1. Overview
Uniform circular motion occurs when an object travels along a circular path at a constant speed. Although the speed is constant, the velocity is not. Velocity is a vector quantity, and because the direction of motion is continuously changing as the object follows the curve, the velocity is continuously changing. According to Newton’s First Law, a change in velocity implies the presence of an acceleration, which in turn requires a resultant force. In this topic, we establish the kinematic language—radians, angular displacement, and angular speed—necessary to describe this motion before moving on to the dynamics of centripetal force.
Key Definitions
- Radian (rad): The angle subtended at the centre of a circle by an arc with a length equal to the radius of the circle.
- Angular Displacement ($\theta$): The angle in radians through which a point or line has been rotated about a specified axis. It is the angular analogue of linear displacement.
- Angular Speed ($\omega$): The rate of change of angular displacement with respect to time.
- Period ($T$): The time taken for an object to complete one full revolution ($2\pi$ radians).
- Frequency ($f$): The number of revolutions completed per unit time.
Content
The Radian and Angular Displacement
In A-Level Physics, degrees are rarely used for circular motion because they are an arbitrary division of a circle ($360^\circ$). The radian is a "natural" unit based on the geometry of the circle itself.
The relationship between the arc length $s$, the radius $r$, and the angular displacement $\theta$ (measured in radians) is defined by:
$$\mathbf{s = r\theta}$$
Where:
- $s$ is the arc length (m)
- $r$ is the radius (m)
- $\theta$ is the angular displacement (rad)
Why use Radians? When $\theta$ is measured in radians, the linear distance moved ($s$) is directly proportional to the angle. If an object completes one full revolution, the arc length is the circumference ($2\pi r$). Substituting this into the formula: $2\pi r = r\theta \implies \theta = 2\pi$ Therefore, $360^\circ = 2\pi \text{ rad}$.
Conversion Factors:
- To convert degrees to radians: Multiply by $\frac{\pi}{180}$
- To convert radians to degrees: Multiply by $\frac{180}{\pi}$
| Degrees ($^\circ$) | Radians (rad) |
|---|---|
| $360^\circ$ | $2\pi$ |
| $180^\circ$ | $\pi$ |
| $90^\circ$ | $\pi/2$ |
| $60^\circ$ | $\pi/3$ |
| $45^\circ$ | $\pi/4$ |
Angular Speed ($\omega$)
Angular speed describes how quickly an object rotates. Unlike linear velocity, which depends on the distance from the center, all points on a rotating rigid body (like a spinning disk) share the same angular speed.
The formula for angular speed is:
$$\mathbf{\omega = \frac{\Delta \theta}{\Delta t}}$$
Where:
- $\omega$ is the angular speed ($\text{rad s}^{-1}$)
- $\Delta \theta$ is the change in angular displacement (rad)
- $\Delta t$ is the time interval (s)
Deriving the Relationship between $\omega$, $T$, and $f$
For an object completing one full circle:
- The angular displacement $\Delta \theta = 2\pi \text{ rad}$.
- The time taken $\Delta t$ is the period $T$.
Substituting these into the angular speed definition:
$$\mathbf{\omega = \frac{2\pi}{T}}$$
Since frequency $f$ is the reciprocal of the period ($f = 1/T$), we can also write:
$$\mathbf{\omega = 2\pi f}$$
Relating Linear Speed ($v$) and Angular Speed ($\omega$)
Consider an object moving at a constant speed $v$ around a circle of radius $r$. In a time interval $t$, it covers an arc length $s$. The linear speed is: $v = \frac{s}{t}$
From the definition of angular displacement, $s = r\theta$. Substituting this: $v = \frac{r\theta}{t}$
Since $\omega = \frac{\theta}{t}$, we arrive at the fundamental link between linear and angular kinematics:
$$\mathbf{v = r\omega}$$
Conceptual Implications of $v = r\omega$:
- For a fixed angular speed $\omega$, the linear speed $v$ is directly proportional to the radius $r$.
- On a rotating platform (like a merry-go-round), a person standing at the outer edge travels at a higher linear speed than someone standing near the center, even though they both complete one revolution in the same amount of time.
Velocity in Circular Motion
It is vital to distinguish between speed and velocity.
- Speed ($v$): The magnitude of the velocity. In uniform circular motion, this is constant.
- Velocity ($\vec{v}$): A vector quantity. At any point in the path, the velocity vector is a tangent to the circle.
- Because the direction of the tangent changes at every point, the velocity is constantly changing. This means the object is accelerating, even if the speedometer reading is constant.
Worked Example 1 — Conversion of Angular Units
A cooling fan rotates at a constant rate of 2400 revolutions per minute (rpm). Calculate its angular speed in $\text{rad s}^{-1}$.
Step 1: Convert revolutions per minute to revolutions per second (Frequency) $$f = \frac{2400 \text{ rev}}{60 \text{ s}} = 40 \text{ Hz}$$
Step 2: Use the relationship between $\omega$ and $f$ $$\omega = 2\pi f$$ $$\omega = 2 \times \pi \times 40$$
Step 3: Calculate the final value $$\omega = 251.327... \text{ rad s}^{-1}$$
Answer: $\omega = 251 \text{ rad s}^{-1}$ (3 s.f.)
Worked Example 2 — Linear Speed on a Rotating Body
A standard vinyl record has a diameter of 30.0 cm and rotates at $33\frac{1}{3}$ revolutions per minute. Calculate the linear speed of a dust particle sitting on the outer edge of the record.
