Circular Motion

CIE A-Level Further Mathematics· 45 min read

1. Fundamentals of Circular Motion★★☆☆☆⏱ 15 min

📘 Definition

Centripetal Acceleration

a_c

Acceleration always directed towards the centre of a circular path, required to change the direction of velocity for uniform circular motion.

Example:

A mass moving at $2 ms^{- 1}$ in a circle of radius $2 m$ has $a_{c} = \frac{2 ^{2}}{2} = 2 ms^{- 2}$ .

In uniform circular motion, speed is constant but velocity direction changes continuously, so acceleration is non-zero. By Newton's second law, a resultant force directed towards the centre is required to produce this acceleration.

a_c = \frac{v^2}{r} = \omega^2 r

F_{\text{net (radial)}} = m a_c = \frac{m v^2}{r} = m \omega^2 r

📐 Worked Example

A 3 kg particle moves in a horizontal circle of radius 0.5 m at 4 revolutions per second. Calculate the resultant centripetal force on the particle.

1
First calculate angular velocity $ω$ . 4 revolutions per second = $4 \times 2 π = 8 π rad s^{- 1}$ .
2
Substitute into the centripetal force formula:
3
$F = m \omega^2 r = 3 \times (8\pi)^2 \times 0.5$
4
Calculate the final value:
5
$F = 96 \pi^2 \approx 947 \text{ N}$

2. Horizontal Circular Motion★★★☆☆⏱ 20 min

Horizontal circular motion describes motion where the circular path lies in a horizontal plane. The most common exam problem is the conical pendulum, where a mass on a string moves in a horizontal circle and the string traces out a cone.

📘 Definition

Conical Pendulum

A mass attached to a fixed point by a light inextensible string, moving in a horizontal circle, so the string makes a constant angle with the vertical.

📐 Worked Example

A conical pendulum has length 1.5 m, and a 4 kg bob moves in a horizontal circle with the string inclined at $4 5^{\circ}$ to the vertical. Find the angular speed of the bob.

1
Resolve forces vertically (no acceleration vertically): $T cos 4 5^{\circ} = m g$
2
Resolve forces horizontally (resultant towards centre = centripetal force): $T sin 4 5^{\circ} = m ω^{2} r$
3
Find radius: $r = l sin 4 5^{\circ} = 1.5 sin 4 5^{\circ}$
4
Divide the horizontal equation by the vertical equation to eliminate $T$ and $m$ :
5
$\tan 45^\circ = \frac{\omega^2 (l \sin 45^\circ)}{g}$
6
Since $tan 4 5^{\circ} = 1$ , rearrange to find $ω^{2}$ :
7
$\omega^2 = \frac{g}{l \cos 45^\circ} = \frac{9.8}{1.5 \times \frac{\sqrt{2}}{2}} \approx 9.25$
8
Take the square root: $ω \approx 3.04 rad s^{- 1}$

3. Vertical Circular Motion: Strings★★★★☆⏱ 25 min

In vertical circular motion, speed is not constant because gravity does work on the particle as it moves up and down the circle. We use conservation of energy to find speed at any point, then calculate tension or reaction force.

📘 Definition

Critical Speed for Strings

For a particle attached by an inextensible string, the string goes slack if tension becomes negative. The minimum speed at the highest point to complete a full circle occurs when tension is zero, so $v^{2} = g r$ .

📐 Worked Example

A 2 kg mass is attached to a light inextensible string of length 1.2 m, projected from the lowest point with speed $u$ . Find the minimum $u$ for the mass to complete a full vertical circle.

1
Let $v$ = speed at the highest point. For minimum $u$ , tension $T = 0$ at the top, so gravity provides all centripetal force:
2
$mg = \frac{m v^2}{r} \implies v^2 = gr = 9.8 \times 1.2 = 11.76$
3
Use conservation of energy: kinetic energy at bottom = kinetic energy at top + gravitational potential energy gained. Height gained = $2 r = 2.4$ m:
4
$\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mg(2r)$
5
Cancel $m$ from all terms and substitute values:
6
$u^2 = v^2 + 4gr = 11.76 + 4(9.8)(1.2) = 58.8$
7
Take the square root: $u = 58.8 \approx 7.67 ms^{- 1}$

4. Vertical Circular Motion: Rods★★★★☆⏱ 25 min

For a particle attached to a rigid rod, or moving on the inside of a fixed circular track, the rod can exert both tension (pull towards the centre) and compression (push away from the centre). This changes the minimum speed condition for completing a full circle.

