# A centrifuge accelerates uniformly from rest to 15,000 rpm in 220 s. Through how many revolutions did it turn in this time?

Oct 20, 2016

I got: $2.75 \times {10}^{4}$revolutions
BUT check my maths!

#### Explanation:

We can say that:
$1 \text{rpm} = \frac{2 \pi}{60} \frac{r a d}{s}$

[1 revolution is $2 \pi$ radians
and 1 min=60 s]

In our case the object changes ANGULAR velocity and goes from ${\omega}_{0} = 0$ to ${\omega}_{f} = 15 , 000 \text{rpm} = 15 , 000 \times \frac{2 \pi}{60} \frac{r a d}{s}$ in $t = 220 s$.
This corresponds to an angular acceleration:
$\alpha = \frac{{\omega}_{f} - {\omega}_{0}}{t} = 15 , 000 \times \frac{2 \pi}{60} \times \frac{1}{220} = 7.14 \frac{r a d}{s} ^ 2$
Let us use a (rotational) kinematic relationship to find tha ANGULAR distance $\theta$ described, as:
$\theta = {\omega}_{0} t + \frac{1}{2} \alpha {t}^{2}$
$\theta = 0 + \frac{1}{2} \cdot 7.14 \cdot \left({220}^{2}\right) \approx 1.72 \times {10}^{5} r a d$ in total (during the $220 s$)

But each revolution corresponds to $2 \pi$ rad giving us:
$\frac{1.7 \times {10}^{5}}{2 \pi} = 2.75 \times {10}^{4}$revolutions