An object moving in a circle of radius $r$ with speed $v$ experiences an acceleration directed toward the centre.
Angular velocity $\omega$ (rad/s): rate of change of angle:
$$\omega = \frac{d\theta}{dt}$$
Relationship to linear speed:
$$v = r\omega$$
Period $T$ (time for one revolution): $T = \dfrac{2\pi}{\omega}$.
Frequency $f$ (revolutions per second): $f = \dfrac{1}{T} = \dfrac{\omega}{2\pi}$.
For uniform circular motion (constant $\omega$), with centre at origin:
$$\mathbf{r}(t) = r\cos(\omega t)\,\mathbf{i} + r\sin(\omega t)\,\mathbf{j}$$
$$\mathbf{v}(t) = -r\omega\sin(\omega t)\,\mathbf{i} + r\omega\cos(\omega t)\,\mathbf{j}$$
Check: $|\mathbf{v}| = r\omega$ and $\mathbf{r}\cdot\mathbf{v} = 0$ (velocity is perpendicular to radius). \checkmark
$$\mathbf{a}(t) = -r\omega^2\cos(\omega t)\,\mathbf{i} - r\omega^2\sin(\omega t)\,\mathbf{j} = -\omega^2\mathbf{r}$$
The acceleration points toward the centre with magnitude:
$$a_c = r\omega^2 = \frac{v^2}{r}$$
By Newton’s second law:
$$F_c = ma_c = \frac{mv^2}{r} = mr\omega^2$$
This force is not a new type of force — it is provided by tension, gravity, friction, or normal force depending on the context.
| Scenario | Centripetal force provided by |
|---|---|
| Ball on a string (horizontal circle) | Tension in the string |
| Car rounding a bend | Friction from the road |
| Satellite orbiting Earth | Gravitational force |
| Ball in a vertical circle | Tension + component of gravity |
A mass $m$ on a string of length $L$ moves in a horizontal circle. The string makes angle $\theta$ with the vertical.
Vertical equilibrium: $T\cos\theta = mg \Rightarrow T = \dfrac{mg}{\cos\theta}$.
Horizontal (centripetal): $T\sin\theta = \dfrac{mv^2}{r}$ where $r = L\sin\theta$.
$$\frac{mg\sin\theta}{\cos\theta} = \frac{mv^2}{L\sin\theta} \Rightarrow v^2 = gL\sin^2\theta/\cos\theta$$
Period: $T_p = \dfrac{2\pi r}{v} = 2\pi\sqrt{\dfrac{L\cos\theta}{g}}$.
For a mass on a string in a vertical circle of radius $r$, at angle $\theta$ from the bottom:
$$T - mg\cos\phi = \frac{mv^2}{r}$$
(where $\phi$ is angle from vertical). String stays taut when $T \geq 0$:
$$v^2 \geq gr\cos\phi$$
At the top of the loop ($\phi = 180^\circ$, $\cos\phi = -1$): minimum speed $v_{\min} = \sqrt{gr}$.
KEY TAKEAWAY: In circular motion, the centripetal acceleration $v^2/r$ always points toward the centre. It is provided by the net inward force, not an extra force.
EXAM TIP: For problems with circular motion, always resolve forces radially (centripetal direction) and axially (tangential or vertical). Identify clearly which force supplies the centripetal component.
COMMON MISTAKE: Writing “centrifugal force” as a force in a free-body diagram. In an inertial frame, there is no centrifugal force. The net inward force IS the centripetal force.
