NBSC Cromer Campus HSC Physics
Module 5 Lesson 05 Tutor Active
Student Isla Nguyen

Lesson

Banked Tracks and Conical Pendulums
Module 5 — Advanced Mechanics · Syllabus 5.2

Banked Tracks and Conical Pendulums

The geometry of the track can replace some or all of the frictional burden. This lesson makes that shift explicit, then prepares the move to orbital motion.

Tier 1 rapid exposition and MCQ consolidation Tier 2 evaluate with quantitative evidence Shared renderer $\ce{^{14}_6C -> ^{14}_7N + e- + \bar{\nu}_e}$
OrientationTier 1 · Understand
Lesson Goal Derive the ideal banking angle for friction-free turning, compare it with a flat-road model, and set up the force logic needed for gravitation and orbits.

Lesson 4 treated uniform circular motion on a flat road, where friction supplied the horizontal inward force. This lesson changes the geometry so the normal force itself can contribute that inward component. The prerequisite ideas are centripetal force, resolution of forces into perpendicular components, and Newton's second law. Lesson 6 then keeps the same centripetal structure but replaces the contact-force setting with gravitation.

The app surface keeps these relationships visible without turning the lesson into a dashboard. The main reading panel remains dense and printable, while the tutoring workflow lives in its own dedicated pane.

By The End
  • Derive the ideal banking relation $ \tan\theta = \dfrac{v^2}{rg} $ from force components.
  • Predict the direction of friction when the vehicle moves above or below the design speed.
  • Transfer the same force geometry to a conical pendulum, including the radius relation $r = L\sin\theta$.
  • Evaluate competing designs using quantitative evidence rather than descriptive commentary.
Tutor Focus
  • Method Force diagram first, resolved equations second, algebraic reduction third.
  • Units Convert km h$^{-1}$ to m s$^{-1}$ before substitution and retain units in prose.
  • Reasoning Turn each numerical result into a physical claim about friction demand or robustness.
  • Writing Finish with a judgement sentence that names the decisive evidence and the practical qualification.
Core ContentTier 1 · Apply

For a frictionless banked curve, vertical balance and horizontal centripetal demand are written separately. The normal force does both jobs once it is resolved into components.

$N \cos \theta = mg$
(1)
$N \sin \theta = \dfrac{mv^2}{r}$
(2)
$\tan \theta = \dfrac{v^2}{rg}$
(3)

Equation (3) is the organising result. It shows that the ideal banking angle depends on speed and radius, not directly on the mass of the vehicle. A conical pendulum has the same geometry: tension replaces the normal force, but the component equations produce the same tangent relationship.

Exam Technique Do not jump directly to Equation (3). The marker needs to see the force components and their directions before the tangent relation is algebraically justified.
Same Geometry, New Context
  • Banked track $N\cos\theta = mg$ and $N\sin\theta = \dfrac{mv^2}{r}$.
  • Conical pendulum $T\cos\theta = mg$ and $T\sin\theta = \dfrac{mv^2}{r}$.
  • Radius link For a conical pendulum, the circular radius is $r = L\sin\theta$.
  • Cancellation Mass disappears because it appears in both vertical-balance and horizontal-inward equations.
When Friction Appears
  • At design speed friction is zero and the bank alone provides the inward component.
  • Above design speed the vehicle tends to slide up the slope, so friction acts down the slope toward the centre.
  • Below design speed the vehicle tends to slide down the slope, so friction acts up the slope away from the centre.
  • Method cue State the direction of friction in words before writing any modified component equations.
Concept CheckTier 1 · Remember / Apply

Attempt the question set directly in the lesson surface. Feedback stays compact, and any question can be pushed into the tutor pane for explanation or writing support.

Q1
State the magnitude of the net centripetal force required to hold a particle of mass $m$ on a circular path of radius $r$ at constant speed $v$.
Remember
Q2
A vehicle rounds a frictionless banked curve of inclination $\theta$. Which component of the contact force on the vehicle supplies the centripetal force?
Understand
Q3
A curve of radius $70\ \text{m}$ is to be negotiated at $25\ \text{m s}^{-1}$ with no reliance on friction. Calculate the ideal banking angle, taking $g = 9.8\ \text{m s}^{-2}$.
Apply
Q4
A conical pendulum of string length $1.2\ \text{m}$ makes a steady cone at $60^\circ$ to the vertical. Its period is nearest to:
Apply
Q5
A vehicle travels slower than the design speed for a frictionless banked curve. What is the direction of the friction force on the vehicle?
Understand
Q6
For a conical pendulum of string length $L$ making an angle $\theta$ with the vertical, the radius of the circular path is:
Remember
Q7
For a fixed curve radius, the design speed is doubled. By what factor does $\tan\theta$ change?
Apply
Q8
Why does vehicle mass cancel from the ideal banking-angle relation?
Reason
Applied PracticeTransition · Apply

1 A 1200 kg car travels around a 70 m radius curve banked at 10°. Find the speed at which no friction is required and the centripetal force at that speed.

