Welcome to Further Circular Motion!
Welcome to one of the most exciting and dynamic topics in Mechanics! Have you ever wondered why high-speed race tracks are tilted on corners, or why a speeding van is at risk of toppling over on a sharp turn? In this chapter, we build upon basic circular motion to explore how friction, track banking, and rigid body mechanics keep objects moving safely along curved paths.
Don't worry if this topic seems a bit daunting at first. We will break every concept down step-by-step, look at the forces clearly, and show you exactly what CCEA examiners look for.
Standard Exam Conventions for CCEA:
• Acceleration due to gravity is taken as \(g = 9.8\text{ m s}^{-2}\) unless stated otherwise.
• Non-exact numerical answers should always be rounded to 3 significant figures.
• Formulae are provided in the CCEA GCE Mathematical Formulae and Tables booklet, but understanding how to derive and apply them is essential for top marks.
1. Kinematics of Uniform Circular Motion: The Foundations
Before we tackle complex banked curves and toppling vehicles, let's quickly review the fundamental building blocks of circular motion.
Key Terms and Definitions
• Angular Velocity (\(\omega\)): The rate at which an angle is swept out with respect to time, measured in radians per second (\(\text{rad s}^{-1}\)):
\(\omega = \frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{2\pi}{T} = 2\pi f\)
where \(T\) is the period of one complete revolution in seconds (\(\text{s}\)), and \(f\) is the frequency in Hertz (\(\text{Hz}\)).
• Linear / Tangential Speed (\(v\)): The actual distance traveled along the circular arc per unit time:
\(v = r\omega\)
where \(r\) is the radius of the circular path.
• Centripetal Acceleration (\(a_c\)): Even when moving at a constant speed, an object in circular motion is constantly changing its direction of travel. This means it is continuously accelerating towards the centre of the circle:
\(a = r\omega^2 = \frac{v^2}{r} = v\omega\)
• Centripetal Force: Remember that "centripetal force" is not a new, mystical physical force! It is simply the resultant real force pointing towards the centre that provides the necessary acceleration according to Newton's Second Law (\(F = ma\)):
\(\Sigma F_{\text{radial}} = m a = \frac{m v^2}{r} = m r \omega^2\)
Quick Review & Takeaway: Whenever an object moves in a horizontal circle of radius \(r\), resolve forces vertically (where acceleration is zero, so \(\Sigma F_y = 0\)) and resolve horizontally towards the centre (where \(\Sigma F_x = \frac{mv^2}{r} = mr\omega^2\)).
---2. Conical Pendulums
A classic circular motion problem involves a particle of mass \(m\) suspended from a fixed point by a light, inextensible string of length \(L\). When set into steady rotation, the string traces out the surface of a cone, while the mass moves in a horizontal circle at a constant height.
Setting Up the Equations
Let \(\theta\) be the angle the string makes with the vertical.
• The radius of the horizontal circle is given by: \(r = L\sin\theta\)
• The height of the cone is given by: \(h = L\cos\theta\)
• The only two forces acting on the particle are the tension \(T\) acting along the string and the gravitational force \(mg\) acting straight down.
1. Vertical Equilibrium:
Because there is no vertical movement, the upward vertical component of tension balances the weight:
\(T\cos\theta = mg \implies T = \frac{mg}{\cos\theta}\)
2. Horizontal Equation of Motion:
The horizontal component of tension provides the necessary centripetal force towards the centre of the circle:
\(T\sin\theta = m r \omega^2\)
Substituting \(r = L\sin\theta\) into the radial equation gives:
\(T\sin\theta = m (L\sin\theta)\omega^2 \implies T = m L \omega^2\)
3. Period of Rotation:
Equating our two expressions for tension \(T\):
\(\frac{mg}{\cos\theta} = m L \omega^2 \implies \omega^2 = \frac{g}{L\cos\theta} = \frac{g}{h}\)
Since the period is \(T_{\text{period}} = \frac{2\pi}{\omega}\), we arrive at the elegant formula:
\(T_{\text{period}} = 2\pi\sqrt{\frac{h}{g}} = 2\pi\sqrt{\frac{L\cos\theta}{g}}\)
Memory Aid: Notice how similar this period formula is to a simple pendulum (\(2\pi\sqrt{\frac{L}{g}}\))! For a conical pendulum, simply replace the full string length \(L\) with the vertical height \(h = L\cos\theta\).
