In my physics class, we have seen experimentally that objects with lower $I$ values (like spheres) will reach the bottom of a ramp sooner and with a higher final velocity than objects with higher $I$ values (like a hoop) assuming total mass and radius are the same. However when looking at energy, I see that gravitational potential = translational KE + rotational KE. Since the sphere has a higher translational velocity, it has more translational kinetic energy. But since it’s moving faster, it must be rotating faster, and therefore have a higher $\omega$. Looking at the equation for rotational KE $= \frac{1}{2} I \omega^2$, I would assume that since $\omega$ is squared it would have a much bigger impact than that of decreasing the $I$ value.

Why does rotational kinetic energy not increase as quickly with objects that have a lower I value?

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    $\begingroup$ Please use MathJax to type equations and mathematical symbols. I edited the question to comply to these guidelines. $\endgroup$ Mar 8 at 5:14

5 Answers 5


I would assume that since $\omega$ is squared it would have a much bigger impact than that of decreasing the $I$ value.

Generalizing ratios between translational and rotational kinetic energy makes sense only when object is rotating without slipping. In that case translational and rotational velocities are proportional $v = \omega r$, and you should compare moment of inertia $I$ against mass $m$.

Total kinetic energy of a rotating body is combination of translational and rotational kinetic energy

$$K = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$$

Moment of inertia for most (simple) rigid objects can be generalized as $I = c m r^2$:

  • solid sphere has $c = 2/5$,
  • hollow sphere has $c = 2/3$,
  • solid cylinder has $c = 1/2$,
  • hollow cylinder has $c = 1$ etc.

Total kinetic energy can then be written as

$$K = \frac{1}{2} m v^2 + \frac{1}{2} c m (\omega r)^2 = \frac{1+c}{2} m v^2$$

From the work-energy theorem $\Delta K = W$ and if we assume that initial kinetic energy is zero, the final translational velocity is then

$$v = \sqrt{\frac{1}{1 + c} \frac{2 W}{m}} = \sqrt{\frac{2 W}{m + I/r^2}}$$

From this we can conclude that translational velocity $v$ will be larger for smaller values of $c$, which also means smaller moment of inertia $I$.


Your generalization is not true. You can change geometry and material density to get whatever result you want but these differences need to be fairly extreme to get results different from what you observed.

With regards to $\frac{1}{2}I\omega^2$, substitute in $\omega = \frac{v}{r}$. You get $\frac{1}{2}I\frac{v^2}{r^2}$. Compare that with $\frac{1}{2}mv^2$. You can see that the linear velocity has a numerically larger influence on the translational kinetic energy than the rotation kinetic energy so long as $I$ is comparable to or less than $m$. However, like I said earlier, things are not equal because m is not I so you can change geometry to achieve whatever result you want.

If you were doing fair comparisons you would need to compare solid cylinders/spheres against thin spherical shells and thin tubes of the same radius. But a thin tube only has double the moment of inertia for the same mass compared to a solid cylinder ($mr^2$ vs $\frac{1}{2}mr^2$), while a thin spherical shell has only 60% higher moment of inertia than a solid sphere of the same mass. But the mathematically ideal forms of the tube and spherical shell have almost infinitely less volume of material to hold that mass. That means in order to have the same mass as their solid counterparts, they have to be made of a much denser material or in the tube's case it could also be much longer. You are probably unlikely to pick a pair of such objects by random chance.

And if you allow the radi of the thin shape be different that doesn't make it easier to find appropriate objects to test either because you can't increase the radius of the shape to fit more mass into the object because increased radius reduces the numerical contribution of the mass to the rotational kinetic energy faster than the increased mass does.

So I guess the answer to your question boils down to: You were probably comparing similar sized objects made of similar density materials when the equations require that the one of the objects be comically denser or comically larger in size than the other.


I understand the reasoning: the rotational kinetic energy is $\frac{1}{2} I \omega^2$, and the squared $\omega$ should have a larger impact than $I$, and it seems that the rotational kinetic energy should increase due to the increased $\omega$ for decreased $I$. It does feel somewhat puzzling.

But actually the above reasoning needs some implicit assumptions. For example, what if actually $\omega \propto \frac{1}{\sqrt{I}}$ or something like that?

But due to energy conservation, $E=\frac{1}{2}m v^2+\frac{1}{2}I \omega^2$, where $E$ the total energy is the same for the sphere and the hoop. Since $v=\omega r$, solving the above equation for $\omega$ leads to $$\omega=\frac{\sqrt{E}}{\sqrt{\frac{1}{2} m r^2+\frac{1}{2} I}},$$ which is basically what we have implicitly assumed not to be.

Using the above formula for $\omega$ one gets the rotational kinetic energy equals $\frac{\frac{1}{2} I E}{\frac{1}{2} m r^2+\frac{1}{2} I}$, which decreases when $I$ decreases, as one observes experimentally, although $\omega$ is higher.

  • $\begingroup$ Reading this answer, I think I totally misunderstood OP's question. $\endgroup$
    – DKNguyen
    Mar 8 at 6:24

There is a finite amount of kinetic energy.
The kinetic energy is to be divided between translational kinetic energy $\frac 12 mv^2$ and rotational kinetic energy $\frac 12 I\omega^2$.
I think that you are assuming the no slipping constraint $v=r\omega$ so the rotational kinetic energy is $\dfrac 12 \dfrac {I}{r^2} v^2= \dfrac{k^2}{r^2}\frac 12mv^2$ where $k$ is the radius of gyration with $I=mk^2$.

So total kinetic energy = $\frac 12 mv^2+\dfrac {k^2}{r^2}\frac 12 m v^2$

If the moment of inertial gets small $k$ gets smaller and so the proportion of rotational kinetic energy decreases whilst the proportion of rotational energy incraeses.

  • $\begingroup$ "whilst the proportion of rotational energy increases". Did you mean translational kinetic energy? $\endgroup$
    – Bob D
    Mar 8 at 13:16

objects with lower I values will reach the bottom of a ramp sooner and with a higher final velocity than objects with higher I values

The total energy is:

$$E=\frac m2\,v^2+\frac I2\omega^2+m\,g\,h$$

with the rolling condition $~\omega=\frac vr~$

$$E=\frac m2\,v^2+\frac I2\frac{v^2}{r^2}+m\,g\,h$$

for $~t=0,v=0~\quad \Rightarrow E_0=m\,g\,h~$ and at the end of the ramp $~h=0\quad \Rightarrow E_F=\frac m2\,v^2+\frac I2\frac{v^2}{r^2}~$ with $~E_F=E_0~$ you obtain the final velocity squar

$$ v_F^2=2\,{\frac {m\,g\,h\,{r}^{2}}{{r}^{2}m+{\it I}}}$$

thus if $~I_1 < I_2\quad\Rightarrow v_{F1}^2 > v_{F2}^2$


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