[tex]Dane:\\m = 200 \ g = 0,2 \ kg\\r = 25 \ cm = 0,25 \ m\\T = 4 \ s\\Szukane:\\F_{d} = ?\\\\Rozwiazanie\\\\F_{d} = \frac{mv^{2}}{r}\\\\ale\\\\v = \frac{2\pi r}{T}\\\\F_{d} = \frac{m}{r}\cdot(\frac{2\pi r}{T})^{2}\\\\\underline{F_{d} = \frac{4\pi ^{2} mr}{T^{2}}}\\\\F_{d} = \frac{4\cdot3,14^{2}\cdot0,2 \ kg\cdot0,25 \ m}{(4 \ s)^{2}}[/tex]
[tex]\boxed{F_{d}\approx0,12 \ N}[/tex]
