# Uniform Circular Motion.

by MetalCut
Tags: circular, motion, uniform
 P: 21 Hi there. I need some help with this question. Can anyone help me...... A centrifuge is a device in which a small container of material is rotated at a high speed on a circular path. Suppose the centripetal acceleration of the sample is 6.25 X 103 times as large as the acceleration due to gravity. How many revolutions per minute is the sample making, if it is located at a radius of 5.00cm from the axis of rotation? Any help would be appreciated. Thanx
 Emeritus Sci Advisor PF Gold P: 9,789 Could you please show some work or thoughts? HINT: What is the equation for centripetal acceleration? ~H
 P: 21 The equation is a = v2/r. But what do they mean when they say the centripetal acceleration is 6.25x10 3 times as large as the acceleration due to gravity????
Emeritus
PF Gold
P: 9,789

## Uniform Circular Motion.

$$a = \left( 6.25\times 10^{3} \right)g$$

~H
 P: 21 So then that probably means that v2/r = (6,25x10 3)g And the circumference of the circle its rotating in is 0,314m or 31,4cm
 Emeritus Sci Advisor PF Gold P: 9,789 But you want to find revolutions per minute, so your next step would be calculating the angular velcoity ($\omega$). You will need to use; $$v = \omega r$$ ~H
 P: 21 But i can get (v) also with v=(2)(pie)(r)\T So i still need T
 Emeritus Sci Advisor PF Gold P: 9,789 They are effectively the same thing, but you don't need to work out v; $$a = \frac{v^2}{r}$$ $$v = \omega r = \frac{2\pi r}{T}$$ $$a = \frac{\omega^2 r^2}{r}$$ $$\omega^{2} = \frac{a}{r}$$ $$\frac{2\pi}{T} = \sqrt{\frac{a}{r}}$$ ~H
 P: 21 Thanx i think i've got it.

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