Mathematics (9660) | Uniform circular motion

M2.7: Uniform Circular Motion (UCM) – Comprehensive Study Notes

Hello future A-Level physicist and mathematician! Welcome to Uniform Circular Motion. Don't worry if the formulas look intense at first; this topic is incredibly satisfying because it links forces (Newton’s Laws) with kinematics (motion). Everything from satellites orbiting the Earth to a car turning a corner relies on these principles!

In this chapter, we explore objects moving in a perfect circle at a steady pace. This might sound simple, but the mathematics required to describe the changing direction of motion is powerful and essential for later mechanics topics.

1. Defining Uniform Circular Motion (UCM)

Uniform Circular Motion describes the movement of a particle along a circular path at a constant speed.

• Uniform: Means the speed (\(v\)) is constant (e.g., always 10 \(m\ s^{-1}\)).
• Circular Motion: Means the path is a perfect circle, defined by a constant radius (\(r\)).

Why Velocity is Not Constant in UCM

This is the single most important conceptual point to grasp!

Even though the speed is constant, the velocity is not. Why? Because velocity is a vector quantity (it has magnitude and direction). As the particle moves around the circle, its direction is constantly changing.

Key Takeaway: Since the velocity is changing, there must be an acceleration, and thus, a net force (Newton's Second Law: F = ma).

Key Definitions

• Radius (\(r\)): The distance from the center of the circle to the particle.
• Period (\(T\)): The time taken for one complete revolution (measured in seconds, \(s\)).
• Frequency (\(f\)): The number of revolutions per unit time (measured in \(s^{-1}\) or Hertz, \(Hz\)).
  \(\hspace{1cm} \displaystyle f = \frac{1}{T}\)

2. Introducing Angular Speed (\(\omega\))

When dealing with circular motion, it’s often easier to measure how fast the object is rotating in terms of angles, rather than linear distance. This is where Angular Speed (\(\omega\)) comes in.

2.1 What is Angular Speed?

Angular speed is the rate of change of the angle swept out by the radius line.

• It is measured in radians per second (\(rad\ s^{-1}\)).
• One complete revolution is \(2\pi\) radians.

Formulae for Angular Speed

If a particle completes one revolution in time \(T\) (the Period):

1. Relating \(\omega\) to Period (\(T\)):
$$\omega = \frac{\text{Total Angle}}{\text{Time}} = \frac{2\pi}{T}$$

2. Relating \(\omega\) to Linear Speed (\(v\)):

We know that for one revolution, the linear distance is the circumference, \(2\pi r\).
$$v = \frac{\text{Distance}}{\text{Time}} = \frac{2\pi r}{T}$$

Since \(\frac{2\pi}{T} = \omega\), we get the fundamental relationship:

$$v = r\omega$$

Analogy: Imagine a record player. A point near the centre and a point on the edge both have the same angular speed (\(\omega\)), but the point on the edge has a much larger linear speed (\(v\)) because it has to cover a greater distance (\(r\) is larger) in the same amount of time.

Converting Units for Angular Speed

Sometimes the rotation rate is given in Revolutions Per Minute (RPM). You must convert this to \(rad\ s^{-1}\):

1. Convert minutes to seconds: Divide RPM by 60 to get revolutions per second (\(Hz\)).
2. Convert revolutions to radians: Multiply by \(2\pi\). (Since 1 revolution = \(2\pi\) radians).

Step-by-Step Conversion Example: A wheel rotates at 300 RPM.
$$ \omega = \frac{300 \text{ rev}}{1 \text{ min}} \times \frac{1 \text{ min}}{60 \text{ s}} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} = 10\pi \text{ rad s}^{-1} $$

3. Centripetal Acceleration (\(a\))

Since velocity is constantly changing direction, there must be an acceleration. In UCM, this acceleration is known as Centripetal Acceleration.

Direction and Name

The acceleration vector always points towards the centre of the circle.

• Centripetal literally means 'centre-seeking'.
• Since acceleration is perpendicular to the instantaneous velocity (which is tangential), this acceleration only changes the direction of the velocity, not its magnitude (speed).

Formulae for Centripetal Acceleration

These formulas are critical and must be used frequently in problems.

In terms of linear speed (\(v\)) and radius (\(r\)): $$a = \frac{v^2}{r}$$

In terms of angular speed (\(\omega\)) and radius (\(r\)):
(Substitute \(v = r\omega\) into the first formula): $$a = \frac{(r\omega)^2}{r} = \frac{r^2\omega^2}{r}$$ $$a = r\omega^2$$

4. Centripetal Force (\(F\))

According to Newton's Second Law (\(F = ma\)), if there is acceleration, there must be a resultant force causing it. This resultant force is called the Centripetal Force (\(F_c\)).

Centripetal Force is the Resultant Force

This is a key concept: Centripetal force is not a new, separate force. It is the name given to the net resultant force acting radially inwards, which is required to keep the body moving in a circle.

The centripetal force is always provided by existing, physical forces, such as:

• Tension (e.g., swinging a mass on a string).
• Friction (e.g., a car turning a corner).
• Gravity (e.g., satellites orbiting Earth).
• Normal Reaction (e.g., a rollercoaster loop).

Formulae for Centripetal Force (\(F_c\))

Using \(F = ma\):

In terms of linear speed (\(v\)): $$F_c = m a = m \frac{v^2}{r}$$

In terms of angular speed (\(\omega\)): $$F_c = m a = m r \omega^2$$

5. Applications of Uniform Circular Motion

5.1 Horizontal Circles (e.g., Car turning a corner)

Consider a particle moving in a horizontal circle (like a flat racetrack). The forces must be resolved so that the resultant force horizontally provides the centripetal force, \(F_c\).

