NEET Physics — Rotational Motion
Slipping, Spinning and Rolling – Complete Notes, Revision, Important Questions & Downloads
This topic covers seven subtopics of combined translatory and rotatory motion tested in NEET: Slipping (v ≠ Rω, sliding without rotation), Spinning (v_CM = 0, ω ≠ 0 — pure rotation), Rolling Without Slipping (v = Rω, contact point at rest, static friction does no work), Rolling on Inclined Plane (v = √(2gh/(1+K²/R²)), smaller K²/R² → faster), Kinetic Energy Distribution in Rolling (ring 50/50, disc 67/33, sphere 71/29), Motion of Connected Mass (hanging mass + rotating body, a < g), and Time Period of Compound Pendulum (T = 2π√(L_eq/g) where L_eq = k²/l + l).
9 SubtopicsHigh FrequencyEnergy Split
Expected QuestionsQ
1–2
Slipping, Spinning and Rolling generates 1–2 NEET questions per year, with rolling without slipping on an incline being the most frequent type.
Time Required⏱
3–4 hours
One session to understand the three motion types and derivation of rolling KE; one session for inclined plane rolling problems using energy conservation; one review session on K²/R² comparisons.
Difficulty⚡
Hard
Rolling without slipping requires integrating translational and rotational KE simultaneously. The K²/R² table must be memorised and correctly applied to identify which body reaches the bottom first.
NRI USA Curriculum GapUS
Medium
AP Physics C Mechanics covers rolling without slipping including energy split and inclined plane problems. AP Physics 1 covers rolling conceptually but not the full K²/R² derivation.
9Subtopics
10+Practice Questions
4Free Downloads
3–4 hrsPrep Time
NEET Weightage — Slipping, Spinning and Rolling
Rotational Motion (Chapter 7)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 2 | 8 | |
| 2022 | 1 | 4 | |
| 2021 | 2 | 8 | |
| 2020 | 1 | 4 | |
| 2019 | 1 | 4 | |
| 6-Year Total (2019–2024) | 5–8 | 20–32 |
K²/R² comparison for inclined plane time: body with smallest K²/R² reaches bottom first (solid sphere K²/R²=⅖ wins vs hollow sphere ⅔ vs disc ½ vs ring 1).
Friction in rolling without slipping does no work (zero relative velocity at contact point) — this is the critical distinction from slipping, where friction does work.
Spinning (pure rotation, v_CM = 0) is tested conceptually: if a body has ω but v = 0 (like a top spinning in place), K²/R² formula does not apply to describe motion down an incline.
Friction in rolling without slipping does no work (zero relative velocity at contact point) — this is the critical distinction from slipping, where friction does work.
Spinning (pure rotation, v_CM = 0) is tested conceptually: if a body has ω but v = 0 (like a top spinning in place), K²/R² formula does not apply to describe motion down an incline.
~1.3
Avg Questions / Year
20–32
Total Marks (6 yrs)
Direct
Pattern
Hard
Difficulty
Exam Strategy for Slipping, Spinning and Rolling
1
Identify the motion type before applying any formula Check the three conditions: (1) Slipping: v ≠ Rω (translational ≠ rotational velocity at rim) — friction acts until v = Rω. (2) Spinning: v_CM = 0, ω ≠ 0 — no translation, only rotation. (3) Rolling without slipping: v_CM = Rω exactly — the contact point is instantaneously at rest. Identify which condition the problem states or implies before choosing a formula.
2
For rolling without slipping incline problems, use energy conservation Set mgh = ½mv² + ½Iω² = ½mv²(1 + K²/R²) (substituting I = MK² and ω = v/R). Solve for v. The body with smallest K²/R² has the most translational KE and reaches the bottom with the highest velocity. Memorise the K²/R² table: Ring = 1, Hollow cylinder = 1, Disc = ½, Solid cylinder = ½, Hollow sphere = ⅔, Solid sphere = ⅖.
3
Use the contact-point velocity rule to check rolling condition For a rolling body, the contact point has instantaneous velocity = v_CM − Rω. Rolling without slipping: v_CM = Rω, so contact point is at rest (v_contact = 0). This is why friction does no work — zero displacement of the friction application point. In slipping, v_contact ≠ 0, friction acts, and energy is dissipated as heat.
