Motion in a circle

The radian 弧度 is the angle made at the centre of a circle by an arc 弧 whose length equals the radius. For an arc of length $s$ on a circle of radius $r$, the angle in radians is

Radians have no unit (a ratio of lengths). A full circle has $s = 2\pi r$, so $\theta = 2\pi\ \text{rad}$. A half-circle is $\pi\ \text{rad}$; a quarter is $\pi/2\ \text{rad}$.

To convert: $1\ \text{rad} = 180°/\pi \approx 57.3°$. Set your calculator to radians for this topic; "degree" mode will give wrong answers.

An object moves in a circle of radius $r$ at constant speed $v$. Define:

• angular displacement 角位移 $\theta$ — the angle (in radians) turned through by the radius from a chosen start line.
• angular speed 角速度 $\omega$ — the rate of change of angular displacement.

For uniform motion $\omega$ is constant and

Unit: $\text{rad s}^{-1}$.

Period and frequency

If the object goes once round ($2\pi\ \text{rad}$, one revolution 圈) in time $T$ (the period 周期), then

where $f = 1/T$ is the frequency 频率 of turning (Hz).

Linear and angular speed

In one period $T$ the object travels a distance $2\pi r$ (the circumference 周长) at constant speed, so

This links the linear (tangential 切向) speed $v$ with the angular speed $\omega$. At a larger radius (for the same angular speed) the linear speed is larger — a child on the edge of a merry-go-round moves faster than one near the centre, even though both go round once in the same time.

Worked example. A fairground ride of radius $4.0\ \text{m}$ completes one turn every $8.0\ \text{s}$. Find its angular speed and the linear speed of a rider on the edge.

An object moving in a circle at constant speed still has a changing velocity 速度 — its direction keeps changing, even though its size stays the same. A changing velocity needs an acceleration 加速度. This acceleration points towards the centre and is the centripetal acceleration 向心加速度.

Size

The two forms are equal because $v = r\omega$. Pick the one with the quantities you have.

The centripetal acceleration is perpendicular 垂直 to the velocity at every instant — never along the direction of motion. (If part of it were along the motion, the speed would change.) Unit: $\text{m s}^{-2}$.

By Newton's second law, the resultant force 力 on a body in circular motion at constant speed is

This is the centripetal force 向心力. It always points towards the centre — perpendicular to the velocity.

The centripetal force is not a new kind of force — it is the net result of the real forces acting (tension, gravity, friction, electric attraction, normal contact force, …). In a problem, work out which real force(s) provide it.

Worked example. A $0.20\ \text{kg}$ ball on a string is whirled in a horizontal circle of radius $0.50\ \text{m}$ at $3.0\ \text{m s}^{-1}$. Find the centripetal force (the tension in the string).

Where the centripetal force comes from

• Ball on a string in a horizontal circle: the tension 张力 in the string.
• Car turning a flat corner: the friction 摩擦力 between tyres and road ($F = m v^{2}/r$). If the car goes too fast, friction is not enough and it skids outwards.
• Banked corner 倾斜 (no friction): the horizontal part of the normal contact force 支持力; $\tan\theta = v^{2}/(rg)$ for the angle that needs no friction.
• Planet or satellite 卫星 in orbit 轨道: the gravitational attraction 引力, $G M m / r^{2} = m v^{2}/r$.
• Electron 电子 in a circular orbit (Bohr-style model): the electrostatic 静电 attraction between the electron and the positive nucleus 原子核:

where $k = 1/(4\pi\varepsilon_{0})$ and $Z$ is the nuclear charge. Solve for $v$ to get the orbital speed; then $T = 2\pi r/v$.

Vertical circles

When the circle is upright, the speed is not constant (gravity does work) — but at each instant the net force towards the centre still equals $m v^{2}/r$:

• at the bottom of a loop: tension up, weight 重力 down, so $T - mg = m v^{2}/r$ — the tension is largest here.
• at the top of a loop: tension and weight both point down (towards the centre), so $T + mg = m v^{2}/r$ — the tension is smallest. For the slowest speed at the top with the string just tight, set $T = 0$: $mg = m v_{\text{min}}^{2}/r$, giving $v_{\text{min}} = \sqrt{gr}$.

Worked example. A car goes round a vertical loop of radius $2.0\ \text{m}$. Find the minimum speed at the top for the car to keep contact with the track (take $g = 9.81\ \text{m s}^{-2}$).

At the slowest speed the track force is zero, so gravity alone provides the centripetal force: $mg = m v_{\text{min}}^{2}/r$, giving $v_{\text{min}} = \sqrt{gr}$:

The constant-speed result ($v = r\omega$, $\omega$ constant) holds for horizontal circles, or where the force only bends the path (orbits in gravity, charges in a magnetic field 磁场).

1. Find the radius $r$ and choose $v$ or $\omega$. Use $v = r\omega$ to switch between them.
2. Find the centripetal acceleration with $a = v^{2}/r$ or $r\omega^{2}$.
3. List the real forces and write Newton's second law in the radial 径向 direction (towards the centre is positive). Set the net inward force equal to $m v^{2}/r$.
4. For period or frequency: use $\omega = 2\pi/T$, or $T = 2\pi r / v$.
5. Check the directions: centripetal force and acceleration point to the centre; the velocity is along the tangent.

• Work in radians; angular speed $\omega = 2\pi/T = v/r$.
• Centripetal acceleration $a = v^2/r = \omega^2 r$; the net force acts towards the centre — it is provided by tension/gravity/friction, not an extra force.
• Always state what provides the centripetal force in the situation given.