TL;DR Circular motion is about applying Newton's second law in a radial direction. Every "centripetal force" question is really asking: which existing force(s) point toward the centre? Master the link between ω, v, and r, learn the four classic setups (flat road, conical pendulum, banked curve, vertical loop), and you will handle any 9478 circular-motion problem methodically.
1 Where circular motion sits in the H2 Physics syllabus
Circular motion sits within the Newtonian Mechanics section of the 2026 H2 Physics syllabus (9478). It builds directly on kinematics and Newton's laws, and feeds into gravitational fields, where orbital motion is analysed as a special case of uniform circular motion with gravity supplying the centripetal force.
Examiners love this topic because it tests whether students can identify forces correctly - a skill that transfers to every other topic.
2 Key definitions and equations
2.1 Angular displacement and angular velocity
When an object moves in a circle of radius r, its position on the circle is described by an angle θ (in radians) measured from a reference line. The rate of change of this angle is the angular velocity:
ω=dtdθ
For uniform circular motion (constant speed around the circle):
ω=T2π=2πf
where T is the period (time for one full revolution) and f is the frequency. The SI unit of ω is rad s−1.
2.2 Relating linear and angular velocity
A point on the rim travels a distance s=rθ along the arc. Differentiating with respect to time:
v=rω
This is the key bridge between translational and rotational descriptions. A larger radius means a larger linear speed for the same ω.
2.3 Centripetal acceleration
Even though the speed v is constant in uniform circular motion, the direction of velocity is continuously changing. This change requires an acceleration directed toward the centre of the circle:
a=rv2=rω2
This is the centripetal acceleration. It has no tangential component in uniform circular motion.
2.4 Centripetal force
By Newton's second law, the net force needed to produce this acceleration is:
F=ma=rmv2=mrω2
Critical point: centripetal force is not a new type of force. It is the resultant of whichever real forces happen to point toward the centre - friction, tension, gravity, normal contact force, or a combination. Every exam answer must identify the real force(s).
Radial equation checkpoint
Before writing a circular-motion equation, build the inward-force statement from the diagram. This avoids adding a fake "centripetal force" arrow.
Step
Question to ask
What goes into the equation
1
Where is the centre of the circle?
Choose the inward direction toward that centre.
2
Which real forces point inward?
Add only inward components of real forces.
3
Which real forces point outward?
Subtract outward components if you use inward as positive.
4
Is the speed given as v or ω?
Set the inward resultant to mv2/r or mrω2.
5
Are there vertical or tangential conditions?
Write separate equations for directions that are not radial.
Worked check: on a flat road, weight and normal contact force cancel vertically. The radial equation is therefore static friction =mv2/r, not weight plus friction =mv2/r.
Common trap: do not draw a separate arrow labelled "centripetal force". Draw friction, tension, weight, or normal contact force, then let their inward resultant supply the centripetal acceleration.
3 Classic setups and worked examples
3.1 Horizontal circular motion - turntable / car on a flat road
Setup: A small block sits on a rotating turntable (or a car rounds a flat bend). No banking, no tilt.
Force analysis: The only horizontal force available is static friction f. It points radially inward toward the centre.
f=rmv2
The block flies off when the required centripetal force exceeds the maximum static friction μsmg:
vmax=μsgr
Worked example 1
A 0.20 kg coin sits 15 cm from the centre of a turntable. The coefficient of static friction is 0.40. Find the maximum angular speed before the coin slides off.
At the point of sliding: μsmg=mrω2
ω=rμsg=0.150.40×9.81=5.1;rad s−1
Notice that the mass cancels - the answer is independent of the coin's mass.
3.2 Conical pendulum
Setup: A bob of mass m is attached to a string of length L and swings in a horizontal circle so that the string traces a cone. The string makes an angle θ with the vertical.
Force analysis: Two forces act on the bob - tension T along the string and weight mg downward.
Resolve tension into components:
Vertical: Tcosθ=mg (no vertical acceleration)
Horizontal (radial): Tsinθ=rmv2
Dividing the second equation by the first:
tanθ=rgv2
Since the radius of the circle is r=Lsinθ, you can express the period as:
T=2πgLcosθ
This shows that a larger cone angle (faster rotation) gives a shorter period - the bob rises.
