Year 12 Physics Module 5 ⏱ ~45 min 5 MC · 3 Short Answer Lesson 8 of 18

Horizontal Circular Motion

A banked bend tilts the normal force so that part of it points toward the centre of the circular path. This lesson uses free-body diagrams to analyse conical pendulums, design speed, and how static friction changes direction above and below that speed.

Today's hook: At one particular speed on a banked bend, no friction is required. Why does tilting the road make the normal force provide the radial resultant, and which way must friction act when the car is slower or faster than that design speed?
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Connect and orient

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response
Before you read, predict

Why do motorcyclists lean into turns? What would happen if they didn't?

Consider what forces are available on a curved road. Could the motorcycle navigate the turn without leaning?

Warm-up, on a flat (unbanked) horizontal curve, which force provides the centripetal force for a car?

Learning Intentions
goals

Know, 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 $T_\text{period}=2\pi\sqrt{L\cos\theta/g}$

Understand, 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

Can Do, Solve Problems with 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
Scan these before reading
vocab
Conical pendulumA bob on a string moving in a horizontal circle with the string at a constant angle $\theta$ to the vertical.
Banked curveA road or track inclined at angle $\theta$ to the horizontal, allowing normal force to contribute a horizontal component toward the centre.
Banking angleThe angle $\theta$ at which a road is tilted; the design angle satisfies $\tan\theta = v^2/(rg)$.
Design speedThe speed at which no friction is needed on a banked curve: $v_\text{design} = \sqrt{rg\tan\theta}$.
Coefficient of friction ($\mu_s$)The ratio of maximum static friction to normal force. $F_{f,\max} = \mu_s N$.
Normal forceThe perpendicular contact force from a surface. On a banked curve: $N = mg/\cos\theta > mg$.
Centripetal forceThe net force directed toward the centre of a circular path: $F_c = mv^2/r$.
Cross-lesson links: L07 applied centripetal force to simple circular motion. L08 extends this to inclined surfaces and three-dimensional systems, banked curves, conical pendulums, and loops require you to resolve forces in two directions simultaneously, which is the highest-difficulty circular motion skill in HSC Physics.
Misconceptions to fix
✗ Wrong: "Banking is only needed for high-speed racing."
✓ Right: Even gentle highway curves are banked for safety at normal speeds. The banking angle is calculated for the expected traffic speed so that vehicles require little or no friction to navigate the curve safely.
✗ Wrong: "Friction always helps on banked curves."
✓ Right: At exactly the design speed, no friction is needed at all. Friction only becomes necessary when the speed differs from the design speed, and it can act either up or down the slope depending on whether you are too fast or too slow.
✗ Wrong: "The normal force equals weight on a banked curve."
✓ Right: $N\cos\theta = mg$, so $N = mg/\cos\theta > mg$. The normal force is greater than weight because the road must support the vehicle against both gravity (upward component) and the centripetal requirement (horizontal component).

On a banked curve at the design speed, friction provides all the centripetal force.

The normal force on a banked curve equals the weight of the vehicle.

Core Content

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Build the conical-pendulum FBD

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Conical Pendulum
+5 XP

A bob on a string moving in a horizontal circle

Horizontal Circular Motion, conical pendulum and banked curve diagrams

Horizontal circular motion, key situations: conical pendulum, banked curve (no friction), and banked curve with friction.

Tie a ball to a string and spin it above your head: the string traces a cone and the ball sweeps a horizontal circle at constant height. The string pulls the ball upward (supporting it against gravity) and inward (providing centripetal force). These two jobs are done by a single tension force at an angle, and resolving that tension into vertical and horizontal components lets you calculate the speed, radius, and period of the orbit.

Draw the free-body diagram first. Two forces act on the bob: string tension $F_T$ along the string and weight $mg$ downward. Resolving $F_T$:

  • Vertical: $F_T\cos\theta = mg$ (zero vertical acceleration)
  • Horizontal radial: $F_T\sin\theta = \dfrac{mv^2}{r}$ (inward resultant)

Dividing the radial equation by the vertical equation eliminates $F_T$:

$$\tan\theta = \frac{v^2}{rg}$$

This is the fundamental relationship for a conical pendulum. Solving for speed: $v = \sqrt{rg\tan\theta}$.

The period (time for one complete revolution) uses $v = 2\pi r/T_\text{period}$ and $r = L\sin\theta$:

$$T_\text{period} = 2\pi\sqrt{\frac{L\cos\theta}{g}}$$
Conical Pendulum Relationships

$r = L\sin\theta$, radius of circular path

$\tan\theta = v^2/(rg)$, angle–speed relationship

$T_\text{period} = 2\pi\sqrt{L\cos\theta/g}$, period

Tension: $F_T = mg/\cos\theta$

Worked Example 1, Conical Pendulum

A conical pendulum has a bob of mass 0.20 kg on a string of length 0.50 m. The string makes 30° with the vertical. Find the speed of the bob, the tension, and the period of motion.

