This MCQ module is based on: Friction Circular Motion
TOPIC 17 OF 33
Friction Circular Motion
🎓 Class 11
Physics
CBSE
Theory
Ch 4 – Laws of Motion
⏱ ~14 min
🌐 Language: [gtranslate]
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Friction Circular Motion
4.9 Common Forces in Mechanics
In mechanics, we encounter several common types of forces:
- Gravitational force (mg, downward, due to Earth)
- Normal force (N) — contact force perpendicular to surface
- Tension (T) — force in a stretched string or rope, along its length
- Spring force (F = −kx) — restoring force in a spring (Hooke's law)
- Friction (f) — opposing force at contact surfaces
Friction
When a body is in contact with another body and tends to slide (or slides), there develops at the surface of contact a force called friction. Friction always opposes relative motion (or its tendency) between the two surfaces.
Static Friction (f_s)
Static friction opposes impending motion. It is self-adjusting: it takes whatever value is needed (up to a maximum) to keep the body at rest.
\[0 \le f_s \le \mu_s N\]
\[(f_s)_{max} = \mu_s N\]
where \(\mu_s\) is the coefficient of static friction, and N is the normal force. The maximum static friction is also called limiting friction.
Kinetic Friction (f_k)
Kinetic friction acts when surfaces slide against each other:
\[f_k = \mu_k N\]
where \(\mu_k\) is the coefficient of kinetic friction. Experimentally:
\[\mu_k < \mu_s\]
Why is kinetic friction LESS than static friction? When two surfaces are at rest, microscopic interlocking has time to "settle." Once sliding, surfaces lose this interlock — they ride over each other on momentary contact peaks, requiring less force to maintain motion than to start it.
Rolling Friction (f_r)
Rolling friction is much smaller than sliding (kinetic) friction. This is why wheels were a revolutionary invention.
| Friction Type | Typical Range of μ | Example |
|---|---|---|
| Static (rubber on dry concrete) | 0.6 – 1.0 | Tires on dry road |
| Kinetic (rubber on dry concrete) | 0.5 – 0.8 | Sliding tires |
| Static (steel on steel) | 0.6 | Industrial machinery |
| Kinetic (steel on ice) | 0.03 | Skater on ice |
| Rolling (steel ball on steel) | 0.001 | Ball bearings |
4.10 Circular Motion (Dynamics)
We have seen in Chapter 3 that acceleration of a body moving in a circle of radius R with uniform speed v is \(v^2/R\) directed towards the centre. According to the second law, the force \(\vec{f}_c\) providing this acceleration is:
\[f_c = \frac{m v^2}{R}\]
This is the centripetal force. Note: it is NOT a new kind of force — it is whatever force(s) act on the body to provide the inward acceleration. It could be gravity, friction, tension, normal force, or a combination.
Application: Motion of a Car on a Level Road
For a car negotiating a curve of radius R on a level road, the centripetal force is provided by friction:
\[\frac{mv^2}{R} \le \mu_s mg\]
\[v^2 \le \mu_s g R\]
\[v_{max} = \sqrt{\mu_s g R}\]
This is the maximum safe speed on a flat curve. Above this, the car skids outward.
Motion on a Banked Road
To allow higher speeds on curves, roads are banked — the outer edge is raised above the inner edge. The normal force then has a horizontal component that helps provide centripetal force.
For a car on a banked road, ignoring friction, the equations are:
\[N\cos\theta = mg \quad \text{(vertical balance)}\]
\[N\sin\theta = \frac{mv^2}{R} \quad \text{(horizontal centripetal)}\]
Dividing:
\[\tan\theta = \frac{v^2}{Rg}\]
\[\boxed{v_{optimum} = \sqrt{Rg\tan\theta}}\]
This is the optimum (no-friction-needed) speed for the given banking angle. With friction included, the maximum speed is:
\[v_{max} = \sqrt{Rg \cdot \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}}\]
🎯 Interactive: Static Friction Limit
Apply force gradually. Observe when the block starts to slide.
Limiting friction: μ_s N = 39.2 N
Status: At rest (static friction = 25 N)
If sliding: a = 0 m/s²
Worked Examples
Worked Example 1 (NCERT Example 4.7): Sliding Box
A wooden block of mass 2 kg rests on a horizontal table. The coefficient of static friction between the block and table is 0.4. A horizontal force is gradually applied. Find the force at which the block just starts to move.
Given: m = 2 kg, μ_s = 0.4, g = 10 m/s².
