Rotational kinetic energy

3.1's formula $K = \tfrac{1}{2}mv^2$ handles anything that translates: a block, a ball, a cart. It doesn't handle a flywheel spinning on a fixed axle. The flywheel's center of mass is at rest, so 3.1 would say $K = 0$, yet the wheel is clearly carrying energy.

The fix is a second kinetic-energy formula for the rotational piece: $K_\text{rot} = \tfrac{1}{2}I\omega^2$. It plays the same role for a rotating body that $\tfrac{1}{2}mv^2$ plays for a translating one. The two are additive: for a rigid body that's doing both at once, like a rolling ball, $K_\text{tot} = K_\text{trans} + K_\text{rot}$.

Two warnings up front. (1) $K_\text{rot}$ scales with $\omega^2$, not $\omega$. Doubling the angular velocity quadruples the rotational KE; this is the headline trap of the topic. (2) $K_\text{rot}$ is a scalar. The direction of rotation matters for angular momentum (Topic 6.3), but for kinetic energy the $\omega^2$ scrubs the sign just like $v^2$ does for translational KE.

For a rigid body rotating with rotational inertia $I$ about a fixed axis at angular speed $\omega$:

Units: kg, m, rad/s, joules. The $\tfrac{1}{2}$ falls out the same way it did in 3.1: every particle in the body has translational $\tfrac{1}{2}m_iv_i^2$, with $v_i = r_i\omega$ for a rigid rotation. Adding those up gives $\tfrac{1}{2}\omega^2 \sum m_i r_i^2 = \tfrac{1}{2}I\omega^2$. So $K_\text{rot}$ isn't a new kind of energy; it's the sum of the translational KEs of all the pieces, repackaged into one formula that uses $I$ instead of tracking each piece by hand.

• $I$: rotational inertia about the axis (kg·m²). Always positive. Depends on both the total mass and how that mass is distributed.
• $\omega$: angular speed about the same axis (rad/s). Signed in calculations (CCW positive in this course), but the sign drops out of $K_\text{rot}$ because $\omega$ is squared.
• $K_\text{rot}$: rotational kinetic energy (J). Scalar, $\geq 0$, no direction.

The structure mirrors $\tfrac{1}{2}mv^2$ exactly: an inertia term times the speed squared, with a half out front. Replace $m$ with $I$ and $v$ with $\omega$ and you've moved from translation to rotation.

A flywheel on a fixed axle has only $K_\text{rot}$. A sliding block has only $K_\text{trans}$. A rolling ball has both at once: its center of mass is moving (translation) and the body is spinning about that center (rotation). The total kinetic energy is the sum:

Both terms, every time, for any rigid body that's doing both at once.

For rolling without slipping, the linear and angular speeds are tied together by $v_\text{cm} = R\omega$, so $\omega = v_\text{cm}/R$. That lets you write $K_\text{rot}$ in terms of $v_\text{cm}$ alone, which is what makes the ramp energy-conservation problems work out cleanly.

Limiting cases. If $I = 0$ (a point particle, or a massless cylinder), $K_\text{rot}$ disappears and you recover $v = \sqrt{2gh}$, the standard frictionless-slide answer. At the other extreme, a hoop has $I = MR^2$, so $K_\text{rot} = \tfrac{1}{2}Mv^2$, $K_\text{tot} = Mv^2$, and $v = \sqrt{gh}$, slower still. The hoop puts the most mass at the rim, so most of the $mgh$ budget goes into rotation and the least into translation.

Each error below mishandles rotational kinetic energy in a different way — how it scales with spin, what motion it counts, and whether it has a direction at all. You'll meet all three in your diagnostic.