A rolling object has two kinds of kinetic energy at the same time: translational kinetic energy because its center
of mass moves, and rotational kinetic energy because it spins about its center of mass.
1. Total kinetic energy
The total kinetic energy of an object rolling without slipping is
\[
K_{\mathrm{total}}
=
K_{\mathrm{trans}}+K_{\mathrm{rot}}.
\]
The translational part is
\[
K_{\mathrm{trans}}=\frac12Mv_{\mathrm{cm}}^2.
\]
The rotational part is
\[
K_{\mathrm{rot}}=\frac12I_{\mathrm{cm}}\omega^2.
\]
Therefore,
\[
K_{\mathrm{total}}
=
\frac12Mv_{\mathrm{cm}}^2
+
\frac12I_{\mathrm{cm}}\omega^2.
\]
2. No-slip condition
Rolling without slipping means that the point touching the ground is instantaneously at rest. This gives the
relation
\[
v_{\mathrm{cm}}=\omega R.
\]
Equivalently,
\[
\omega=\frac{v_{\mathrm{cm}}}{R}.
\]
3. Inertia factor \(k\)
Many rolling objects can be written in the form
\[
I_{\mathrm{cm}}=kMR^2.
\]
Substituting \(\omega=v_{\mathrm{cm}}/R\) gives
\[
K_{\mathrm{rot}}
=
\frac12(kMR^2)\left(\frac{v_{\mathrm{cm}}}{R}\right)^2.
\]
\[
K_{\mathrm{rot}}=\frac12kMv_{\mathrm{cm}}^2.
\]
Thus, the total rolling kinetic energy becomes
\[
K_{\mathrm{total}}
=
\frac12Mv_{\mathrm{cm}}^2(1+k).
\]
4. Common rolling objects
| Object |
\(I_{\mathrm{cm}}\) |
\(k\) |
Energy implication |
| Solid sphere |
\(I_{\mathrm{cm}}=\frac25MR^2\) |
\(k=\frac25\) |
Less energy goes into rotation, so it rolls faster down a ramp. |
| Solid cylinder / disk |
\(I_{\mathrm{cm}}=\frac12MR^2\) |
\(k=\frac12\) |
More rotational energy than a solid sphere. |
| Thin spherical shell |
\(I_{\mathrm{cm}}=\frac23MR^2\) |
\(k=\frac23\) |
Even more energy is stored in rotation. |
| Hoop / thin ring |
\(I_{\mathrm{cm}}=MR^2\) |
\(k=1\) |
Largest rotational share among these examples. |
5. Rolling down from height
If a rolling object starts from rest at height \(h\), and losses are ignored, gravitational potential energy becomes
total kinetic energy:
\[
Mgh=K_{\mathrm{total}}.
\]
Using the simplified rolling form,
\[
Mgh=\frac12Mv^2(1+k).
\]
The mass cancels:
\[
gh=\frac12v^2(1+k).
\]
Solving for final speed,
\[
v=\sqrt{\frac{2gh}{1+k}}.
\]
This shows why objects with different shapes roll down the same ramp with different speeds. The object with the
smaller \(k\) has less rotational inertia and a larger final translational speed.
6. Required height for a target speed
The same equation can be rearranged to solve for the height needed to reach a target speed:
\[
h=\frac{v^2(1+k)}{2g}.
\]
7. Contact-point interpretation
At any instant, a rolling object can also be treated as rotating about the contact point. By the parallel axis
theorem,
\[
I_P=I_{\mathrm{cm}}+MR^2.
\]
Then the kinetic energy can be written as
\[
K=\frac12I_P\omega^2.
\]
This gives the same result as the translation-plus-rotation form:
\[
\frac12I_P\omega^2
=
\frac12(I_{\mathrm{cm}}+MR^2)\omega^2.
\]
\[
K=
\frac12I_{\mathrm{cm}}\omega^2
+
\frac12M(\omega R)^2.
\]
\[
K=
K_{\mathrm{rot}}+K_{\mathrm{trans}}.
\]
8. Worked example: solid sphere from \(2.5\ \mathrm{m}\)
For a solid sphere,
\[
k=\frac25=0.4.
\]
If it rolls from rest through a vertical drop of
\[
h=2.5\ \mathrm{m},
\]
then
\[
v=\sqrt{\frac{2gh}{1+k}}.
\]
Using \(g=9.80665\ \mathrm{m/s^2}\),
\[
v
=
\sqrt{\frac{2(9.80665)(2.5)}{1+0.4}}.
\]
\[
v
\approx
5.92\ \mathrm{m/s}.
\]
The total kinetic energy at the bottom is
\[
K_{\mathrm{total}}=Mgh.
\]
Key idea: rolling without slipping splits energy into translation and rotation. A larger \(k\) means a larger
rotational share and a smaller final speed for the same drop height.
