Analyse Conical Pendulum Motion
- Resolve tension into vertical and horizontal components
- Relate the conical pendulum angle to speed and radius
- Derive and use the period formula for a conical pendulum
Derive and Use the Banked Curve Formula
- Analyse forces on a banked curve with and without friction
- Derive the design speed formula from force components
- Solve for maximum safe speed on banked and flat curves
Solve Problems Involving Friction on Curved Paths
- Determine the direction of friction on banked curves
- Calculate maximum speed on flat and banked curves
- Evaluate the safety implications of banking angle choices
Why do motorcyclists lean into turns? What would happen if they didn't?
Horizontal Circular Motion
Analyse conical pendulums, banked curves, and situations where friction, tension, or normal force provide the centripetal force. Derive the banking angle formula.
Conical Pendulum
A bob on a string moving in a horizontal circle
Horizontal Circular Motion
A conical pendulum consists of a bob of mass $m$ attached to a string of length $L$, moving in a horizontal circle with the string at constant angle $\theta$ to the vertical. The bob traces out a circular path of radius $r = L\sin\theta$.
Two forces act on the bob: tension $T$ along the string, and weight $mg$ downward. The tension has two components:
- Vertical: $T\cos\theta$ balances the weight: $T\cos\theta = mg$
- Horizontal: $T\sin\theta$ provides the centripetal force: $T\sin\theta = \dfrac{mv^2}{r}$
Dividing the horizontal equation by the vertical equation:
$$\frac{T\sin\theta}{T\cos\theta} = \frac{mv^2/r}{mg}$$ $$\tan\theta = \frac{v^2}{rg}$$This is the fundamental relationship for a conical pendulum. Solving for speed:
$$v = \sqrt{rg\tan\theta}$$The period $T_{\text{period}}$ (time for one complete revolution) can be found from $v = \dfrac{2\pi r}{T_{\text{period}}}$. Substituting and simplifying:
$$T_{\text{period}} = 2\pi\sqrt{\frac{L\cos\theta}{g}}$$Worked Example — Conical Pendulum
A conical pendulum has a bob of mass 0.20 kg on a string of length 0.50 m. The string makes an angle of 30° with the vertical. Find the speed of the bob, the tension in the string, and the period of motion.
$m = 0.20 \text{ kg}$, $L = 0.50 \text{ m}$, $\theta = 30°$, $g = 9.8 \text{ m s}^{-2}$
Radius: $r = L\sin\theta = 0.50 \times \sin 30° = 0.25 \text{ m}$
$v$ (speed), $T$ (tension), $T_{\text{period}}$ (period)
Use $T\cos\theta = mg$ for tension, $\tan\theta = v^2/(rg)$ for speed, and $T_{\text{period}} = 2\pi\sqrt{L\cos\theta/g}$ for period.
Tension: $T = \dfrac{mg}{\cos\theta} = \dfrac{0.20 \times 9.8}{\cos 30°} = \dfrac{1.96}{0.866} = 2.26 \text{ N}$
Speed: $v = \sqrt{rg\tan\theta} = \sqrt{0.25 \times 9.8 \times \tan 30°} = \sqrt{0.25 \times 9.8 \times 0.577} = \sqrt{1.41} = 1.19 \text{ m s}^{-1}$
Period: $T_{\text{period}} = 2\pi\sqrt{\dfrac{L\cos\theta}{g}} = 2\pi\sqrt{\dfrac{0.50 \times \cos 30°}{9.8}} = 2\pi\sqrt{0.0442} = 2\pi \times 0.210 = 1.32 \text{ s}$
Banked Curves (No Friction)
When the normal force alone provides the centripetal force component
A banked curve is a road or track inclined at angle $\theta$ to the horizontal. When a vehicle travels at exactly the design speed, the normal force $N$ from the road surface provides precisely the required centripetal force, and no friction is needed.
Resolving the normal force:
- Vertical: $N\cos\theta = mg$ (balances weight)
- Horizontal: $N\sin\theta = \dfrac{mv^2}{r}$ (provides centripetal force)
Dividing the horizontal by the vertical equation:
$$\tan\theta = \frac{v^2}{rg}$$Solving for the design speed:
$$v_{\text{design}} = \sqrt{rg\tan\theta}$$At the design speed, no friction is required. The normal force alone provides exactly the right centripetal force. This is why icy banked curves can still be navigated safely at the correct speed.
Worked Example — Design Speed on a Banked Curve
A highway curve of radius 80 m is banked at 12°. Calculate the design speed at which no friction is required.
