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How do we describe rotation and the force needed to keep an object moving in a circle?

Angular displacement, velocity and acceleration, the angular equations of motion, the link between angular and linear quantities, and central (centripetal) force.

An SQA Advanced Higher Physics answer on angular motion, covering angular displacement, velocity and acceleration, the angular equations of motion, the link to linear quantities, radial acceleration and the central (centripetal) force needed for circular motion.

Generated by Claude Opus 4.814 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this key area is asking
  2. Angular kinematics
  3. Linking angular and linear quantities
  4. Radial acceleration and central force
  5. Examples in context
  6. Try this

What this key area is asking

The SQA wants you to describe rotation with angular displacement, angular velocity and angular acceleration, apply the angular equations of motion (the rotational twins of the linear ones), convert between angular and linear quantities through the radius, and calculate the central (centripetal) force that keeps an object moving in a circle.

Angular kinematics

Angles are measured in radians: one full turn is 2π2\pi radians, so a steady rotation rate gives ω=2πT=2πf\omega = \frac{2\pi}{T} = 2\pi f. Because ω\omega and α\alpha are defined by the same calculus as their linear counterparts, the angular equations of motion have exactly the same form for constant α\alpha:

ω=ω0+αt,θ=ω0t+12αt2,ω2=ω02+2αθ.\omega = \omega_0 + \alpha t, \qquad \theta = \omega_0 t + \tfrac{1}{2}\alpha t^2, \qquad \omega^2 = \omega_0^2 + 2\alpha\theta.

You solve rotational kinematics problems exactly as you solve linear ones, just with the angular symbols.

Linking angular and linear quantities

This is why the rim of a wheel travels faster than a point near the hub, and why a longer spanner turns a bolt through a larger arc. The conversion v=rωv = r\omega links the rotational description to ordinary speed and is needed constantly in this area.

Radial acceleration and central force

Even at constant speed, an object going round a circle is accelerating, because its velocity is continually changing direction. This acceleration points towards the centre.

By Newton's second law a central acceleration needs a central (centripetal) force of the same direction:

F=mar=mv2r=mrω2.F = ma_r = \frac{mv^2}{r} = mr\omega^2.

This is not a new kind of force; it is the name for whatever real force (friction, tension, gravity, the normal force) points towards the centre and supplies the needed inward pull. If that force is removed, the object flies off tangentially, in a straight line, not outward.

Examples in context

A fairground ride presses you towards the centre through the seat, supplying the central force; remove the wall and you would continue tangentially. Satellites are held in circular orbit because gravity provides exactly the central force mv2r\frac{mv^2}{r}. A centrifuge spins samples so fast that the large rω2r\omega^2 separates components by density. A car on a banked track uses a component of the normal force, not just friction, to supply the inward force, letting it corner faster.

Try this

Q1. State the unit of angular velocity. [1 mark]

  • Cue. Radians per second (rad s−1\text{rad s}^{-1}).

Q2. A point lies 0.30 m0.30\ \text{m} from the axis of a wheel turning at ω=10 rad s−1\omega = 10\ \text{rad s}^{-1}. Find its linear speed. [2 marks]

  • Cue. v=rω=0.30×10=3.0 m s−1v = r\omega = 0.30 \times 10 = 3.0\ \text{m s}^{-1}.

Q3. State the direction of the central force on an object moving in a circle. [1 mark]

  • Cue. Towards the centre of the circle (radially inward).

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA AH style5 marksA wheel starting from rest reaches an angular velocity of 24extrads−124 ext{rad s}^{-1} in 6.06.0 s with constant angular acceleration. Calculate the angular acceleration and the total angle turned through in that time.
Show worked answer →

Use the angular equations of motion, which mirror the linear ones.

Angular acceleration from ω=ω0+αt\omega = \omega_0 + \alpha t: 24=0+α(6.0)24 = 0 + \alpha(6.0), so α=4.0 rad s−2\alpha = 4.0\ \text{rad s}^{-2}.

Angle turned from θ=ω0t+12αt2\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2: θ=0+12(4.0)(6.0)2=72 rad\theta = 0 + \tfrac{1}{2}(4.0)(6.0)^2 = 72\ \text{rad}.

Markers reward selecting the correct angular equation, substituting consistently, and both answers with units (rad per second squared, and radians). Starting from rest means ω0=0\omega_0 = 0.

SQA AH style4 marksA car of mass 1200extkg1200 ext{kg} rounds a bend of radius 50extm50 ext{m} at 15extms−115 ext{m s}^{-1}. Calculate the central force required and state what provides it.
Show worked answer →

The central (centripetal) force is F=mv2rF = \frac{mv^2}{r}.

Substitute: F=1200×15250=1200×22550=5400 NF = \frac{1200 \times 15^2}{50} = \frac{1200 \times 225}{50} = 5400\ \text{N}.

It is provided by friction between the tyres and the road, directed towards the centre of the bend.

Markers reward the correct relationship, the value with unit, and naming friction as the source. A common error is to think a force acts outward; the net force is inward (centripetal).

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