  1. Given. $m = 0.20\ \text{kg}$, $L = 0.50\ \text{m}$, $\theta = 30°$, $g = 9.8\ \text{m s}^{-2}$. Radius: $r = L\sin\theta = 0.50 \times \sin 30° = 0.25\ \text{m}$.
  2. Find. $v$, $F_T$ (tension), $T_\text{period}$ (period).
  3. Method. Use $F_T = mg/\cos\theta$, then $\tan\theta = v^2/(rg)$, then $T_\text{period} = 2\pi\sqrt{L\cos\theta/g}$.
  4. Tension. $F_T = \dfrac{0.20 \times 9.8}{\cos 30°} = \dfrac{1.96}{0.866} = \mathbf{2.26\ \text{N}}$
  5. Speed. $v = \sqrt{rg\tan\theta} = \sqrt{0.25 \times 9.8 \times \tan 30°} = \sqrt{0.25 \times 9.8 \times 0.577} = \sqrt{1.41} = \mathbf{1.19\ \text{m s}^{-1}}$
  6. Period. $T_\text{period} = 2\pi\sqrt{\dfrac{0.50 \times \cos 30°}{9.8}} = 2\pi\sqrt{\dfrac{0.50 \times 0.866}{9.8}} = 2\pi\sqrt{0.0442} = 2\pi \times 0.210 = \mathbf{1.32\ \text{s}}$

Conical pendulum: $r=L\sin\theta$; vertical $F_T\cos\theta=mg$; radial $F_T\sin\theta=mv^2/r$; hence $\tan\theta=v^2/(rg)$. Period $T_\text{period}=2\pi\sqrt{L\cos\theta/g}$. Keep $F_T$ distinct from period.

Pause, copy the highlighted conical pendulum equations into your book before moving on.

For a conical pendulum, increasing the speed of the bob will:

Explore the conical pendulum
interactive

Vary the string angle and connect the free-body diagram to radius, speed, tension and period.

Open fullscreen ↗
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Build the banked-curve FBD

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Banked Curves (No Friction)
+5 XP

When the normal force alone provides the centripetal force component

We just saw that the conical pendulum uses $\tan\theta = v^2/(rg)$ from resolving tension. That raises a question: do roads use the same idea, can banking replace the need for friction? This card answers it → yes; at design speed, $N\sin\theta$ provides all centripetal force and $v_\text{design} = \sqrt{rg\tan\theta}$.

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, no friction is needed.

Draw the free-body diagram first: only weight and the normal force act in the frictionless design-speed model. Resolve the normal force:

  • Vertical: $N\cos\theta = mg$ (balances weight)
  • Horizontal: $N\sin\theta = \dfrac{mv^2}{r}$ (provides centripetal force)

Dividing horizontal by vertical eliminates $N$:

$$\tan\theta = \frac{v^2}{rg}$$

This is mathematically identical to the conical pendulum equation. Solving for the design speed:

$$v_\text{design} = \sqrt{rg\tan\theta}$$
Key Insight

At the design speed, 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, no friction is required. This is why steep banking allows very high-speed cornering in motorsport.

Worked Example 2, 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.

  1. Given. $r = 80\ \text{m}$, $\theta = 12°$, $g = 9.8\ \text{m s}^{-2}$.
  2. Find. $v_\text{design}$.
  3. Method. Use $v_\text{design} = \sqrt{rg\tan\theta}$.
  4. Solve. $v_\text{design} = \sqrt{80 \times 9.8 \times \tan 12°} = \sqrt{80 \times 9.8 \times 0.213} = \sqrt{166.9} = \mathbf{12.9\ \text{m s}^{-1}}$ (46.4 km/h).

Banked curve (no friction): $N\cos\theta = mg$; $N\sin\theta = mv^2/r$; dividing → $\tan\theta = v^2/(rg)$. Design speed $v_\text{design} = \sqrt{rg\tan\theta}$. Normal force $N = mg/\cos\theta > mg$.

Add the highlighted banking equations and the $N > mg$ result to your notes before the check below.

The formula $\tan\theta = v^2/(rg)$ applies to both a conical pendulum and a frictionless banked curve.

On a banked curve at the design speed, the normal force is greater than the weight of the vehicle.