1
Use $\tan \theta = \dfrac{v^2}{rg}$.
$v = \sqrt{rg \tan \theta}$
$v = \sqrt{(70)(9.8)\tan 10^\circ} = 11.0\ \text{m s}^{-1}$
Required speed: $11.0\ \text{m s}^{-1}$
2
$F_c = \dfrac{mv^2}{r}$
$F_c = \dfrac{(1200)(11.0)^2}{70} = 2.07 \times 10^3\ \text{N}$
Centripetal force: $2.1 \times 10^3\ \text{N}$
Trace Check 1 speed complete    2 force complete ~2 min writing time

2 A 0.40 kg stopper on a 1.40 m string forms a conical pendulum at 35° to the vertical. Find the radius of the circular path, the period, and the speed.

1
Radius from geometry: $r = L\sin\theta$
$r = 1.40\sin35^\circ = 0.80\ \text{m}$
Radius: $0.80\ \text{m}$
2
Period from the component model: $T = 2\pi\sqrt{\dfrac{L\cos\theta}{g}}$
$T = 2\pi\sqrt{\dfrac{1.40\cos35^\circ}{9.8}} = 2.15\ \text{s}$
Period: $2.15\ \text{s}$
3
Then $v = \dfrac{2\pi r}{T}$
$v = \dfrac{2\pi(0.80)}{2.15} = 2.35\ \text{m s}^{-1}$
Speed: $2.35\ \text{m s}^{-1}$
Transfer Check The same horizontal-inward / vertical-balance split is still doing the work. Only the force name changes from $N$ to $T$.

3 Use the 70 m, 10° bank to stress-test the evaluation. Find the minimum friction coefficient required at 80 km h$^{-1}$ and 100 km h$^{-1}$, then compare with the flat-road case.

1
For speeds above the design speed, friction acts down the slope, so
$\mu = \dfrac{k\cos\theta - \sin\theta}{k\sin\theta + \cos\theta}$, where $k = \dfrac{v^2}{rg}$.
Direction first, algebra second.
2
At $80\ \text{km h}^{-1}$, $v = 22.2\ \text{m s}^{-1}$ and $k = 0.72$.
$\mu_{\text{banked}} = 0.48$ and $\mu_{\text{flat}} = \dfrac{v^2}{rg} = 0.72$.
80 km h$^{-1}$: banked demand lower by $0.24$
3
At $100\ \text{km h}^{-1}$, $v = 27.8\ \text{m s}^{-1}$ and $k = 1.13$.
$\mu_{\text{banked}} = 0.79$ and $\mu_{\text{flat}} = 1.13$.
100 km h$^{-1}$: banked still marginally feasible, flat road is beyond the stated dry-road range
Deep PracticeTier 2 · Evaluate
Evaluate Prompt Evaluate whether a banked or flat track is preferable for a 70 m radius curve navigated at 80–100 km h$^{-1}$, given $\mu_s = 0.9$–$1.0$ for a dry flat road, banking at 10° with $\mu_s = 0.6$–$0.8$ for the banked design. Consider sensitivity to surface condition and practical construction implications.

The judgement needs quantitative evidence. A descriptive answer is insufficient. The decisive comparison is whether geometry reduces dependence on friction over the stated operating band, while the final sentence still acknowledges construction and maintenance implications.

Evaluate Structure Open with a judgement, support it with two quantitative comparisons, then explain why the rejected design is less robust under the stated conditions.
Judgement Path
  • Step 1 Establish the ideal banked speed to show that the operating band sits above the friction-free case.
  • Step 2 Compare required friction at 80 km h$^{-1}$ for banked and flat designs.
  • Step 3 Repeat the comparison at 100 km h$^{-1}$ because that is the stress-test point.
  • Step 4 Conclude with the more robust design, then acknowledge cost, construction, or speed-specific limitations.
What Must Be Visible
  • Equation trail Define the controlling relationship before substitution.
  • Numerical evidence Quote at least two friction coefficients or threshold speeds.
  • Rejected option Explain why the weaker design is less reliable under the stated conditions.
  • Qualification Keep one practical caveat, but do not let it replace the physics comparison.
Evidence BankTier 2 · Quantify

This is the compact numerical bank for the evaluation. In a strong response, each number must do argumentative work rather than sit as an isolated calculation.

