[tex]dane:\\r = 150 \ cm = 1,5 \ m\\T = 8 \ s\\m = 35 \ kg\\szukane:\\v = ?\\F_{d} = ?\\\\Rozwiazanie\\\\v = \frac{2\pi r}{T}\\\\v = \frac{2\cdot3,14\cdot1,5 \ m}{8 \ s} = 1,1775\frac{m}{s} \approx1,18\frac{m}{s}\\\\\\F_{d} = \frac{mv^{2}}{r}\\\\F_{d} = \frac{35 \ kg\cdot(1,1775\frac{m}{s})^{2}}{1,5 \ m}\approx32,4 \ N[/tex]
