Physics • Year 12 • Module 5 • Lesson 8
Horizontal Circular Motion
Apply force-diagram reasoning and formula selection to real data sets, cause-and-effect problems, and a comparative analysis of flat vs banked curves.
1. Interpret experimental data — conical pendulum survey
A student set up five different conical pendulums and recorded the data below. The string length was 0.60 m in every trial. 8 marks
| Trial | Angle θ (degrees) | Period T (s) — measured | Period T (s) — calculated using 2π√(L cos θ/g) | % difference |
|---|---|---|---|---|
| 1 | 10 | 1.52 | ||
| 2 | 20 | 1.49 | ||
| 3 | 30 | 1.42 | ||
| 4 | 45 | 1.29 | ||
| 5 | 60 | 1.07 |
1.1 Complete the “calculated” column using T = 2π√(L cos θ/g) with L = 0.60 m and g = 9.8 m s−2. Show one full working for Trial 3. 6 marks (1 per row, including working shown for Trial 3)
1.2 Describe the trend in period as the angle increases, and explain it physically in terms of the vertical component of tension. 2 marks
2. Interpret graph — maximum speed vs radius on wet and dry roads
The graph below shows how the maximum safe speed on a flat unbanked curve changes with radius for two road conditions. 7 marks
Figure 2. Maximum safe speed on a flat unbanked curve vs curve radius for dry road (μs = 0.70) and wet road (μs = 0.30). Calculated using vmax = √(μsgr). Illustrative data.
2.1 Describe the shape of each curve and explain, using the formula vmax = √(μsgr), why the graph is not a straight line. 2 marks
2.2 Using the graph, estimate the maximum safe speed (in km/h) on a wet road for a curve of radius 200 m. Show how you read the graph. 2 marks
2.3 A driver approaches a flat curve of radius 80 m on a dry road at 80 km/h. Using vmax = √(μsgr), calculate whether this is safe and explain the consequence of exceeding vmax. 3 marks
3. Compare flat and banked curves across five features
Complete the two-column table for a curve of radius 100 m with μs = 0.40, comparing a flat unbanked curve with a banked curve at θ = 15°. 10 marks (1 per cell)
| Feature | Flat curve (unbanked) | Banked curve (θ = 15°) |
|---|---|---|
| Forces providing Fc | ||
| Normal force magnitude | ||
| Direction of friction when v = vdesign | ||
| Maximum safe speed (km/h) — calculated | ||
| Safety on an icy surface (μs ≈ 0) |
4. Predict and justify — Daytona banking scenario
Daytona International Speedway has turns banked at 31° with a turn radius of approximately 300 m. Under race conditions the track surface can be treated as having μs ≈ 0.90 (rubber on asphalt). 5 marks
4.1 Calculate the design speed in km/h for Daytona’s banked turns. 2 marks
4.2 A car travels at 200 km/h (55.6 m s−1) through the Daytona turn. Predict the direction friction acts and justify your answer with reference to the design speed. 2 marks
4.3 If the track became perfectly frictionless (μs = 0), predict the only speed at which a car could safely navigate the turn. 1 mark
Q1.1 — Calculated periods (6 marks — 1 per row + working)
Formula: T = 2π√(L cos θ/g) with L = 0.60 m, g = 9.8 m s−2.
Trial 1 (θ = 10°): T = 2π√(0.60 × cos 10° / 9.8) = 2π√(0.60 × 0.985 / 9.8) = 2π√(0.06031) = 2π × 0.2456 = 1.54 s
Trial 2 (θ = 20°): T = 2π√(0.60 × 0.940 / 9.8) = 2π√(0.05755) = 2π × 0.2399 = 1.51 s
Trial 3 (θ = 30°) [full working required]: T = 2π√(0.60 × cos 30° / 9.8) = 2π√(0.60 × 0.8660 / 9.8) = 2π√(0.05306) = 2π × 0.2303 = 1.45 s
Trial 4 (θ = 45°): T = 2π√(0.60 × 0.7071 / 9.8) = 2π√(0.04330) = 2π × 0.2081 = 1.31 s
Trial 5 (θ = 60°): T = 2π√(0.60 × 0.5000 / 9.8) = 2π√(0.03061) = 2π × 0.1750 = 1.10 s
Q1.2 — Trend explanation (2 marks)
As the angle θ increases, the period decreases [1]. Physically, increasing θ reduces L cos θ (the effective “pendulum length”). A shorter effective length means the bob completes each revolution faster, so the period is smaller. The vertical component of tension T cos θ must equal mg; as θ increases, the string tilts more, the vertical component decreases for a given T, so to maintain balance the bob must orbit faster and thus the period shortens [1].
