Rotational kinetic energy

A flywheel spinning on a fixed axle still carries kinetic energy: it can do work as it slows down. That stored energy is its rotational kinetic energy, $K = \tfrac{1}{2}I\omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular speed in rad/s.

Three moves carry the topic. First, use the rotational formula, not the linear one: pair $I$ with $\omega$, never the plain mass $m$ with a speed $v$. Second, keep the formula whole: the leading $\tfrac{1}{2}$ and the square on $\omega$ are both required. Third, choose the moment of inertia that matches the body and its axis, since $I$ is what makes $K$ depend on shape.

If you took algebra-based Physics 1, none of this needs calculus: the formula and the moment-of-inertia shapes carry straight over. The work is keeping the linear and rotational sides apart, which is where the three mistakes below live.

Linear kinetic energy is $\tfrac{1}{2}mv^2$. Rotational kinetic energy follows the same pattern, with every linear quantity replaced by its rotational partner:

• mass $m$ becomes the moment of inertia $I$ (how the body resists changes in spin),
• linear speed $v$ becomes the angular speed $\omega$ (in rad/s),
• so $\tfrac{1}{2}mv^2$ becomes $K = \tfrac{1}{2}I\omega^2$.

The units are still joules: $I$ is in $\text{kg}\cdot\text{m}^2$ and $\omega$ is in rad/s, so $\tfrac{1}{2}I\omega^2$ comes out in $\text{kg}\cdot\text{m}^2/\text{s}^2 = \text{J}$. For example, a disk with $I = 4$ and $\omega = 3$ rad/s has $K = \tfrac{1}{2}(4)(3^2) = 18$ J.

Rotational kinetic energy is $K = \tfrac{1}{2}I\omega^2$: the linear formula with $I$ in place of $m$ and $\omega$ in place of $v$. Pair $I$ with $\omega$, and the answer comes out in joules.

$K$ is only as right as the $I$ you put in it. The moment of inertia depends on how the mass is spread around the axis, so each shape has its own formula:

• solid disk or cylinder, central axis: $I = \tfrac{1}{2}MR^2$,
• thin hoop or ring, central axis: $I = MR^2$,
• solid sphere, about a diameter: $I = \tfrac{2}{5}MR^2$,
• thin spherical shell, about a diameter: $I = \tfrac{2}{3}MR^2$.

Mass farther from the axis counts more, so a hoop ($MR^2$) has a larger moment of inertia than a solid disk of the same mass and radius ($\tfrac{1}{2}MR^2$). For example: a solid sphere with $M = 5$ kg and $R = 2$ m has $I = \tfrac{2}{5}(5)(2^2) = 8$. Spinning at $\omega = 2$ rad/s, its energy is $K = \tfrac{1}{2}(8)(2^2) = 16$ J. Using a disk's $\tfrac{1}{2}MR^2$ by mistake would give $I = 10$ and $K = 20$ J: a wrong shape, a wrong answer.

Match $I$ to the shape and axis before computing $K$. The same $M$ and $R$ give different $I$ (and different $K$) for a disk, a hoop, a solid sphere, and a shell.

The formula $K = \tfrac{1}{2}I\omega^2$ has three parts that all matter: the $\tfrac{1}{2}$, the moment of inertia $I$, and the square of the angular speed. Dropping the $\tfrac{1}{2}$ doubles the answer; writing $\tfrac{1}{2}I\omega$ instead of $\tfrac{1}{2}I\omega^2$ drops the square.

Because $\omega$ is squared, the energy is very sensitive to angular speed. Double $\omega$ and $K$ goes up by $2^2 = 4$, not $2$. Triple $\omega$ and $K$ goes up by $3^2 = 9$. Cut $\omega$ to a third and $K$ drops to a ninth. A flywheel spun to twice its speed stores four times the energy, not twice. That is why flywheels store so much energy at high spin rates.

Keep the $\tfrac{1}{2}$ and the square. Since $K$ depends on $\omega^2$, scaling the angular speed by a factor $k$ scales the energy by $k^2$: double $\omega$ and $K$ quadruples.

The most common slip is reaching for $\tfrac{1}{2}mv^2$ when a body is spinning. A spinning rigid body has no single linear speed: the rim moves fast, points near the axis barely move, and the axis itself does not move at all. There is no one $v$ to plug in. The moment of inertia $I$ rolls that whole spread of mass and distance into one number, and $\omega$ is the one speed every point shares. That is why a spinning body needs $K = \tfrac{1}{2}I\omega^2$.

A tempting trap is the rim speed $v = \omega R$. It is a real speed, but only the rim has it, so putting it into $\tfrac{1}{2}mv^2$ or $\tfrac{1}{2}Iv^2$ gives the wrong energy. The rim speed never appears in the rotational formula. In the same way, the plain mass $m$ is not the moment of inertia: writing $\tfrac{1}{2}m\omega^2$ puts mass where $I$ belongs.

For a purely spinning body, do not use $\tfrac{1}{2}mv^2$. There is no single $v$. Use $K = \tfrac{1}{2}I\omega^2$, with the moment of inertia and the one angular speed all points share.

Nearly every rotational-energy slip traces back to treating a spin like a slide, mangling the $\tfrac{1}{2}I\omega^2$ structure, or plugging in the wrong $I$.

• Linear swapUsing $\tfrac{1}{2}mv^2$ for a spinning body, dropping a rim speed $v = \omega R$ in for $\omega$, or putting mass $m$ where $I$ belongs. Fix: pair $I$ with $\omega$.
• Broken structureDropping the leading $\tfrac{1}{2}$, forgetting to square $\omega$, or assuming $K$ scales with $\omega$ rather than $\omega^2$. Fix: keep the $\tfrac{1}{2}$ and the square.
• Wrong $I$Grabbing the moment of inertia of the wrong shape or axis. Fix: match $I$ to the body before computing $K$.

Pair $I$ with $\omega$, keep the $\tfrac{1}{2}$ and the square, and pick the right $I$. Everything in this topic reduces to those three habits.