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.


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

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

  • $\begingroup$ The time taken for each revolution is not constant; it gets longer as the centrifuge slows down. $\endgroup$ – leongz Nov 23 '12 at 23:54
  • $\begingroup$ @leongz Well, how would I find the time interval during which the centrifuge slows down? $\endgroup$ – Mack Nov 24 '12 at 0:29
  • $\begingroup$ It may be helpful to consider the graph of angular velocity against time. The slope of the graph gives the constant angular acceleration, while the area under the graph gives the angular distance traveled. $\endgroup$ – leongz Nov 24 '12 at 1:17

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.

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  • $\begingroup$ Thank you very much! I completely forgot about the rotational kinematic equations, until you had refreshed my mind. $\endgroup$ – Mack Nov 24 '12 at 12:54

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