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I arrived at the same answer, which I posted on his YouTube video, this morning. As of 3:30 PM central time today, he has not released it to the public view. So I'm going to assume it is probably correct.person123 said:Alright. I finally think I have the answer... They both follow the format of ##IΔω=F_{f}rt## . I divided one by the other, leaving me with:
##\frac {r_{2}^4(ω_{2}-ω_{1})} {r_{1}^4ω_{1}}=-\frac {r_{2}} {r_{1}}##
. When simplifying and substituting the values for r1 and r2, it gave me:
##ω_{2}=\frac {ωr_{2}^2} {r_{1}^2+r_{2}^2}##
##ω_{1}=\frac {r_{2}^3ω} {r_{1}^3+r_{1}r_{2}^2}##