[tex]dane:\\m = 2 \ kg\\r = 300 \ cm = 3 \ m\\T = 0,1 \ min = 0,1\cdot60 \ s = 6 \ s\\szukane:\\F_{d} = ?\\\\Rozwiazanie\\\\v = \frac{2\pi r}{T}\\\\v^{2} = \frac{4\pi^{2}r^{2}}{T^{2}}\\\\F_{d} = \frac{mv^{2}}{r} = \frac{\frac{m\cdot4\pi^{2}r^{2}}{T^{2}}}{r}\\\\F_{d} = \frac{4\pi^{2}mr}{T^{2}}\\\\F_{d} = \frac{4\cdot3,14^{2}\cdot2 \ kg\cdot3 \ m}{(6 \ s)^{2}}\\\\F_{d}\approx6,57 \ N[/tex]
