📚 Study Notes: Circular Motion (Further Mechanics 9231, Paper 3)
Hello future Further Mathematicians! Welcome to the exciting chapter of Circular Motion. While you studied forces and straight-line motion in standard A-Level Mechanics, here we deal with particles moving in curves—specifically, perfect circles.
This chapter is crucial because it combines concepts of kinematics, forces, and (in vertical circles) energy. Don't worry if it seems challenging; by breaking it down into linear and angular components, you’ll master it in no time!
1. Understanding Angular Speed (\(\omega\))
When an object moves in a straight line, we use linear speed ($v$). When it moves in a circle, we often find it easier to use Angular Speed.
What is Angular Speed?
Angular speed, denoted by \(\omega\) (the Greek letter omega), measures how fast the angle of rotation changes. It tells you the rate at which the particle sweeps out an angle around the centre of the circle.
- Definition: The rate of change of the angle \(\theta\).
- Formula: \(\omega = \frac{\Delta\theta}{\Delta t}\) (or \(\omega = \frac{v}{r}\))
- Standard Units: Radians per second (\(\text{rad s}^{-1}\)). Always use radians for \(\omega\).
🔧 The Key Link: Linear Speed (\(v\)) and Angular Speed (\(\omega\))
The speed you are used to (linear speed, $v$) is directly linked to angular speed (\(\omega\)) and the radius ($r$):
$$ v = r\omega $$
Quick Review Box:
- $v$ is linear speed (m/s).
- $r$ is the radius (m).
- $\omega$ is angular speed ($\text{rad s}^{-1}$).
Analogy: Imagine a CD spinning. A dust particle near the centre (small $r$) and a particle near the edge (large $r$) have the same angular speed ($\omega$); they both complete a rotation in the same time. But the particle on the edge has a much greater linear speed ($v$) because it has to travel a larger distance ($2\pi r$) in that same time.
Key Takeaway for Section 1
Circular motion uses two speeds: linear ($v$) and angular ($\omega$). They are linked by the simple formula $v = r\omega$.
2. Centripetal Acceleration and Force
For an object to move in a circle, its velocity vector must constantly change direction. According to Newton’s laws, changing velocity means there must be an acceleration, and thus a net force.
Acceleration in Circular Motion
A particle moving in a circle, even at constant speed, is always accelerating. This acceleration is called Centripetal Acceleration.
- Direction: Always directed towards the centre of the circle.
- Formulas (You must know and use both):
$$ a = \frac{v^2}{r} \quad \text{or} \quad a = r\omega^2 $$
(Note: The syllabus states that the proof of these formulae is not required, but applying them is essential.)
The Centripetal Force (\(F\))
The net force required to cause this centripetal acceleration is called the Centripetal Force.
- Formula (Newton's Second Law, \(F = ma\)):
$$ F = \frac{mv^2}{r} \quad \text{or} \quad F = mr\omega^2 $$
🚨 Crucial Concept Alert!
The centripetal force ($F$) is not a separate, new type of force. It is the *resultant* force acting on the particle, always pointing towards the centre. The force is provided by familiar mechanisms like tension in a string, friction, gravity, or the normal contact force.
In any circular motion problem, the solution involves identifying the forces acting on the particle and then setting the resultant force component towards the centre equal to the centripetal force requirement:
$$ \sum F_{\text{towards centre}} = mr\omega^2 $$
Key Takeaway for Section 2
The net force towards the centre of the circle must always equal $mr\omega^2$ or $mv^2/r$. This is the foundation of all circular motion problems.
3. Modelling Motion in a Horizontal Circle (H.C.M.)
Problems involving horizontal circles usually involve steady motion where the particle is moving on a flat plane or is suspended as a conical pendulum.
Steps for Solving H.C.M. Problems
- Draw a clear force diagram: Show all forces acting on the particle (Tension $T$, Weight $mg$, Normal Reaction $R$, etc.).
- Establish the radius ($r$): This is the radius of the horizontal circle, which might not be the length of the string ($l$). If the string makes an angle \(\alpha\) with the vertical, $r = l\sin\alpha$.
- Resolve Vertically (Balance): Since the particle is not accelerating vertically (it’s moving horizontally), the vertical forces must balance (\(\sum F_y = 0\)).
- Resolve Horizontally (Centripetal Force): The net force directed horizontally (towards the centre) provides the centripetal force (\(\sum F_x = mr\omega^2\)).
Example: The Conical Pendulum
A particle of mass $m$ is attached to a string of length $l$ and moves in a horizontal circle of radius $r$. The string makes an angle \(\alpha\) with the vertical.
1. Forces: Tension $T$ (up the string), Weight $mg$ (down).
2. Vertical Resolution (\(\sum F_y = 0\)):
$$ T\cos\alpha = mg $$
3. Horizontal Resolution (\(\sum F_x = mr\omega^2\)):
$$ T\sin\alpha = m r\omega^2 $$
You can then solve these two simultaneous equations for unknowns like $T$, $\omega$, or $\alpha$. (Remember $r = l\sin\alpha$ if $l$ is given.)
💡 Common Mistake: Confusing the radius of the circle ($r$) with the length of the string ($l$). Always ensure you use the radius of the horizontal path in the $mr\omega^2$ formula.
Key Takeaway for Section 3
In H.C.M., the motion is steady. Vertical forces balance, and the horizontal resultant force provides the centripetal requirement $mr\omega^2$.
