# Moment of Inertia

This may sound like a stupid question and probably is, but I'll take my chances anyway.

If the moment of inertia of an N particle distribution is defined as

$$I=\sum_{n=1}^N m_ir_i^2,$$

where $r_i$ are the perpendicular distances repect to the axis of rotation, then if I consider a solid disk of radius R with its mass concentrated in two points in the circunference, then the moment of inertia is $I=m_1 r_1^2+m_2 r_2^2=(m_1+m_2)R^2$, right?

If I consider the center of mass of those two points, shouldn't the moment of inertia would be the same? If so,

$$r_{cm}=\frac{m_1r_1+m_2r_2}{m_1+m_2},$$

so

$$I=\frac{\|m_1r_1+m_2r_2\|^2}{m_1+m_2}=\frac{(m_1^2+m_2^2)R^2+2m_1m_2 r_1\cdot r_2}{m_1+m_2}.$$

So, what am I doing wrong?

What trows me out about the first calculation, is that no matter how the points are distributed along the circumference, the moment of inertia is the same. This can't be right, or can it?

Any help will be much appretiated.

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D H
Staff Emeritus
Your first calculation of the moment of inertia implicitly assumes the axis of rotation passes through the center of the disk. The moment of inertia about the axis normal to the disk and passing through the center of the disk will be the same so long as the two masses are placed on the edge of the disk.

The second calculation (which you are not doing correctly) uses a different axis of rotation (normal to the disk, but passing through the center of mass). This result will of course differ from the first because the two axes of rotation are not the same.

daniel_i_l
Gold Member
This may sound like a stupid question and probably is, but I'll take my chances anyway.
Rule: Never be afraid to ask a question!

Well, maybe I forgot to tell you that the center of the disk is the axis of rotation (the disk is rotating because of the friction), and yes, the masess are constrained to the edge of the disk.

So you are telling me that the first calculation is the correct one?
If so, why it doesn't matter how the two points are distributed (they could be in oposit diameters or very close to each other)?

On the other hand, the center of mass measured from the center of the disk is indeed

$$\vec{r}_{cm}=\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}.$$

Then how come I am calculating the moment of inertia with repect to an axis passing trough the center of mass, if the $r$ in the definition is the distance of a point to the axis of rotation (wich in this case is the origin)?

Finally, could you tell me why my calculation is wrong?

$$\begin{array}{rcl}I & = & (m_1+m_2)r_{cm}^2=(m_1+m_2)\left\|\dfrac{m_1\vec{r_1}+m_2\vec{r_2}}{m_1+m_2}\right\|^2 \\ \\ & = &\dfrac{m_1^2 \|\vec{r}_1\|^2+m_2^2\|\vec{r}_2\|^2+2m_1m_2\vec{r}_1\cdot\vec{r}_2} {m_1+m_2}\\ \\&=&\dfrac{(m_1^2 +m_2^2)R^2+2m_1m_2 \vec{r}_1\cdot \vec{r}_2}{m_1+m_2}.\end{array}$$

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D H
Staff Emeritus
Finally, could you tell me why my calculation is wrong?
Here.

$$\begin{array}{rcl}I & = & (m_1+m_2)r_{cm}^2\end{array}$$
This equation is treating the masses as if they were concentrated at the center of mass. They aren't concentrated there, so this equation is wrong.

Isn't that the point of the center of mass?

D H
Staff Emeritus
No. You need to read up on inertia some more.

Think about it this way. Suppose you have a whole bunch of masses, all the same. You distribute them evenly around the circumference of the disk. This configuration will have quite a bit of rotational inertia. The center of mass of the configuration will be at the center of the disk (i.e., $r_{cm}=0$). Now consider another configuration, with all of the masses piled at the center of the disk. The center of mass of this configuration is the same ($r_{cm}=0$) but this configuration has no rotational inertia.

Oh, now I see... But why the moment of inertia does not depend on the distributions of the masses (in this particular example, of course)?

---EDIT---

the moment of inertia only depends on the geometry of the object (even if the mass distribution is not homogeneous)?

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