Example: Car on a flat road.

The forces acting on the car are:

1. Weight (\(mg\)) (downwards)
2. Normal Reaction (\(R\)) (upwards)
3. Friction (\(F_{friction}\)) (horizontally, towards the centre of the turn)

In the vertical plane, the forces balance: \(R = mg\).
In the horizontal plane, the Friction Force provides the necessary centripetal force:
$$F_{friction} = F_c = m \frac{v^2}{r}$$

The maximum speed the car can take the corner at is limited by the maximum static friction, \(F_{max} = \mu R\). If \(m \frac{v^2}{r} > \mu R\), the car skids!

5.2 Satellite Orbits (Horizontal orbits)

For a satellite maintaining a circular orbit around a large body (like Earth), the Gravitational Force is the only force acting towards the centre of the circle.

Therefore, the Gravitational Force is the Centripetal Force:
$$F_{gravitational} = F_c$$

(Note: In M2, you will primarily deal with situations where the gravitational force is constant or acting towards the centre, as specified in the syllabus, rather than using the inverse square law of gravity.)

6. The Conical Pendulum

The conical pendulum is a classic example of UCM. It involves a particle attached to a string, swinging in a horizontal circle, creating a cone shape (hence the name).

Step-by-Step Analysis

Let the string have length \(L\), and the angle the string makes with the vertical be \(\theta\). The radius of the horizontal circle is \(r = L \sin \theta\).

Forces acting on the particle \(m\):

1. Weight (\(mg\)) acting vertically downwards.
2. Tension (\(T\)) acting along the string.

We resolve the Tension force into vertical and horizontal components:

1. Vertical Resolution (Equilibrium): The particle is not accelerating vertically, so the forces balance.
$$T \cos \theta = mg$$

2. Horizontal Resolution (Centripetal Force): The horizontal component of tension provides the centripetal force, \(F_c\).
$$T \sin \theta = F_c = m r \omega^2$$

By dividing the horizontal equation by the vertical equation, we can eliminate the Tension \(T\):
$$\frac{T \sin \theta}{T \cos \theta} = \frac{m r \omega^2}{mg}$$
$$\tan \theta = \frac{r \omega^2}{g}$$

Since \(r = L \sin \theta\), this relationship allows you to find the angular speed \(\omega\) if you know the angle \(\theta\) and the length \(L\).

7. Vector Description of Circular Motion (i, j)

For A-Level Mechanics (M2), you must be able to describe the position, velocity, and acceleration of a particle in UCM using \(\mathbf{i}\) and \(\mathbf{j}\) unit vectors, where motion occurs in the \(x-y\) plane.

Let the circle have radius \(r\) and constant angular speed \(\omega\). We typically assume the centre of the circle is at the origin \((0, 0)\). The position of the particle at time \(t\) can be described by an angle \(\theta = \omega t\).

Position Vector (\(\mathbf{r}\))

The vector pointing from the origin to the particle:
$$\mathbf{r} = (r \cos(\omega t))\mathbf{i} + (r \sin(\omega t))\mathbf{j}$$

Students may be required to show that motion is circular by demonstrating that the magnitude of \(\mathbf{r}\) is constant (i.e., its distance from the origin is constant: \(|\mathbf{r}| = \sqrt{(r \cos(\omega t))^2 + (r \sin(\omega t))^2} = r\)).

Velocity Vector (\(\mathbf{v}\))

Velocity is the derivative of position with respect to time, \(\mathbf{v} = \frac{d\mathbf{r}}{dt}\).
$$\mathbf{v} = \frac{d}{dt} \left( (r \cos(\omega t))\mathbf{i} + (r \sin(\omega t))\mathbf{j} \right)$$
$$\mathbf{v} = (-r \omega \sin(\omega t))\mathbf{i} + (r \omega \cos(\omega t))\mathbf{j}$$

Note that the magnitude of the velocity is \(|\mathbf{v}| = r\omega\), which confirms the constant speed. The velocity vector is tangential (perpendicular to \(\mathbf{r}\)).

Acceleration Vector (\(\mathbf{a}\))

Acceleration is the derivative of velocity with respect to time, \(\mathbf{a} = \frac{d\mathbf{v}}{dt}\).
$$\mathbf{a} = \frac{d}{dt} \left( (-r \omega \sin(\omega t))\mathbf{i} + (r \omega \cos(\omega t))\mathbf{j} \right)$$
$$\mathbf{a} = (-r \omega^2 \cos(\omega t))\mathbf{i} + (-r \omega^2 \sin(\omega t))\mathbf{j}$$

We can factor out \(- \omega^2\):
$$\mathbf{a} = - \omega^2 \left( (r \cos(\omega t))\mathbf{i} + (r \sin(\omega t))\mathbf{j} \right)$$
Since the bracketed term is just the position vector \(\mathbf{r}\):
$$\mathbf{a} = - \omega^2 \mathbf{r}$$

This mathematical result is very elegant! It proves two things:

• Magnitude: \(|\mathbf{a}| = \omega^2 |\mathbf{r}| = r\omega^2\). (Matches our formula!)
• Direction: The negative sign (\(-\omega^2\)) shows that the acceleration vector \(\mathbf{a}\) points in the exact opposite direction to the position vector \(\mathbf{r}\). Since \(\mathbf{r}\) points outwards, \(\mathbf{a}\) points inwards, towards the centre!