$r = 80 \text{ m}$, $\theta = 12°$, $g = 9.8 \text{ m s}^{-2}$
$v_{\text{design}}$
Use $v_{\text{design}} = \sqrt{rg\tan\theta}$
$v_{\text{design}} = \sqrt{80 \times 9.8 \times \tan 12°} = \sqrt{80 \times 9.8 \times 0.213} = \sqrt{166.9} = 12.9 \text{ m s}^{-1}$
$v_{\text{design}} = 12.9 \times 3.6 = 46.4 \text{ km/h}$
Banked Curves (With Friction)
Friction acts up or down the slope depending on speed
When a vehicle travels at a speed different from the design speed, friction is required to prevent sliding. The direction of friction depends on whether the vehicle is travelling faster or slower than $v_{\text{design}}$:
- $v > v_{\text{design}}$ (too fast): The vehicle tends to slide up the slope. Friction acts down the slope, adding to the centripetal force.
- $v < v_{\text{design}}$ (too slow): The vehicle tends to slide down the slope. Friction acts up the slope, reducing the net centripetal force.
Maximum Speed (Friction Acting Down the Slope)
At maximum speed, friction acts at its limiting value $F_f = \mu_s N$ down the slope. Resolving forces parallel and perpendicular to the surface:
Perpendicular to surface: $N = mg\cos\theta + \dfrac{mv^2_{\max}}{r}\sin\theta$
Parallel to surface (down the slope, friction adds to the horizontal component):
$$N\sin\theta + \mu_s N\cos\theta = \frac{mv^2_{\max}}{r}$$After algebraic manipulation (dividing through by $\cos\theta$ and substituting):
$$v_{\max} = \sqrt{rg\left(\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}\right)}$$You must derive this formula in exam questions unless told otherwise. Do not memorise the final form — memorise the force diagram and the method of resolving.
Worked Example — Maximum Speed with Friction
A banked curve has radius 100 m and banking angle 8°. The coefficient of static friction between tyres and road is 0.25. Calculate the maximum safe speed.
$r = 100 \text{ m}$, $\theta = 8°$, $\mu_s = 0.25$, $g = 9.8 \text{ m s}^{-2}$
$v_{\max}$
Use $v_{\max} = \sqrt{rg\left(\dfrac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}\right)}$ with $\tan 8° = 0.141$
$v_{\max} = \sqrt{100 \times 9.8 \times \left(\dfrac{0.141 + 0.25}{1 - 0.25 \times 0.141}\right)}$
$v_{\max} = \sqrt{980 \times \left(\dfrac{0.391}{0.965}\right)} = \sqrt{980 \times 0.405} = \sqrt{397} = 19.9 \text{ m s}^{-1}$
$v_{\max} = 19.9 \times 3.6 = 71.6 \text{ km/h}$
Friction on Flat Curves
The limiting case: no banking, friction only
On a flat (unbanked) curve, the only horizontal force available to provide centripetal force is friction between the tyres and the road. At the maximum speed before slipping:
$$F_c = F_{\text{friction}} = \mu_s N = \mu_s mg$$Since $F_c = \dfrac{mv^2_{\max}}{r}$:
$$\frac{mv^2_{\max}}{r} = \mu_s mg$$$$v_{\max} = \sqrt{\mu_s gr}$$
This is why icy curves are so dangerous. When $\mu_s$ drops to near zero, $v_{\max}$ drops to near zero — any speed causes the car to slide outward. Notice that $v_{\max}$ does not depend on the mass of the vehicle.
Worked Example — Flat Curve Maximum Speed
A car travels around a flat unbanked curve of radius 30 m. The coefficient of static friction between the tyres and the road is 0.60. Calculate the maximum speed without slipping.
$\mu_s = 0.60$, $r = 30 \text{ m}$, $g = 9.8 \text{ m s}^{-2}$
$v_{\max}$
Use $v_{\max} = \sqrt{\mu_s gr}$
$v_{\max} = \sqrt{0.60 \times 9.8 \times 30} = \sqrt{176.4} = 13.3 \text{ m s}^{-1}$
$v_{\max} = 13.3 \times 3.6 = 47.9 \text{ km/h}$
Essential Formulae — Horizontal Circular Motion
Common Misconceptions
"Banking is only needed for high-speed racing" — No. Even gentle highway curves are banked for safety at normal speeds. The banking angle is calculated for the expected traffic speed.
