Question

The question is, "A centrifuge in a medical laboratory rotates at an angular speed of 3500 rev/min. When switched off, it rotates 46.0 times before coming to rest. Find the constant angular acceleration of the centrifuge."

So, I said that the centrifuge makes 58.3 revolutions every second $(58.3~rev./s)$, and one revolution takes 0.0172 s $(0.0172~s/rev.)$

$\omega_i=\Large\frac{58.3~rev.}{s} \cdot \frac{2\pi}{1~rev.}\small=366~rad/s$

$\Large\frac{0.0172~s}{1~rev.}\cdot\frac{46.0~rev.}{1}\small=0.7912$ This is how long it continues to rotate after the centrifuge stops applying a force.

$\omega_f=0$

$\alpha=-463~rad/s^2$ However, the answer $-223~rad/s^2$

What did I do incorrectly? Is any of my analysis erroneous?

Answer 1

Let's look at the given informations. We have an initial angular velocity, and we know how many revolutions it takes to get to zero angular velocity. So we have $\omega_i$, $\omega_f$, and $\theta$. Looking at that list of givens I would guess to use $$ \omega_f = \alpha t+\omega_i \quad\text{and}\quad \theta_f=\frac{1}{2}\alpha t^2 + \omega_i t + \theta_i $$ This has all the items we are interested in, and all the givens, and only time as an extra piece. Then we have two equations and two unknowns. Solving for $t$ in the first one we get $$ t=\frac{\omega_f -\omega_i}{\alpha} $$ Then the second equations becomes after substituting in $$ \theta_f = \frac{1}{2}\alpha \left( \frac{\omega_f -\omega_i}{\alpha} \right)^2 + \omega_i \left( \frac{\omega_f -\omega_i}{\alpha} \right) $$ $$ \alpha=\frac{1}{2\theta_f}(\omega_f -\omega_i)^2+\frac{\omega_i}{\theta_f}(\omega_f -\omega_i) $$ now we know that the final speed is zero, so put that in and simplify. $$ \alpha=\frac{1}{2\theta_f}(\omega_i)^2-\frac{\omega_{i}^{2}}{\theta_f}=-\frac{1}{2}\frac{\omega_{i}^{2}}{\theta_f} $$ Hope this helps.