Difference between revisions of "Talk:162: Angular Momentum"

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The issue date is not given, as i don't have a clue about it. Could someone fix this? [[User:Rikthoff|Rikthoff]] ([[User talk:Rikthoff|talk]]) 19:30, 3 August 2012 (EDT)
 
The issue date is not given, as i don't have a clue about it. Could someone fix this? [[User:Rikthoff|Rikthoff]] ([[User talk:Rikthoff|talk]]) 19:30, 3 August 2012 (EDT)
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:When the page was updated to the new comic template by [[User:Bpothier]] he fixed the date. [[User:Lcarsos|lcarsos]] ([[User talk:Lcarsos|talk]]) 20:48, 28 August 2012 (UTC)
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That actually is a neat physics puzzle, which has probably (i.e. certainly) been addressed somewhere on the net. I may incorporate that some day. --[[User:Quicksilver|Quicksilver]] ([[User talk:Quicksilver|talk]]) 05:58, 24 August 2013 (UTC)
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I tried to calculate the change in Earth's period, assuming that she was standing in the north pole (latitude = 90º N), where her spinning would have more effect. I either did something wrong, or my TI-84 Plus is not capable of detecting the very small effect her spinning would have on the Earth's rotation. I assumed the Earth had a period of exactly 24 hours, and got the same value to the second, even if she was spinning at 1000 turns per second, which seems like a lot.
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Here's the formula:
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L_Earth_i = L_Earth_f + L_spinner <=>
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I_Earth * (2*PI)/T_Earth_i = I_Earth * (2*PI)/T_Earth_f + I_spinner* (2*PI) * f_spinner <=>
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(1/T_Earth_f) = (1/T_Earth_i) - (I_spinner/I_Earth)*f_spinner <=>
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T_Earth_f = 1/((1/T_Earth_i) - (I_spinner/I_Earth)*f_spinner)
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Where the variables have names in the format:
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[variable name]_[object it refers to]_[situation (i or f stand for initial and final)]
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L = Angular Moment
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I = Moment of Inertia
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T = Period of rotation about one's axis
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f = frequency
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I used as values:
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T_Earth_i = 86400 seconds (24 hours exactly)
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I_spinner = 62,04 Kg.m^2 (Found on Wolfram|Alpha, for a 62Kg adult human being)
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I_Earth = 8,03e+37 Kg.m^2 (http://scienceworld.wolfram.com/physics/MomentofInertiaEarth.html)
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f_spinner = the frequency of the woman's spinning in complete turns per second. {{unsigned ip|2.82.142.28}}
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:Taking that a bit further, the relative decrease is:
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  (T_Earth_f - T_Earth_i)/T_Earth_i = 1 / (I_Earth/(I_spinner*T_Earth_i*f_spinner) - 1)
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      = 1 / ( 1.5 e+28 - 1) ~= 67 e-30
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:Fwiw, the absolute value is 5.767 yocto-seconds. If the ''entire'' world population would spin at that 1000 turns per second (and at favourable locations as in your assumptions), the effect will still be a measly 0.041 pico-seconds. So T_Earth_f = 86 399.999 999 999 999 958 ... But the TI-84 only has about 14 digits precision, i believe, so even that won't show up. -- [[Special:Contributions/173.245.51.210|173.245.51.210]] 22:46, 30 October 2013 (UTC)
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Is it possible for someone to write an equation that factors in latitude (and, if relevant, longitude) that we could plug our locations into and get a value from? That would be awesome. Thanks. [[Special:Contributions/108.162.250.208|108.162.250.208]] 02:48, 23 February 2014 (UTC)
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The visual style and theme of this comic is clearly referencing the 'Spinning Ballerina Optical Illusion' (evidenced by the grey-to-white gradient 'glow', as well as her arm and leg positions).
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[[Special:Contributions/108.162.249.231|108.162.249.231]] 03:03, 30 September 2014 (UTC)
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The 'Spinning Ballerina' optical illusion does not apply here, Megan clearly stated that she was spinning 'counterclockwise' and due to the fact that she is drawn with hair (not a silhouette) lets you know where she is facing. Therefore the bent leg on the right of the image is her left leg.  Plus there aren't many ways to draw a stick figure 'spinning'.  Now if it were Cueball doing the spinning THEN I would agree with you because there would not be a reference point to make any type of judgement and therefore a point could be made that could be a reference Randall was trying to make.
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[[User:Nexxuz|Nexxuz]] ([[User talk:Nexxuz|talk]])
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The momentum of Megan is in fact one order of magnitude smaller than what appeared in above calculation, so the dilation effect is still smaller.
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http://www.wolframalpha.com/input/?i=moment+of+inertia+of+solid+cylinder+for+a%3D0.25m%2Cmass%3D62kg {{unsigned ip|173.245.50.90}}

