Rotational Dynamics & Rolling (IPhO Mechanics)

IPhO mechanics lesson on torque, angular momentum, moments of inertia, and rolling without slipping.

  • International Physics Olympiad preparation
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Rigid body questions become straightforward once you separate (1) translation of the centre of mass from (2) rotation about a chosen axis, and treat rolling as a constraint rather than “a friction force you guess”.

Learning objectives

By the end, you should be able to:

  • IPHO-M-ROT-01: combine ∑ F = ma_cm, ∑τ_cm = I_cmα and the correct contact constraint;
  • IPHO-M-ROT-02: infer static-friction direction from incipient slip and verify |fₛ| ≤ μₛN; and
  • check the chosen moment of inertia, sign convention and limiting behaviour.
Rolling body on an incline with force and torque informationA rigid body rolls down an incline. Weight acts vertically through the centre, the normal force is perpendicular to the surface, and static friction acts up the slope for passive rolling down the incline. The no-slip relation couples centre-of-mass acceleration and angular acceleration.mgNfaa = αRthen test |f| ≤ μsNα
Scroll diagram horizontally to read all labels.
For passive rolling down a fixed incline, translation, rotation, and a = αR must be solved together. Afterwards verify |f| ≤ μsN.
Prerequisites (quick refresh)

1. Definitions

  • Torque about point O: vector τ_O = vector r × vector F, and in general ∑ vector τ_O = d vector L_O/dt.
  • Moment of inertia about an axis: I = ∫ r_⊥² dm (distance perpendicular to the axis).
  • Parallel axis theorem: I_O = I_cm + md².
  • Angular momentum about the centre (fixed axis through centre): L_cm = I_cmω.
  • Rolling without slipping (radius R on a stationary surface):
    v_cm = ω R, a_cm = α R
    along the direction of motion.
  • Static friction fₛ: adjusts to enforce no slip, with |fₛ| ≤ μₛ N.
  • Instantaneous contact inertia: for a rolling body, I_P = I_cm + mR² about the contact point P.

2. Key ideas

  • Default template:
    ∑ Fₓ = m a_cm, ∑τ_cm = I_cmα, a_cm = α R.
  • On a stationary surface with pure rolling, static friction does no work at the contact (instantaneous contact speed is zero). Energy methods are often the fastest.
  • The rolling acceleration down an incline has a memorisable form:
    a = (g sin α)/(1 + I_cm/(mR²)).
  • Friction direction is set by the tendency to slip at the contact, not by the direction of the centre-of-mass velocity.
Three fast routes you should recognise
  1. Newton + torque about centre, then use a = α R.
  2. Energy + rolling: mgh = 1/2 mv² + 1/2 Iω² with v = ω R.
  3. One-coordinate energy or Lagrangian method, with x = Rθ built in from the start.

3. Detailed explanation

3.1 Translation plus rotation (why it is reliable)

For planar rolling with one translational coordinate x and one rotation angle θ, you can nearly always write:

  • translation of the centre of mass,
  • rotation about the centre of mass,
  • the no-slip constraint relating x and θ.

This avoids guessing friction and gives you a closed system of equations.

3.2 Why torque about contact is not the default

The contact point is an instantaneous centre of zero velocity, but it is generally not a fixed inertial pivot and may have nonzero acceleration. Therefore the fixed-axis equation ∑τ_P = I_Pα cannot be invoked solely because the contact point is instantaneously at rest.

Use centre-of-mass translation plus centre-of-mass torque as the reliable dynamics route. An angular-momentum calculation about a moving point needs the full transport relation and an explicit justification; it is rarely the shortest safe competition solution.

3.3 Static friction: direction and size

A clean approach:

  1. Decide the likely direction of slip if the surface were frictionless.
  2. Set friction opposite that slip tendency.
  3. Solve for fₛ from dynamics.
  4. Verify |fₛ| ≤ μₛ N. If not, rolling without slipping is impossible and the motion must involve slipping.

4. Common mistakes

  • Using v = ω R even though the body is slipping (constraint broken).
  • Using the wrong axis for I or forgetting the parallel axis theorem.
  • Assuming friction always opposes motion of the centre of mass.
  • Writing ∑τ = Iα about a point that is accelerating without justification.

5. Problem-solving tips

  1. State your sign convention early (clockwise positive or anticlockwise positive).
  2. If you compute friction, always check it against μₛ N.
  3. Use the limit I_cm → 0 as a quick checksum: you should recover particle-like motion.
  4. If a surface is moving (belt), do not assume static friction does no work in the lab frame.

6. Worked examples

1) Solid cylinder rolling down an incline (find a and minimum mu_s)

A solid cylinder (mass m, radius R) rolls without slipping down an incline of angle α.

For a solid cylinder, I_cm = 1/2 mR².

Translation along the slope (down-slope positive):

mg sin α - f = ma.

Rotation about the centre:

fR = I_cmα, a = α R

so

f = (I_cm/R²)a = 1/2 ma.

Substitute into translation:

mg sin α - 1/2 ma = ma ⇒ a = (2/3)g sin α.

Then

f = 1/2 ma = (1/3)mg sin α.

No-slip requires |f| ≤ μₛ N = μₛ mg cos α, hence

μₛ ≥ f/(mg cos α) = 1/3 tan α.
2) Pulling a cylinder: friction direction can flip (centre pull vs top pull)

A cylinder (mass m, radius R, I_cm = β mR²) rolls without slipping on rough horizontal ground. Compare two cases with the same horizontal force F to the right.

(A) Force applied through the centre. Translation:

F - f = ma.

Rotation about centre:

fR = I_cmα, a = α R ⇒ f = (I_cm/R²)a = β ma.

So

F-β ma = ma ⇒ a = F/(m(1 + β)), f = (β/(1 + β))F (left).

(B) Force applied tangentially at the top. Translation:

F + f = ma.

Torque about centre (clockwise positive for rolling right):

FR - fR = I_cmα, a = α R.

Solve the pair:

F-f = β ma, F + f = ma.

Add them:

2F = m(1 + β)a ⇒ a = 2F/(m(1 + β)).

Then

f = ma - F = F(1-β)/(1 + β) (right if β ≤ 1).

Takeaway: static friction enforces no slip, and its direction depends on the torque you apply.

7. Extension

1) Rolling on a moving belt: static friction can add energy

A cylinder (mass m, radius R, moment of inertia I_cm) is gently placed on a belt moving right at constant speed u. Initially v_cm = 0 and ω = 0. It slips at first (kinetic friction), then eventually reaches rolling without slipping relative to the belt.

During slipping, friction accelerates translation and produces angular acceleration:

a = F_f/m, α = (F_f R)/I_cm.

Starting from rest:

v = (F_f/m)t, ω = ((F_f R)/I_cm)t.

No slip relative to the belt means the contact point matches belt speed:

v + ω R = u.

Eliminate t to get final speeds (independent of F_f):

v_f = uI_cm/(I_cm + mR²), ω_f = umR/(I_cm + mR²).

Why this is a mind stretcher: even when slipping stops, the belt’s motor is the energy source, so mechanical energy of the cylinder can increase.

8. Practice and evidence

Practice
  • Do one incline rolling problem where you must compute fₛ and check |fₛ| ≤ μₛ N.
  • Do one pulled-cylinder or spool problem and justify the friction direction from slip tendency.
  • Treat IPHO-M-ROT-01/02 as passed only when both solutions include a declared angular sign convention and a no-slip feasibility check.
  • Then continue via the IPhO Mechanics Hub.
Syllabus and review details

No official syllabus alignment is listed for this lesson.

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