4. Modelling Motion in a Vertical Circle (V.C.M.)
Vertical circular motion is significantly more complex because the force of gravity constantly changes the particle’s speed and the direction in which gravity acts relative to the path.
The syllabus requires solving V.C.M. problems without loss of energy, meaning we use the Principle of Conservation of Mechanical Energy.
🌟 Step 1: Using Conservation of Energy
To find the speed $v$ at any point in the circle, we compare its energy to a known point (usually the bottom, where speed is maximum).
Let $v_B$ be the speed at the bottom, and $v_P$ be the speed at point P, which is height $h$ above the bottom. Potential Energy (P.E.) is taken as zero at the bottom.
$$ \text{K.E.}_B + \text{P.E.}_B = \text{K.E.}_P + \text{P.E.}_P $$
$$ \frac{1}{2}m v_B^2 + 0 = \frac{1}{2}m v_P^2 + mgh $$
You can use geometry (usually involving the radius $r$ and angle $\theta$ from the vertical) to express $h$.
🌟 Step 2: Using Centripetal Force
At any point P, we resolve forces towards the centre of the circle. The net force must equal $mv^2/r$.
Let $T$ be the tension (or $R$ the normal force) and $\theta$ be the angle measured from the vertical radius.
$$ T + (mg \cos\theta) = \frac{m v^2}{r} $$
(Note: At the top point, \(\theta = 180^\circ\), so \(\cos\theta = -1\). The equation becomes: \(T - mg = mv^2/r\) or \(T = mg + mv^2/r\).)
Procedure: Use the Energy equation (Step 1) to find $v^2$ at P, substitute $v^2$ into the Force equation (Step 2), and then solve for $T$ or $R$.
🔰 Critical Conditions for V.C.M.
A key focus of V.C.M. problems is determining the minimum conditions required for the particle to complete the circle or remain in contact with a track.
1. Conditions for a Particle on a String (Tension T)
A string can only pull; it cannot push. If the tension $T$ becomes zero, the string goes slack, and the particle stops moving in a circle.
- Critical condition: $T = 0$.
- This occurs at the top of the circle or somewhere in the top half (i.e., when $v$ is minimum).
To find the minimum speed $v_{\text{top}}$ required to complete the loop, set $T=0$ at the highest point (H) where $h = 2r$ (relative to the bottom):
At the top, the force equation is (resolving towards the centre): $T + mg = mv_{\text{top}}^2/r$.
Setting $T=0$: $$ mg = \frac{m v_{\text{top}}^2}{r} $$ $$ v_{\text{top}}^2 = gr $$ $$ v_{\text{top}} = \sqrt{gr} $$
This is the minimum speed at the top required to just keep the string taut.
2. Conditions for a Particle on a Track (Normal Force R)
If a particle is sliding inside a circular tube or loop, $R$ is the normal contact force. If $R$ becomes zero, the particle loses contact with the track.
- Critical condition: $R = 0$.
The calculation is identical to the tension case. For a particle moving inside a vertical loop, the minimum speed at the top to maintain contact is also $v_{\text{top}} = \sqrt{gr}$.
🧩 Calculating the Minimum Speed at the Bottom
Often, the question asks for the minimum speed at the bottom ($v_{\min}$) required to complete the circle. We use energy conservation, relating the critical speed at the top ($v_{\text{top}} = \sqrt{gr}$) back to the bottom (height difference $h=2r$):
$$ \frac{1}{2}m v_{\min}^2 = \frac{1}{2}m v_{\text{top}}^2 + mg(2r) $$
Substituting $v_{\text{top}}^2 = gr$:
$$ \frac{1}{2}m v_{\min}^2 = \frac{1}{2}m (gr) + 2mgr $$
Dividing by $m$ and multiplying by 2:
$$ v_{\min}^2 = gr + 4gr = 5gr $$
$$ v_{\min} = \sqrt{5gr} $$
This is the minimum speed at the bottom needed for a particle attached to a string or moving on a track to complete the vertical circle.
Key Takeaway for Section 4
V.C.M. requires using Conservation of Energy to link speeds and then resolving forces towards the centre. The critical condition for completing the circle is maintaining $T \geq 0$ (or $R \geq 0$) at the very top, which requires a top speed of $v_{\text{top}} = \sqrt{gr}$.
💭 Quick Chapter Review: Circular Motion Checklist
- Angular Speed: \(\omega\) (in \(\text{rad s}^{-1}\)).
- Linear/Angular Link: \(v = r\omega\).
- Centripetal Acceleration: \(a = r\omega^2 = v^2/r\). Always towards the centre.
- Centripetal Force: \(F = mr\omega^2 = mv^2/r\). This is the resultant force towards the centre.
- Horizontal Circles: Forces balance vertically (\(\sum F_y = 0\)). Net horizontal force is \(mr\omega^2\).
- Vertical Circles: Use Conservation of Energy (\(\frac{1}{2}m v^2 + mgh = \text{constant}\)). Resolve forces towards the centre.
- Critical Speed (V.C.M.): Minimum speed at the top to complete a loop is \(v = \sqrt{gr}\).
- Critical Speed (V.C.M. Bottom): Minimum speed at the bottom to complete a loop is \(v = \sqrt{5gr}\).
You’ve covered the core concepts of Circular Motion! Practice applying these energy and force resolutions, and remember to define your direction for resolving forces (usually towards the centre).