Explicit physics instruction
Use K_total=½Mv²+½Iω² and v=rω. Rolling without slipping constrains v and ω so they cannot be chosen independently.
For a rolling body with M=2 kg, v=3 m/s, I=0.50 kg·m², and ω=6 rad/s, K_trans=9 J and K_rot=9 J, so total KE=18 J.
If slipping occurs, v=rω no longer applies exactly.
Core relations: K_total=½Mv²+½Iω² and v=rω.
Worked example
At r=0.50 m and v=2.0 m/s, rolling without slipping requires ω=v/r=4.0 rad/s.
Guided question
A wheel rolls at 3 m/s with radius 0.30 m. Find ω.
Independent work
- Solve three rotational-energy or rolling calculations.
- Draw an energy bar chart.
- State one condition where v=rω fails.
Misconception check
Rolling does not mean all kinetic energy is translational; rotation stores part of the total.
Formative check
What extra energy term must be included for a rolling object?
Reveal response
Rotational kinetic energy ½Iω².
Mentor evidence conference
Ask Archaemenes to audit rolling constraints, moment-of-inertia reasoning, and energy accounting.
Teacher guidance
Foundation: Calculate K_rot and v=rω.
Core: Combine translational and rotational energy.
Extended: Compare shapes using I and conservation residuals.