What is the velocity of the point of contact in pure rolling motion?

In pure rolling without slipping, the point of contact with the ground is instantaneously at rest: v_contact = v_cm - R * omega = 0. Therefore, static friction does zero work during pure rolling on a stationary surface.

Which body reaches the bottom of an inclined plane first when released from rest?

The body with the smallest ratio of (k^2 / R^2) has the greatest acceleration and reaches first. The race order is: Solid Sphere (k^2/R^2 = 0.4) wins first, followed by Solid Disc / Cylinder (0.5), then Hollow Sphere (0.67), and finally Ring / Hollow Cylinder (1.0) reaches last.

What is the minimum coefficient of friction required for pure rolling down an incline of angle theta?

The minimum coefficient of static friction is mu_min = (tan theta) / (1 + R^2 / k^2). If the actual friction coefficient is less than this value, the body slips while rolling.

§ 6.6NCERT Class 11 · Physics · Chapter 6

Rotational Dynamics & Rolling Motion

QCC Notes · Class 11 Physics NCERT-aligned Updated 26 Sep 2026 9 min read

Real-world motion frequently blends translational translation with rotational spin. A bicycle wheel, a bowling ball, or a rolling marble combines both worlds. Rolling Motion is a composite motion: rotation about the moving centre of mass coupled with the translational motion of the centre of mass.

1. Equations of Rotational Kinematics & Work-Energy Theorem

For constant angular acceleration α, rotational kinematics mirrors linear kinematics directly:

Linear Kinematics Equation (a = const)	Rotational Kinematics Equation (α = const)
v = u + a t	ω = ω₀ + α t
s = u t + ½ a t²	θ = ω₀ t + ½ α t²
v² = u² + 2 a s	ω² = ω₀² + 2 α θ
s_nth = u + ½ a (2n − 1)	θ_nth = ω₀ + ½ α (2n − 1)

Work-Energy Theorem in Pure Rotation

The work done by a net torque τ equals the change in rotational kinetic energy:

W_rot = ∫ τ dθ = ½ I ω_f² − ½ I ω_i² = ΔK_rot

2. Kinematics of Pure Rolling (Rolling Without Slipping)

Consider a circular body (cylinder, disc, or sphere) of radius R rolling on a stationary flat surface:

Translational Component: Every point of the body translates forward with velocity v_cm.
Rotational Component: The body rotates with angular velocity ω about the centre of mass. A point at the top moves forward at +Rω, while the lowest point of contact moves backward at −Rω.
No-Slip Condition: For pure rolling, the instantaneous velocity of the contact point with the surface must be ZERO:

v_contact = v_cm − R ω = 0 ⇒ v_cm = R ω and a_cm = R α

Velocities at Different Points on a Rolling Wheel

The velocity of any point P at angle θ from the lowest contact point is given by:

v_P = 2 v_cm sin(θ / 2)

Bottom Contact Point (θ = 0): v = 0 (Instantaneous center of zero velocity).
Center of Wheel (CM): v = v_cm.
Top Point (θ = 180°): v = 2 v_cm (Moves at twice the car's speed!).

3. Total Kinetic Energy of a Rolling Body

By König's theorem, the total kinetic energy K of a rolling body is the sum of its translational kinetic energy of the centre of mass and its rotational kinetic energy about the centre of mass:

K_total = K_trans + K_rot = ½ M v_cm² + ½ I_cm ω²

Using I_cm = M k² (radius of gyration) and ω = v_cm / R:

K_total = ½ M v_cm² [ 1 + (k² / R²) ]

Fractional breakdown of energy:

Translational Fraction: K_trans / K_total = 1 / [1 + (k²/R²)]
Rotational Fraction: K_rot / K_total = (k²/R²) / [1 + (k²/R²)]

4. Rolling Down an Inclined Plane & The Great Incline Race

When a circular body of mass M, radius R, and radius of gyration k rolls down an incline of angle θ and height h from rest:

Acceleration: a = (g sin θ) / [ 1 + (k² / R²) ]

Velocity at Bottom: v = √[ (2 g h) / (1 + (k² / R²)) ]

Time to Reach Bottom: t = (1 / sin θ) √[ (2 h / g) (1 + k²/R²) ]

The Famous Incline Race Comparison

Because acceleration a is inversely proportional to [1 + (k²/R²)], the body with the smallest k²/R² value has the largest acceleration and finishes the race FIRST:

Rolling Body	k² / R² Value	Acceleration (a)	Velocity at Bottom (v)	Finishing Order in Race
Solid Sphere	2 / 5 = 0.40	(5/7) g sin θ ≈ 0.714 g sin θ	√(10gh / 7) ≈ 1.195 √(gh)	1st (Fastest!)
Solid Disc / Cylinder	1 / 2 = 0.50	(2/3) g sin θ ≈ 0.667 g sin θ	√(4gh / 3) ≈ 1.155 √(gh)	2nd
Spherical Shell (Hollow)	2 / 3 ≈ 0.67	(3/5) g sin θ = 0.600 g sin θ	√(6gh / 5) ≈ 1.095 √(gh)	3rd
Ring / Hollow Cylinder	1.00	(1/2) g sin θ = 0.500 g sin θ	√(gh) = 1.000 √(gh)	4th (Slowest!)

SURPRISING RESULT: The acceleration and race outcome are completely independent of the mass M and radius R of the bodies! A tiny marble of mass 10 g and radius 5 mm will beat a massive solid disc of 100 kg and radius 1 m every single time, because race speed depends solely on the geometric mass distribution parameter (k²/R²).

