UY1: More About Rolling Motion

1) At a glance

• Prerequisite: pure rolling relations, torque, I_CM, static/kinetic friction.
• Outcomes:
  • Test when pure rolling is valid.
  • Compare rolling objects with different moments of inertia.
• Solve slipping-to-rolling transition under kinetic friction.
• Key result (general incline, no slip): a_CM = (g sin θ)/(1 + I_CM/MR²)

2) Setup

We use two regimes:

• Pure rolling: contact point instantaneously at rest, static friction.
• Slipping + rolling: contact point moves relative to surface, kinetic friction.

For a rigid body of radius R on an incline:

• Along slope: Mg sin θ downhill.
• Normal: N = Mg cos θ.
• Friction opposes relative motion at contact.

Friction-direction quick test (1D along the surface):

• Define vᵣₑₗ = v_CM - Rω for the contact point’s velocity relative to the ground along the surface.
• If vᵣₑₗ > 0, the contact point tends to slip forward, so kinetic/static friction points backward (opposes that slip).
• Pure rolling means vᵣₑₗ = 0.

3) Core method and key derivations

A) No-slip condition for a solid sphere on incline

From the sphere result:

f = (2/7)Mg sin θ

No slip requires f ≤ μₛN = μₛMg cos θ, so:

tan θ ≤ (7/2)μₛ

If this fails, sliding occurs and v_CM ≠ Rω.

B) Which object arrives first?

For objects released from same height h, no slip:

v_bottom = square root of (2gh/(1 + I/MR²))

Smaller I/(MR²) means larger acceleration and larger bottom speed.

Typical values:

• Solid sphere: I/(MR²) = 2/5
• Solid cylinder: I/(MR²) = 1/2
• Hoop (thin hollow cylinder): I/(MR²) = 1

Order from fastest to slowest: solid sphere, solid cylinder, hoop.

C) Slipping to rolling on a horizontal floor (corrected)

A uniform sphere gets initial v₀ at its centre, with ω₀ = 0. While slipping:

• Translation: v dot = -μₖ g so v(t) = v₀-μₖgt.
• Rotation about CM: τ = fₖR = μₖMgR, so

α = τ/I = μₖMgR/((2/5)MR²) = (5μₖ g)/2R

Hence

ω(t) = ω₀ + α t = ((5μₖ g)/2R)t

Rolling starts at t = tᵣ when v = Rω:

v₀-μₖgtᵣ = R(((5μₖ g)/2R)tᵣ)

v₀ = (7/2)μₖgtᵣ ⇒ μₖ = 2v₀/7gtᵣ

This fixes the common coefficient error in ω(t).

D) Checks

• Units: μₖ = 2v₀/(7gtᵣ) is dimensionless.
• Sign consistency in slipping case above: v decreases, ω increases until matching.
• Limit μₖ → 0: no rotational spin-up, so transition time diverges, which is physically consistent.

4) Worked examples

Example 1: race down incline

Three bodies start from rest at same height and roll without slipping:

• Solid sphere
• Solid cylinder
• Hoop

Using a = g sin θ/(1 + I/(MR²)), the one with smallest I/(MR²) has largest a.

Answer: solid sphere first, hoop last.

Example 2: find μₖ from transition time

Given v₀ = 1.0 m s⁻¹ and tᵣ = 0.50 s for a uniform sphere,

μₖ = 2v₀/7gtᵣ = 2(1.0)/7(9.81)(0.50) ≈ 5.8 × 10⁻²

So μₖ ≈ 0.058.

5) Practice set (with hints + answers)

1. For a rolling solid sphere on incline, derive the minimum μₛ to prevent slip. Hint: use f = (2/7)Mg sin θ and f ≤ μₛMg cos θ. Answer: μₛ ≥ 2/7 tan θ.

2. A hoop and a sphere start from same height. Which has larger rotational energy fraction at bottom? Hint: compare Kᵣₒₜ/K using K = (1/2)Mv² + (1/2)Iω². Answer: hoop has larger rotational fraction.

3. A sphere has v₀ = 2.0 m s⁻¹, reaches rolling in 0.40 s. Find μₖ. Hint: use μₖ = 2v₀/(7gtᵣ). Answer: μₖ ≈ 0.146.

6) Summary and next steps

Rolling problems split cleanly into two cases: no-slip (constraint valid) and slipping (constraint invalid until transition). Always check which regime you are in before writing equations.

Next steps:

• Angular Momentum
• Sphere On An Incline
• Moment Of Inertia (why β = I/(MR²) controls the motion)
• Back To Mechanics (UY1)