AP Physics C: Mechanics / Unit 6: Energy and Momentum of Rotating Systems
6.5 Kinetic Energy of a System with Translational and Rotational Motion
Notes
Rolling without slipping: $v_{cm}=R\omega$.
- Total KE = translational + rotational: $\displaystyle \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$.
- Objects with smaller $I$ (relative to $mR^2$) roll down ramps faster.
Practice
Released together on a ramp, which reaches the bottom first?
Smallest $\displaystyle \frac{I}{mR^2}$.
For rolling without slipping, $v=$
No slipping at the contact point.
A rolling ball has kinetic energy that is:
It moves and spins.
Total KE of a solid sphere rolling at speed $v$:
$\displaystyle \frac{1}{2}mv^2+\frac{1}{2}\cdot\frac{2}{5}mR^2\frac{v^2}{R^2}$.
Ready for more? Take the Unit 6 test.