Question

A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point located 8.00 cm from the axis of rotation of the centrifuge moves with a speed of 150 m/s when the centrifuge is at full speed. (a) What is the angular acceleration (in rad/s2) of the centrifuge as it spins up? (b) How many revolutions does the centrifuge make as it goes from rest to its final angular speed?

Answer 1

Answer:

a) $\alpha = 18.75\,\frac{rad}{s^{2}}$, b) $n \approx 14920.776\,rev$

Explanation:

a) The final angular speed is:

$\omega = \frac{v}{R}$

$\omega = \frac{150\,\frac{m}{s} }{0.08\,m}$

$\omega = 1875\,\frac{rad}{s}$

The angular acceleration experimented by the centrifuge is determined by means of the following kinematic expression:

$\alpha = \frac{\omega - \omega_{o}}{\Delta t}$

$\alpha = \frac{1875\,\frac{rad}{s} - 0\,\frac{rad}{s} }{100\,s}$

$\alpha = 18.75\,\frac{rad}{s^{2}}$

b) The change in angular position is:

$\Delta \theta = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \alpha}$

$\Delta \theta = \frac{\left(1875\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(18.75\,\frac{rad}{s^{2}} \right)}$

$\Delta \theta = 93750\,rad$

Lastly, the total amount of revolutions made by the centrifuge is:

$n = (93750\,rad)\cdot \left(\frac{1}{2\pi}\,\frac{rev}{rad} \right)$

$n \approx 14920.776\,rev$