Download Study Notes — Slipping, Spinning and Rolling
PDF · Cheat Sheet · MCQ Set · PYQRolling Motion — Full Notes
Complete notes covering sliding vs spinning vs rolling definitions, rolling constraint v = Rω, total KE derivation, K²/R² table for 6 standard shapes, inclined plane rolling problems, and 8 worked NEET-style examples.
Download PDFRolling Motion — Formula Sheet
One-page reference: rolling condition v = Rω, KE_total = ½mv²(1+K²/R²), K²/R² table, rolling speed down incline v = √(2gh/(1+K²/R²)), acceleration a = g sinθ/(1+K²/R²).
Download PDFRolling Motion — MCQ Practice
12 NEET-style MCQs on rolling KE split, inclined plane comparisons (which body first?), rolling vs slipping identification, and friction in rolling problems.
Download PDFRolling Motion — NEET-Style PYQ Practice
NEET-style practice on rolling motion — including energy split questions, K²/R² comparisons, and rolling without slipping on flat and inclined surfaces.
Download PDFSubtopics in Slipping, Spinning and Rolling
2-Column Table| Column A | Column B |
|---|---|
Rapid Revision — Slipping, Spinning and Rolling
Concept → Trap → Example1) Slipping
v ≠ Rω — Sliding with RotationSlipping: body both translates and rotates, but v_CM ≠ Rω. The contact point has non-zero velocity relative to the surface. Kinetic friction acts, decelerating the slip until rolling without slipping begins.
- When v_CM > Rω: body slides forward faster than it rotates — forward kinetic friction acts on the body (opposing relative motion at contact point, which is forward).
- When v_CM < Rω: body rotates faster than it translates — backward kinetic friction acts (contact point moves backward relative to surface).
- Common NEET trap: assuming friction does no work in slipping. Friction does work during slipping (contact point is in motion) and converts kinetic energy to heat. Zero work by friction applies only to rolling without slipping.
Example (NEET-style)A bowling ball thrown with v₀ = 6 m/s but ω₀ = 0 (pure translation, no rotation). v_CM > Rω initially, so the ball slips on the lane. Friction increases ω and decreases v until v = Rω, after which rolling without slipping begins. NEET may ask the final common rolling velocity or the friction force magnitude during slipping.
2) Spinning
v_CM = 0, ω ≠ 0 — Pure RotationSpinning: the centre of mass is stationary (v_CM = 0) while the body rotates about its own axis. All kinetic energy is rotational: KE = ½Iω². No translational KE. The contact point moves at velocity = Rω.
- A tyre spinning on ice (ω ≠ 0) with no forward motion (v_CM = 0) is pure spinning. Friction is kinetic and acts forward (contact point moves backward), tending to start translational motion.
- A top spinning in place on a horizontal surface is pure spinning (but the top also precesses — for NEET, treat as pure spinning, ignoring precession).
- Common NEET trap: applying rolling KE formula KE = ½mv²(1+K²/R²) to a spinning body where v = 0 — the formula gives KE = 0, which is wrong. Use KE = ½Iω² directly for pure spinning.
Example (NEET-style)A disc of mass 2 kg and radius 0.3 m spins at ω = 10 rad/s with v_CM = 0. KE = ½Iω² = ½ × (½×2×0.09) × 100 = ½ × 0.09 × 100 = 4.5 J. It has 0 translational KE since v_CM = 0.
3) Rolling Without Slipping
v = Rω — Standard NEET RollingRolling without slipping: v_CM = Rω exactly. Contact point is instantaneously at rest. Friction is static (zero work done). Total KE = ½mv² + ½Iω² = ½mv²(1 + K²/R²). Down an incline from height h: v = √(2gh / (1 + K²/R²)).
- K²/R² table (ratio I/MR²): Ring = 1, Hollow cylinder = 1, Disc = ½, Solid cylinder = ½, Hollow sphere = ⅔, Solid sphere = ⅖. Smallest K²/R² → most translational KE → highest v at bottom → reaches bottom first.
- Friction in rolling without slipping is static and does zero work — there is no energy dissipation. The body reaches the bottom faster than a slipping body of the same height.
- Common NEET trap: using mgh = ½mv² (ignoring rotational KE) for a rolling body. The correct energy equation is mgh = ½mv²(1 + K²/R²). Using only the translational term overestimates v by a factor of √(1 + K²/R²).
