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

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

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Answer 1 · 2016-10-20T13:48:04

I got: $2.75 \times 10^{4}$ $2.75xx10^4$ revolutions
BUT check my maths!

Explanation:

We can say that:
$1 rpm = \frac{2 π}{60} \frac{r a d}{s}$ $1"rpm"=(2pi)/60(rad)/s$

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

In our case the object changes ANGULAR velocity and goes from $ω_{0} = 0$ $omega_0=0$ to $ω_{f} = 15, 000 rpm = 15, 000 \times \frac{2 π}{60} \frac{r a d}{s}$ $omega_f=15,000"rpm"=15,000xx(2pi)/60(rad)/s$ in $t = 220 s$ $t=220s$ .
This corresponds to an angular acceleration:
$α = \frac{ω_{f} - ω_{0}}{t} = 15, 000 \times \frac{2 π}{60} \times \frac{1}{220} = 7.14 \frac{r a d}{s^{2}}$ $alpha=(omega_f-omega_0)/t=15,000xx(2pi)/60xx1/220=7.14(rad)/s^2$
Let us use a (rotational) kinematic relationship to find tha ANGULAR distance $θ$ $theta$ described, as:
$θ = ω_{0} t + \frac{1}{2} α t^{2}$ $theta=omega_0t+1/2alphat^2$
$θ = 0 + \frac{1}{2} \cdot 7.14 \cdot (220^{2}) \approx 1.72 \times 10^{5} r a d$ $theta=0+1/2*7.14*(220^2)~~1.72xx10^5rad$ in total (during the $220 s$ $220s$ )

But each revolution corresponds to $2 π$ $2pi$ rad giving us:
$\frac{1.7 \times 10^{5}}{2 π} = 2.75 \times 10^{4}$ $(1.7xx10^5)/(2pi)=2.75xx10^4$ revolutions

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

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