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Infer friction direction and limits

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Banked Curves (With Friction)
+5 XP

Friction acts up or down the slope depending on speed

We just saw that at the design speed, no friction is needed. That raises a question: what happens when the car goes faster or slower than design speed? This card answers it → friction acts down the slope (too fast) or up the slope (too slow), giving $v_{\max} = \sqrt{rg(\tan\theta + \mu_s)/(1 - \mu_s\tan\theta)}$.

When a vehicle travels at a speed different from the design speed, friction is required to prevent sliding. The direction of friction depends on the relationship between actual speed and design speed:

Speed regimeImpending motion without frictionStatic friction direction
$v<v_\text{design}$Down the bank, toward the inner edgeUp the bank, outward
$v=v_\text{design}$No relative slipNo friction required
$v>v_\text{design}$Up the bank, toward the outer edgeDown the bank, inward

Do not memorise friction as “always inward” or “always up”. It opposes the impending relative motion between tyre and road.

Maximum Speed (Friction Acting Down the Slope)

At maximum speed, friction is at its limiting value $F_f = \mu_s N$ acting down the slope. Resolving forces:

Perpendicular to surface: $N = mg\cos\theta + \dfrac{mv^2_{\max}}{r}\sin\theta$

Along the horizontal (centripetal direction):

$$N\sin\theta + \mu_s N\cos\theta = \frac{mv^2_{\max}}{r}$$

After algebraic manipulation:

$$v_{\max} = \sqrt{rg\left(\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}\right)}$$

Domain check: this maximum-speed form requires $1-\mu_s\tan\theta>0$. If the denominator is zero or negative, the rearranged limiting formula is not physically applicable as written.

Exam Tip

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 forces parallel and perpendicular to the road surface. The derivation is worth marks.

Worked Example 3, Maximum Speed with Friction

A banked curve has radius 100 m and banking angle 8°. The coefficient of static friction is 0.25. Calculate the maximum safe speed.

  1. Given. $r = 100\ \text{m}$, $\theta = 8°$, $\mu_s = 0.25$, $g = 9.8\ \text{m s}^{-2}$, $\tan 8° = 0.141$.
  2. Find. $v_{\max}$.
  3. Method. $v_{\max} = \sqrt{rg\left(\dfrac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}\right)}$
  4. Solve. $v_{\max} = \sqrt{100 \times 9.8 \times \left(\dfrac{0.141 + 0.25}{1 - 0.25 \times 0.141}\right)} = \sqrt{980 \times \dfrac{0.391}{0.965}} = \sqrt{980 \times 0.405} = \sqrt{397} = \mathbf{19.9\ \text{m s}^{-1}}$ (71.6 km/h).

Too fast ($v > v_\text{design}$): friction acts DOWN the slope. Too slow ($v < v_\text{design}$): friction acts UP. Maximum speed: $v_{\max} = \sqrt{rg(\tan\theta + \mu_s)/(1 - \mu_s\tan\theta)}$, derive from force diagrams; do not just memorise.

Pause, write the highlighted friction-direction rules and the $v_{\max}$ formula into your book before moving on.

A car drives slower than the design speed on a banked curve. In which direction does friction act?

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Apply the flat-curve limiting case

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Friction on Flat Curves
+5 XP

The limiting case: no banking, friction only

We just saw how banking raises the maximum safe speed through the normal force. That raises a question: what is the situation on an ordinary flat road with no banking at all? This card answers it → only friction available; $v_{\max} = \sqrt{\mu_s g r}$, which is independent of mass.

On a flat (unbanked) horizontal curve, the only horizontal force available to provide centripetal force is static 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 $F_c = mv^2_{\max}/r$, substituting:

$$\frac{mv^2_{\max}}{r} = \mu_s mg \quad\Rightarrow\quad v_{\max} = \sqrt{\mu_s g r}$$
Safety Warning

This is why icy curves are so dangerous. When $\mu_s$ drops to near zero, $v_{\max} = \sqrt{\mu_s g r}$ drops to near zero, any speed causes the car to slide outward. Crucially, $v_{\max}$ does not depend on the mass of the vehicle, which is why speed limits apply equally to all vehicles.

Worked Example 4, Flat Curve Maximum Speed

A car travels around a flat unbanked curve of radius 30 m. The coefficient of static friction is 0.60. Calculate the maximum speed without slipping.

  1. Given. $\mu_s = 0.60$, $r = 30\ \text{m}$, $g = 9.8\ \text{m s}^{-2}$.
  2. Find. $v_{\max}$.
  3. Method. $v_{\max} = \sqrt{\mu_s g r}$.
  4. Solve. $v_{\max} = \sqrt{0.60 \times 9.8 \times 30} = \sqrt{176.4} = \mathbf{13.3\ \text{m s}^{-1}}$ (47.9 km/h).

