A simple mathmatical problem..

Discussion in 'Physics & Math' started by ZMacZ, Jan 26, 2015.

  1. ZMacZ Registered Senior Member

    Messages:
    123
    Well it's like so..
    v.v
    a= ------
    r

    with v being

    v= 2.pi.r

    I tried looking at it through simple web stuff..but...

    How come they always get a G force factor that's higher than 10/r ?

    I used a circle radius 1 meter..and the spinning was 1 RPS..(rotation per second)

    And the answer was 39.4784 m/ss

    All the webstuffs show a different answer above 40...like 40.15 and 40.25..

    Am I right or am I wrong ?

    (thanx for any helpful answer in advance..)
     
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  3. ZMacZ Registered Senior Member

    Messages:
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    How do I add a pic ?
    Inserting the image only allows from another site ??
     
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  5. ZMacZ Registered Senior Member

    Messages:
    123
    anyways the first formula is acceleration (centrifugal acceleration) = v squared divided by r..

    and my answer would acceleration = 4 times pi squared times r...

    and with an r of 1 that would yield 39.4784...

    (or rather 39.4784 per r)

    (making my life easy by simplifying to 40 m/ss per meter for r..)

    but still why do they come up with numbers larger than 40 ?
     
    Last edited: Jan 26, 2015
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  7. mathman Valued Senior Member

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    My guess - webstuff is doing something different. Your answer is correct for the formula you are using.
     
  8. rpenner Fully Wired Valued Senior Member

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    4,833
    For rotation about a fixed point with constant angular velocity, \(v = \omega r, \quad a = \frac{v^2}{r} = \frac{\omega^2 r^2}{r} = \omega^2 r\).

    If \(\omega = 1 \, \textrm{rotations per second} = 2 \pi \, \textrm{radians per second} = 2 \pi \, \textrm{second}^{-1}\) and \(r = 1 \, \textrm{meter}\) it follows that \(a = \omega^2 r = 4 \pi^2 \, \textrm{meter} \cdot \textrm{second}^{-2} \approx \frac{10975}{278} \frac{\textrm{m}}{\textrm{s}^2} \approx 39.4784 \frac{\textrm{m}}{\textrm{s}^2} \)

    Quick check: Since \( 3.14 \lt \pi < 3.15 \lt \sqrt{10}\) it follows that \(a \lt 40 \frac{\textrm{m}}{\textrm{s}^2}\).

    If there is some sort of perverse multiple rounding procedure that gives 40.15 or 40.25, it's not an obvious one. That's why half or more of your physics homework is to show your work.
     
    Last edited: Jan 27, 2015
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  9. ZMacZ Registered Senior Member

    Messages:
    123
    so..my formula is ok then ?..guud...I thought I was getting brainfarted...

    btw..it's not about homework..

    I was looking for an easy converter...but doing the math by hand I found that discrepancy...

    Anyways...I found a simple way to stay at 1G on the moon..

    An 100metre arm above a maglev...and spinning at once every 25 seconds roughly..I'd say a lil more than that..since you actually need
    to counter the low level gravity time with a lil higher level gravity to avoid bone bone decalcification...
    And it's not like a wild ride...unless people have a gyroscopic sense that can tell we're actually moving around...
    (take motion sickness tablets

    Please Register or Log in to view the hidden image!

    ?..)

    Anyways..it would still require a big underground excavation...but with 0.1 G moving the soil out can be as simple as using a hoover..

    (like a brick can weigh like 2500 grams..but in 0.1 G it would be 250 grams...very hoover able..)

    I'd say using chemical plastic that hardens quickly through exothermic reaction...not needing anything but to mix it...
    then a nice deep shaft...
    and beneath all that the same process but in a dome like shape..the surface soil would be useful to shield from micrometeorites

    yes..it's a moonbase..easy construction of vessel to mars...

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    (and SOOO cool...)
     
  10. Jason.Marshall Banned Banned

    Messages:
    654
    I think you might be on to something although I disapprove of your persuits of masor weaponry did you not see what happened in the James Bond movie ? and besides Tesla had that all figured out decades ago and purposely dismantled the technology so it wouldn't fall in the wrong hands, imaging an angry lunatic gets access to this imaging what he would do to his neighbors the guy that cut him off coming out his driveway, hope you get my drift.

    I would encourage you to write a letter to the author of the source that gave you the quantity "40.15 and 40.25" and ask them two questions how many digits of accuracy did they define pi at before they made their calculation and what was the circumstantial basis they were using that specific quantity for, I have noticed some of these inconsistencies myself and currently investigating to get to the bottom of it so your input would be helpful thanks in advance because it seems like you are happy with your formula.
     
    Last edited: Feb 4, 2015
  11. Dr_Toad It's green! Valued Senior Member

    Messages:
    2,527
    Where do you come up with this garbage? He never said anything (in this thread, at least) about MASER weaponry, James Bond, Tesla, or even the Smurfs.

    You seem to have a cognition problem.
     
  12. ZMacZ Registered Senior Member

    Messages:
    123
    Ugh..the mass does not change..inertia would stay the same ofc.but the force required to move it would be less..so..hooverable..
    .
    Then again..no atmosphere ?..umm.. tent ? (yes..one capable to hold enough atmosphere to allow for suitless work..and with that I mean the bulky suits, not the thin life bubble ones, that you'd wear
    instead of the bulky ones, but without solid helmet..a close-able foldable one..)
    (like a minibubble for temporary atmosphere till you can get the atmosphere back..)
     
    Last edited by a moderator: Mar 24, 2016

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