Ideal banked speed: $v_0 = \sqrt{rg\tan\theta} = 11.0\ \text{m s}^{-1} = 39.6\ \text{km h}^{-1}$
bank only
Flat road demand: $\mu_{\text{flat}} = \dfrac{v^2}{rg}$
all friction
Banked road demand above $v_0$: $\mu = \dfrac{k\cos\theta - \sin\theta}{k\sin\theta + \cos\theta}$
$k = \dfrac{v^2}{rg}$
Numbers To Quote
  • Design speed $11.0\ \text{m s}^{-1}$ or $39.6\ \text{km h}^{-1}$.
  • At 80 km h$^{-1}$ banked surface requires $\mu \approx 0.48$; flat road requires $\mu \approx 0.72$.
  • At 100 km h$^{-1}$ banked surface requires $\mu \approx 0.79$; flat road requires $\mu \approx 1.13$.
  • Decision cue The banked design keeps the friction burden lower at both comparison speeds.
What Those Numbers Mean
  • Low design speed The 10° bank is tuned for a much lower friction-free speed than the operating band.
  • 80 km h$^{-1}$ Both designs are still possible, but the banked design uses substantially less friction reserve.
  • 100 km h$^{-1}$ The banked design is at the upper edge of its stated range, while the flat road exceeds the stated dry-road band.
  • Final judgement Prefer the banked design for robustness, then mention its construction complexity and speed specificity.
Argument Rule Two numbers plus one qualification are enough. More arithmetic is not better unless it sharpens the verdict.
Scientific WritingTier 2 · Communicate

Tomorrow’s priority is not just solving the physics correctly but expressing the judgement in precise scientific prose. The tutor is configured to critique structure, units, variable definitions, and claim-evidence-reasoning flow.

Writing Moves Keep variables symbolic before substitution, attach units to every numerical quantity, and translate the result into an explicit physical claim before moving to the next sentence.
  • Sentence 1 Name the preferred design and the operating range.
  • Sentence 2 State the controlling relationship and the relevant speed thresholds.
  • Sentence 3 Compare the friction demand for the banked and flat designs.
  • Sentence 4 Close with the decisive evidence and the practical qualification.
Band 6 Rubric
  • R1Labelled force representation for each case3 marks
  • R2Formula → substitution → units → result4 marks
  • R3Quantitative comparison of both designs3 marks
  • R4Explicit verdict naming decisive evidence2 marks
Sentence Bank
  • Opening judgement The banked design is preferable across the stated operating band because the geometry shares the centripetal requirement with the normal force.
  • Evidence sentence 1 Using $ \tan\theta = \dfrac{v^2}{rg} $, the 10° bank has an ideal friction-free speed of only $11.0\ \text{m s}^{-1}$, so the 80–100 km h$^{-1}$ band must be evaluated with friction included.
  • Evidence sentence 2 At $80\ \text{km h}^{-1}$ the banked curve requires $\mu \approx 0.48$, whereas the flat road requires $\mu \approx 0.72$.
  • Closing qualification Although a flat road may be simpler to construct, the banked surface is the more robust design under the stated conditions because it maintains a lower friction demand at both comparison speeds.
Common Fixes
  • Avoid calling centripetal force an extra force; it is the net inward result of real forces.
  • Avoid substituting km h$^{-1}$ directly into SI equations.
  • Avoid writing a result without saying what that result implies for the design choice.
  • Avoid ending with “therefore it is better” without naming why it is better.
Draft Workshop
Draft is saved locally in this browser.
Common ErrorsReview · Refine

The fastest way to lift the quality of a response is to remove the repeated errors that flatten otherwise correct physics. These checks are also good tutor prompts when a draft feels almost correct but not yet sharp.

Misconceptions To Catch
  • Extra-force mistake Treating centripetal force as an extra named force rather than the inward resultant.
  • Component swap Writing $N\sin\theta = mg$ and $N\cos\theta = \dfrac{mv^2}{r}$ without checking the geometry.
  • Radius miss Forgetting that the conical-pendulum radius is horizontal, so $r = L\sin\theta$.
  • Friction shortcut Using the same friction direction above and below the design speed.
Recovery Checks
  • Axis first Draw a vertical axis and an inward horizontal axis before you label forces.
  • Direction first State whether friction is up or down the slope before you substitute $f = \mu N$.
  • SI first Convert speed units before entering the equation line.
  • Verdict last End with a sentence that names the better design and the numerical evidence that decided it.
Syllabus MappingNESA Stage 6 · Module 5.2

The lesson and the tutor are aligned to the Stage 6 outcomes below. The same section ids are used in the live tutor calls so feedback can stay grounded in the relevant syllabus lane.

  • Conduct investigations to explain and evaluate the variables that affect objects executing uniform circular motion.Evaluate
  • Analyse the forces acting on an object executing uniform circular motion in a variety of situations, including cars on flat or banked roads and conical pendulums.Analyse
  • Solve problems and make quantitative predictions using $F_{\text{net}} = \dfrac{mv^2}{r}$ and $\tan\theta = \dfrac{v^2}{rg}$.Apply
Section Trace Orientation and core content build the model. Concept check consolidates recall and application. Applied practice secures exam method. Deep practice and scientific writing move the student into judgement and communication.
Next LessonContinuity
06
Newton's Law of Universal Gravitation

The inward-force structure used here becomes the bridge to orbital motion once the contact normal is replaced by an inverse-square gravitational force.