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

Then teach the underlying concepts
Don't copy without citing sources
preview
?

Write a one sentence answer...

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

22
Gió Share
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

##### Just asked! See more
• 8 minutes ago
• 8 minutes ago
• 9 minutes ago
• 11 minutes ago
• A minute ago
• A minute ago
• 3 minutes ago
• 3 minutes ago
• 6 minutes ago
• 7 minutes ago
• 8 minutes ago
• 8 minutes ago
• 9 minutes ago
• 11 minutes ago