Rolling friction is the resistive force that opposes the motion of a body rolling on a surface. When a wheel, ball, or cylinder rolls without slipping, it deforms slightly at the contact patch — and the surface deforms too. This deformation dissipates energy and creates a net horizontal retarding force: rolling friction.
The key NCERT definition (Class 11 Physics Chapter 4, page 60) places rolling friction alongside static and kinetic friction as one of the three friction types. The critical hierarchy for NEET is:
μ_rolling ≪ μ_kinetic < μ_static (for the same surface pair)
This is why wheels replaced sledges: converting sliding into rolling dramatically reduces friction. Rolling friction is typically 2–3 orders of magnitude smaller than sliding (kinetic) friction for hard surfaces.
What rolling friction depends on:
What it does NOT depend on (at NEET level):
The trap that costs marks: confusing the friction hierarchy. When a question contrasts a block sliding down a rough incline versus a cylinder rolling down the same incline, the sliding block experiences kinetic friction f_k = μ_k N, which is far larger than the rolling friction on the cylinder. Students who use the same μ for both cases get the comparison wrong. On a rough incline, the deceleration for a sliding block is g(sin θ − μ_k cos θ). For a rolling body, the friction is much smaller, and the analysis involves rotational inertia — but the foundational point is that rolling friction is NOT kinetic friction, and you cannot substitute one coefficient for the other.
Watch-out: NEET occasionally tests whether you know that rolling friction exists as a distinct, much smaller quantity — not as a calculation, but as a conceptual ranking question.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following correctly ranks the three types of friction for the same pair of surfaces?
Answer: A. A is correct. For the same surface pair, the coefficient of static friction is greatest, kinetic friction is intermediate, and rolling friction is the smallest — as stated in NCERT Class 11 Physics Chapter 4, page 60.
Why B is wrong: B is wrong because rolling friction is the smallest of the three, not the largest. This reverses the entire hierarchy.
Why C is wrong: C is wrong because static friction coefficient is always greater than kinetic for the same surface pair, not the other way around.
Why D is wrong: D is wrong because rolling friction is smaller than kinetic friction, not larger. Placing μ_rolling between μ_static and μ_kinetic contradicts the established hierarchy.
Rolling friction is significantly less than sliding friction because:
Answer: C. C is correct. Rolling friction arises from slight deformation at the contact patch. This deformation — and the associated energy loss — is far smaller than the surface interactions during sliding, which is why rolling friction is much less than kinetic friction.
Why A is wrong: A is wrong because a rolling body does have a finite contact area (the contact patch). The area may be small, but it is not zero.
Why B is wrong: B is wrong because the rolling body's weight acts downward and the surface provides a normal reaction. The normal force is present and determines the magnitude of rolling friction.
Why D is wrong: D is wrong because kinetic friction applies when surfaces slide against each other. Rolling without slipping avoids sliding, so kinetic friction does not govern the interaction — but it does not become 'zero'; instead, rolling friction (a different, smaller quantity) takes over.
A steel ball rolls on a steel rail. A rubber ball of the same mass rolls on soft sand. Which experiences greater rolling friction, and why?
Answer: D. D is correct. Rolling friction depends on the deformability of both surfaces. Rubber on sand deforms far more at the contact patch than steel on a steel rail, dissipating more energy per revolution and producing greater rolling friction.
Why A is wrong: A is wrong because the relevant factor for rolling friction is deformability, not the general 'friction coefficient' label. Steel on steel has very low deformation, so rolling friction is very small despite both surfaces being metal.
Why B is wrong: B is wrong because the normal force depends on weight, not surface hardness. Both balls have the same mass, so both have the same normal force. The difference lies in deformation, not normal force.
Why C is wrong: C is wrong because equal mass only ensures equal weight and normal force. Rolling friction also depends on the deformability of the surfaces, which differs drastically between the two cases.
A block of mass 4 kg rests on a rough plane inclined at 20° to the horizontal. The incline angle has been increased until the block is on the verge of sliding (limiting equilibrium). Take g = 10 m/s², sin 20° = 0.342, cos 20° = 0.940. What is the coefficient of static friction between the block and the plane?