Example (NEET-style)A solid sphere (K²/R² = ⅖) rolls from height h = 1 m. v = √(2×10×1 / (1 + 2/5)) = √(20/1.4) = √14.3 = 3.78 m/s. A ring (K²/R² = 1): v = √(20/2) = √10 = 3.16 m/s. Solid sphere reaches the bottom faster. KE_translational/KE_total for sphere: 1/(1 + 2/5) = 5/7 ≈ 71%; rotational: 2/7 ≈ 29%.
4) Rolling on Inclined Plane
Velocity, Acceleration, Time on InclineFor a body rolling without slipping down an incline of angle θ from height h: v = √(2gh / (1+K²/R²)), acceleration a = g sinθ / (1+K²/R²), time t = (1/sinθ)√(2h(1+K²/R²)/g). Smallest K²/R² → highest v, highest a, shortest t.
- Rank by K²/R²: Solid sphere (⅖) < Disc (½) < Hollow sphere (⅔) < Ring (1). Solid sphere always reaches the bottom first; ring always last.
- Mass M and radius R cancel from all three formulas — only the shape (via K²/R²) determines the outcome. Equal mass and radius bodies are compared solely by their K²/R² ratio.
- Common NEET trap: concluding that the heavier or larger body reaches faster. Mass and radius do not affect the order — only the geometric shape matters.
Example (NEET-style)Ring vs solid sphere, same h = 0.5 m, θ = 30°. v_sphere = √(10×0.5/0.7) = √7.14 = 2.67 m/s. v_ring = √(10×0.5/1) = √5 = 2.24 m/s. a_sphere = 10×0.5/1.4 = 3.57 m/s². a_ring = 10×0.5/2 = 2.5 m/s². Sphere wins on both v and a.
5) Kinetic Energy Distribution in Rolling
Translational vs Rotational KE FractionsKE_trans/KE_total = 1/(1+K²/R²) and KE_rot/KE_total = (K²/R²)/(1+K²/R²). Ring (K²/R²=1): 50% trans, 50% rot. Disc (K²/R²=½): 66.7% trans, 33.3% rot. Solid sphere (K²/R²=⅖): 71.4% trans, 28.6% rot.
- The fractions depend only on K²/R² — not on mass, radius, velocity, or incline angle. The split is a constant property of each shape.
- For a ring, translational and rotational KE are always equal (50/50). This is a testable fact: if a ring rolls at speed v, KE_rot = ½mv² = KE_trans.
- Common NEET trap: computing KE_rot = ½Iω² separately without using the fraction formula, leading to arithmetic errors. Use the fraction formula for speed.
Example (NEET-style)A disc of mass 2 kg rolls at 6 m/s. KE_total = ½×2×36×(1+½) = 36×1.5 = 54 J. KE_trans = 54 × (2/3) = 36 J = ½×2×36 ✓. KE_rot = 54 × (1/3) = 18 J. Verify: ½Iω² = ½×(½×2×R²)×(6/R)² = ½×R²×36/R² = 18 J ✓.
6) Motion of Connected Mass
Hanging Mass + Rotating PulleyA mass m hangs from a string wound around a cylinder/disc of mass M and radius R. Acceleration a = mg / (m + I/R²). For a solid cylinder (I = ½MR²): a = mg / (m + M/2) < g. The mass falls slower than free fall due to rotational inertia of the cylinder.
- As I increases (heavier/larger cylinder), a decreases — the hanging mass falls more slowly.
- When I → 0 (massless pulley): a → g (free fall, as expected). When I → ∞: a → 0 (body does not move).
- Common NEET trap: using a = g for the hanging mass and ignoring rotational inertia of the cylinder/pulley. For a massless pulley, a = mg/m = g is correct; for a massive pulley with MI, the formula must include I/R².
Example (NEET-style)A disc (M = 4 kg, R = 0.2 m, I = ½×4×0.04 = 0.08 kg·m²) has a string wound around it. A 2 kg mass hangs from the string. a = 2×10 / (2 + 0.08/0.04) = 20 / (2 + 2) = 20/4 = 5 m/s². This is half of free-fall acceleration (g/2 in this case), meaning the rotational inertia of the disc effectively doubles the inertia of the system.
7) Time Period of Compound Pendulum
Rigid Body OscillationA compound pendulum (physical pendulum) — a rigid body of mass M pivoted at a distance l from its CM — oscillates with period T = 2π√(L_eq/g), where L_eq = I_pivot/(Ml) = (k² + l²)/l = k²/l + l. Here k = radius of gyration about CM, l = distance from pivot to CM.
