Circular Motion

Circular motion is described most naturally using the radian rather than the degree. One radian is the angle subtended at the centre of a circle by an arc whose length equals the radius:

$\theta = \frac{s}{r}$

where $s$ is the arc length and $r$ the radius. Because a full circle has circumference $2\pi r$, a complete turn is $\dfrac{2\pi r}{r}=2\pi$ radians $=360°$. This gives the conversions you will use constantly:

• $360° = 2\pi\ \text{rad}$, so $180°=\pi\ \text{rad}$ and $1\ \text{rad}=\dfrac{180}{\pi}\approx57.3°$.
• To convert degrees → radians, multiply by $\dfrac{\pi}{180}$; radians → degrees, multiply by $\dfrac{180}{\pi}$.

Angular displacement $\theta$ is the angle swept out by the radius line, and angular velocity $\omega$ is the rate at which that angle changes:

$\omega = \frac{\Delta\theta}{\Delta t}$

measured in radians per second ($\text{rad s}^{-1}$). For one complete revolution the object sweeps $2\pi$ rad in one period $T$, so

$\omega = \frac{2\pi}{T} = 2\pi f$

where $f=\dfrac{1}{T}$ is the frequency (revolutions per second, in hertz).

Linking angular and linear speed. A point at radius $r$ travels a distance $s=r\theta$ along the arc. Dividing by time gives $v=r\omega$, i.e.

$\omega = \frac{v}{r} \qquad\Longleftrightarrow\qquad v = \omega r$

so points further from the centre move faster (larger $v$) even though every point shares the same $\omega$. Here $v$ is the tangential (linear) speed in $\text{m s}^{-1}$.

Calculator check. Put your calculator in radian mode for these problems. A frequent slip is leaving it in degrees, which quietly wrecks any answer that uses an angle.

An object going round a circle at constant speed is nevertheless accelerating, because acceleration is the rate of change of velocity, and velocity is a vector. The velocity is always tangential — it points along the direction of travel, at a tangent to the circle — so as the object moves round, the direction of the velocity is constantly changing even though its magnitude (the speed) is not.

The change in velocity, and therefore the acceleration, points towards the centre of the circle. This is called the centripetal ("centre-seeking") acceleration:

$a = \frac{v^2}{r} = \omega^2 r$

(the two forms are equivalent because $v=\omega r$, so $\dfrac{v^2}{r}=\dfrac{(\omega r)^2}{r}=\omega^2 r$).

Because the centripetal acceleration is always perpendicular to the velocity, it changes the direction of motion without changing the speed. If the centripetal force were suddenly removed (e.g. a string snaps), the object would fly off along the tangent — a straight line in the direction it was travelling at that instant — not radially outwards.

By Newton's second law, an acceleration requires a resultant force in the same direction. For circular motion this resultant acts towards the centre and is called the centripetal force:

$F = ma = \frac{mv^2}{r} = m\omega^2 r$

The single most important idea in this whole topic is that the centripetal force is not a new or separate force. It is the name for the resultant of the real forces already acting, resolved towards the centre. In different situations that real force is supplied by:

• Tension — a ball whirled on a string, a conical pendulum;
• Gravity — a satellite or the Moon orbiting the Earth (linked forward to Unit 5, Gravitational Fields);
• Friction — a car cornering on a flat road (friction between tyres and road points inward);
• Normal (reaction) contact force — the horizontal component on a banked track, or the seat pushing a passenger inward;
• Electrostatic or magnetic force — an electron circling a nucleus, or a charged particle in a magnetic field.

So when you draw a free-body diagram you show only the real forces (weight, tension, normal reaction, friction). You do not add an extra arrow labelled "centripetal force" — instead you set the resultant of the real forces towards the centre equal to $\dfrac{mv^2}{r}$.

Worked routine for any horizontal-circle problem

1. Draw the free-body diagram with the real forces only.
2. Resolve those forces towards the centre (horizontal) and perpendicular to it (vertical).
3. Set the resultant towards the centre $=\dfrac{mv^2}{r}$ (or $m\omega^2 r$).
4. Use any perpendicular equilibrium condition (e.g. vertical: $T\cos\theta=mg$ for a conical pendulum) to eliminate unknowns.

For a car on a flat bend, the inward force is friction: $f=\dfrac{mv^2}{r}$. The maximum cornering speed before skidding is set by the maximum available friction — increase $r$ or reduce $v$ to corner safely. A banked track tilts the road so that a component of the normal reaction also points inward, reducing the reliance on friction (examined qualitatively at IAL).

In a vertical circle — a bucket of water swung on a rope, a ball on a string, a loop-the-loop — the speed is not constant, because gravity does work on the object as it rises and falls. Two tools solve every such problem:

1. Energy conservation to find the speed at a given point.
2. Newton's second law towards the centre to find the tension or contact force there.

Speed from energy. Taking the bottom of the circle as the reference height, the object rises by $2r$ (a diameter) to reach the top. Conservation of energy gives

$\tfrac12 m v_{\text{bottom}}^2 = \tfrac12 m v_{\text{top}}^2 + mg(2r)$

so the object is always slowest at the top and fastest at the bottom.

Force from Newton's second law. At any point the resultant force towards the centre equals $\dfrac{mv^2}{r}$. The two special points:

• At the top, both the weight $mg$ and the tension $T$ point downwards — towards the centre: $T + mg = \frac{mv_{\text{top}}^2}{r}\quad\Rightarrow\quad T = \frac{mv_{\text{top}}^2}{r} - mg$
• At the bottom, the tension points up (towards the centre) but the weight points down (away): $T - mg = \frac{mv_{\text{bottom}}^2}{r}\quad\Rightarrow\quad T = \frac{mv_{\text{bottom}}^2}{r} + mg$

So the tension is greatest at the bottom and least at the top.

The minimum-speed condition at the top. As the object slows near the top, the tension falls. The slowest it can travel and still complete the circle is when the string just goes slack, i.e. tension $T=0$. Then gravity alone supplies the centripetal force:

$mg = \frac{mv_{\text{top}}^2}{r}\quad\Rightarrow\quad v_{\text{top,min}} = \sqrt{gr}$

Below this speed the object cannot maintain contact/tension and leaves the circular path — the water falls out of the bucket. (For an object on the inside of a track the same condition applies with the normal reaction $N$ in place of $T$: minimum speed when $N=0$.)

A satellite or the Moon orbiting the Earth is circular motion in which the real force supplying the centripetal force is gravity. There is no string and no friction — the gravitational pull of the Earth on the satellite is entirely used to keep it turning:

$\underbrace{\frac{GMm}{r^2}}_{\text{gravitational force (Unit 5)}} = \underbrace{\frac{mv^2}{r}}_{\text{centripetal requirement}}$

You are not asked to derive this at Unit 4, but recognising that the centripetal force for an orbit is provided by gravity is exactly the idea that Unit 5 (Gravitational Fields) builds on to derive orbital speed, period and geostationary orbits. The mass $m$ of the satellite cancels, so the orbital speed depends only on the central mass $M$ and the orbital radius $r$ — every object at a given orbital radius must move at the same speed, whatever its mass.

The general point to carry forward: in every context — string, bend, orbit — you identify the real inward force and equate it to $\dfrac{mv^2}{r}=m\omega^2 r$. Circular motion supplies the geometry; the real force supplies the physics.