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$v=33 800rpm$

Giancoli 6th Edition, Chapter 5, Problem 17

(2:54)

Chapter 5, Problem 17 is solved.

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Transcript for this Giancoli solution

So we have a centrifuge of radius point zero nine metres, that’s nine centimetres, so let’s draw a centrifuge with radius ‘r’ equals zero point zero nine zero zero metres and it has a centripetal acceleration of particle at that point. It’s going to be fifteen thousand ‘g’s’. So we’ll take a hundred and fifteen thousand times nine point eight and that gives us one point one two seven times ten raised to power six metres per second squared. So now that we know its centripetal acceleration we can find its velocity since we also know the radius of its curvature. Centripetal acceleration ‘ac=’ is ‘v’ squared divided by the radius ‘r’, meaning that velocity is the square root of radius times centripetal acceleration, multiply both sides by ‘r’ and we take the square root, this is the square root of zero point zero nine times one point one two seven times ten raised to power six and that speed is three hundred and eighteen point four eight one metres per second. Keeping lots of digits because they want the answer expressed in revolutions per minute. So let’s do that unit conversion: velocity is three hundred and eighteen point four eight one metres per second times sixty seconds per minute, this gets rid of our seconds leaving us with minutes, so that’s half done, we have the minutes now on the bottom times one revolution, we want revolutions on top because it’s ‘rpm’ revolutions per minute that we are looking for. At the bottom, we’ll put metres so we have one revolution every time it goes a full circumference, so we need two pi ‘r’ here, ‘r’ being point zero nine, so we have revolutions per minute and calculating gives us three three seven nine one RPM which we should write with three significant figures as thirty three thousand eight hundred ‘rpm’