$$v = \omega r \qquad \omega = \frac{2\pi}{T} = 2\pi f \qquad T = \frac{1}{f}$$ $$\text{Centripetal acceleration } a = \frac{v^2}{r} = \omega^2 r \qquad \text{Centripetal force } F = \frac{mv^2}{r} = m\omega^2 r$$

• $\omega$ is measured in rad/s; $1$ revolution $= 2\pi$ rad $= 360^\circ$.
• The centripetal force points towards the centre of the circle.
• Vertical circle: at the top, $T + mg = \dfrac{mv^2}{r}$; at the bottom, $T - mg = \dfrac{mv^2}{r}$.