Talk:162: Angular Momentum

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.