# Circular Motion

> CIE A-Level Further Mathematics · Further Mechanics Unit 3
> Source: https://www.owlsprep.com/study/cie-9231-u3-circular-motion/

This subtopic covers applying Newton's laws and conservation of energy to motion along circular paths. You will learn to solve problems for horizontal motion (conical pendulums) and vertical motion (strings and rods), including finding minimum speed conditions for full loops.

**Prerequisites:** Newton's laws of motion; Force resolution in 2D; Conservation of mechanical energy

## Learning objectives

- Apply Newton's second law to uniform circular motion
- Solve problems involving horizontal circular motion (conical pendulums, banked tracks)
- Use conservation of energy to solve vertical circular motion problems
- Distinguish between motion on strings/rods and find reaction forces at different points
- Determine minimum speed conditions for completing full vertical circles

## Fundamentals of Circular Motion

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

*Notation:* a_c

*Example:* A mass moving at $2 \text{ ms}^{-1}$ in a circle of radius $2 \text{ m}$ has $a_c = \frac{2^2}{2} = 2 \text{ ms}^{-2}$.

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.

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

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

> **warning**
>
> Centripetal force is **not** an extra force added to tension, gravity or reaction. It is the resultant of all forces acting towards the centre of the circle.

**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 $\omega$. 4 revolutions per second = $4 \times 2\pi = 8\pi \text{ rad s}^{-1}$.
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. $$F = 96 \pi^2 \approx 947 \text{ N}$$

## Horizontal Circular Motion

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.

**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 $45^\circ$ to the vertical. Find the angular speed of the bob.

1. Resolve forces vertically (no acceleration vertically): $T \cos 45^\circ = mg$
2. Resolve forces horizontally (resultant towards centre = centripetal force): $T \sin 45^\circ = m \omega^2 r$
3. Find radius: $r = l \sin 45^\circ = 1.5 \sin 45^\circ$
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 $\tan 45^\circ = 1$, rearrange to find $\omega^2$:
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: $\omega \approx 3.04 \text{ rad s}^{-1}$

> **Exam tip**
>
> Always draw a clear diagram to avoid mixing up $l \sin\theta$ and $l \cos\theta$ for the radius of a conical pendulum.

## Vertical Circular Motion: Strings

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.

**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 $v^2 = gr$.

**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 = $2r = 2.4$ 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: $u = \sqrt{58.8} \approx 7.67 \text{ ms}^{-1}$

## Vertical Circular Motion: Rods

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.

**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 $\sqrt{gr}$.

**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 $6 \text{ ms}^{-1}$. 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 = $2r = 1.6$ 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: $mg + R = \frac{m v^2}{r}$:
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.

> **Exam tip**
>
> Always check the sign of your reaction force. A negative value means it acts opposite to your assumed direction (towards the centre), which indicates compression for rods.

## Common pitfalls

- **Wrong:** Adding centripetal force as an extra force alongside tension, gravity and reaction.
  - Why it fails: Centripetal force is not a separate force, it is the resultant of existing forces acting towards the centre.
  - Correct: Find the sum of all forces acting along the radial line towards the centre, then set this equal to $\frac{mv^2}{r}$.
- **Wrong:** Using $r$ as the height difference between the lowest and highest point of a vertical circle.
  - Why it fails: The highest point is 2 times the radius above the lowest point, not 1 times $r$.
  - Correct: Always use a height difference of $2r$ for lowest to highest point in vertical circle energy calculations.
- **Wrong:** Using the critical speed condition $v \geq \sqrt{gr}$ for a particle on a rigid rod.
  - Why it fails: Students memorize the string condition and incorrectly apply it to rods, which can support weight with zero speed at the top.
  - Correct: For strings: $v \geq \sqrt{gr}$ at the top; for rigid rods: $v \geq 0$ at the top.
- **Wrong:** Calculating the radius of a conical pendulum as $l \cos\theta$ instead of $l \sin\theta$.
  - Why it fails: Confusion between the sides of the right triangle formed by the string, vertical axis and radius.
  - Correct: Draw a clear diagram: if $\theta$ is the angle between the string and the vertical, $r = l \sin\theta$.
- **Wrong:** Resolving forces tangentially instead of radially when calculating centripetal force.
  - Why it fails: Acceleration for uniform circular motion is only in the radial direction, so resultant force must be calculated radially.
  - Correct: Always resolve forces along the radial line (towards the centre of the circle) for centripetal force calculations.

## Cheatsheet

| Concept | Formula/Rule | Key Note |
| --- | --- | --- |
| Centripetal acceleration | $a_c = \frac{v^2}{r} = \omega^2 r$ | Always directed to centre |
| Centripetal force | $F_{net} = \frac{mv^2}{r}$ | Resultant force, not an extra force |
| Conical pendulum | $\omega^2 = \frac{g}{l \cos\theta}$ | Radius $r = l \sin\theta$ |
| String vertical circle | $v_{top} \geq \sqrt{gr}$ | Min projection speed: $u = \sqrt{5gr}$ |
| Rod vertical circle | $v_{top} \geq 0$ | Min projection speed: $u = \sqrt{4gr}$ |
| Vertical circles | $\Delta KE = -\Delta PE$ | Always use conservation of energy |

## 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.

- [Equilibrium of rigid bodies](https://www.owlsprep.com/study/cie-9231-u3-equilibrium-of-rigid-bodies/)
- [Elasticity and simple harmonic motion](https://www.owlsprep.com/study/cie-9231-u3-elasticity-and-simple-harmonic-motion/)
- [Further Probability & Statistics](https://www.owlsprep.com/study/cie-9231-u4-overview/)

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