2.1 D=1,1; r=D/2= 1,1/2=0,55 m m=3,5 t=3,5*10^3 kg; n= 20*10^3 min^-1= 20*10^3/60=333,34 s^-1 Energia kinetyczna koła żyroskopu Ek=Iω^2/2 ω=2πf f=333,33Hz I=mr^2/2 Ek=mr^2/2*(2πf)^2/2=mr^2*(2πf)^2/4 Ek= 3,5*10^3*0,55^2*(2*π*333,34)^2/4 Ek=1,161*10^9 J Ek= 1,161 GJ 2.2 n= 10^4 min^-1= 10^4/60=166,67 s^-1[Hz] mas ta sama Ek= 3,5*10^3*0,55^2*(2*π*166,67)^2/4 Ek=2,9*10^8 J Ek= 0,29 GJ Energia wykorzystana ΔE=E20-E10= 1,161-0,29=0,871 zużyto 0,871 GJ albo f1=2f2=333,34 Hz f2=f1/2=0,5f1 ΔE=E20-E10 ΔE= mr^2*(2πf1)^2/4-mr^2*(2π0,5f1)^2/4 ΔE=mr^2*(2π)^2*(f1^2-(0,5f1)^2) ΔE=(1/4)*mr^2*(2πf1)^2(1-0,5^2) ΔE= 3,5*10^3*0,55^2*(2*π*333,34)^2*(1-0,5^2)/4 ΔE=8,708*10^8 ΔE=0,871 GJ
[latex]E_k= frac{Jw^2}{2} \ E_{k1}= frac{Jw_1^2}{2} \ E_{k2}= frac{Jw_2^2}{2} \ czesc= frac{E_{k1}-E_{k2}}{E_{k1}} = frac{ frac{Jw_1^2}{2}-frac{Jw_2^2}{2} }{frac{Jw_1^2}{2}} = frac{w_1^2-w_2^2}{w_1^2}=1- frac{w_2^2}{w_1^1} =1-( frac{w_2}{w_1} )^2= \ =1-( frac{10000}{20000} )^2=3/4[/latex]
