Physics - Circular Motion Question with Solution | TestHub

The centripetal force needed to keep the stone moving at $5\mathrm{m}{\mathrm{s}}^{-1}$ is $$F_c=\frac{m{v}^2}{r}=\frac{(1\mathrm{kg}){\left(5{\mathrm{ms}}^{-1}\right)}^2}{0.5\mathrm{m}}=50\mathrm{N}$$ The weight of the stone is $W=mg=(1\mathrm{kg})\left(9.8\mathrm{m}{\mathrm{s}}^{-2}\right)=9.8\mathrm{N}$. At the top of the circle, $$T=F_c-W=50\mathrm{N}-9.8\mathrm{N}=40.2\mathrm{N}$$