---3. Further Circular Motion on Banked Tracks
When a car turns a corner on a flat road, only friction between the tyres and road prevents it from sliding outwards. If the track is banked (tilted at an angle \(\theta\) to the horizontal), the normal reaction force \(R\) is tilted inward, contributing to the required centripetal acceleration. This makes cornering much safer and faster!
Case A: Smooth Banked Track (No Friction)
Imagine an icy track or a track specifically designed so that vehicles do not rely on friction at a particular "design speed" \(v_0\).
• Vertical forces balance: \(R\cos\theta = mg\)
• Horizontal centripetal force: \(R\sin\theta = \frac{mv_0^2}{r}\)
Dividing the horizontal equation by the vertical equation gives:
\(\frac{R\sin\theta}{R\cos\theta} = \frac{\frac{mv_0^2}{r}}{mg} \implies \tan\theta = \frac{v_0^2}{rg}\)
Design Speed (\(v_0\)): \(v_0 = \sqrt{rg\tan\theta}\). At this exact speed, a car negotiates the bend with zero friction needed!
Case B: Rough Banked Track (With Friction \(F \le \mu R\))
What happens if the vehicle travels faster or slower than the ideal design speed \(v_0\)? Friction comes to the rescue!
1. High Speed (\(v > v_0\)) — Tendency to slip UP the slope:
When traveling very fast, the car tends to skid up the banking. Therefore, the frictional force \(F\) acts downwards along the slope (towards the inside of the track).
• Resolving Vertically (\(\uparrow = \downarrow\)):
\(R\cos\theta - F\sin\theta = mg\)
• Resolving Horizontally towards centre (\(\rightarrow\)):
\(R\sin\theta + F\cos\theta = \frac{mv_{\text{max}}^2}{r}\)
At the maximum speed before slipping occurs, friction reaches its limiting value: \(F = \mu R\).
2. Low Speed (\(v < v_0\)) — Tendency to slip DOWN the slope:
If the car travels too slowly, it tends to slide down the icy incline under its own weight. Friction acts upwards along the slope to prevent this.
• Resolving Vertically (\(\uparrow = \downarrow\)):
\(R\cos\theta + F\sin\theta = mg\)
• Resolving Horizontally towards centre (\(\rightarrow\)):
\(R\sin\theta - F\cos\theta = \frac{mv_{\text{min}}^2}{r}\)
At the minimum speed before sliding down occurs, friction reaches its limiting value: \(F = \mu R\).
Crucial Examiner Warning: Always resolve horizontally and vertically for circular motion on banked tracks! Do not resolve perpendicular to the inclined track. Because the acceleration is strictly horizontal, resolving perpendicular to the slope often incorrectly leads students to write \(R = mg\cos\theta\), which is wrong here!
---4. Sliding vs Overturning (Toppling) on Bends
In advanced mechanics, vehicles are no longer treated as simple point particles. A vehicle is a rigid body with dimensions: a track width of \(2d\) (distance between left and right wheels) and a centre of mass located at a height \(h\) above the road surface.
Why Do Vehicles Topple?
When a vehicle negotiates a horizontal curve of radius \(r\) at speed \(v\):
• The normal reactions at the inner wheels (\(R_1\)) and outer wheels (\(R_2\)) are not equal.
• As the speed increases, the vehicle tries to roll outwards. The inner wheels lift, meaning \(R_1\) decreases while \(R_2\) increases.