3.3 Banked curves
Without friction (ideal banking):
On a frictionless banked road at angle θ, the normal force N is perpendicular to the road surface. Its horizontal component supplies the centripetal force:
Nsinθ=rmv2,Ncosθ=mg
tanθ=rgv2
This defines the design speed. At exactly this speed, no friction is needed.
With friction:
If the car goes faster than the design speed, friction acts down the slope (preventing the car from sliding up). If slower, friction acts up the slope. The algebra becomes longer but the method is the same: resolve all forces radially and vertically, then solve simultaneously.
3.4 Vertical circular motion
Setup: A ball on a string (or a bucket of water) swings in a vertical circle of radius r.
At the top of the loop, both the weight mg and the tension T point downward (toward the centre):
T+mg=rmv2
The minimum speed at the top occurs when T=0 (string just goes slack):
vmin=gr
At the bottom of the loop, tension acts upward and weight acts downward:
T−mg=rmv2
So the tension at the bottom is always greater than at the top - this is why strings and tracks are most likely to break at the bottom.
Worked example 2
A 0.50 kg ball swings on a 0.80 m string in a vertical circle. At the top, its speed is 4.0;m s−1. Find the tension in the string at the top.
For a satellite of mass m orbiting the Earth (mass M) at radius r from the Earth's centre, the gravitational force provides the centripetal force:
r2GMm=rmv2
This gives the orbital speed:
v=rGM
A larger orbital radius means a slower orbital speed - counterintuitive but important. This result bridges directly into the gravitational fields topic.
4 Common exam traps
Trap 1 - "Centripetal force acts on the body"
Never write this without immediately stating which real force supplies it. Examiners will withhold marks if you treat centripetal force as a stand-alone force on a free-body diagram.
✗ "The centripetal force keeps the car on the road." ✓ "The frictional force between the tyres and the road surface provides the centripetal force."
Trap 2 - Centripetal vs centrifugal
There is no outward "centrifugal force" in an inertial reference frame. It only appears in a rotating (non-inertial) frame. In H2 Physics, always work in an inertial frame and use centripetal (inward) force only.
Trap 3 - Forgetting unit conversions
Turntable problems often give speed in revolutions per minute (rpm). Convert to rad s−1:
ω=602π×n
where n is rpm. Forgetting this conversion is one of the most common arithmetic errors.
Trap 4 - Vertical loop force direction at the top
At the top of a vertical loop, both the weight and the normal contact force (or tension) point toward the centre. Students often mistakenly draw the normal force pointing outward. Remember: for a ball on the inside of a loop, the surface pushes inward.
5 Practice questions
Question 1 - Flat curve
A car of mass 1200 kg travels at constant speed around a flat, circular bend of radius 50 m. The coefficient of static friction between the tyres and the road is 0.60.
(a) Calculate the maximum speed at which the car can round the bend without skidding. (b) Explain what happens if the road is wet and μs drops to 0.30.
Hint: Set friction equal to the required centripetal force. For part (b), compare the new vmax with the original.
Question 2 - Vertical loop
A toy car of mass 0.10 kg travels around the inside of a vertical circular loop of radius 0.25 m.
(a) Find the minimum speed at the top of the loop for the car to maintain contact with the track. (b) If the speed at the top is 2.5 m s−1, find the normal contact force at the top.
Hint: At minimum speed, the normal contact force is zero. For part (b), use N+mg=mv2/r.
Question 3 - Conical pendulum
A 0.30 kg bob is attached to a 0.60 m string and swings in a horizontal circle. The string makes an angle of 30° with the vertical.
(a) Find the tension in the string. (b) Determine the speed of the bob. (c) Calculate the period of revolution.
Hint: Resolve the tension vertically to find T, then use the horizontal component for the centripetal force.
6 What to study next
Gravitational fields (Topic 8): Orbital motion is circular motion with gravity as the centripetal force - see our H2 Physics notes hub for the full topic list.
Simple harmonic motion: The projection of uniform circular motion onto a diameter gives SHM - understanding circular motion deeply helps with oscillations.
Electromagnetic force on charged particles: In Topic 17, a charge moving in a magnetic field follows a circular path because qvB provides the centripetal force.
For structured revision with a tutor, visit our H2 Physics tuition page.
