Fccc = mv2r\; \frac{mv^2}{r}rmv2
mass = 200g = 0.2kg, v = 5 m/s, r = 1.5 m
Fccc = 0.2×5×51.5\; \frac{0.2 \times 5 \times 5}{1.5}1.50.2×5×5 ≈ 3.33N
At the bottom of the circle, the tension T in the string must counteract the weight of the stone as well as provide the centripetal force:
T = Fccc + mg
T = 3.33 + 0.2 x 10 = 5.3N