Minimum Friction for Pure Rolling

To prevent slipping down the incline, static friction f_s must satisfy f_s ≤ μ_s N:

μ_s,min = (tan θ) / [ 1 + (R² / k²) ]

5. Visualising Pure Rolling Kinematics

Figure 6.6 — Vector Velocity Addition in Pure Rolling Motion

  [Pure Translation]        [Pure Rotation]             [Combined Pure Rolling]
      v_cm                      +Rω                           v_top = 2 v_cm
       -->                       -->                              ====>
    +-------+                 +-------+                         +-------+
    |   O-->| v_cm     +      |   O   |                =        |   O-->| v_cm
    +-------+                 +-------+                         +-------+
       -->                       <--                                o
      v_cm                      -Rω                           v_bottom = 0
                                                       (Instantaneous Rest Point!)
      

Exam Read-out: Because v_bottom = 0, the contact point never slides relative to the ground. Consequently, static friction does ZERO work on a stationary plane, and total mechanical energy is conserved!

Step-by-Step Solved Numerical Examples

Solved Example 1: Kinetic Energy Fractions of a Rolling Solid Sphere

Problem: A uniform solid sphere of mass M rolls without slipping along a horizontal floor with velocity v. Find (1) the translational kinetic energy, (2) the rotational kinetic energy, and (3) the percentage of total energy stored in rotational motion.

K_rot / K_total = 2 / 7 ≈ 28.57%

Step 1: Write Individual Kinetic Energies:
Translational KE: K_trans = ½ M v².
Rotational KE: K_rot = ½ I ω².
For a solid sphere, I = (2/5) M R² and pure rolling condition gives ω = v/R.
K_rot = ½ [(2/5) M R²] (v/R)² = (1/5) M v².

Step 2: Total Kinetic Energy:
K_total = K_trans + K_rot = (½ + 1/5) M v² = (7/10) M v².

Step 3: Rotational Percentage:
Fraction = K_rot / K_total = (1/5) / (7/10) = 2 / 7 ≈ 28.57%.
(Translational fraction is 5 / 7 ≈ 71.43%).

Solved Example 2: Incline Velocity Comparison (Disc vs Ring)

Problem: A solid disc and a thin ring of the same mass and radius are released simultaneously from rest from the top of an inclined plane of height h = 3 m. What are their velocities upon reaching the bottom? (Take g = 9.8 m/s²).

v_disc = 6.26 m/s · v_ring = 5.42 m/s

Step 1: Formula for Velocity at Bottom:
v = √[ (2 g h) / (1 + k²/R²) ].

Step 2: Solid Disc (k²/R² = 1/2):
v_disc = √[ (2 × 9.8 × 3) / (1 + 0.5) ] = √[ 58.8 / 1.5 ] = √39.2 ≈ 6.26 m/s.

Step 3: Thin Ring (k²/R² = 1):
v_ring = √[ (2 × 9.8 × 3) / (1 + 1.0) ] = √[ 58.8 / 2.0 ] = √29.4 ≈ 5.42 m/s.
The disc reaches the bottom faster and with higher velocity.

6. Solved NEET & JEE Practice Questions

Q1. A solid cylinder of mass M and radius R rolls without slipping down an inclined plane of angle θ. Its linear acceleration is:

A. g sin θ
B. (2/3) g sin θ
C. (1/2) g sin θ
D. (5/7) g sin θ

Answer: B

Explanation: For a solid cylinder, k²/R² = 1/2. Acceleration a = (g sin θ) / [1 + (k²/R²)] = (g sin θ) / [1 + 1/2] = (2/3) g sin θ.

Q2. In pure rolling motion on a flat horizontal road, the work done by the force of friction is:

A. Positive
B. Negative
C. Zero
D. Infinite

Answer: C

Explanation: In pure rolling without slipping, the instantaneous velocity of the contact point is zero (v_contact = 0). Since there is zero displacement of the point of application during contact, the work done by static friction is identically zero.

Q3. A solid sphere, a hollow sphere, a solid cylinder, and a circular ring of the same mass are rolled down an incline from the same height. Which one arrives at the bottom LAST?

A. Solid sphere
B. Hollow sphere
C. Solid cylinder
D. Circular ring

Answer: D

Explanation: Time taken t ∝ √(1 + k²/R²). The ring has the largest k²/R² value (1.0), meaning it has the lowest acceleration and will reach the bottom last.

Q4. A wheel starts from rest and rotates with a constant angular acceleration of 3 rad/s². The total angular displacement in the first 4 seconds is:

A. 12 rad
B. 24 rad
C. 36 rad
D. 48 rad

Answer: B

Explanation: θ = ω₀ t + ½ α t² = 0 + ½ (3) (4)² = ½ (3)(16) = 24 rad.

7. Frequently Asked Questions (FAQs)

Why does friction act uphill when a body rolls down an inclined plane?

As the body tries to slide down under the component of gravity (Mg sin θ), static friction acts uphill to oppose relative motion at the contact point. This uphill friction produces a counter-clockwise torque about the center of mass (τ = f · R), which provides the angular acceleration α required to maintain the pure rolling constraint a_cm = R·α.

What happens if the incline is frictionless (μ = 0)?

Without friction, there is no torque about the center of mass (τ = 0), so angular acceleration is zero (α = 0). The body does not roll at all; it simply slips/slides down as a frictionless block with acceleration a = g sin θ, converting all potential energy into purely translational kinetic energy.

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