Flat curve: $F_c = \mu_s mg$, so $v_{\max} = \sqrt{\mu_s g r}$. Mass cancels, $v_{\max}$ is independent of vehicle mass. When $\mu_s \approx 0$ (ice), $v_{\max} \approx 0$; a banked curve always permits a higher $v_{\max}$ for the same $r$.

Add the highlighted flat-curve formula and the mass-independence result to your notes before the check below.

Fill the gap. For a car on a flat curve, the maximum speed formula is $v_{\max} = \sqrt{\_\_\_ \cdot g \cdot r}$. The missing symbol is the coefficient of _____ friction.

Essential formulae, horizontal circular motion

Conical pendulum: $r = L\sin\theta$ · $\tan\theta = v^2/(rg)$ · $T = 2\pi\sqrt{L\cos\theta/g}$

Banked curve (no friction): $\tan\theta = v^2/(rg)$ · $v_\text{design} = \sqrt{rg\tan\theta}$

Banked curve (with friction): $v_{\max} = \sqrt{rg(\tan\theta + \mu_s)/(1 - \mu_s\tan\theta)}$

Flat curve: $v_{\max} = \sqrt{\mu_s g r}$ (mass-independent)

Three of these statements about a banked curve at the design speed are correct. Pick the odd one out.

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Apply, diagnose and consolidate

Real World, Daytona International Speedway

Daytona International Speedway features 31° banking in the turns, with a turn radius of approximately 300 m. The design speed is:

$v_\text{design} = \sqrt{300 \times 9.8 \times \tan 31°} = \sqrt{300 \times 9.8 \times 0.601} = \sqrt{1767} \approx 42.0\ \text{m s}^{-1}$ (151 km/h)

At this speed, no friction is needed. The normal force from the banked surface alone provides exactly the required centripetal force. Cars could theoretically maintain circular paths even on a frictionless surface at this speed, which is why steep banking enables extremely high-speed cornering.

A banked curve is designed for 90 km/h. A driver takes the curve at 60 km/h. Compared to the design speed, friction must act:

Activities

Activity 1, Calculation Drills
ApplyBand 3–4

Apply the derived formulae to solve practical problems. Show all working.

  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.

For the same radius and same coefficient of friction, how does the maximum safe speed on a banked curve compare to a flat curve?

Activity 2, Direction of Friction on Banked Curves
UnderstandBand 4

Explain the role of friction at different speeds on a banked curve.

A car is driving on a banked curve at a speed slower than the design speed. Explain which way friction acts and why the vehicle tends to slide. Then repeat for a car driving faster than the design speed. Use a force diagram in your explanation.

Wrap-up, Misconceptions & Summary

Misconceptions, final check

✗ "Increasing the banking angle always increases safety."
✓ Very steep banking raises the design speed, which benefits fast traffic, but slow vehicles (lorries, cars braking to turn) may slide down the slope if going below the minimum safe speed. Highway engineers choose angles (typically 3–8°) that balance the full speed distribution of expected traffic.
✗ "On a flat curve, a heavier car can go faster before sliding."
✓ $v_{\max} = \sqrt{\mu_s g r}$, mass cancels completely. A 10 tonne truck and a 1 tonne car have the same $v_{\max}$ on the same flat curve with the same coefficient of friction.

Copy into your books

Conical Pendulum

  • $F_T\cos\theta = mg$ (vertical)
  • $F_T\sin\theta = mv^2/r$ (radial)
  • $\tan\theta = v^2/(rg)$
  • $T_\text{period} = 2\pi\sqrt{L\cos\theta/g}$

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 DOWN slope
  • $v < v_\text{design}$: friction UP slope
  • $v_{\max} = \sqrt{rg(\tan\theta + \mu_s)/(1 - \mu_s\tan\theta)}$

Flat Curve

  • $F_c = \mu_s mg$
  • $v_{\max} = \sqrt{\mu_s g r}$
  • Independent of mass
  • Icy roads: $\mu_s \approx 0 \Rightarrow v_{\max} \approx 0$
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Apply horizontal circular motion independently

Quick recall, horizontal circular motion
+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 10 marks
+5 XP

ApplyBand 5(3 marks) 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.

1 mark: correct radius; 1 mark: correct tension using $T = mg/\cos\theta$; 1 mark: correct speed using $v = \sqrt{rg\tan\theta}$

ApplyBand 5(3 marks) 2. A highway curve of radius 120 m is banked at 8°. Calculate the design speed. If the coefficient of static friction is 0.30, calculate the maximum safe speed.

