KAHIBARO
Table of Contents
Rotation as Kinetic Energy
When an object rotates, its parts move with different linear speeds. A point farther from the axis moves faster than a point closer to the axis. Because kinetic energy depends on speed, a rotating object stores kinetic energy in the motion of all its parts. This energy is called rotational kinetic energy.
For a single particle of mass $m$ moving with speed $v$, the kinetic energy is
$$
K = \frac{1}{2}mv^2
$$
In a rotating rigid body, each small mass element has its own speed. If the body rotates with angular speed $\omega$, and a mass element is at distance $r$ from the axis, then its linear speed is
$$
v = r\omega
$$
So the kinetic energy of that small mass element is
$$
dK = \frac{1}{2}dm \, v^2 = \frac{1}{2}dm \, (r\omega)^2
$$
Adding the contributions from all parts of the object gives the total rotational kinetic energy:
$$
K_{\text{rot}} = \frac{1}{2}\omega^2 \int r^2 \, dm
$$
The integral $\int r^2 \, dm$ is the moment of inertia $I$, so the formula becomes
$$
K_{\text{rot}} = \frac{1}{2}I\omega^2
$$
Important formula:
$$
K_{\text{rot}} = \frac{1}{2}I\omega^2
$$
Rotational kinetic energy depends on both the moment of inertia $I$ and the angular speed $\omega$.
Why the Moment of Inertia Matters
Two objects rotating with the same angular speed do not necessarily have the same rotational kinetic energy. The object with the larger moment of inertia has more rotational kinetic energy. This is because more of its mass is effectively farther from the axis, so more energy is needed to keep it rotating at the same angular speed.
This is similar to translational motion, where larger mass means larger kinetic energy at the same speed. In rotation, the role of mass is played by moment of inertia.
The comparison is shown clearly in the table below.
| Translational motion | Rotational motion |
|---|---|
| Mass $m$ | Moment of inertia $I$ |
| Speed $v$ | Angular speed $\omega$ |
| $K = \frac{1}{2}mv^2$ | $K_{\text{rot}} = \frac{1}{2}I\omega^2$ |
Rotational Kinetic Energy of a Rigid Body
A rigid body is an object whose shape does not change while it rotates. Every point in the body has the same angular speed $\omega$, but different points have different linear speeds because $v = r\omega$.
This means that points farther from the axis contribute more strongly to the energy, since the energy contains $r^2$:
$$
K_{\text{rot}} = \frac{1}{2}\omega^2 \int r^2 \, dm
$$
So mass far from the axis increases rotational kinetic energy a lot more than mass near the axis.
Example with Particles
Suppose two small masses rotate together about the same axis. Let one mass be $m_1$ at distance $r_1$, and the other be $m_2$ at distance $r_2$. Both rotate with angular speed $\omega$.
The total rotational kinetic energy is
$$
K_{\text{rot}} = \frac{1}{2}m_1(r_1\omega)^2 + \frac{1}{2}m_2(r_2\omega)^2
$$
Factoring out $\omega^2$ gives
$$
K_{\text{rot}} = \frac{1}{2}(m_1r_1^2 + m_2r_2^2)\omega^2
$$
Since
$$
I = m_1r_1^2 + m_2r_2^2
$$
we again get
$$
K_{\text{rot}} = \frac{1}{2}I\omega^2
$$
This shows how the general formula comes naturally from adding the kinetic energies of all the particles.
Rolling Objects
If an object rolls without slipping, it often has two kinds of kinetic energy at the same time. The center of mass moves forward, and the object also rotates about its center.
Then the total kinetic energy is
$$
K_{\text{total}} = \frac{1}{2}Mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2
$$
Here, $M$ is the total mass, $v_{\text{cm}}$ is the speed of the center of mass, and $I_{\text{cm}}$ is the moment of inertia about the center of mass.
This chapter focuses only on the rotational part, but it is important to remember that rotating objects can also translate.
For rolling motion, do not forget that there may be both translational and rotational kinetic energy:
$$
K_{\text{total}} = \frac{1}{2}Mv_{\text{cm}}^2 + \frac{1}{2}I_{\text{cm}}\omega^2
$$
Units
The SI unit of rotational kinetic energy is the joule, just like any other form of energy.
From the formula
$$
K_{\text{rot}} = \frac{1}{2}I\omega^2
$$
the units are
$$
[I] = \text{kg}\cdot\text{m}^2
$$
and
$$
[\omega] = \text{rad/s}
$$
Radians are dimensionless, so
$$
[I\omega^2] = \text{kg}\cdot\text{m}^2/\text{s}^2 = \text{J}
$$
Common Cases
Using known moments of inertia, we can write rotational kinetic energy for common objects.
| Object | Moment of inertia | Rotational kinetic energy |
|---|---|---|
| Particle at distance $r$ | $I = mr^2$ | $K = \frac{1}{2}mr^2\omega^2$ |
| Solid disk about center | $I = \frac{1}{2}MR^2$ | $K = \frac{1}{4}MR^2\omega^2$ |
| Solid sphere about center | $I = \frac{2}{5}MR^2$ | $K = \frac{1}{5}MR^2\omega^2$ |
| Hoop about center | $I = MR^2$ | $K = \frac{1}{2}MR^2\omega^2$ |
| Rod about center, perpendicular to rod | $I = \frac{1}{12}ML^2$ | $K = \frac{1}{24}ML^2\omega^2$ |
These formulas show again that the distribution of mass changes the energy even when mass and angular speed are the same.
A Simple Numerical Example
Consider a wheel with moment of inertia
$$
I = 2.0\ \text{kg}\cdot\text{m}^2
$$
rotating with angular speed
$$
\omega = 3.0\ \text{rad/s}
$$
Then its rotational kinetic energy is
$$
K_{\text{rot}} = \frac{1}{2}I\omega^2
$$
$$
K_{\text{rot}} = \frac{1}{2}(2.0)(3.0)^2
$$
$$
K_{\text{rot}} = 1.0 \times 9.0 = 9.0\ \text{J}
$$
So the wheel has $9.0\ \text{J}$ of rotational kinetic energy.
Physical Meaning
Rotational kinetic energy is the energy associated with spinning motion. If the angular speed increases, the energy increases as $\omega^2$. This means doubling the angular speed makes the rotational kinetic energy four times larger.
Also, if the moment of inertia increases while angular speed stays the same, the energy increases in direct proportion.
Key relationships:
If $\omega$ doubles, then
$$
K_{\text{rot}} \to 4K_{\text{rot}}
$$
If $I$ doubles, then
$$
K_{\text{rot}} \to 2K_{\text{rot}}
$$
Visualizing the Energy Distribution
Because each part of the object contributes according to $r^2$, the outer regions are especially important in rotation. This is why a hoop and a solid disk of the same mass and radius do not have the same rotational kinetic energy at the same angular speed. The hoop has more of its mass at larger radius, so it has larger $I$, and therefore larger $K_{\text{rot}}$.
Final Formula to Remember
Rotational kinetic energy is the rotational form of kinetic energy for a rigid body. The central result is
$$
K_{\text{rot}} = \frac{1}{2}I\omega^2
$$
This is the rotational analogue of
$$
K = \frac{1}{2}mv^2
$$
This formula is used whenever an object spins about an axis and you know its moment of inertia and angular speed.