Wahadło - polecenie w załączniku.   Wg. odpowiedzi wynik to 0,128m

[latex]x= sqrt{ l^{2}-R^{2}} \ tg alpha = frac{R}{ sqrt{l^{2}-R^{2}}} \ tg alpha = frac{F_{b}}{mg} \ frac{R}{sqrt{ l^{2}-R^{2}}} = frac{m omega ^{2} R}{mg} \ [/latex]   [latex]g= omega^{2} sqrt{ l ^{2}-R^{2}} \ \ R= sqrt{ l^{2} - frac{g^{2}}{ omega ^{4}}} = sqrt{l^{2} - frac{g^{2}}{( frac{2 pi }{T})^{4}}} = sqrt{l^{2} - frac{g^{2}T^{4}}{16 pi ^{4}}} \ [/latex]   [latex]R=sqrt{l^{2} - frac{g^{2}( frac{t}{n})^{4}}{16 pi ^{4}}} approx sqrt{0,0625m^{2} - frac{100 ( frac{18,4}{20})^{4}}{16 cdot 3,14 cdot 3,14}m^{2}} approx 0,128m[/latex]

Narysuj sobie siły działające na tę kulkę. (rys) Z rysunku widać: [latex]sinalpha=frac{R}{l}=frac{F_d}{sqrt{F_d^2+Q^2}}\ F_d=momega^2R\ Q=mg\ frac{R^2}{l^2}=frac{F_d^2}{F_d^2+Q^2}\ R^2(F_d^2+Q^2)=F_d^2l^2\ R^2m^2omega^4R^2+R^2m^2g^2=m^2omega^4R^2l^2\[/latex] [latex]R^2omega^4+g^2=omega^4l^2\ R^2omega^4=omega^4l^2-g^2\ R=sqrt{frac{omega^4l^2-g^2}{omega^4}}[/latex] [latex]omega=frac{2pi}{T}\ T=frac{t}{n}=frac{18,4s}{20}=0,92s\ omega=frac{2*3,14}{0,92s}=6,83frac{rad}{s}\ Podstawiamy do wzoru na promien[/latex] [latex]R=sqrt{frac{(6,83frac{rad}{s})^4*(0,25m)^2-(10frac{m}{s^2})^2}{(6,83frac{rad}{s})^4}}\ R=sqrt{frac{2176,12frac{rad^4}{s^4}*0,0625m^2-100frac{m^2}{s^4}}{2176,12frac{rad^4}{s^4}}}\ [/latex] [latex]R=sqrt{frac{136,0075frac{rad^4m^2}{s^4}-100frac{m^2}{s^4}}{2176,12frac{rad^4}{s^4}}}\ R=sqrt{frac{36,0075frac{rad^4m^2}{s^4}}{2176,12frac{rad^4}{s^4}}}=sqrt{0,0165m^2}=0,128m[/latex]