Q2.1 — Curve shape (2 marks)
Both curves rise steeply at small radii and flatten out at larger radii — each has the shape of a square-root function [1]. Because vmax = √(μsgr), v is proportional to √r, not r. Doubling r increases v by a factor of √2 (≈ 1.41), so equal increases in radius give diminishing increases in maximum speed, producing the characteristic concave-down curve [1].
Q2.2 — Reading graph for wet road, r = 200 m (2 marks)
From the graph, reading up from r = 200 m to the wet-road line and across to the y-axis gives approximately 87 km/h [1]. Calculated check: vmax = √(0.30 × 9.8 × 200) × 3.6 = √(588) × 3.6 = 24.25 × 3.6 = 87.3 km/h (graph reading consistent) [1].
Q2.3 — 80 km/h on dry road r = 80 m (3 marks)
vmax = √(0.70 × 9.8 × 80) = √(548.8) = 23.4 m s−1 = 84.3 km/h [1].
Since 80 km/h < 84.3 km/h, the speed is within the safe limit on a dry road [1]. If the driver exceeded vmax, the required centripetal force would exceed the maximum friction force available. The car would skid outward, continuing on a path closer to a straight line rather than following the curve — it would leave the road [1].
Q3 — Compare flat vs banked (10 marks)
Forces providing Fc: Flat: friction only. Banked: horizontal component of normal force N sin θ, plus friction if v ≠ vdesign.
Normal force: Flat: N = mg. Banked: N = mg / cos 15° = mg / 0.966 ≈ 1.035 mg (greater than mg).
Direction of friction when v = vdesign: Flat: friction always acts inward (there is no design speed concept). Banked: no friction at all at design speed.
Maximum safe speed: Flat: vmax = √(0.40 × 9.8 × 100) = √(392) = 19.8 m s−1 = 71.3 km/h. Banked: vmax = √{100 × 9.8 × (tan 15° + 0.40) / (1 − 0.40 × tan 15°)} = √{980 × (0.268 + 0.40) / (1 − 0.40 × 0.268)} = √{980 × 0.668 / 0.893} = √{733.6} = 27.1 m s−1 = 97.6 km/h. Banked is higher.
Safety on icy surface (μs ≈ 0): Flat: vmax ≈ 0; any speed causes skidding. Banked: vdesign = √(rg tan θ) = √(100 × 9.8 × 0.268) = √(262.6) = 16.2 m s−1 = 58.3 km/h; the car can be navigated safely at exactly the design speed even on ice.
Q4.1 — Daytona design speed (2 marks)
vdesign = √(rg tan θ) = √(300 × 9.8 × tan 31°) = √(300 × 9.8 × 0.6009) [1] = √(1766.6) = 42.0 m s−1 = 151 km/h [1].
Q4.2 — Friction direction at 200 km/h (2 marks)
The car is travelling faster than the design speed (200 km/h > 151 km/h) [1]. When v > vdesign, the car tends to slide up and outward on the banked surface, so friction acts down the slope (toward the inside of the turn) to provide additional centripetal force and prevent the car from sliding up [1].
Q4.3 — Frictionless speed (1 mark)
On a frictionless banked surface, the only safe speed is the design speed: vdesign = 151 km/h (42.0 m s−1). At any other speed, the car would slide either up or down the banking [1].