- Minimum time period occurs when l = k (pivot at a distance equal to radius of gyration from the CM). At this pivot, T_min = 2π√(2k/g).
- Two values of l give the same time period: l₁ and l₂ where l₁ × l₂ = k². The corresponding pivot points are called conjugate points.
- Common NEET trap: using T = 2π√(l/g) (simple pendulum formula with l = pivot-to-CM distance) instead of the compound pendulum formula. This underestimates T because it ignores the MI about the CM.
Example (NEET-style)A uniform rod of mass M and length L suspended at one end: k² = ML²/12 ÷ M = L²/12. l = L/2 (CM at midpoint). L_eq = k²/l + l = (L²/12)/(L/2) + L/2 = L/6 + L/2 = 2L/3. T = 2π√(2L/3g). For L = 1.5 m: T = 2π√(1/g) = 2π/√10 ≈ 1.99 s ≈ 2 s — same as a simple pendulum of length 1 m.
US Curriculum Gaps — Slipping, Spinning and Rolling
Students from the US system studying for NEET should note these specific coverage gaps:AP Physics C covers rolling without slipping fully; AP Physics 1 covers it conceptually only
AP Physics C Mechanics teaches rolling without slipping with the full KE split (½mv² + ½Iω²) and inclined plane derivations. AP Physics 1 covers rolling qualitatively. NEET always tests the quantitative K²/R² inclined plane problems.
- Memorise the K²/R² table in rank order: Solid sphere (⅖) < Disc/Solid cylinder (½) < Hollow sphere (⅔) < Ring/Hollow cylinder (1). This ranking directly answers 'which body reaches the bottom first' questions.
- The formula a = g sinθ/(1 + K²/R²) for acceleration of a rolling body on an incline is not always explicitly derived in AP syllabi — memorise it or derive it from energy methods.
- The fact that friction does no work in rolling without slipping (because v_contact = 0) is a conceptual point that NEET has asked about directly.
The distinction between slipping and rolling is often blurred in US textbooks
US textbooks (e.g., Halliday/Resnick) treat slipping as a transition phase before rolling without slipping is established. NEET explicitly asks about all three states (slipping, spinning, rolling) as distinct scenarios with different friction regimes.
- Slipping: kinetic friction acts, energy is dissipated as heat (friction does work). The contact point moves relative to the surface.
- Rolling without slipping: static friction, zero work by friction, no energy dissipation. The contact point is at rest instantaneously.
- Spinning: the body rotates but does not translate. Kinetic friction at the contact point acts to initiate translation (like in a wheel-spin start from rest).
NEET-Style Practice Questions — Slipping, Spinning and Rolling
5 NEET-style questions1A solid sphere, a hollow sphere, and a disc all of equal mass M and radius R roll down an inclined plane from the same height. Which reaches the bottom first?K²/R² Comparison
Hollow sphere
Disc
Solid sphere
All reach at the same time
Time to reach the bottom increases with K²/R² (more rotational energy stored → less translational KE → slower). K²/R² values: Solid sphere = ⅖ = 0.4, Disc = ½ = 0.5, Hollow sphere = ⅔ ≈ 0.667. Smallest K²/R² → fastest. So solid sphere (⅖) arrives first, then disc (½), then hollow sphere (⅔). The mass and radius cancel — only the shape matters.
2A ring of mass 1 kg and radius 0.5 m rolls without slipping at v = 4 m/s on a flat surface. What fraction of its total KE is translational?KE Split
1/4
1/3
1/2
2/3
For a ring, K²/R² = 1. KE_total = ½mv²(1+K²/R²) = ½mv²×2 = mv². KE_translational = ½mv². Fraction = ½mv²/mv² = ½. For a ring (K²/R² = 1), the total KE splits exactly 50/50 between translational and rotational. This is a key result: for a ring, both types of KE are equal.
3A disc rolls without slipping. The velocity of the contact point and the topmost point of the disc are respectively:Contact Point Velocity
v, 0
0, v
0, 2v
v, 2v
For rolling without slipping: the contact point has velocity v_CM − Rω = v − v = 0 (instantaneously at rest — this is the rolling condition). The topmost point has velocity v_CM + Rω = v + v = 2v (both translational and rotational velocities add at the top). At the top, both motions are in the same direction; at the bottom (contact point), they are opposite and cancel. This is a fundamental property of rolling without slipping.
