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RPM required to simulate Earth's gravity (Centrifugal acceleration)

  1. Mar 22, 2014 #1
    1. The problem statement, all variables and given/known data

    Here's the problem: "On a journey to Mars, one design is to have a section of the spacecraft rotate to simulate gravity. If the radius of this section is 30 meters, how many RPMs must it rotate to simulate one Earth gravity (1 g = 9.8 meters/sec^2)?"

    2. Relevant equations

    [itex]A = \frac{V^2}{R}[/itex]

    3. The attempt at a solution

    I completed this problem, getting an answer of 5.5 RPM. Apparently I was wrong, and the answer should have been 14 RPM. I asked my instructor about it, and he showed me the calculations used to get 14 RPM, but it still didn't quite make sense to me.

    Here's the process my instructor used to solve the problem:
    Circumference is C = 2∏(30) = 188 meters.

    1 rpm is like rotating 1 full circular rotation every minute, so 188/60 = 3.1 meters/sec.

    Then V = 3.1 meters/sec x RPM.

    Then A = (3.1 RPM)^2/188 = 0.05 x rpm^2.

    So: 9.8 = 0.05rpm^2

    RPM = 14

    --

    Here's my process:
    A = 9.8 (m)/(s^2)
    R = 30 m
    V = √(AR)
    V = 17.15 m/s

    1 revolution = circumference = 2∏30 m
    [1 revolution/(2∏30 m)] * [60s/1 min] (1 rev is equal to 2π30m and sixty seconds is equal to one minute)
    V = (17.15 m/s)/∏ RPM
    V = 5.5 RPM

    --

    I realize, of course, that we both used different methods to get the answer. I thought I did it wrong, so I solved the problem again, using his method, to try to get the correct answer. He used 188m for the radius component when solving for A=V^2/R, but when I used 30m I got an answer of 5.5RPM. The radius, 30m, should be used in that equation, not 188m, right?

    I would greatly appreciate any help in understanding this problem. I can't tell where/how I went wrong in solving it.
     
  2. jcsd
  3. Mar 22, 2014 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Why is he dividing by 188 (circumference)?? Should be 30.

    You are right and the instructor is wrong.

    It is easier to use Ac = ω2r; so [itex]\omega = \sqrt{A/r}[/itex] where ω is the angular speed in radians/sec.

    Letting RPM = 60ω/2∏:

    [itex]RPM = \frac{60}{2\pi}*\sqrt{A/r} = \frac{60}{2\pi}\sqrt{9.8/30} = 5.5[/itex]

    AM
     
    Last edited: Mar 22, 2014
  4. Mar 23, 2014 #3
    I also agree, it must be divided by radius = 30 not circumference.
     
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