Rotational kinetic energy is the energy a rigid body has because it is spinning about an axis.
It is the rotational analogue of translational kinetic energy.
1. Main formula
For a rigid body rotating with angular speed \(\omega\), the rotational kinetic energy is
\[
K_{\mathrm{rot}}=\frac12 I\omega^2.
\]
Here \(I\) is the moment of inertia about the rotation axis. Angular speed must be in radians per second.
2. Where the formula comes from
A small mass element \(m_i\) at distance \(r_i\) from the axis has linear speed
\[
v_i=\omega r_i.
\]
Its kinetic energy is
\[
K_i=\frac12 m_i v_i^2
=\frac12 m_i(\omega r_i)^2.
\]
Adding all mass elements gives
\[
K_{\mathrm{rot}}
=
\sum_i \frac12 m_i\omega^2 r_i^2
=
\frac12\omega^2\sum_i m_i r_i^2.
\]
Since
\[
I=\sum_i m_i r_i^2,
\]
the result is
\[
K_{\mathrm{rot}}=\frac12 I\omega^2.
\]
3. Angular-speed conversions
The calculator accepts several ways to describe the rotation rate. Each one is converted to \(\omega\):
| Input |
Meaning |
Conversion to \(\omega\) |
| Angular speed |
\(\omega\) in rad/s |
\(\omega=\omega\) |
| Frequency |
\(f\) revolutions per second |
\(\omega=2\pi f\) |
| Period |
\(T\) seconds per revolution |
\(\omega=\frac{2\pi}{T}\) |
| Rotational speed |
\(n\) revolutions per minute |
\(\omega=2\pi\frac{n}{60}\) |
4. Rolling objects
A rolling object has both translational and rotational kinetic energy. The total kinetic energy is
\[
K_{\mathrm{total}}
=
K_{\mathrm{trans}}+K_{\mathrm{rot}}.
\]
The translational part is
\[
K_{\mathrm{trans}}=\frac12 Mv_{\mathrm{cm}}^2.
\]
The rotational part is
\[
K_{\mathrm{rot}}=\frac12 I_{\mathrm{cm}}\omega^2.
\]
For rolling without slipping,
\[
v_{\mathrm{cm}}=\omega r.
\]
Therefore,
\[
K_{\mathrm{total}}
=
\frac12 M(\omega r)^2+\frac12 I_{\mathrm{cm}}\omega^2.
\]
5. Moment of inertia connection
Moment of inertia measures how mass is distributed around the axis. A larger \(I\) means more energy is stored
at the same angular speed.
| Object |
Axis |
Moment of inertia |
| Solid disk / cylinder |
Central symmetry axis |
\(I=\frac12 MR^2\) |
| Thin hoop / ring |
Central symmetry axis |
\(I=MR^2\) |
| Solid sphere |
Diameter |
\(I=\frac25 MR^2\) |
| Thin spherical shell |
Diameter |
\(I=\frac23 MR^2\) |
| Thin rod |
Center axis, perpendicular to rod |
\(I=\frac{1}{12}ML^2\) |
| Thin rod |
End axis, perpendicular to rod |
\(I=\frac13 ML^2\) |
6. Units
In SI units,
\[
I\rightarrow \mathrm{kg\,m^2},
\qquad
\omega\rightarrow \mathrm{rad/s}.
\]
Then
\[
K_{\mathrm{rot}}\rightarrow \mathrm{J}.
\]
Radians are dimensionless, so \(\mathrm{kg\,m^2/s^2}\) is the joule.
7. Worked example
A flywheel has
\[
I=2.5\ \mathrm{kg\,m^2},
\]
and spins with angular speed
\[
\omega=120\ \mathrm{rad/s}.
\]
Use
\[
K_{\mathrm{rot}}=\frac12 I\omega^2.
\]
Substitute the values:
\[
K_{\mathrm{rot}}
=
\frac12(2.5)(120)^2.
\]
\[
K_{\mathrm{rot}}
=
1.25(14400)
=
18000\ \mathrm{J}.
\]
Therefore,
\[
\boxed{K_{\mathrm{rot}}=1.8\times10^4\ \mathrm{J}}.
\]
Key idea: rotational kinetic energy grows with the square of angular speed. If \(\omega\) doubles,
\(K_{\mathrm{rot}}\) becomes four times larger.