"Friction always helps on banked curves" — No. At exactly the design speed, no friction is needed at all. Friction only becomes necessary when the speed differs from the design speed.
Daytona International Speedway
Daytona International Speedway features 31° banking in the turns, with a turn radius of approximately 300 m. The design speed is:
$$v_{\text{design}} = \sqrt{300 \times 9.8 \times \tan 31°} = \sqrt{300 \times 9.8 \times 0.601} = \sqrt{1767} = 42.0 \text{ m s}^{-1}$$
$$v_{\text{design}} = 42.0 \times 3.6 = 151 \text{ km/h}$$
At this speed, no friction is needed between the tyres and the track. The normal force from the banked surface alone provides exactly the required centripetal force. Cars can theoretically maintain their circular path even on a perfectly smooth, frictionless track surface. This is why steep banking allows such high-speed cornering.
Conical Pendulum and Banking Calculations
1 A conical pendulum has string length 0.40 m and makes an angle of 25° with the vertical. Calculate the period of motion.
2 A highway curve of radius 150 m is banked at 5°. Calculate the design speed in km/h.
3 A car travels around a flat unbanked curve of radius 20 m. The coefficient of static friction is 0.50. Calculate the maximum safe speed in km/h.
Direction of Friction on Banked Curves
A car is driving on a banked curve at a speed slower than the design speed. Explain which way friction acts and why. Then repeat for a car driving faster than the design speed. Use a diagram in your explanation.
📝 Copy into Books
▾Conical Pendulum
- $T\cos\theta = mg$ (vertical balance)
- $T\sin\theta = mv^2/r$ (horizontal)
- $\tan\theta = v^2/(rg)$
- $T_{\text{period}} = 2\pi\sqrt{L\cos\theta/g}$
- $r = L\sin\theta$
Banked Curve (No Friction)
- $N\cos\theta = mg$
- $N\sin\theta = mv^2/r$
- $\tan\theta = v^2/(rg)$
- $v_{\text{design}} = \sqrt{rg\tan\theta}$
Banked Curve (With Friction)
- $v > v_{\text{design}}$: friction acts DOWN slope
- $v < v_{\text{design}}$: friction acts UP slope
- $v = v_{\text{design}}$: no friction needed
- $v_{\max} = \sqrt{rg(\tan\theta + \mu_s)/(1 - \mu_s\tan\theta)}$
Flat Curve
- $F_c = F_{\text{friction}} = \mu_s mg$
- $v_{\max} = \sqrt{\mu_s gr}$
- Independent of mass
- Icy roads: $\mu_s \approx 0$, so $v_{\max} \approx 0$
Revisit Your Prediction
Now that you have studied horizontal circular motion, revisit your prediction about motorcyclists leaning into turns.
When a motorcyclist leans into a turn, the normal force from the road tilts. The horizontal component of this normal force provides the centripetal force needed for circular motion. Leaning also lowers the centre of mass, increasing stability. If the rider did not lean, the only horizontal force would be friction, which may be insufficient — the motorcycle would slide outward (continue in a straight line due to inertia). At low speeds, a rider can turn without much lean because friction alone can provide the small centripetal force needed. At high speeds, a large lean angle is required to generate sufficient horizontal force component.
Write a revised explanation:
1. For a conical pendulum, increasing the speed of the bob will:
2. On a banked curve at the design speed:
3. A car drives slower than the design speed on a banked curve. Friction acts:
4. For the same radius and coefficient of friction, a banked curve allows:
5. The maximum speed on a flat unbanked curve depends on:
1. A conical pendulum has a bob of mass 0.50 kg on a string of length 0.80 m. The bob moves in a horizontal circle with the string at 35° to the vertical. Calculate the tension in the string and the speed of the bob.
2. A highway curve of radius 120 m is banked at 8°. Calculate the design speed. If the coefficient of static friction between the road and tyres is 0.30, calculate the maximum safe speed.
3. Assess whether increasing the banking angle always increases safety on a highway curve. Consider the effects on vehicles travelling at different speeds.