1. The Condition for Overturning (Toppling Outwards)
At the point of overturning, the inner wheels lose contact with the road entirely:
Condition for toppling: \(R_1 = 0\)
To find the critical overturning speed \(v_{\text{topple}}\) on a flat road:
• Take moments about the point of contact of the outer wheels (where reaction \(R_2\) and friction act):
Clockwise restoring moment from weight = Counter-clockwise overturning moment from effective inertia
\(mg \cdot d = \left(\frac{mv^2}{r}\right) \cdot h \implies v_{\text{topple}} = \sqrt{\frac{grd}{h}}\)
2. The Condition for Sliding (Skidding Outwards)
The vehicle will slide sideways when the required centripetal force exceeds the total maximum static friction available:
\(F_{\text{max}} = \mu (R_1 + R_2) = \mu mg\)
Setting \(\frac{mv^2}{r} = \mu mg\) yields the critical sliding speed:
\(v_{\text{slide}} = \sqrt{\mu g r}\)
3. Which Happens First?
To determine whether a vehicle skids or topples first as it speeds up, compare the two critical speeds:
• Vehicle will SLIDE first if: \(v_{\text{slide}} < v_{\text{topple}} \implies \mu < \frac{d}{h}\)
• Vehicle will TOPPLE first if: \(v_{\text{topple}} < v_{\text{slide}} \implies \mu > \frac{d}{h}\)
Real-World Insight: Tall vehicles like double-decker buses or transit vans have a large height \(h\) and a relatively small half-width \(d\), making \(\frac{d}{h}\) small. On high-grip roads (large \(\mu\)), they are prone to toppling over before they skid!
---5. Step-by-Step Worked Example
Question: A car of mass \(1200\text{ kg}\) travels around a horizontal circular track of radius \(r = 50\text{ m}\) banked at an angle of \(\theta = 20^\circ\) to the horizontal. The coefficient of friction between the tyres and the track is \(\mu = 0.35\). Calculate the maximum speed at which the car can travel without slipping up the track. Take \(g = 9.8\text{ m s}^{-2}\).
Step 1: Identify the limiting state and draw a force diagram.
The car is traveling at maximum speed, so it tends to slip UP the incline. Therefore, limiting friction \(F = \mu R\) acts down the incline.
Step 2: Set up the vertical equilibrium equation.
\(\Sigma F_y = 0 \implies R\cos(20^\circ) - F\sin(20^\circ) = mg\)
Since \(F = 0.35R\):
\(R\cos(20^\circ) - 0.35R\sin(20^\circ) = 1200(9.8)\)
\(R\left(\cos(20^\circ) - 0.35\sin(20^\circ)\right) = 11760\)
\(R(0.93969 - 0.35 \times 0.34202) = 11760\)
\(R(0.93969 - 0.11971) = 11760\)
\(R(0.81998) = 11760 \implies R \approx 14341.8\text{ N}\)
Step 3: Set up the horizontal radial equation of motion.
\(\Sigma F_x = \frac{mv^2}{r} \implies R\sin(20^\circ) + F\cos(20^\circ) = \frac{mv_{\text{max}}^2}{r}\)
Substitute \(F = 0.35R\):
\(R\left(\sin(20^\circ) + 0.35\cos(20^\circ)\right) = \frac{1200 v_{\text{max}}^2}{50}\)
\(14341.8 \times \left(0.34202 + 0.35 \times 0.93969\right) = 24 v_{\text{max}}^2\)
\(14341.8 \times (0.34202 + 0.32889) = 24 v_{\text{max}}^2\)
\(14341.8 \times 0.67091 = 24 v_{\text{max}}^2\)
\(9622.07 = 24 v_{\text{max}}^2 \implies v_{\text{max}}^2 \approx 400.92\)
\(v_{\text{max}} = \sqrt{400.92} \approx 20.023\text{ m s}^{-1}\)
Step 4: State final answer with correct units and significant figures.
The maximum speed is \(20.0\text{ m s}^{-1}\) (to 3 s.f.).
6. Summary of Common Pitfalls to Avoid
• Inventing Centripetal Force: Never draw a force labeled "\(F_c\)" or "Centripetal Force" on your Free-Body Diagram! Centripetal force is merely the vector sum of real forces (friction, normal reaction, tension) pointing towards the centre.
• Wrong Direction for Friction: Remember: at maximum speed, friction points down the slope; at minimum speed, friction points up the slope.
• Horizontal Radius vs Slanted Length: In conical pendulum questions, make sure to use the horizontal radius \(r = L\sin\theta\), not the string length \(L\), when calculating \(mr\omega^2\).
• Toppling Equilibrium Condition: When testing for overturning, always set the normal reaction of the inner wheel to zero (\(R_{\text{inner}} = 0\)).