Answer:
7500 revolutions
Explanation:
Given:
Time taken for the centrifuge (t) = 3 min
Initial angular speed of the centrifuge (ω₁) = 0 rad/min (Initially at rest)
Final rotational speed (N₂) = 5000 rpm
So, final angular speed of the centrifuge is given as:
[tex]\omega_2=2\pi N_2=2\times \pi\times 5000=10000\pi\ rad/min[/tex]
Now, using the concept of angular motion, the angular acceleration is given as:
[tex]\alpha =\dfrac{\omega_2-\omega_1}{t}\\\\\alpha =\frac{10000\pi-0}{3}\\\\\alpha=\frac{10000\pi}{3}\ rad/min^2[/tex]
Now, angular displacement of the centrifuge is determined using the angular equation of motion which is given as:
[tex]\theta=\omega_1t+\frac{1}{2}\alpha t^2\\\\\theta=0+\frac{1}{2}\times \frac{10000\pi}{3}\times (3)^2\\\\\theta=15000\pi\ rad[/tex]
Now, we know that, 1 revolution corresponds to a angular displacement of 2π radians.
So, 2π rad = 1 revolution
∴ 15000π rad = [tex]\frac{15000\pi}{2\pi}=7500\ revolutions[/tex]
Therefore, the centrifuge makes 7500 revolutions.
