I'm trying to calculate the moment of inertia for a cylinder about a longitudinal axis, but I don't know where I went wrong with my approach.

$$I=\int r^2 dm$$

Assuming constant density:

$$\frac{M}{V} = \frac{dm}{dv} $$

Then in order to find $dv$ I found the volume by summing all the chords in the base of the cylinder from R to -R and multiplying my the length of the cylinder. (Where $R$ is the radius and $r$ is the distance from the axis of rotation, which I kept at the origin.

$$V = 2L\int \sqrt{R^2-r^2} dr$$


$$dv = 2L\sqrt{R^2-r^2} dr$$

And I know that the volume of the cylinder is:

$$V = πR^2L$$

So then...

$$dm = \frac{2M}{πR^2}\sqrt{R^2-r^2}$$

Substituting the original definition of moment of inertia from $-R$ to $R$ gives me:

$$\frac{2M}{πR^2}\int_{-R}^{R}r^2\sqrt{R^2-r^2}dr = \frac{1}{4}MR^2$$

However, the moment of inertia I looked up in a physics textbook is exactly two times this (the factor is $1/2,$ not $1/4$). I also solved for the moment of inertia of a sphere and similarly got exactly half of the accepted answer. I have looked over this thoroughly and don't know where I went wrong, but I suspect it is something to do with my integrating bounds?

  • $\begingroup$ how about $\text d m = 2\pi h\rho r\text dr$, to integrate on $r\in [0,R]$? $\endgroup$
    – Phoenix87
    Dec 24, 2015 at 23:53
  • $\begingroup$ The moment of inertia of one of your chords of mass $dm$ is not $dm R^2$ because the chord is not a point mass. Its spatial extent gives it a larger moment of inertia. To find how much larger than $dm R^2$ the chord's moment of inertia is, I would use the parallel axis theorem and the moment of inertia of a rod. Then you should get the right answer. Feel free to right up a complete answer for yourself if this works. $\endgroup$ Dec 25, 2015 at 0:01
  • $\begingroup$ @Phoenix87 Yup, this is an excellent solution (and probably a better one!) but I was just looking for the flaw in my solution. $\endgroup$
    – somil
    Dec 25, 2015 at 14:38
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs Maybe I'm missing something, but when I solve for dv, I'm solving for the point volume for every point change in r. This doesnt have anything to do with rotational inertia, correct me if I'm wrong $\endgroup$
    – somil
    Dec 25, 2015 at 14:51

2 Answers 2


What is wrong with your answer?

I am going to write an answer explaining why your solution is wrong, because I don't think a comment would be enough. First, I am going to make a change of variable. You used the variable $r$ to refer to the distance from the axis of the cylinder. I feel more comfortable using the symbol $x$ for this variable, so that is what I am going to. The reason I fee more comfortable with this choice is because you don't integrate over cylindrical shells of radius $r$; you integrate over surfaces of constant $x$.

Anyway, let's look at what you did. Your equation $$dV = 2L \sqrt{R^2-x^2}dx$$ is correct. This is great so far.

The next equation I want to look at is

$$dm = \frac{2M}{\pi R^2}\sqrt{R^2-x^2}dx.$$

Notice you forgot the $dx$ in your original question, but its obvious that is what you meant. Now let's think about what this equation means. It means that the mass in the surface of constant $x$ of width $dx$ is the $dm$ given by the equation. This equation is also totally fine, but it isn't as useful as you think.

I do have a problem with your next equation. Your next equation is

$$I = \frac{2M}{\pi R^2}\int x^2 \sqrt{R^2-x^2}dx.$$

The reason I have a problem with this equation is that it is really saying $I=\int x^2 dm /M$, but of course we know that it should be an $r$ instead of an $x$: $I=\int r^2 dm /M$. The reason this distinction is important is because your surfaces of constant $x$ are not surfaces of constant $r$. Now we can't fix this problem by just writing

$$I = \frac{2M}{\pi R^2}\int r^2 \sqrt{R^2-x^2}dx$$

because each surface of constant $x$ is not at a well-defined $r$: the part of the surface near the surface of the cylinder has $r=R$, but the part in the middle of the surface has $r=x$. So the integral above doesn't make sense.

Correct way to get the answer

There are two ways to find the answer then. One way is just to write out the integral in rectangular coordinates and the other way is to use the parallel axis theorem to find the moment of inertia of each surface of constant $x$ and the integrate over constant $x$.

