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UY1: Rolling Motion

Learn pure rolling kinematics and energy decomposition, with derivations, checks, and worked UY1-level examples.

  • University Physics Year 1
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Learning objectives

  • Use vector and differential or integral calculus to express physical change and accumulation.
  • Use complex numbers, linear algebra, and differential equations to solve coupled physical models.
  • Choose an efficient mathematical method and validate the result physically.
Cluster role + pathway links

This page is the core concept page for pure rolling constraints in the UY1 mechanics pathway.

Scope boundary: pure rolling first

This page is for pure rolling without slipping. If the contact condition breaks, do not force v_CM = Rω.

For mixed slip/roll problems:

  • start with force/torque equations and friction direction checks,
  • then test whether pure rolling is satisfied.

1) At a glance

  • Prerequisite: translational kinematics, angular speed/acceleration, moment of inertia, kinetic energy.
  • Outcomes:
    • Relate translation and rotation in pure rolling.
    • Use v_CM = Rω and a_CM = Rα correctly.
    • Write total kinetic energy of a rolling rigid body.
  • Key result: K = (1/2)Mv_CM² + (1/2)I_CMω², v_CM = Rω
Common traps (constraints + friction)
  • The rolling constraint v_CM = Rω only holds for pure rolling (no slip).
  • Static friction can be non-zero in pure rolling and still do no work (because the contact point has zero instantaneous velocity on a rigid surface).
  • “Instantaneous axis at the contact point” is a velocity/energy trick; don’t blindly treat the contact point as a fixed pivot for dynamics.

2) Setup

Model and assumptions:

  • Rigid body of radius R, mass M.
  • Motion on a fixed surface with pure rolling (no slipping).
  • Symmetric body (sphere, cylinder, hoop), so one principal axis is enough.

Coordinates/signs:

  • Positive x is forward translation of the centre of mass (CM).
  • Positive rotation is the sense consistent with forward rolling.
  • ω and α are about the CM unless stated otherwise.

Rolling constraint:

s = Rθ ⇒ v_CM = ds/dt = Rω, a_CM = dv_CM/dt = Rα

Direction note:

  • The constraint links the component of CM motion along the rolling direction to the signed ω about the symmetry axis. If you flip your sign convention for ω, v_CM = Rω must flip with it.

3) Core method and derivation

A) Why the contact point can be used as an instantaneous axis

At one instant, the point in contact with the ground has zero velocity relative to the ground for pure rolling. So the motion is equivalent to instantaneous rotation about that contact point P.

This gives an alternative kinetic-energy form:

K = (1/2)I_Pω²

Using parallel-axis theorem:

I_P = I_CM + MR²

So

K = (1/2)(I_CM + MR²)ω²; = (1/2)I_CMω² + (1/2)M(Rω)²; = (1/2)I_CMω² + (1/2)Mv_CM²

This is the standard translation + rotation decomposition.

Exam tip: instantaneous axis (what it does and doesn’t do)

Use the contact point as an instantaneous axis to relate velocities and to rewrite kinetic energy using I_P = I_CM + MR².

Don’t automatically use it as a pivot for ∑ τ = Iα unless you have checked the conditions for that equation (fixed axis / appropriate point choice).

B) Checks

  • Units:
    • Mv² has unit kg m²s⁻² = J.
    • Iω² has unit (kg m²)(s⁻²) = J.
  • Consistency limit:
    • If ω → 0, then v_CM → 0 (from rolling constraint), so K → 0.
  • Sign check:
    • Forward rolling with our sign convention gives v_CM,ω > 0, so energy terms are positive.

4) Worked example

A solid cylinder (I_CM = (1/2)MR²) rolls without slipping at v_CM = 4.0 m s⁻¹. Find:

  1. ω
  2. Fraction of total kinetic energy that is rotational.

Solution:

ω = v_CM/R

Total kinetic energy:

K = (1/2)Mv² + (1/2)((1/2)MR²)(v/R)²; = (1/2)Mv² + (1/4)Mv²; = (3/4)Mv²

Rotational part is Kᵣₒₜ = (1/4)Mv², so

Kᵣₒₜ/K = (1/4)/(3/4) = 1/3

Answer:

  • ω = v/R
  • Rotational fraction = 1/3.

5) Practice set (with hints + answers)

  1. A hoop rolls at speed v. Write total kinetic energy in terms of M and v. Hint: for a hoop, I_CM = MR². Answer: K = (1/2)Mv² + (1/2)Mv² = Mv².

  2. A sphere has ω = 12 rad s⁻¹ and radius 0.10 m in pure rolling. Find v_CM. Hint: use the rolling constraint directly. Answer: v_CM = Rω = 1.2 m s⁻¹.

  3. True/false: In pure rolling, static friction must be zero at all times. Hint: think of rolling with acceleration. Answer: False. Static friction can be non-zero and still do no net work at the instantaneous contact point.

6) Summary and next steps

Pure rolling links translation and rotation through v_CM = Rω and a_CM = Rα. The kinetic energy is exactly the sum of translational and rotational parts, or equivalently (1/2)I_Pω² about the contact point.

MOI links (common rolling bodies)

Next steps:

If you only have 20-30 minutes, do one question each on:

  1. kinematic constraint usage,
  2. energy split by body type (hoop/cylinder/sphere),
  3. incline acceleration comparison.