Step 1: Identify the radius in SI units $$r = \frac{\text{diameter}}{2} = \frac{0.300 \text{ m}}{2} = 0.150 \text{ m}$$
Step 2: Calculate the angular speed $\omega$ First, find frequency: $f = \frac{33.33}{60} = 0.5555 \text{ Hz}$ $$\omega = 2\pi f = 2 \times \pi \times 0.5555 = 3.491 \text{ rad s}^{-1}$$
Step 3: Calculate linear speed $v$ $$v = r\omega$$ $$v = 0.150 \times 3.491$$ $$v = 0.5236... \text{ m s}^{-1}$$
Answer: $v = 0.524 \text{ m s}^{-1}$ (3 s.f.)
Worked Example 3 — Geostationary Satellite Kinematics
A geostationary satellite remains above the same point on the Earth's equator at all times. The radius of its orbit is $4.22 \times 10^7 \text{ m}$. Calculate the magnitude of its velocity.
Step 1: Determine the period $T$ For a geostationary satellite, the period must match the Earth's rotation: $$T = 24 \text{ hours} = 24 \times 3600 \text{ s} = 86,400 \text{ s}$$
Step 2: Calculate the angular speed $\omega$ $$\omega = \frac{2\pi}{T}$$ $$\omega = \frac{2\pi}{86,400} = 7.272 \times 10^{-5} \text{ rad s}^{-1}$$
Step 3: Calculate the linear speed $v$ $$v = r\omega$$ $$v = (4.22 \times 10^7) \times (7.272 \times 10^{-5})$$ $$v = 3068.7... \text{ m s}^{-1}$$
Answer: $v = 3.07 \times 10^3 \text{ m s}^{-1}$ (3 s.f.)
Worked Example 4 — Comparing Two Points
A rigid rod of length 2.0 m rotates about one end at a rate of 5.0 rad s⁻¹. Point A is 0.5 m from the pivot, and Point B is at the far end of the rod. Calculate the difference in linear speeds between Point A and Point B.
Step 1: Identify that $\omega$ is the same for both points $$\omega = 5.0 \text{ rad s}^{-1}$$
Step 2: Calculate $v_A$ $$v_A = r_A \omega = 0.5 \times 5.0 = 2.5 \text{ m s}^{-1}$$
Step 3: Calculate $v_B$ $$v_B = r_B \omega = 2.0 \times 5.0 = 10.0 \text{ m s}^{-1}$$
Step 4: Find the difference $$\Delta v = v_B - v_A = 10.0 - 2.5 = 7.5 \text{ m s}^{-1}$$
Answer: $\Delta v = 7.5 \text{ m s}^{-1}$
Key Equations
| Equation | Symbols & Units | Data Sheet? |
|---|---|---|
| $\theta = \frac{s}{r}$ | $\theta$: angle ($\text{rad}$), $s$: arc length ($\text{m}$), $r$: radius ($\text{m}$) | No |
| $\omega = \frac{\Delta \theta}{\Delta t}$ | $\omega$: angular speed ($\text{rad s}^{-1}$), $t$: time ($\text{s}$) | No |
| $\omega = \frac{2\pi}{T}$ | $T$: period ($\text{s}$) | Yes |
| $\omega = 2\pi f$ | $f$: frequency ($\text{Hz}$) | Yes |
| $v = r\omega$ | $v$: linear speed ($\text{m s}^{-1}$) | Yes |
Common Mistakes to Avoid
- ❌ Wrong: Using degrees in the formula $v = r\omega$ or $s = r\theta$.
- ✅ Right: These equations are derived specifically for radians. Always convert degrees to radians first.
- ❌ Wrong: Confusing diameter and radius in exam questions.
- ✅ Right: Always check if the question provides the "diameter" or "radius." Divide the diameter by 2 before calculating $v$ or $\omega$.
- ❌ Wrong: Stating that an object in uniform circular motion has "constant velocity."
- ✅ Right: The speed is constant, but the velocity is changing because the direction is changing. This is a common trap in multiple-choice questions.
- ❌ Wrong: Forgetting to convert "revolutions per minute" (rpm) to "revolutions per second" (Hz).
- ✅ Right: Divide rpm by 60 to get frequency in Hz before using $\omega = 2\pi f$.
- ❌ Wrong: Assuming $\omega$ changes with the radius on a rigid rotating object.
- ✅ Right: For a solid object (like a wheel), $\omega$ is the same for every point on the object. Only the linear speed $v$ changes with $r$.
Exam Tips
- Precision in Definitions: When defining the radian, you must state it is the angle subtended at the centre of the circle. Omitting "at the centre" is a frequent reason for losing the mark.
- The "Show That" Question: If asked to show that $v = r\omega$, start from the very beginning:
- Define $v = \frac{\Delta s}{\Delta t}$
- State the definition of the radian: $\Delta s = r \Delta \theta$
- Substitute: $v = \frac{r \Delta \theta}{\Delta t}$
- Identify $\omega = \frac{\Delta \theta}{\Delta t}$, therefore $v = r\omega$.
- Significant Figures: Cambridge 9702 is strict on significant figures. Look at the data provided. If the radius is $0.50 \text{ m}$ (2 s.f.) and the period is $2.1 \text{ s}$ (2 s.f.), your final answer should be given to 2 s.f.
- Calculator Mode: While most circular motion calculations are done manually, if you are using trigonometric functions in related topics (like resolving forces in circular motion), ensure your calculator is in RAD mode.
- Vector Diagrams: When drawing the velocity vector on a circular path, use a ruler to ensure the line is a tangent. Draw a small square at the intersection of the radius and the velocity vector to indicate they are at $90^\circ$ to each other.