Unit 6 · Topic 07

Rolling motion and moving axes

Rolling combines center-of-mass translation with rotation about the center of mass.

Unit 6RollingOpen navigation

Concept 01

Instantaneous contact

Essential The minimum you should retain

For rolling without slipping, |v_cm|=R|ω|.

UnderstandInterpret and connect

Signs depend on axis orientation and angular convention.

DeepenFormulation and conditions

The contact point is instantaneously at rest relative to an ideal ground.

ExploreConnections for further study

The constraint fails during slipping.

Mathematical relation

Rolling without slipping

vcm=Rω,acm=Rα,apend=gsinβ1+IcmMR2 v_{cm}=R\omega,\quad a_{cm}=R\alpha,\quad a_{pend}=\frac{g\sin\beta}{1+I_{cm}/(MR^2)}
Represents

Rolling constraint and slope acceleration.

Physical interpretation

Mass distribution changes acceleration.

DeepenVariables, conditions, and checks

Variables

R
radius; usual unit: m
β
slope angle; usual unit: rad or °

Conditions of application

  • Rolling without slipping; compatible signs.

Dimensional check

m/s and m/s².

A schematic wheel moves over the ground and marks its center, contact, and the axial direction of ω.v_cmRCMv_contact=0ω ⊗

In rolling without slipping, the CM moves with v_cm=Rω v_{cm}=R\omega , contact is instantaneously at rest, and clockwise rotation means ω points into the plane.

Concept 02

Translating and rotating

Essential The minimum you should retain

Total K adds (1/2)Mv_cm² and (1/2)I_cmω².

UnderstandInterpret and connect

At equal v_cm, a ring and disk partition energy differently.

DeepenFormulation and conditions

Using contact-point I and also adding translation may double-count.

ExploreConnections for further study

Energy compares speeds down a slope.

Worked example

Energy of a rolling disk

A solid 2.0 kg, 0.30 m disk rolls without slipping at 4.0 m/s.

Given
  • M=2.0 kg
  • R=0.30 m
  • v_cm=4.0 m/s
Target

Find total K.

  1. Constraint

    ω=v_cm/R=13.33 rad/s.

  2. Inertia

    I_cm=(1/2)MR²=0.090 kg·m².

  3. Translation

    K_trans=16 J.

  4. Rotation

    K_rot=8 J.

  5. Total

    K=24 J.

Conclusion

K_total=24 J.

Conceptual stacked bars compare translation and rotation.K_transring K_rotK_transdisk K_rot

At equal v_cm, the ring has a larger rotational share than the disk because I_cm is larger.

Concept 03

Distribution and acceleration

Essential The minimum you should retain

On a slope, a_cm=g sinβ/[1+I_cm/(MR2)] a_{cm}=g\sin\beta/[1+I_{cm}/(MR^2)] .

UnderstandInterpret and connect

Larger I_cm/(MR²) reduces acceleration at the same angle.

DeepenFormulation and conditions

The relation assumes rolling without slipping and negligible losses.

ExploreConnections for further study

Mass cancels for geometrically similar bodies.

Concept 04

Enforcing the constraint

Essential The minimum you should retain

Static friction may supply needed torque without dissipating energy at ideal contact.

UnderstandInterpret and connect

Its direction depends on the slipping tendency caused by other forces.

DeepenFormulation and conditions

It does not always point up or down the slope.

ExploreConnections for further study

If required friction exceeds μ_sN, slipping begins and the model changes.

Concept review

Common errors

Each warning includes a concrete way to review the reasoning, not only an incorrect-answer marker.

Imposing f_s=μ_sN in every rolling case.

Static friction adjusts up to its maximum when needed.

Concluding the whole wheel is at rest because instantaneous contact is at rest.

The CM translates while the wheel rotates.