Circular motion

What the exam asks

• Convert between degrees, revolutions and radians, and use $\omega = \dfrac{v}{r} = 2\pi f = \dfrac{2\pi}{T}$.
• Use $a = \dfrac{v^2}{r} = \omega^2 r$ and . Edexcel and IAL also ask you to .
• Identify the real force providing the centripetal force: tension, friction, a normal reaction, gravity or lift.
• Solve horizontal-circle problems (bends, banking, conical pendulums, aircraft) and vertical-circle problems (loops, buckets, bridges), usually combined with energy conservation.
• (OCR A) Describe an experiment to investigate circular motion, such as the whirling bung.

Core ideas

Radians and angular speed

An angle in radians is $\theta = \dfrac{\text{arc length}}{\text{radius}}$, so one full turn is $2\pi$ rad and $1\ \text{rad} \approx 57.3^\circ$. is in rad s⁻¹.

For one revolution, $\omega = \dfrac{2\pi}{T} = 2\pi f$, and the linear (tangential) speed is $v = \omega r$. Every point on a rigid rotating object has the . Points further out have larger and larger acceleration.

Why constant speed still means acceleration

Velocity is a vector. In uniform circular motion the speed is constant, but the direction of the velocity changes continuously, so the velocity changes and the object accelerates. The acceleration points towards the centre (centripetal). By Newton’s second law there must be a resultant force towards the centre.

That force is always perpendicular to the velocity, so it does no work and the kinetic energy stays constant.

Derivation (Edexcel): in a short time $\Delta t$ the velocity vector turns through $\Delta\theta$. The velocity-change triangle has two sides of length $v$, so $|\Delta v| \approx v\,\Delta\theta$, directed towards the centre. Then

Centripetal force is not a new force

“Centripetal” describes a direction. It is not an extra force to add to your free-body diagram. Always name the real force or forces:

Horizontal circles with an angled force

Resolve vertically (the forces balance, because there is no vertical acceleration) and horizontally (the resultant equals $mv^2/r$). For a banked track with no friction, or an aircraft:

For a conical pendulum of string length $L$ at angle $\theta$ to the vertical, the radius is $r = L\sin\theta$. Then $T\cos\theta = mg$ and , which gives .

Vertical circles

The speed changes around a vertical circle, because gravity does work, so combine energy conservation with the centripetal equation at the point in question:

• Top (the string pulls, or the track pushes, downwards towards the centre): $T + mg = \dfrac{mv^2}{r}$.
• Bottom: , so the tension or reaction is greatest here.

Worked examples

Exam technique

• Draw a free-body diagram first, then write “resultant force towards centre $= mv^2/r$”. This earns the method mark even if the arithmetic slips.
• Convert rpm immediately: $\omega = \text{rpm} \times \dfrac{2\pi}{60}$.

Common mistakes

Quick recap

• $\omega = \dfrac{v}{r} = 2\pi f$, and radians are arc length ÷ radius.
• $a = \dfrac{v^2}{r} = \omega^2 r$ towards the centre. Changing direction means changing velocity, so there is an acceleration.