Further Mathematics | Motion in a circle

Motion in a Circle: Essential Study Notes (M3 Mechanics)

Hello Future Engineer! Welcome to one of the most dynamic and fascinating topics in Mechanics: Motion in a Circle. Don't worry if this chapter seems tricky at first—it connects concepts you already know (forces, acceleration) but applies them in a circular path. By the end of these notes, you'll understand why roller coasters feel the way they do and how satellites stay in orbit!

Why is this important? When an object moves in a circle, its direction is constantly changing. A changing velocity means there must be acceleration, and according to Newton’s Second Law, acceleration requires a force. This study explores the specific nature of that acceleration and the forces responsible.

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1. Defining Angular Motion: Radians and Angular Velocity

When dealing with circles, we need to shift from measuring distance (metres) to measuring rotation (angles).

Key Definitions

• Angular Displacement (\(\theta\)): This is the angle turned through. In Further Maths Mechanics, this must be measured in radians (rad).
• Angular Velocity (\(\omega\)): This measures how fast the angle is changing, or the rate of rotation. It is the angular displacement per unit time.
  $$ \omega = \frac{\text{d}\theta}{\text{d}t} $$

Units Check: The standard unit for angular velocity is radians per second (rad s\(\text{}^{-1}\)).

Did you know? One complete revolution is \(2\pi\) radians. If a particle completes \(f\) revolutions per second (frequency), then its angular velocity is \(\omega = 2\pi f\).

Quick Review: Radian Conversion

Remember: \(180^{\circ} = \pi\) radians. If a question gives you angles in degrees, you must convert them to radians before using the mechanics formulas involving \(\omega\)!

Key Takeaway: Circular motion requires us to use angular measure (\(\theta\)) and its rate of change (\(\omega\)).

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2. Relating Linear Speed (v) and Angular Speed (\(\omega\))

An object moving in a circle has both linear speed (\(v\)) and angular speed (\(\omega\)). These two are directly related by the radius of the circle, \(r\).

The Fundamental Relationship

Imagine you and a friend are on a carousel. You stand near the edge (large \(r\)) and your friend stands near the centre (small \(r\)). You both complete one rotation in the same amount of time (same \(\omega\)), but you have to cover a much larger circumference, so your linear speed (\(v\)) must be greater.

The relationship is defined as:

$$ v = r\omega $$

Where:

• \(v\) is the linear speed (m s\(\text{}^{-1}\))
• \(r\) is the radius of the circular path (m)
• \(\omega\) is the angular velocity (rad s\(\text{}^{-1}\))

Accessibility Tip: If the question gives you \(v\) and \(r\), use \(v = r\omega\) to find \(\omega\). If it gives you \(\omega\) and \(r\), use it to find \(v\). Simple rearrangement!

Key Takeaway: Linear speed is proportional to the radius for a fixed angular velocity: \(v \propto r\).

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3. Centripetal Acceleration

This is perhaps the single most important concept in this chapter. Because the direction of motion is always changing (even if speed is constant), the object is accelerating.

Definition and Direction

This acceleration is called Centripetal Acceleration, which means "centre-seeking" acceleration.

• Magnitude: The magnitude of the acceleration is constant if \(v\) and \(r\) are constant.
• Direction: The acceleration always acts towards the centre of the circle.

Formulas for Centripetal Acceleration (a)

We have two main formulas, depending on whether we know \(v\) or \(\omega\):

1. Using linear speed (\(v\)): $$ a = \frac{v^2}{r} $$

2. Using angular speed (\(\omega\)): (Substitute \(v = r\omega\) into the first equation) $$ a = r\omega^2 $$

Memory Aid: If you forget the formulas, remember that \(v = r\omega\) links them. If you know one, you can derive the other!

Key Takeaway: Acceleration is always directed towards the centre. Use \(a = v^2/r\) or \(a = r\omega^2\).

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4. Centripetal Force (The Net Force)

Newton’s Second Law (\(F=ma\)) must still apply. Since there is acceleration towards the centre (\(a\)), there must be a resultant force causing it. This resultant force is the Centripetal Force (\(F_c\)).

The Centripetal Force Formula

Since \(F = ma\), we substitute the acceleration formulas:

1. Using linear speed (\(v\)): $$ F_c = \frac{mv^2}{r} $$

2. Using angular speed (\(\omega\)): $$ F_c = mr\omega^2 $$

Crucial Concept: What is Centripetal Force?

Centripetal Force is NOT a separate, additional force.

It is the NET RESULTANT FORCE acting towards the centre, provided by existing forces like tension, gravity, friction, or the normal reaction force.

Example Analogy: Imagine spinning a bucket of water overhead. The tension in the rope (T) is providing the necessary centripetal force (\(F_c\)) to keep the bucket moving in a circle.

Common Mistake to Avoid: Never include a fictitious "centrifugal force" (the outward feeling you experience) in your free-body diagrams when analyzing the forces acting on the object. Only include actual physical forces (T, R, mg, F). The centripetal force is the sum of these forces in the radial direction.

Key Takeaway: \(F_c\) is the net force towards the centre, required by \(F_c = mr\omega^2\).

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5. Motion in a Horizontal Circle (Conical Pendulum)

In many scenarios, the object moves in a circle in a horizontal plane. Gravity is present but usually does not change the speed, it only affects the vertical equilibrium.

