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$T=10s$

Giancoli 6th Edition, Chapter 5, Problem 45

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Chapter 5, Problem 45 is solved.

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Transcript for this Giancoli solution

We have a cylindrical space ship of diameter thirty two meters but in or formulas we’re always working with ‘r’ so let’s make sure we convert that right away. So we have our cylindrical space ship with a radius 'r' of sixteen meters which is half the diameter. We want to figure out how fast this cylinder has to rotate in order to give a simulated gravity of zero point six ‘g’s. We know that the centripetal acceleration is ‘v2’ over 'r' and that has to be zero point six zero ‘g’s. Solving for ‘v’: multiply both sides by 'r' and take the square root we have: ‘v’ equals the square root of zero point six zero times nine point eight times sixteen meters which is nine point six nine nine five meters per second which is not our answer because it wants to know what the periods of the rotation is, so how much time elapses with one complete rotation. What we’ve found so far is the tangential speed of a point on the rim of this cylinder. So: ‘v’ is two times pi times 'r' over‘t’ which is to say that a point travels the full circumference of the cylinder in a period called ‘t’. ‘t’ is what we’re looking for which is the amount of time for one rotation: ‘t’ equals two times pi times 'r' over ‘v’, substituting in numbers our answer will be two times pi times sixteen meters divided by nine point six nine nine five meters per second giving us an answer of ten seconds. For the 5th Edition we have zero point five ‘g’s and the speed in the 5th Edition is eight point eight five four meters per second and substituting gives us an answer of eleven seconds.