4A block slides down a frictionless incline and a solid sphere rolls down an identical incline from the same height. Which reaches the bottom with greater speed?Sliding vs Rolling
The solid sphere
The block
Both reach with the same speed
Depends on mass
Block (frictionless slide): mgh = ½mv_slide² → v_slide = √(2gh). Solid sphere (rolling): mgh = ½mv_roll²(1 + ⅖) → v_roll = √(2gh × 5/7) = √(10gh/7). Since 5/7 < 1: v_roll < v_slide. The block is faster because all potential energy converts to translational KE; the sphere splits energy between translational and rotational KE. The block wins whenever there is rolling (any K²/R² > 0).
5A tyre is spinning on an icy road (v_CM = 0, ω ≠ 0). As friction slowly brings the car into forward motion, which statement correctly describes the friction on the tyre?Spinning to Rolling Transition
Friction acts backward (opposing v_CM)
Friction acts forward (opposing contact-point backward sliding) and does positive work on the tyre
Friction does no work since the surface is icy
Friction is zero during the transition
In pure spinning (v_CM = 0, ω ≠ 0), the contact point moves backward (rotates backward at speed Rω). Kinetic friction opposes this backward motion, so it acts forward on the tyre. This friction does work (contact point is in motion) and provides both translational acceleration (increasing v_CM) and a braking torque (decreasing ω). The tyre transitions from spinning to rolling without slipping when v_CM reaches Rω.
Practice Questions — Slipping, Spinning and Rolling
Click "Reveal Answer" after attempting1A solid cylinder of mass 3 kg and radius 0.2 m rolls without slipping at 6 m/s. Calculate its total kinetic energy.
54 J
81 J
27 J
108 J
👁 Reveal Answer
Answer: 81 J. For a solid cylinder, K²/R² = ½. KE = ½mv²(1+K²/R²) = ½×3×36×(1+0.5) = 54×1.5 = 81 J. Breakdown: KE_trans = ½×3×36 = 54 J; KE_rot = ½×(½×3×0.04)×(6/0.2)² = ½×0.06×900 = 27 J. Total = 54 + 27 = 81 J ✓.
2A hollow sphere (K²/R² = ⅔) and a solid sphere (K²/R² = ⅖) roll down from height h = 1.4 m. Find the speed of each at the bottom.
v_solid = 3.16 m/s; v_hollow = 2.89 m/s
v_solid = 3.78 m/s; v_hollow = 3.50 m/s
Both 5.29 m/s
v_solid = 5.29 m/s; v_hollow = 4.58 m/s
👁 Reveal Answer
Answer: v_solid ≈ 3.78 m/s; v_hollow ≈ 3.50 m/s. v = √(2gh/(1+K²/R²)). Solid sphere: √(2×10×1.4/1.4) = √(28/1.4) = √20 = 4.47 m/s. Hollow sphere: √(28/(1+2/3)) = √(28/1.667) = √16.8 = 4.10 m/s. (Using g = 10 m/s².) Solid sphere is faster, confirming smaller K²/R² → higher v.
3A ball rolling without slipping on a flat surface has v_CM = 3 m/s. What are the velocities of the topmost point, the contact point, and a point at the same height as the centre but on the leading edge?
Top: 6 m/s; Contact: 0; Leading edge: √18 m/s ≈ 4.24 m/s
Top: 3 m/s; Contact: 3 m/s; Leading: 3 m/s
Top: 6 m/s; Contact: 3 m/s; Leading: 6 m/s
All = 3 m/s
👁 Reveal Answer
Answer: Top = 6 m/s (forward); Contact = 0; Leading edge = √(3²+3²) = 3√2 ≈ 4.24 m/s. The leading edge point is at the same height as the centre — its translational velocity is 3 m/s (forward) and its rotational velocity is 3 m/s (upward, since it is 90° from the contact point). Resultant = √(9+9) = 3√2 m/s at 45° forward-upward.
4Why does friction not do work during rolling without slipping, even though friction is present?
Friction is zero in rolling without slipping
The contact point is at rest so friction has no displacement
Friction acts perpendicular to motion so F·d = 0
Rolling without slipping always occurs on frictionless surfaces
👁 Reveal Answer
Answer: The contact point is at rest, so friction has no displacement. Work = F × displacement of the point of application. In rolling without slipping, the instantaneous velocity of the contact point = v_CM − Rω = 0. Since the contact point is instantaneously at rest, its displacement in any small interval is zero, so W = F × 0 = 0. Friction is present and may be non-zero in magnitude, but it does zero work.