The block starts moving when applied force just exceeds limiting friction: \[F = (f_s)_{max} = \mu_s N = \mu_s mg = 0.4 \times 2 \times 10 = \boxed{8 \text{ N}}\] For F < 8 N, static friction adjusts to balance F (no motion). For F > 8 N, the block accelerates.
The block starts moving when applied force just exceeds limiting friction: \[F = (f_s)_{max} = \mu_s N = \mu_s mg = 0.4 \times 2 \times 10 = \boxed{8 \text{ N}}\] For F < 8 N, static friction adjusts to balance F (no motion). For F > 8 N, the block accelerates.
Worked Example 2 (NCERT Example 4.10): Banked Road
A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the (a) optimum speed of the racecar to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid slipping?
Given: R = 300 m, θ = 15°, μ_s = 0.2, g = 9.8 m/s². tan 15° ≈ 0.268.
(a) Optimum speed (no friction needed): \[v_o = \sqrt{Rg\tan\theta} = \sqrt{300 \times 9.8 \times 0.268}\] \[v_o = \sqrt{787.9} \approx \boxed{28.07 \text{ m/s} \approx 101 \text{ km/h}}\]
(b) Maximum permissible speed: \[v_{max} = \sqrt{\frac{Rg(\mu_s + \tan\theta)}{1 - \mu_s\tan\theta}} = \sqrt{\frac{300 \times 9.8 \times (0.2+0.268)}{1 - 0.2 \times 0.268}}\] \[v_{max} = \sqrt{\frac{1376}{0.946}} = \sqrt{1455} \approx \boxed{38.14 \text{ m/s} \approx 137 \text{ km/h}}\]
(a) Optimum speed (no friction needed): \[v_o = \sqrt{Rg\tan\theta} = \sqrt{300 \times 9.8 \times 0.268}\] \[v_o = \sqrt{787.9} \approx \boxed{28.07 \text{ m/s} \approx 101 \text{ km/h}}\]
(b) Maximum permissible speed: \[v_{max} = \sqrt{\frac{Rg(\mu_s + \tan\theta)}{1 - \mu_s\tan\theta}} = \sqrt{\frac{300 \times 9.8 \times (0.2+0.268)}{1 - 0.2 \times 0.268}}\] \[v_{max} = \sqrt{\frac{1376}{0.946}} = \sqrt{1455} \approx \boxed{38.14 \text{ m/s} \approx 137 \text{ km/h}}\]
Worked Example 3: Conical Pendulum
A stone of mass 0.25 kg tied to a string of length 1.5 m moves in a horizontal circle. If the string makes 30° with the vertical, find: (a) the tension in the string, (b) the speed of the stone. (g = 10 m/s²)
The string traces out a cone. The horizontal component of tension provides centripetal force.
Vertical balance: T cos θ = mg \[T = \frac{mg}{\cos\theta} = \frac{0.25 \times 10}{\cos30°} = \frac{2.5}{0.866} \approx \boxed{2.89 \text{ N}}\]
Horizontal (centripetal): T sin θ = mv²/r, where r = L sin θ = 1.5 × 0.5 = 0.75 m. \[v^2 = \frac{T \sin\theta \cdot r}{m} = \frac{2.89 \times 0.5 \times 0.75}{0.25} = 4.34\] \[v = \sqrt{4.34} \approx \boxed{2.08 \text{ m/s}}\]
Vertical balance: T cos θ = mg \[T = \frac{mg}{\cos\theta} = \frac{0.25 \times 10}{\cos30°} = \frac{2.5}{0.866} \approx \boxed{2.89 \text{ N}}\]
Horizontal (centripetal): T sin θ = mv²/r, where r = L sin θ = 1.5 × 0.5 = 0.75 m. \[v^2 = \frac{T \sin\theta \cdot r}{m} = \frac{2.89 \times 0.5 \times 0.75}{0.25} = 4.34\] \[v = \sqrt{4.34} \approx \boxed{2.08 \text{ m/s}}\]
📐 Activity 4.4 — Inclined Plane Friction Coefficient
Setup: Place a wooden block on a hinged board (inclined plane). Slowly raise the angle until the block just begins to slide. Measure the angle.
Predict: What does this critical angle tell you?
At the critical (sliding) angle θ_s, the component of gravity along the incline equals limiting friction:
\[mg\sin\theta_s = \mu_s mg\cos\theta_s\]
\[\boxed{\mu_s = \tan\theta_s}\]
This is called the angle of repose — a beautiful experimental method to measure μ_s using only a ruler/protractor and a board!
Example: if block slips at θ = 27°, then μ_s = tan(27°) ≈ 0.51.