Answers
▾Activity 1: Conical Pendulum Period
$T_{\text{period}} = 2\pi\sqrt{\dfrac{L\cos\theta}{g}} = 2\pi\sqrt{\dfrac{0.40 \times \cos 25°}{9.8}} = 2\pi\sqrt{\dfrac{0.40 \times 0.906}{9.8}}$
$T_{\text{period}} = 2\pi\sqrt{0.0370} = 2\pi \times 0.192 = \mathbf{1.21 \text{ s}}$ (accept 1.20–1.22 s)
Activity 2: Highway Design Speed
$v_{\text{design}} = \sqrt{rg\tan\theta} = \sqrt{150 \times 9.8 \times \tan 5°} = \sqrt{150 \times 9.8 \times 0.0875}$
$v_{\text{design}} = \sqrt{128.6} = 11.3 \text{ m s}^{-1} = \mathbf{40.8 \text{ km/h}}$
Activity 3: Flat Curve Maximum Speed
$v_{\max} = \sqrt{\mu_s gr} = \sqrt{0.50 \times 9.8 \times 20} = \sqrt{98} = 9.90 \text{ m s}^{-1}$
$v_{\max} = 9.90 \times 3.6 = \mathbf{35.6 \text{ km/h}}$
Multiple Choice Answers
1. B ($\tan\theta = v^2/(rg)$, so increasing $v$ increases $\theta$)
2. B (At design speed, normal force alone provides centripetal force)
3. B (When $v < v_{\text{design}}$, friction acts up the slope)
4. C (Banking adds the normal force component, increasing $v_{\max}$)
5. B ($v_{\max} = \sqrt{\mu_s gr}$; mass cancels out)
SA1: Conical Pendulum Tension and Speed (3 marks)
Radius: $r = L\sin\theta = 0.80 \times \sin 35° = 0.80 \times 0.574 = 0.459 \text{ m}$ (1 mark)
Vertical: $T\cos\theta = mg$
$T = \dfrac{mg}{\cos\theta} = \dfrac{0.50 \times 9.8}{\cos 35°} = \dfrac{4.90}{0.819} = \mathbf{5.98 \text{ N}}$ (1 mark)
Horizontal: $\tan\theta = \dfrac{v^2}{rg}$
$v = \sqrt{rg\tan\theta} = \sqrt{0.459 \times 9.8 \times \tan 35°} = \sqrt{0.459 \times 9.8 \times 0.700}$
$v = \sqrt{3.15} = \mathbf{1.77 \text{ m s}^{-1}}$ (1 mark)
Alternative: Use $T\sin\theta = mv^2/r$ directly: $v = \sqrt{Tr\sin\theta/m} = \sqrt{5.98 \times 0.459 \times \sin 35°/0.50} = 1.77 \text{ m s}^{-1}$
SA2: Highway Curve Design and Maximum Speed (3 marks)
Design speed:
$v_{\text{design}} = \sqrt{rg\tan\theta} = \sqrt{120 \times 9.8 \times \tan 8°} = \sqrt{120 \times 9.8 \times 0.141}$ (1 mark)
$v_{\text{design}} = \sqrt{165.8} = 12.9 \text{ m s}^{-1} = \mathbf{46.4 \text{ km/h}}$ (1 mark)
Maximum speed with friction:
$v_{\max} = \sqrt{rg\left(\dfrac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}\right)} = \sqrt{120 \times 9.8 \times \left(\dfrac{0.141 + 0.30}{1 - 0.30 \times 0.141}\right)}$
$v_{\max} = \sqrt{1176 \times \left(\dfrac{0.441}{0.958}\right)} = \sqrt{1176 \times 0.460} = \sqrt{541} = 23.3 \text{ m s}^{-1}$
$v_{\max} = \mathbf{83.8 \text{ km/h}}$ (1 mark)
SA3: Assessing Banking Angle and Safety (4 marks)
Increasing the banking angle increases the design speed ($v_{\text{design}} = \sqrt{rg\tan\theta}$), which benefits fast-moving traffic by reducing their reliance on friction. (1 mark)
However, a steeper banking angle can be hazardous for slow-moving traffic (lorries, vehicles in wet conditions, or those slowing to turn). When $v < v_{\text{design}}$, friction must act up the slope to prevent sliding down. If the angle is too steep, slow vehicles may slide inward. (1 mark)
Very steep banking (like a velodrome at 42° or Daytona at 31°) requires vehicles to travel at minimum speeds to avoid sliding down the slope. This is unsuitable for public roads where traffic speeds vary widely. (1 mark)
Therefore, the optimal banking angle balances the speed distribution of expected traffic. Highway engineers choose angles (typically 3–8°) that provide safety for the majority of vehicles at normal highway speeds, rather than maximising the angle. (1 mark)
Horizontal Circular Motion Challenge
Test your knowledge of conical pendulums, banked curves, and friction under time pressure. Beat the boss to unlock your mastery badge.
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