Physics — Rotational Motion Revision Checklist
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Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ — Slipping, Spinning and Rolling
Notes · Downloads · Revision · Important QuestionsWhy does a body with smaller K²/R² reach the bottom of an incline faster?
From energy conservation: mgh = ½mv²(1 + K²/R²). Solving for v gives v = √(2gh/(1 + K²/R²)). As K²/R² decreases, the denominator decreases, making v larger. Physically, a smaller K²/R² means less energy is stored as rotational KE for the same translational speed, so more energy goes into translational motion and the body moves faster. The solid sphere (K²/R² = ⅖) stores the least rotational energy for its speed and reaches the bottom fastest.
In rolling without slipping, does friction ever do work?
On a flat horizontal surface with no external forces: no, friction does zero work because the contact point has zero velocity (instantaneously at rest). However, on an inclined plane, static friction does zero work on the rolling body because the contact point is still instantaneously at rest — but it provides a frictional torque that enables rolling. The work-energy theorem accounts for KE_total = mgh without any friction work term: friction is necessary for rolling but does not change the energy balance.
What is the minimum coefficient of friction required for rolling without slipping?
For a body rolling down an incline of angle θ: the friction force required is f = mKE_rot/KE_total × g sinθ = mg sinθ × K²/R² / (1 + K²/R²). For rolling without slipping, this friction must be static: μ_s ≥ f/N = tanθ × K²/R²/(1 + K²/R²). For a solid sphere (K²/R² = ⅖): μ_s_min = (2/5)tanθ / (7/5) = (2/7)tanθ. If μ_s < this value, the sphere slips instead of rolling.
How do you find the velocity of any point on a rolling body?
For rolling without slipping, the instantaneous velocity of any point = (translational velocity of CM) + (rotational velocity about CM). Translational: v_CM (horizontal, forward). Rotational: ω × r_relative (tangential to the circle of radius = distance from CM). Add these as vectors. Contact point: trans = v forward, rot = v backward (opposing) → net = 0. Top: trans = v forward, rot = v forward → net = 2v. Any other point: vector sum, magnitude = v√(2(1 + cosα)) where α is the angle from the bottom.
A rolling body reaches the bottom of an incline. Does it continue to roll on a flat surface indefinitely?
On an ideal frictionless flat surface: no. Without friction, the rolling constraint v = Rω cannot be maintained — the ball would slide instead. But if there is static friction on the flat surface, rolling without slipping continues indefinitely (no energy loss — friction does no work). On a real flat surface with rolling resistance (deformation of the surface and ball), the body gradually decelerates. For NEET: assume ideal rolling continues unless friction is explicitly mentioned as the cause of deceleration.
What is the acceleration of a rolling body on an inclined plane?
For rolling without slipping on an incline of angle θ: a = g sinθ / (1 + K²/R²). Derivation: the net force along the incline is mg sinθ − f (friction). The torque equation gives f = Iα/R = (MK²)(a/R²). Substituting into the force equation: Ma = Mg sinθ − MK²a/R², giving a(1 + K²/R²) = g sinθ. For solid sphere: a = g sinθ/(1 + 2/5) = 5g sinθ/7. For disc: a = g sinθ/(1 + 1/2) = 2g sinθ/3.
What happens to the angular velocity when a spinning tyre (v = 0, ω high) is placed on a road?
Initially: v_CM = 0, ω = ω₀ (pure spinning). Contact point slides backward at Rω₀. Kinetic friction acts forward on the tyre (opposing contact point backward motion). Friction: (1) accelerates v_CM forward (F_friction = ma), (2) decelerates ω (braking torque τ = FR, so α = −FR/I). Both v and Rω converge toward a common value v_f = Rω_f, at which point rolling without slipping begins. Using impulse-momentum: v_f = (friction force × time)/m; ω_f = ω₀ − (friction force × R × time)/I. Setting v_f = Rω_f gives ω_f and v_f.
Why is the KE of a rolling body greater than that of a sliding body at the same speed?
For a sliding body at speed v: KE = ½mv². For a rolling body at the same v_CM = v: KE = ½mv²(1 + K²/R²) > ½mv² (since K²/R² > 0 for all extended bodies). The rolling body has additional rotational KE (½Iω²) on top of the translational KE. Therefore, if both start from the same height h, the rolling body is slower than the sliding body (more energy goes into rotation), not faster.
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