The moment of inertia of a solid cylinder is $\frac{1}{2}mr^2$, and the moment of inertia of a composite object is the sum of the moments of inertia of the objects it consists of minus the moment of inertia of any air space. That would imply that the moment of inertia of a cylindrical tube is $\frac{1}{2}m(R^2-r^2)$, but it is in fact $\frac{1}{2}m(R^2+r^2)$. Why? Is this a situation in which the rule I described cannot be applied, and if so why?

  • 14
    $\begingroup$ In short: you used $m$ to denote three different masses. $\endgroup$ Commented Oct 1, 2023 at 7:48

5 Answers 5


In the formula $I_{\rm solid}= \frac 12 mr^2$ the "$m$" is $m_{\rm solid}=\rho \pi r^2L$, where $L$ is the length and $\rho$ the density.

So $$ I_{\rm solid}[r]= \frac 12 \rho \pi L r^4. $$ In the hollow cylinder the "$m$" is $m_{\rm hollow}= \pi \rho (R^2-r^2)L$.

So $$ I_{\rm hollow} = m_{\rm hollow}\frac 12 (R^2+r^2)\\= \pi \rho (R^2-r^2)L\frac 12 (R^2+r^2)\\ = \frac 12 \pi \rho L(R^4-r^4)\\ = I_{\rm solid}[R]- I_{\rm solid} [r], $$
as you expected.

  • $\begingroup$ D-d-d-d-diiference of squares! (well, 4th powers). $\endgroup$
    – Yakk
    Commented Oct 2, 2023 at 15:03

But it is.

As Kamil Maciorowski notes in the comments, your problem is that you're using the letter $m$ to denote three different masses. So let's fix that.

Specifically, let $M$ be the mass of the bigger cylinder (with radius $R$), $m$ the mass of the smaller cylinder (with radius $r$) and $m_t = M - m$ the mass of the tube formed by removing the smaller cylinder from the bigger one.

We'll also need to assume that all the cylinders and tubes have the same length and are made of the same homogeneous substance (since the formulas you used are only valid for homogeneous objects). In particular, that implies that the masses of the cylinders are proportional to their cross section areas, and thus to their radii squared. That is to say, $M = cR^2$, $m = cr^2$ and $m_t = M - m = c(R^2 - r^2)$, where $c$ is some constant.

(For cylinders / tubes of length $L$ and made of a homogeneous material of density $\rho$, $c = \rho \pi L$, but that doesn't really matter here. All that matters is that $c$ is the same for all the objects.)

In particular, this means that the moments of inertia of the two cylinders and the tube, as given by the formulas you quoted, are:

  • Bigger cylinder: $\frac12 M R^2 = \frac12 c R^4$
  • Smaller cylinder: $\frac12 m r^2 = \frac12 c r^4$
  • Tube: $\frac12 m_t (R^2 + r^2) = \frac12 c (R^2 - r^2) (R^2 + r^2)$

Now, to show that the moment of inertia of the tube is indeed the difference between the moments of inertia of the two cylinders, all we need to do is some simple algebra:

$$\begin{aligned} \tfrac12 m_t (R^2 + r^2) &= \tfrac12 c(R^2 - r^2) (R^2 + r^2) \\ &= \tfrac12 c(R^4 - r^4) \\ &= \tfrac12 c R^4 - \tfrac12 c r^4 \\ &= \tfrac12 M R^2 - \tfrac12 m r^2. \end{aligned}$$


Let's put your argument in more explicit terms:

The moment of inertia of a solid cylinder of radius R is $\frac{1}{2}mR^2$. If we take the portion of that cylinder that is within radius $r$ of the center line, the moment of inertia of that cylinder is $\frac{1}{2}mr^2$. So the moment of inertia of the portion of the cylinder that is between $r$ and $R$ should be $\frac{1}{2}m(R^2-r^2)$, but it is in fact $\frac{1}{2}m(R^2+r^2)$.

Your error is in simply putting "$m$" everywhere. If you have one object, then $m$ can represent its mass, but if you are adding and subtracting the moments of inertia for several different objects with different masses, you can't just use $m$ to represent all of their different masses. The moment of inertia of the inner cylinder is indeed one half the mass of the inner cylinder times the square of its radius. But if $m$ is the mass of the entire cylinder, then clearly the inner circle will not also have mass $m$! Its mass will be $\frac {mr^2}{R^2}$. So the moment of inertia will be $\frac{m(R^2)}2-\frac{mr^4}{2R^2} =\frac{m(R^4-r^4)}{2R^2}=\frac{m(R^2-r^2)(R^2+r^2)}{2R^2} = \frac 1 2 m\left(1-\frac{r^2}{R^2}\right)(R^2+r^2)$ and $m_{h} = m\left(1-\frac{r^2}{R^2}\right)$ is the mass of the hollow cylinder, so the moment of inertia of the hollow cylinder is $\frac 1 2 m_h(R^2+r^2)$.