Answer: B. At limiting equilibrium, resolving the weight along and perpendicular to the incline gives f_s,max = mg sin θ = 4 × 10 × 0.342 = 13.68 N and N = mg cos θ = 4 × 10 × 0.940 = 37.60 N. Since f_s,max = μ_s N, μ_s = 13.68 / 37.60 ≈ 0.364. (This is the same tan θ_max = μ_s relation used to find the angle of repose, NCERT Class 11 Physics Chapter 4, page 61.)
Why A is wrong: A (0.342) comes from treating the normal force as N = mg instead of mg cos θ, so it reports sin θ itself instead of f_s,max/N (trap: forgetting to resolve the normal force along the incline).
Why C is wrong: C (2.75) inverts the ratio, computing cos θ / sin θ instead of sin θ / cos θ = f_s,max / N (trap: dividing the wrong way round).
Why D is wrong: D (0.940) reports cos θ itself — the normal-force component per unit weight — mistaking it directly for the coefficient of static friction (trap: confusing a trigonometric component with μ_s).
A solid cylinder and a block of the same mass are released from rest at the top of the same rough incline (angle θ, coefficient of kinetic friction μ_k). The cylinder rolls without slipping; the block slides. Which statement is correct?
Answer: B. B is correct. The sliding block experiences kinetic friction f_k = μ_k N = μ_k mg cos θ opposing its motion. The rolling cylinder does not slide, so kinetic friction does not apply — instead, it experiences rolling friction, which is much smaller in magnitude. Confusing the two friction types leads to incorrect velocity or time-of-descent comparisons.
Why A is wrong: A is wrong because it reverses the assignment: the block slides (kinetic friction) and the cylinder rolls (rolling friction), not the other way around.
Why C is wrong: C is wrong because the kinetic friction formula f_k = μ_k N applies only to sliding surfaces. The cylinder rolls without slipping, so it does not experience kinetic friction — it experiences rolling friction, a much smaller force.
Why D is wrong: D is wrong because on a rough incline, friction is present for both sliding and rolling bodies. The mg sin θ component is the gravitational pull along the incline; friction opposes motion in addition to this.
At the NEET level, rolling friction for a rigid body on a hard surface does NOT significantly depend on:
Answer: A. A is correct. Like static and kinetic friction, rolling friction at the NEET level is treated as independent of the apparent area of contact (Coulomb-Amontons model). The other three factors — normal force, surface deformability, and roller radius — all influence rolling friction.
Why B is wrong: B is wrong because the very origin of rolling friction is surface deformation at the contact patch. More deformable surfaces produce greater rolling friction.
Why C is wrong: C is wrong because rolling friction is proportional to the normal force, just as static and kinetic friction are. It does depend on normal force.
Why D is wrong: D is wrong because the radius of the rolling body affects how much deformation occurs per revolution. Larger radii generally produce less rolling friction for rigid bodies on hard surfaces.
A student claims: 'When a car tyre rolls on a road without skidding, kinetic friction acts on the tyre.' This claim is:
Answer: C. C is correct. In pure rolling (no slipping), the contact point of the tyre is instantaneously at rest relative to the road. Since there is no relative sliding at the contact, static friction (not kinetic) acts. Kinetic friction would apply only if the tyre skidded.
Why A is wrong: A is wrong because 'motion relative to the road' refers to the tyre as a whole translating. At the contact point specifically, there is no relative sliding in pure rolling — so the friction there is static, not kinetic.
Why B is wrong: B is wrong because kinetic friction requires relative sliding between the two surfaces in contact. A tyre rolling without slipping has no sliding at the contact patch.
Why D is wrong: D is wrong because friction does act on a rolling tyre — static friction at the contact point is what enables rolling without slipping and provides the necessary torque.
A body rests on a horizontal surface. A gradually increasing horizontal force is applied. Before the body starts to move, the friction force on the body:
Answer: D. D is correct. Static friction is self-adjusting: it matches the applied tangential force exactly, preventing motion, until the applied force reaches the ceiling f_s,max = μ_s N = μ_s mg. Only at that threshold does motion begin. Using f_s = μ_s mg before the limit is reached overestimates friction (a common mistake documented in the corpus).
Why A is wrong: A is wrong because μ_s mg is the MAXIMUM static friction, not the actual value at every instant. Below the threshold, static friction equals the applied force, which can be much less than μ_s mg. This is the classic 'static friction always at maximum' mistake.