First way of getting the answer

Let's look at the first method first. We get the following expression for the moment of inertia:

$$I = \frac{M}{\pi R^2}\int^R_{-R} \int^\sqrt{R^2-x^2}_{-\sqrt{R^2-x^2}} (x^2+y^2) dy\,dx $$

Doing the inner integral, you get \begin{equation} \begin{aligned} I &= \frac{2M}{\pi R^2}\int^R_{-R} \left(\frac{4}{3} R^2 + \frac{2}{3}x^2 \right) \sqrt{R^2-x^2} dx \\ &=\frac{1}{3} \frac{2M}{\pi R^2}\int^R_{-R}R^2\sqrt{R^2-x^2} dx+\frac{2}{3} \frac{2M}{\pi R^2}\int^R_{-R}x^2\sqrt{R^2-x^2} dx \end{aligned} \end{equation}

Now you have already evaluated the integral in the second term in your question. The integral in the first term is easy because it is just the area of a semicircle. So we get

\begin{equation} \begin{aligned} I &= \frac{1}{3} \frac{2M}{\pi R^2} R^2 \pi R^2/2 +\frac{2}{3} \frac{M R^2}{4} \\ &= MR^2/2 \end{aligned} \end{equation}

Second way of getting the answer

Now let's look at the second way of getting the answer. This time we will just calculate the moment of inertia of each surface of constant $x$ and add them up. We could find in a table that the moment of inertia of a rectangle of width $2\sqrt{R^2-x^2}$ and mass $dm$ about its center of mass is $\frac{1}{3} dm \left(R^2-x^2\right).$ But we are more interested in the moment of inertia about the origin when this surface of constant $x$ is displaced a distance $x$ from the origin. Using the parallel axis theorem, we find that the moment of inertia is $\frac{1}{3} dm \left(R^2-x^2\right) + dm x^2$. This is the moment of inertia of each surface of constant $x$. Adding these up we get the total moment of inertia:

$$I = \int dI = \int \frac{1}{3} dm \left(R^2-x^2\right) + dm x^2 = \int \frac{1}{3} R^2 dm + \frac{2}{3} x^2 dm.$$

Now plugging in our expression for $dm$, we get that $$dI = \frac{2M}{\pi R^2} \int_{-R}^R \frac{1}{3} R^2 \sqrt{R^2-x^2} + \frac{2}{3} x^2 \sqrt{R^2-x^2} dx.$$

This is the same expression we got from the first way of computing the moment of intertia. So this will give us the correct answer. Also, we can see that the $y$ integral of the first method just gave us the moment of inertia of each surface of constant $x$.

I should add that the easiest way to get the moment of inertia is to integrate over surfaces of constant $r$ (where $r$ is the distance from the axis of the cylinder. You get that $dI = r^2 dm$ where $dm = \frac{M}{\pi R^2} 2 \pi r dr.$ So then you get $I = \int dI = \frac{M}{\pi R^2}\int_0^R r^2 2 \pi r dr = \frac{2 M}{R^2} \int_0^R r^3 dr = MR^2 /2$

  • $\begingroup$ Wow, awesome! My issue is that I made a huge conceptual mistake by equating r with x. I assumed that r and x would always be the same, but I now realize that r is much greater towards the top of the chord even when x remains the same. Your solutions are excellent, thanks again. $\endgroup$
    – somil
    Dec 26, 2015 at 18:59

Suppose that the length of the cylinder is $L$, the radius of the cross section is $R$. We choose cylindrical coordinates to solve the question.

The momentum of inertia is that, $$ I=\int s^2dm, $$ and we assume the density is constant, we have, $$ I=\int s^2 \rho dv. $$ In cylindrical coordinates, $dv=sds\;dz\;d\theta$, so we get, \begin{align} I&=\rho\cdot \int\int\int s^2\cdot sds\;dz\;d\theta \\& =\rho\int_0^R s^3ds\int_0^Ldz\int_0^{2\pi}d\theta \\ & =\rho\cdot 2\pi \cdot L\cdot \frac{R^4}{4} \\ & =\frac{ \rho \pi R^4L}{2} .\end{align}

Then we substitute $\rho=\frac{M}{V}=\frac{M}{\pi R^2 L}$ into the above equation, we have \begin{align} I&=\frac{M}{\pi R^2 L} \cdot \frac{\pi R^4 L}{2} \\ & =\frac{1}{2}MR^2. \end{align}

  • $\begingroup$ Awesome! Although, I agree this is an excellent valid solution, I was actually looking for the flaw in my solution rather than another viable one... $\endgroup$
    – somil
    Dec 25, 2015 at 14:36
  • $\begingroup$ @drsom So give my answer an upward vote? $\endgroup$
    – Wang Yun
    Dec 25, 2015 at 15:25
  • $\begingroup$ I did, but I dont have enough points on physics stack for it to show up yet sorry ... $\endgroup$
    – somil
    Dec 25, 2015 at 15:48
  • $\begingroup$ @drsom It doesn't matter. $\endgroup$
    – Wang Yun
    Dec 25, 2015 at 15:50

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