It might be easier to understand with specific numbers. Suppose that $R=3, r=2, m=18$. If the density and length of the cylinder are fixed, and only the radius is variable, then the mass is proportional to the radius squared. So if a cylinder with radius $3$ has mass $18$, then the inner cylinder has mass $8$, and the portion between them has mass $10$. The moment of inertia of the total cylinder is $\frac 1 2 18 \cdot 3^3 = 81$. The moment of inertia of the inner cylinder is $\frac 1 2 8 \cdot 2^2 = 16$. The moment of inertia of the hollow cylinder is $\frac 1 2 10 (9+4)=65$. And if we check $81-65$, we indeed get $16$.

If you start with a cylinder with moment of inertia $\frac{1}{2}mr^2$ and make it hollow and while keeping its mass the same, then you can't be making it hollow by simply removing the center. You must be taking the mass in the center and squishing it into the outside, increasing the density. Since you're pushing the mass towards the outside, the moment of inertia must be increasing, not decreasing.


Mass moment of inertia of a composite body can only be added/subtracted when presented in relation to density and volume.

$$ m_{\rm sum} = \sum_i \underbrace{ \rho_i V_i }_{m_i}$$

$$ I_{\rm sum} = \sum_i \underbrace{ \rho_i V_i r_i^2}_{I_i} $$

where $r_i$ is the individual radius of gyration.

As a matter of convenience in the end we divide the above to get

$$ I_{\rm sum} = \frac{m_{\rm sum}}{ \sum_i \rho_i V_i} \sum_i \rho_i V_i r_i^2 = m_{\rm sum} r_{\rm sum}^2 $$

For a homogeneous part with uniform density $\rho$ you need to split the above into $i$ additive volumes, and $j$ subtractive volumes

$$m_{\rm sum} =\sum_i \underbrace{ \rho V_i}_{m_i} - \sum_j \underbrace{ \rho V_j}_{m_j} $$

$$ I_{\rm sum} = \sum_i \underbrace{\ \rho V_i r_i^2}_{I_i} -\sum_j \underbrace{ \rho V_j r_j^2}_{I_j} $$

In your example, we have the following parts

Component Part Volume Rad. Gyration Part Mass Part MMOI
Outer Cylinder $V_1 = \ell \pi R^2$ $r_1 = \tfrac{1}{\sqrt{2}} R $ $m_1 = \rho \ell \pi R^2$ $I_1 = \rho \tfrac{\pi}{2} \ell R^4 $
Inner Cylinder $V_2 = \ell \pi r^2$ $r_2 = \tfrac{1}{\sqrt{2}} r $ $m_2 = -\rho \ell \pi r^2 $ $I_2 = -\rho \tfrac{\pi}{2} \ell r^4 $
Combined $V = \ell \pi \left( R^2 - r^2 \right)$ $$r^2 = \tfrac{1}{2} \left(R^2 + r^2\right)$$ $$m_1+m_2 = \rho \ell \pi \left( R^2 - r^2 \right)$$ $$I_1 + I_2 = \rho \tfrac{\pi}{2} \left( R^4 -r^4\right)$$

Now the combined radius of gyration is found by

$$ r^2 = \frac{I_1+I_2}{m_1+m_2} = \frac{ \rho \tfrac{\pi}{2} \left( R^4 -r^4\right)} { \rho \ell \pi \left( R^2 - r^2 \right) } = \tfrac{1}{2} \left(R^2 + r^2\right) $$

And here is the source of the $+$ sign which confuses you. It comes from $ \frac{R^4-r^4}{R^2-r^2} = \frac{ (R^2-r^2)(R^2+r^2)}{R^2-r^2} = R^2+r^2$


The Moment of Inertia of cylinder x is proportional to the mass of the object $m_x$ and the 2nd power of the radius $r_x$.

$$I_x \propto m_x {r_x}^2$$ However, $m_x \propto {r_x}^2$, so

$$I_x \propto {r_x}^4$$

If z is shape x minus y, we get:

$$I_z = I_x - I_y \propto {r_x}^4 - {r_y}^4$$

Using the difference of squares (well, quartics), we get:

$$I_z \propto ({r_x}^2 - {r_y}^2)({r_x}^2 + {r_y}^2)$$

Now, lets find the mass of z:

$$m_z = m_x - m_y \propto {r_x}^2 - {r_y}^2$$ which we can factor out of $I_z$: $$I_z \propto m_z ({r_x}^2 + {r_y}^2)$$ and all that is left is to determine the constant.

The core trick, as observed by others, is that the inertia of a tube is proportional to the 4th power of the radius. And that the $m$ in the equation refers to different things in the smaller, larger and hollow cylinder cases.

We can also sanity check this. If we have a tube of fixed mass and we make it thinner, we are moving the same mass further away from the axis. This should increase the moment of inertia.


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