Why B is wrong: B is wrong because friction begins opposing the applied force immediately, not only above mg. The threshold for motion is μ_s mg (for a horizontal surface), not mg.
Why C is wrong: C is wrong because kinetic friction applies only after sliding begins. Before the body moves, static friction (not kinetic) acts, and it self-adjusts below μ_s mg.
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Given
A block and a solid cylinder, each of mass m = 2.0 kg, are placed at the top of a rough incline of length L = 5.0 m at angle θ = 30°. The coefficient of kinetic friction between the block and incline is μ_k = 0.20. The cylinder rolls without slipping; rolling friction is negligible compared to kinetic friction. Take g = 10 m/s² (exact, problem-defined).
Required
Find the velocity of each body at the bottom of the incline.
Concept
The sliding block experiences kinetic friction opposing motion along the incline. The rolling cylinder, with negligible rolling friction, behaves approximately as if on a smooth incline (for force analysis), but its kinetic energy splits between translational and rotational modes.
Formula
For the sliding block: a_block = g(sin θ − μ_k cos θ)
Final velocity: v² = 2 a L (starting from rest)
For the rolling solid cylinder (negligible rolling friction): v² = (2gL sin θ) / (1 + I/(mR²)) = (2gL sin θ) / (1 + 1/2) = (4gL sin θ) / 3
Substitution
Block:
a_block = 10(sin 30° − 0.20 × cos 30°) = 10(0.500 − 0.20 × 0.866) = 10(0.500 − 0.173) = 10 × 0.327 = 3.27 m/s²
v_block² = 2 × 3.27 × 5.0 = 32.7 m²/s²
Cylinder:
v_cyl² = (4 × 10 × 5.0 × sin 30°) / 3 = (4 × 10 × 5.0 × 0.500) / 3 = 100/3 = 33.3 m²/s²
Calculation
v_block = √32.7 ≈ 5.72 m/s
v_cyl = √33.3 ≈ 5.77 m/s
Note: g = 10 m/s² is an exact problem-defined value. The angles sin 30° = 0.500 and cos 30° = 0.866 are standard exact/tabulated values. These do not limit the significant figures of the answer.
Final answer
The sliding block reaches the bottom at approximately 5.7 m/s. The rolling cylinder reaches the bottom at approximately 5.8 m/s. Despite kinetic friction acting on the block and essentially no friction on the cylinder, the velocities are close because the cylinder diverts energy into rotation.
Common trap
The high-frequency trap here is using the same friction coefficient for both the sliding block and the rolling cylinder (trap: treating rolling friction as equal to kinetic friction). On this incline, the block loses energy to kinetic friction (f_k = μ_k mg cos θ), while the rolling cylinder loses negligible energy to friction. Students who apply μ_k to the rolling case incorrectly reduce the cylinder's velocity.
A second trap: forgetting the μ_k cos θ term entirely for the sliding block and writing a = g sin θ (the smooth-incline answer). This overestimates the block's speed.
Similar NEET-style question
Two identical solid spheres are released simultaneously from rest at the top of two inclines of the same angle and length. One incline is smooth; the other is rough with μ_k = 0.15. On the smooth incline the sphere rolls without slipping. On the rough incline the sphere slides without rolling. Which sphere reaches the bottom first? (Answer: compare v² = 2gL sin θ/1.4 = 10gL sin θ/7 for the rolling sphere vs v² = 2gL(sin θ − μ_k cos θ) for the sliding sphere; the result depends on θ and μ_k. For a solid sphere I/mR² = 2/5, so the denominator is 1 + 2/5 = 1.4 — note that 4gL sin θ/3 is the CYLINDER result, where I/mR² = 1/2.)
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Static friction f_s opposes impending motion between surfaces in contact and is self-adjusting up to a maximum f_s ≤ μ_s N (where N is normal force, μ_s coefficient of static friction). Kinetic friction f_k = μ_k N opposes actual sliding motion (μ_k < μ_s typically). Rolling friction is much smaller than sliding friction.
-- NCERT Class 11 Physics, Ch. 4, p. 60More in Laws of Motion: 16 exam traps and mistakes · 9 formulas · 10 question patterns from its other lessons.
No question in our NEET 2020–2025 set targets this topic directly.
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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