Energy and Momentum of Rotating Systems

A rotating object has rotational kinetic energy $K = \frac{1}{2}I\omega^2$, even when its center of mass isn't moving, because every piece of it is moving. If an object both moves and spins, its total kinetic energy is its translational kinetic energy plus its rotational kinetic energy about the center of mass.

A torque that acts while an object turns through an angle does work on it: $W = \int \tau\,d\theta$, or $W = \tau\Delta\theta$ for a constant torque. This work is the area under a graph of torque versus angular position. The net work done by all the torques equals the change in the object's rotational kinetic energy.

Angular momentum is the rotational version of momentum: $L = I\omega$ for a rigid object and $\vec{L} = \vec{r} \times \vec{p}$ for a particle, which has angular momentum about a point even when it moves in a straight line. A net torque changes it, $\tau_{\text{net}} = \frac{dL}{dt}$, and the angular impulse $\int \tau\,dt$ (the area under a torque–time graph) equals $\Delta L$.

If the net external torque on a system is zero, its total angular momentum stays constant. A spinning skater who pulls in their arms lowers their rotational inertia, so their angular velocity goes up, and when a lump of clay sticks to a pivoted rod, the angular momentum about the pivot is the same before and after.

An object rolling without slipping has $v_{\text{cm}} = r\omega$ and $a_{\text{cm}} = r\alpha$. Its kinetic energy is split between translation and rotation, which is why a hoop rolls down a ramp more slowly than a solid disk. Static friction helps it roll but does no work on it. If the object slips, kinetic friction takes mechanical energy out, and $v_{\text{cm}} = r\omega$ no longer holds.

For a satellite orbiting a much more massive object, the gravitational potential energy is $U = -\frac{GMm}{r}$, which is zero when they're infinitely far apart. In a circular orbit the speed, kinetic energy and potential energy all stay constant. In an elliptical orbit only the total energy and angular momentum stay constant, so the satellite speeds up as it gets closer. Escape velocity, $v_{\text{esc}} = \sqrt{\frac{2GM}{r}}$, is the launch speed that makes the total mechanical energy zero.