KAHIBARO
Table of Contents
Kinetic Energy in Rolling Motion
When an object rolls, its motion is a combination of two kinds of motion at the same time. The whole object moves forward, and the object also spins about its center. Because of this, a rolling object has two parts of kinetic energy, translational kinetic energy and rotational kinetic energy.
If the center of mass moves with speed $v$, the translational kinetic energy is
$$
K_{\text{trans}} = \frac{1}{2}mv^2
$$
If the object rotates with angular speed $\omega$ about its center of mass, the rotational kinetic energy is
$$
K_{\text{rot}} = \frac{1}{2}I\omega^2
$$
Here, $m$ is the mass of the object and $I$ is its moment of inertia about the axis through its center.
So the total kinetic energy of a rolling object is
$$
K_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
$$
For a rolling object, the total kinetic energy is the sum of translational and rotational kinetic energy:
$$
K_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
$$
Rolling Without Slipping and Energy
In rolling without slipping, the translational speed and angular speed are related by
$$
v = \omega R
$$
where $R$ is the radius of the object. This lets us rewrite the rotational part in terms of the forward speed.
Substitute $\omega = \frac{v}{R}$ into the rotational kinetic energy:
$$
K_{\text{rot}} = \frac{1}{2}I\left(\frac{v}{R}\right)^2
$$
Then the total energy becomes
$$
K_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\frac{v^2}{R^2}
$$
or
$$
K_{\text{total}} = \frac{1}{2}v^2\left(m + \frac{I}{R^2}\right)
$$
This form is very useful because it shows how both the mass and the distribution of mass affect the motion.
For rolling without slipping, always use
$$
v = \omega R
$$
This relation connects translational motion and rotational motion.
Why Different Objects Roll Differently
Two objects can have the same mass and radius, but if their mass is distributed differently, they have different moments of inertia. This changes how much of their total kinetic energy is rotational.
An object with a larger moment of inertia puts more of its energy into rotation. That leaves less for translational motion if the total energy is fixed. This is why some objects roll down a slope faster than others.
For example, a hoop has more of its mass far from the center than a solid disk. So the hoop has a larger moment of inertia, and more of its kinetic energy is rotational.
Energy Partition
It is often helpful to compare how the total kinetic energy is divided.
Using $v = \omega R$, the rotational energy becomes
$$
K_{\text{rot}} = \frac{1}{2}\frac{I}{R^2}v^2
$$
So the ratio between rotational and translational kinetic energy is
$$
\frac{K_{\text{rot}}}{K_{\text{trans}}} = \frac{I/R^2}{m} = \frac{I}{mR^2}
$$
This ratio depends only on the shape of the object.
For many common rolling objects:
| Object | Moment of inertia $I$ | Total kinetic energy |
|---|---|---|
| Hoop | $mR^2$ | $K = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$ |
| Solid disk or solid cylinder | $\frac{1}{2}mR^2$ | $K = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$ |
| Solid sphere | $\frac{2}{5}mR^2$ | $K = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$ |
From this table, the hoop has the greatest fraction of energy in rotation, while the solid sphere has less.
Rolling Down a Height
If a rolling object starts from rest at height $h$ and rolls without slipping, gravitational potential energy changes into both translational and rotational kinetic energy.
The energy equation is
$$
mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
$$
Using $v = \omega R$,
$$
mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\frac{v^2}{R^2}
$$
Solving for $v$ gives
$$
v^2 = \frac{2gh}{1 + \frac{I}{mR^2}}
$$
and therefore
$$
v = \sqrt{\frac{2gh}{1 + \frac{I}{mR^2}}}
$$
This shows clearly that the speed depends on the moment of inertia.
For an object rolling from height $h$ without slipping,
$$
mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
$$
and
$$
v = \sqrt{\frac{2gh}{1 + \frac{I}{mR^2}}}
$$
Comparison of Common Shapes
If several objects start from the same height and roll without slipping, the one with the smallest value of $\frac{I}{mR^2}$ reaches the bottom first.
| Object | $\frac{I}{mR^2}$ | Speed at bottom |
|---|---|---|
| Solid sphere | $\frac{2}{5}$ | Largest |
| Solid disk | $\frac{1}{2}$ | Smaller |
| Hoop | $1$ | Smallest |
So a solid sphere rolls faster than a solid disk, and a solid disk rolls faster than a hoop.
Physical Picture
The total motion of a rolling object can be imagined as forward motion of the center plus spinning around the center.
The arrow $v$ shows the translational motion of the center of mass. The curved arrow shows the rotation. Both contribute to the total kinetic energy.
A Useful Special Case
If an object slides without rotating, only translational kinetic energy appears:
$$
K = \frac{1}{2}mv^2
$$
If an object spins in place without moving forward, only rotational kinetic energy appears:
$$
K = \frac{1}{2}I\omega^2
$$
But for rolling motion, both are present together. That is what makes rolling motion different from simple sliding or simple spinning.
Final Insight
Rolling motion stores energy in two ways. The object moves forward as a whole, and it also rotates about its center. Because of this, the same amount of gravitational potential energy can lead to different speeds for different shapes. The key reason is the moment of inertia, which tells how strongly the object resists rotational motion.