A.4.12 (HL)—Rotational kinetic energy

Rotational energy

For a rigid body rotating about a fixed axis,

$$E_k=\frac12I\omega^2=\frac{L^2}{2I}$$

Combine forms of motion

A rolling object may have translational kinetic energy of its centre of mass and rotational kinetic energy about its centre. Include both when accounting for total kinetic energy.

Worked example from local Question Bank row 31644

A rod of weight $36.0\,\mathrm{N}$ lowers its centre of mass by $5.00/2=2.50\,\mathrm{m}$ and has $I=30.6\,\mathrm{kg\,m^2}$. If the gravitational transfer becomes rotational kinetic energy,

$$E_k=(36.0)(2.50)=90.0\,\mathrm{J}$$
$$90.0=\frac12(30.6)\omega^2\Rightarrow\omega=2.43\,\mathrm{rad\,s^{-1}}$$

The energy result is in joules; angular speed is in radians per second.

Common trap

Do not use $\frac12mv^2$ alone for a rotating body when its rotational motion contributes energy.