AQA 7408 · Section 3.6.1.1 · Paper 1
Circular motion: angular speed, acceleration and force
Learn why constant speed can still mean acceleration, connect radians and angular speed to linear motion, and identify the real force that supplies the inward resultant.
Content diagnostic
Commit before you calculate
Active revision
Use three passes
Draw directions
Mark the tangent velocity and the inward acceleration before choosing an equation.
Link quantities
Move between frequency, angular speed, linear speed, acceleration and resultant force.
Name the force
Identify tension, friction, gravity or a contact force as the real inward resultant.
Core model
Velocity changes even when speed does not
Velocity is a vector. In uniform circular motion its magnitude stays constant, but its direction changes at every point. That changing velocity means the object accelerates.
The acceleration is always directed towards the centre. It is perpendicular to the instantaneous velocity.
Newton’s second law then requires an inward resultant force:
Centripetal means centre-seeking. It describes the direction and role of the resultant; it is not a separate type of force.
Radians and angular speed
Connect one turn to the motion at the rim
Radian measure
One full turn is 2π radians. Convert degrees using θ(rad) = θ(°)π/180.
Angular speed
Every point on a rigid rotating object has the same angular speed and frequency.
Linear speed
At the same angular speed, a point farther from the axis travels faster because it covers a larger circumference each turn.
Worked example
A platform turns at 120 revolutions per minute. A marker is 0.30 m from the axis. The frequency is 120/60 = 2.0 Hz, so ω = 2πf = 12.6 rad s⁻¹. Its linear speed is v = ωr = 12.6 × 0.30 = 3.8 m s⁻¹ to 2 significant figures.
Useful simulation
Predict, vary one quantity, then explain
Before moving a control, predict what happens to acceleration if frequency doubles while radius stays fixed. The model reports numerical values as well as displaying the directions.
Force identification
Ask which real interaction points inwards
Parcel on a turntable
Static friction supplies the inward resultant. If the required force exceeds the maximum available friction, the parcel slips.
Car on a level bend
Tyre–road friction supplies the horizontal inward resultant. Weight and normal contact force balance vertically.
Satellite in circular orbit
Gravity supplies the inward resultant. Do not add a second “centripetal force” arrow to gravity.
Misconception repair
Four ideas that commonly cost marks
- “Constant speed means zero acceleration.” False: velocity direction changes.
- “Centripetal force is an extra force.” False: it is the inward resultant of real forces.
- “The velocity points towards the centre.” False: velocity is tangential; acceleration is inward.
- “Degrees can be substituted into radian equations.” False: convert to radians first.
Original exam-style assessment
20 marks with worked feedback
Attempt every question before opening its answer. Show substitutions, retain units and state a direction whenever the quantity is a vector.
1. At the right-hand point in the turntable figure, state the direction of the parcel’s velocity and acceleration. Explain why it accelerates even if its speed is constant.[3 marks]
Velocity is tangential to the circle, upwards in the figure. Acceleration is radially inwards, towards the centre. The velocity changes because its direction changes continuously, so there is acceleration even though the speed is constant.
2. A point moves in a circle of radius 0.45 m at a frequency of 1.8 Hz. Calculate its angular speed and linear speed.[3 marks]
ω = 2πf = 2π × 1.8 = 11 rad s⁻¹ to 2 significant figures. Then v = ωr = 11.3 × 0.45 = 5.1 m s⁻¹.
3. A 0.65 kg object travels at 7.2 m s⁻¹ in a circle of radius 1.4 m. Calculate its centripetal acceleration and the resultant force.[4 marks]
a = v²/r = 7.2²/1.4 = 37.0 m s⁻². F = ma = 0.65 × 37.0 = 24.1 N, directed towards the centre.
4. Convert an angle of 135° into radians.[2 marks]
θ = 135 × π/180 = 3π/4 = 2.36 rad.
5. A car follows a level circular road. Identify the horizontal force that keeps it on the circular path and explain why “centripetal force” is not an additional force.[3 marks]
Static friction from the road acts towards the centre. “Centripetal” describes the inward resultant force required for circular motion; here that resultant horizontal force is supplied by friction rather than by a separate new force.
6. Two equal masses rotate with the same angular speed. One is twice as far from the axis as the other. Compare their centripetal accelerations and forces.[2 marks]
Using a = ω²r, doubling r at fixed ω doubles the acceleration. Since the masses are equal, F = mω²r also doubles.
7. An object is being whirled in a horizontal circle on a string. Describe its motion immediately after the string breaks.[2 marks]
It moves in a straight line tangent to the circle at the release point. The inward tension has disappeared, so there is no longer an inward resultant force to curve the path.
8. Which statement is correct: A) velocity and resultant force are both inward; B) velocity is tangent and resultant force is inward; C) both are tangent; D) resultant force is outward?[1 mark]
B. Instantaneous velocity is tangent to the path and the resultant force is radially inwards.
Exam checklist
Before moving on
- I can distinguish constant speed from constant velocity.
- I can draw tangent velocity and inward acceleration directions.
- I can convert degrees to radians and use ω = 2πf.
- I can choose between v²/r and ω²r from the supplied data.
- I can identify the real force or forces providing the inward resultant.
- I never add a separate “centripetal force” to a force diagram.
Specification coverage
AQA 7408 section 3.6.1.1
- 3.6.1.1(a) Explain that motion in a circular path at constant speed involves acceleration and requires a centripetal force.
- 3.6.1.1(b) Use angular speed ω = v/r = 2πf.
- 3.6.1.1(c) Use radian measure of angle.
- 3.6.1.1(d) Recognise that direction of angular velocity is not considered.
- 3.6.1.1(e) Use centripetal acceleration a = v²/r = ω²r.
- 3.6.1.1(f) Recognise that derivation of centripetal acceleration is not examined.
- 3.6.1.1(g) Use centripetal force F = mv²/r = mω²r.
Continue the AQA sequence
Connect circular motion with simple harmonic motion and resonance. Revisit mechanics and materials for resultant forces and Newton’s laws, or waves for frequency, phase and radians. Use the AQA Paper 1 revision hub for the wider assessment route.
Written against AQA Physics 7408 section 3.6.1.1. All questions are original. The static figure is an original generated PNG and contains no assessed text.