📘 Definition

Minimum Speed for Rods

A rigid rod can support the weight of the particle even if speed is zero, so the minimum speed at the highest point is 0, not $g r$ .

📐 Worked Example

A 1 kg mass is attached to a light rigid rod of length 0.8 m, pivoted at one end. It is projected from the lowest point with speed $6 ms^{- 1}$ . Find the force exerted by the rod at the highest point.

1
Use conservation of energy to find $v^{2}$ at the top, height gain = $2 r = 1.6$ m:
2
$\frac{1}{2} (1)(6)^2 = \frac{1}{2} (1) v^2 + (1)(9.8)(1.6)$
3
Rearrange to get $v^{2}$ :
4
$18 = 0.5 v^2 + 15.68 \implies v^2 = 4.64$
5
Let $R$ be the force from the rod towards the centre. Resultant force towards centre: $m g + R = \frac{m v ^{2}}{r}$ :
6
$R = \frac{(1)(4.64)}{0.8} - (1)(9.8) = 5.8 - 9.8 = -4 \text{ N}$
7
The negative sign means $R$ acts away from the centre: the rod exerts an upward compression force of 4 N on the mass.

5. Common Pitfalls

Wrong move:

Adding centripetal force as an extra force alongside tension, gravity and reaction.

Why:

Centripetal force is not a separate force, it is the resultant of existing forces acting towards the centre.

Correct move:

Find the sum of all forces acting along the radial line towards the centre, then set this equal to $\frac{m v ^{2}}{r}$ .

Wrong move:

Using $r$ as the height difference between the lowest and highest point of a vertical circle.

Why:

The highest point is 2 times the radius above the lowest point, not 1 times $r$ .

Correct move:

Always use a height difference of $2 r$ for lowest to highest point in vertical circle energy calculations.

Wrong move:

Using the critical speed condition $v \geq g r$ for a particle on a rigid rod.

Why:

Students memorize the string condition and incorrectly apply it to rods, which can support weight with zero speed at the top.

Correct move:

For strings: $v \geq g r$ at the top; for rigid rods: $v \geq 0$ at the top.

Wrong move:

Calculating the radius of a conical pendulum as $l cos θ$ instead of $l sin θ$ .

Why:

Confusion between the sides of the right triangle formed by the string, vertical axis and radius.

Correct move:

Draw a clear diagram: if $θ$ is the angle between the string and the vertical, $r = l sin θ$ .

Wrong move:

Resolving forces tangentially instead of radially when calculating centripetal force.

Why:

Acceleration for uniform circular motion is only in the radial direction, so resultant force must be calculated radially.

Correct move:

Always resolve forces along the radial line (towards the centre of the circle) for centripetal force calculations.

6. Quick Reference Cheatsheet

Concept	Formula/Rule	Key Note
Centripetal acceleration	$a_{c} = \frac{v ^{2}}{r} = ω^{2} r$	Always directed to centre
Centripetal force	$F_{n e t} = \frac{m v ^{2}}{r}$	Resultant force, not an extra force
Conical pendulum	$ω^{2} = \frac{g}{l c o s θ}$	Radius $r = l sin θ$
String vertical circle	$v_{t o p} \geq g r$	Min projection speed: $u = 5 g r$
Rod vertical circle	$v_{t o p} \geq 0$	Min projection speed: $u = 4 g r$
Vertical circles	$Δ K E = - Δ P E$	Always use conservation of energy

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Vertical circle on a string problem
2023 · 2
Conical pendulum calculation
2024 · 1
Vertical circle on a rod problem

Going deeper

syllabusCIE 9231 Further Mechanics Syllabus

What's Next

Circular motion is a core topic in further mechanics that underpins more advanced concepts including rotational motion, orbital mechanics, and simple harmonic motion. The skills you have developed here, including radial force resolution, combining Newton's laws with conservation of energy, and adapting conditions for different constraints (strings vs rods), are essential for solving complex multi-topic dynamics problems common in CIE A-Level exams. Circular motion is frequently combined with work, energy, collisions, and connected particles in exam questions, so mastering these fundamentals will make it easier to tackle harder problems in the rest of further mechanics.