Rotational kinetic energy

Rotational kinetic energy is the energy a system has because it is spinning, equal to one half the rotational inertia times the square of the angular velocity. It is a scalar measured in joules, and it adds to translational kinetic energy rather than replacing it.

K = \frac{1}{2}I\omega^2

EK 6.1.A.1 ties it to the rotational inertia and angular velocity of the rigid system, and the AP Physics 1 equation sheet prints the equation. The structure mirrors $K = \frac{1}{2}mv^2$ , with $I$ in place of $m$ and $\omega$ in place of $v$ .

It is a scalar. EK 6.1.A.3 says so outright. There is no direction and no sign: a wheel spinning clockwise and the same wheel spinning counterclockwise at the same rate carry identical rotational kinetic energy.

A rolling object has two terms. EK 6.1.A.1.ii states that the total kinetic energy of a rigid system is the sum of the rotational kinetic energy due to rotation about its center of mass and the translational kinetic energy due to the linear motion of that center of mass:

K_{\text{tot}} = \tfrac{1}{2}Mv_{\text{cm}}^2 + \tfrac{1}{2}I\omega^2

Dropping the rotational half is the standard error in a rolling-down-a-ramp energy problem. It is also why a hoop and a sphere released together do not arrive together: the hoop puts a larger share of the same energy into spin.

Spinning in place still counts. EK 6.1.A.2 notes that a rigid system can have rotational kinetic energy while its center of mass is at rest, because the individual points within it are moving and each has kinetic energy.

$\omega$ must be in radians per second. The constraint that links the two terms for a rolling body is rolling without slipping.

Back to the physics glossary