🎯 Competency-Based Questions
Q1. A block of mass 5 kg is placed on a horizontal surface (μ_s = 0.3). What minimum horizontal force is needed to just start motion? (g = 10) L3 Apply
Answer: (b) 15 N. F_min = μ_s mg = 0.3 × 5 × 10 = 15 N.
Q2. A car negotiates a flat curve of radius 50 m with μ_s = 0.5. Find the maximum safe speed. L3 Apply
Answer: \(v_{max} = \sqrt{\mu_s g R} = \sqrt{0.5 \times 9.8 \times 50} = \sqrt{245} \approx \boxed{15.65 \text{ m/s} \approx 56.3 \text{ km/h}}\). Above this, the car will skid outward.
Q3. A road is banked at angle 30°. Find optimum speed for radius 100 m. L3 Apply
Answer: \(v_o = \sqrt{Rg\tan\theta} = \sqrt{100 \times 9.8 \times \tan30°} = \sqrt{100 \times 9.8 \times 0.577} = \sqrt{565.6} \approx \boxed{23.78 \text{ m/s}}\). At this speed, no friction is needed for circular motion.
Q4. Analyse: Why are highways banked instead of being kept flat? Why is rolling motion preferred over sliding? L4 Analyse
Answer:
Banking: A banked road allows higher safe speeds without depending purely on friction. The horizontal component of the normal force provides centripetal force "for free." This is essential because friction is unreliable (rain, oil, ice can drastically reduce μ).
Rolling vs. sliding: Rolling friction (μ_r ~ 0.001) is ~100× smaller than kinetic friction (μ_k ~ 0.5). So rolling wheels lose almost no energy to friction, making transport vastly more efficient.
Banking: A banked road allows higher safe speeds without depending purely on friction. The horizontal component of the normal force provides centripetal force "for free." This is essential because friction is unreliable (rain, oil, ice can drastically reduce μ).
Rolling vs. sliding: Rolling friction (μ_r ~ 0.001) is ~100× smaller than kinetic friction (μ_k ~ 0.5). So rolling wheels lose almost no energy to friction, making transport vastly more efficient.
Q5. HOT (Create): Design a protocol to safely test the coefficient of friction between a new tire compound and an asphalt surface. List equipment, procedure, and safety precautions. L6 Create
Sample Protocol:
- Equipment: Tire-shaped block of new compound, asphalt slab, force gauge (or spring balance), ruler, weights, safety barriers, protective gear.
- Static μ test: Place block + known weight on slab. Pull horizontally with force gauge until block just starts to slide. F_max / (mg) = μ_s.
- Kinetic μ test: Once moving, pull at constant velocity. Steady force F / (mg) = μ_k.
- Inclined plane method: Use a hinged asphalt slab; raise until block slips. tan(angle) = μ_s.
- Repeats: Multiple trials to average; vary normal load (verify F ∝ N independence).
- Safety: Securely clamp setup; wear gloves; ensure no overhead hazards if block flies; test in dry+wet conditions.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: It is harder to start moving a stationary heavy box than to keep it moving.
R: The coefficient of static friction is greater than the coefficient of kinetic friction.
Answer: (A). Both true; R correctly explains A. μ_s > μ_k means starting (limiting friction = μ_s N) requires more force than maintaining sliding (kinetic friction = μ_k N).
A: Centripetal force is a special new kind of force.
R: All forces in nature are: gravitational, electromagnetic, strong nuclear, weak nuclear.
Answer: (D). A is FALSE — "centripetal" is a description (toward center), not a separate force. Could be friction, gravity, tension, normal force, etc. R is TRUE.
A: A coin placed on a rotating turntable will eventually fly off if rotation speed is high enough.
R: Static friction can only provide a limited centripetal force. Beyond a critical ω, friction is insufficient.
Answer: (A). Both true; R explains A. f_s ≤ μ_s mg = mω²r requires ω ≤ √(μ_s g/r). Beyond this, coin flies off.
Frequently Asked Questions - Friction Circular Motion
What is the main concept covered in Friction Circular Motion?
In NCERT Class 11 Physics Chapter 4 (Laws of Motion), "Friction Circular Motion" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Friction Circular Motion useful in real-life applications?
Real-life applications of Friction Circular Motion from NCERT Class 11 Physics Chapter 4 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Friction Circular Motion?
Key formulas in Friction Circular Motion (NCERT Class 11 Physics Chapter 4 Laws of Motion) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 4?
NCERT Class 11 Physics Chapter 4 (Laws of Motion) is structured so each part builds on the previous one. Friction Circular Motion connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Friction Circular Motion?
CBSE board questions from Friction Circular Motion typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Friction Circular Motion lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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