1 mark: correct design speed; 1 mark: correct formula setup for $v_{\max}$; 1 mark: correct $v_{\max}$

EvaluateBand 6(4 marks) 3. Assess whether increasing the banking angle always increases safety on a highway curve. Consider the effects on vehicles travelling at different speeds.

1 mark each for: benefit for fast vehicles; risk for slow vehicles; minimum safe speed concept; overall balanced judgement about optimal angle choice

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Activity Answers

Activity 1 Q1 (Conical pendulum period): $T_\text{period}=2\pi\sqrt{L\cos\theta/g}=2\pi\sqrt{0.40\cos25°/9.8}=\mathbf{1.21\ \text{s}}$

Activity 1 Q2 (Design speed): $v_\text{design} = \sqrt{150 \times 9.8 \times \tan 5°} = \sqrt{150 \times 9.8 \times 0.0875} = \sqrt{128.6} = 11.3\ \text{m s}^{-1} = \mathbf{40.8\ \text{km/h}}$

Activity 1 Q3 (Flat curve): $v_{\max} = \sqrt{0.50 \times 9.8 \times 20} = \sqrt{98} = 9.90\ \text{m s}^{-1} = \mathbf{35.6\ \text{km/h}}$

Short Answer, Model Answers

Q1 (3 marks): Radius: $r = 0.80 \times \sin 35° = 0.80 \times 0.574 = 0.459\ \text{m}$ (1 mark). Tension: $T = mg/\cos\theta = 0.50 \times 9.8/\cos 35° = 4.90/0.819 = \mathbf{5.98\ \text{N}}$ (1 mark). Speed: $v = \sqrt{rg\tan\theta} = \sqrt{0.459 \times 9.8 \times \tan 35°} = \sqrt{0.459 \times 9.8 \times 0.700} = \sqrt{3.15} = \mathbf{1.77\ \text{m s}^{-1}}$ (1 mark).

Q2 (3 marks): Design speed: $v_\text{design} = \sqrt{120 \times 9.8 \times \tan 8°} = \sqrt{120 \times 9.8 \times 0.141} = \sqrt{165.8} = 12.9\ \text{m s}^{-1} = \mathbf{46.4\ \text{km/h}}$ (1 mark). Setup: $v_{\max} = \sqrt{120 \times 9.8 \times (0.141 + 0.30)/(1 - 0.30 \times 0.141)}$ (1 mark). $v_{\max} = \sqrt{1176 \times 0.441/0.958} = \sqrt{1176 \times 0.460} = \sqrt{541} = 23.3\ \text{m s}^{-1} = \mathbf{83.8\ \text{km/h}}$ (1 mark).

Q3 (4 marks): Increasing $\theta$ raises $v_\text{design} = \sqrt{rg\tan\theta}$, meaning fast vehicles rely less on friction, improving safety for high-speed traffic (1 mark). However, slow-moving vehicles (e.g., lorries stopping, vehicles at low speed) may slide down the bank if $v \ll v_\text{design}$. A minimum safe speed exists below which the vehicle slides inward (1 mark). Very steep banking (e.g., 31° at Daytona) requires vehicles to maintain minimum speeds, unsuitable for public roads where traffic speeds vary widely (1 mark). Highway engineers select angles (typically 3–8°) that balance the full speed distribution of expected traffic, steep enough to help at highway speeds, shallow enough to remain safe for all vehicles (1 mark).

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Retrieve and reflect

Check what actually stuck
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quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

Start the module quiz →
Arcade practice · Asteroid Blaster

Quick-fire questions on banked curves and conical pendulums, locks in the $\tan\theta = v^2/(rg)$ relationship and the direction of friction before the next lesson.

How did your thinking change?

At the start you were asked about Daytona International Speedway's 31° banking, engineered in 1959 so that NASCAR cars travelling at ~320 km/h (88.9 m/s) require no lateral friction to navigate the turns.

Verify: $\tan\theta = v^2/(rg)$. From $\theta = 31°$: $\tan 31° = 0.601$. So $r = v^2/(g\tan\theta) = (88.9)^2 / (9.8 \times 0.601) = 7903 / 5.89 \approx 1340 \text{ m}$. The Daytona turns have a radius of approximately 310 m, which means at 320 km/h the car is above design speed and friction is still involved. The design speed for 310 m radius and 31° banking is $v = \sqrt{rg\tan\theta} = \sqrt{310 \times 9.8 \times 0.601} \approx 42.8 \text{ m/s} \approx 154 \text{ km/h}$. Race speed requires friction on top of the normal-force component. Has your understanding of banked curves held up?

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