Revision as of 15:40, 8 December 2014

The issue date is not given, as i don't have a clue about it. Could someone fix this? Rikthoff (talk) 19:30, 3 August 2012 (EDT)

When the page was updated to the new comic template by User:Bpothier he fixed the date. lcarsos (talk) 20:48, 28 August 2012 (UTC)

That actually is a neat physics puzzle, which has probably (i.e. certainly) been addressed somewhere on the net. I may incorporate that some day. --Quicksilver (talk) 05:58, 24 August 2013 (UTC)

I tried to calculate the change in Earth's period, assuming that she was standing in the north pole (latitude = 90º N), where her spinning would have more effect. I either did something wrong, or my TI-84 Plus is not capable of detecting the very small effect her spinning would have on the Earth's rotation. I assumed the Earth had a period of exactly 24 hours, and got the same value to the second, even if she was spinning at 1000 turns per second, which seems like a lot.

Here's the formula:

L_Earth_i = L_Earth_f + L_spinner <=>

I_Earth * (2*PI)/T_Earth_i = I_Earth * (2*PI)/T_Earth_f + I_spinner* (2*PI) * f_spinner <=>

(1/T_Earth_f) = (1/T_Earth_i) - (I_spinner/I_Earth)*f_spinner <=>

T_Earth_f = 1/((1/T_Earth_i) - (I_spinner/I_Earth)*f_spinner)


Where the variables have names in the format:

[variable name]_[object it refers to]_[situation (i or f stand for initial and final)]


L = Angular Moment

I = Moment of Inertia

T = Period of rotation about one's axis

f = frequency


I used as values:

T_Earth_i = 86400 seconds (24 hours exactly)

I_spinner = 62,04 Kg.m^2 (Found on Wolfram|Alpha, for a 62Kg adult human being)

I_Earth = 8,03e+37 Kg.m^2 (http://scienceworld.wolfram.com/physics/MomentofInertiaEarth.html)

f_spinner = the frequency of the woman's spinning in complete turns per second. 2.82.142.28 (talk) (please sign your comments with ~~~~)

Taking that a bit further, the relative decrease is:
  (T_Earth_f - T_Earth_i)/T_Earth_i = 1 / (I_Earth/(I_spinner*T_Earth_i*f_spinner) - 1)
     = 1 / ( 1.5 e+28 - 1) ~= 67 e-30
Fwiw, the absolute value is 5.767 yocto-seconds. If the entire world population would spin at that 1000 turns per second (and at favourable locations as in your assumptions), the effect will still be a measly 0.041 pico-seconds. So T_Earth_f = 86 399.999 999 999 999 958 ... But the TI-84 only has about 14 digits precision, i believe, so even that won't show up. -- 173.245.51.210 22:46, 30 October 2013 (UTC)

Is it possible for someone to write an equation that factors in latitude (and, if relevant, longitude) that we could plug our locations into and get a value from? That would be awesome. Thanks. 108.162.250.208 02:48, 23 February 2014 (UTC)

The visual style and theme of this comic is clearly referencing the 'Spinning Ballerina Optical Illusion' (evidenced by the grey-to-white gradient 'glow', as well as her arm and leg positions). 108.162.249.231 03:03, 30 September 2014 (UTC)

The 'Spinning Ballerina' optical illusion does not apply here, Megan clearly stated that she was spinning 'counterclockwise' and due to the fact that she is drawn with hair (not a silhouette) lets you know where she is facing. Therefore the bent leg on the right of the image is her left leg. Plus there aren't many ways to draw a stick figure 'spinning'. Now if it were Cueball doing the spinning THEN I would agree with you because there would not be a reference point to make any type of judgement and therefore a point could be made that could be a reference Randall was trying to make. Nexxuz (talk)

The momentum of Megan is in fact one order of magnitude smaller than what appeared in above calculation, so the dilation effect is still smaller.

http://www.wolframalpha.com/input/?i=moment+of+inertia+of+solid+cylinder+for+a%3D0.25m%2Cmass%3D62kg 173.245.50.90 (talk) (please sign your comments with ~~~~)