Application: The Conical Pendulum

A particle of mass \(m\) is attached to a string of length \(L\) and whirls around so that the string makes a constant angle \(\alpha\) with the vertical, tracing out a horizontal circle of radius \(r\).

Step-by-Step Force Analysis

1. Draw a free-body diagram showing the forces acting on the particle:

• Weight (\(mg\)): Acts vertically downwards.
• Tension (T): Acts along the string, upwards and inwards.

2. Resolve the Tension (T) into horizontal and vertical components, using the angle \(\alpha\).

• Vertical Component: \(T \cos \alpha\). Since there is no vertical acceleration, this component balances the weight. $$ T \cos \alpha = mg \quad \quad (1) $$
• Horizontal Component: \(T \sin \alpha\). This component acts directly towards the centre of the circle and provides the necessary centripetal force (\(F_c\)). $$ T \sin \alpha = F_c = mr\omega^2 \quad \quad (2) $$

3. Solve simultaneously: Divide equation (2) by equation (1): $$ \frac{T \sin \alpha}{T \cos \alpha} = \frac{mr\omega^2}{mg} $$ $$ \tan \alpha = \frac{r\omega^2}{g} \quad \text{or} \quad \tan \alpha = \frac{v^2}{rg} $$

Note on Radius (r): If the string length \(L\) is given, the radius \(r\) of the horizontal circle is \(r = L \sin \alpha\). You must substitute this into the equations if \(L\) is used instead of \(r\).

Application: Banking of Tracks

Roads and race tracks are often banked (tilted at an angle \(\alpha\)) to help vehicles turn safely. The normal reaction force (R) is used to provide the centripetal force.

If a car moves at the optimal speed \(v\) on a banked curve (meaning there is no tendency to slip, so friction \(F\) is zero):

• Vertical Equilibrium: \(R \cos \alpha = mg\)
• Horizontal Net Force (Centripetal): \(R \sin \alpha = \frac{mv^2}{r}\)

Dividing these equations gives the same relationship as the conical pendulum: $$ \tan \alpha = \frac{v^2}{rg} $$

Key Takeaway: In horizontal motion, the vertical forces balance (equal to zero), and the horizontal component of the tension or reaction provides the required centripetal force.

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6. Motion in a Vertical Circle

This section is more complex because gravity constantly opposes or aids the motion, meaning the speed \(v\) and the tension (or reaction force) \(T\) are NOT constant.

You will usually need to use the conservation of energy (from M2/M3) to find the speed at different points.

Energy Review (Prerequisite)

If the object is moving freely (not driven by a motor), Total Mechanical Energy is conserved: $$ KE_1 + PE_1 = KE_2 + PE_2 $$ $$ \frac{1}{2}mv_1^2 + mgh_1 = \frac{1}{2}mv_2^2 + mgh_2 $$

Force Analysis at Key Points

We define the net force towards the centre as \(F_c = mr\omega^2\) or \(mv^2/r\). This net force is calculated by considering all real forces (Tension/Reaction and Weight) acting in the radial line.

Let \(r\) be the radius.

A. The Bottom of the Circle (Maximum Tension/Reaction)

At the bottom, Tension (T) acts up (towards the centre) and Weight (\(mg\)) acts down (away from the centre). $$ F_{c, \text{bottom}} = T_{\text{max}} - mg $$ $$ T_{\text{max}} = \frac{mv^2}{r} + mg $$

The tension (or reaction force) is maximum here because it must counteract gravity AND provide the centripetal force.

B. The Top of the Circle (Minimum Tension/Reaction)

At the top, Tension (T) and Weight (\(mg\)) both act down (towards the centre). $$ F_{c, \text{top}} = T_{\text{min}} + mg $$ $$ T_{\text{min}} = \frac{mv^2}{r} - mg $$

The tension (or reaction) is minimum here because gravity is helping to provide the required centripetal force.

The Critical Speed (\(v_{\text{min}}\))

A common question involves finding the minimum speed required to complete the loop or circle.

The particle will just complete the circle if, at the highest point, the string becomes slack (or the particle just loses contact with the track). This means the tension (T) or normal reaction (R) must be exactly zero at the top.

Set \(T_{\text{min}} = 0\) in the Top Point equation: $$ 0 + mg = \frac{m(v_{\text{min}})^2}{r} $$

Cancel \(m\) and solve for \(v_{\text{min}}\): $$ v_{\text{min}} = \sqrt{gr} $$

If the particle's speed at the top is less than \(\sqrt{gr}\), it will not complete the circle; it will fall or the string will go slack before reaching the top.

Step-by-Step for Vertical Circle Problems:

1. Identify the highest/lowest points.
2. Use Conservation of Energy (KE + PE) to relate the speeds (\(v_1\) and \(v_2\)) at different points.
3. Apply Newton's Second Law (\(F_c = mv^2/r\)) at the specific point, noting whether Weight helps or hinders the centripetal force.

Key Takeaway: In a vertical circle, speed and tension vary. The critical minimum speed at the top needed to maintain circular motion is \(v_{min} = \sqrt{gr}\).

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Chapter Summary and Study Checklist

You've just tackled some of the core concepts of M3 Mechanics! Here’s a quick review of the absolute necessities:

Final Encouragement: Motion in a circle combines resolving forces (M1) and energy conservation (M2/M3). Practice drawing clear free-body diagrams and labelling the forces correctly. Once you identify which real force (Tension, Reaction, or Gravity) provides the centripetal force, the rest is just algebra! Keep practicing, you've got this!